Organic Chemistry MCAT Flashcards

Question

What is the difference between relative and absolute configuration and how are they determined?

Answer 1

configuration spatial arrangement of atoms or groups of a stereoisomer, relative configuration- configuration in relation to another chiral molecule through chemical interconversion to determine enantio/diasteromers. Absolute configuration- exact spatial arrangement of atoms or groups, independent of other molecules. Step 1) prioritize four substituents looking at atoms directly attached to chiral center relative to highest atomic number if equality priority of combination of atoms attached double bonds count as two bonds. Step 2) atoms with lowest priority/hydrogen back of molecule, and switch the other neighboring groups. 3) If the circle is drawn clockwise, the asymmetric atom is called R (from Latin rectus, meaning “ right”). If it is counterclockwise, it is called S (from Latin sinister, meaning “ left”), 4) Name the compound S&R in parentheses separated by a dash, if more than one location is specified by preceding number with R or S within parentheses without dash. Differences in chirality (handedness) can produce isomers such as enantiomers (non-superimposable mirror images) as well as diasteromers (non- mirror image stereoisomers). Meso compounds contain chiral centers but are not optically active because they have an internal mirror plane of symmetry.

Answer 2

- horizontal lines indicate bonds that project out from the plane of the page, whereas vertical lines indicate bonds going into (or behind) the plane of the page. The point of intersection of the lines represents a carbon atom. With lowest priority projected into the page can be on the top or bottom, another advantage is that, we can manipulate Fischer projections without changing the compound. By interchanging any two pairs of substituents, or by rotating the projection on the plane of the page by 180°, we still have the same compound, if only one pair of substituents is interchanged, or if the molecule is rotated by 90°, the mirror image of the original compound is obtained(R instead of S, or S instead of R). Lowest priority pointed into the page Option 1: Go ahead and determine the order of substituents as normal, number 1 → number 2-→ number 3. Remember, number 4 doesn’t’ t count, so just skip right over it when determining the order. Then obtain the designation(R or S). The true designation will be the opposite of what you just obtained. Option 2: Make a single switch: Just swap the lowest priority group with one of the groups on the vertical axis. Obtain the designation (R or S), and once again, the true designation will be the opposite of what you just found. Option 3: The final option is to make two switches. Basically, just use option 2, moving the lowest priority group into the correct position. Then, switch the other two groups as well. Because we made two switches, this molecule will have the same designation as the initial molecule. This is the same, as holding one substituent in place and rotating the other three in order— like the rotation of a methyl group. Meso compounds are made up of two halves that are mirror images. Thus, they are not optically active. M eS o compounds have a Mirror plane of Symmetry.

Answer 3

• . At the molecular level, one enantiomer will rotate plane-polarized light to the same extent but in the opposite direction of its mirror image (if the concentration and path lengths are equal). Returning to the molecular level, if plane-polarized light is passed through an optically active compound, the orientation of the plane is rotated by an angle α. The enantiomer of this compound will rotate light by the same amount but in the opposite direction. A compound that rotates the plane of polarized light to the right, or clockwise (from the point of view of an observer watching the light approach), is dextrorotatory and is labeled (+). A compound that rotates light toward the left, or counterclockwise, is levorotatory and is labeled (−). The direction of rotation cannot be determined from the structure of a molecule and must be determined experimentally. That is, it is not related to the absolute configuration of the molecule. Beer’s law- α = [α] • concentration • length

Answer 4

* (nonsuperimposable mirror images) are molecules that have the opposite configuration at their one chiral center. Or, if there are multiple chiral centers, they must have the opposite configuration at every one of their chiral centers to be enantiomers. Enantiomers have identical physical properties and most of the same chemical properties. However, the exceptions are optical activity and how they react in chiral environments. A compound is optically active if it has the ability to rotate plane-polarized light. Ordinary light is unpolarized, which means it consists of waves vibrating in all possible planes perpendicular to its direction of motion. A polarizer allows light waves oscillating only in a particular direction to pass through, producing plane-polarized light. Since optical activity occurs when plane-polarized light is shifted in one direction by a chiral molecule. The amount of rotation depends on the number of molecules that a light wave encounters. This depends on two factors: the concentration of the optically active compound and the length of the tube through which the light passes. 1 g/mL for concentration and 1 dm for length. Rotations measured at different concentrations and tube lengths can be converted to a standardized specific rotation ([α]) * When both (+) and (-) enantiomers are present in equal concentrations, we call the mixture a racemic mixture. In these mixtures, the rotations cancel each other out and no optical activity is observed. A racemic mixture displays no optical activity. Because they have different arrangements in space, they will have different physical properties.

Answer 5

has different chemical properties, but because they have the same functional groups, they might behave in a similar fashion. Diastereomers are non-mirror-image configurational isomers. For example, cis and trans-alkenes are diasteromers. Diasteromers are also possible when a molecule has two or more stereogenic centers and differ at some, but not all, of these centers. This means that diasteromers are required to have multiple chiral centers. For any molecule with chiral centers, there are 2n possible stereoisomers. Thus, if a compound has two chiral carbon atoms, it has a maximum of four possible stereoisomers

Answer 6

the criterion for optical activity of a molecule containing any number of chiral centers is that it has no plane of symmetry. If an internal plane of symmetry exists, the molecule is not optically active, even though it possesses chiral centers. This plane of symmetry can occur either through the chiral center or between chiral centers. As long as there is a plane of symmetry, the compound is not chiral. A molecule with an internal plane of symmetry is called a meso compound

Answer 7

Carbon is tetravalent, which means that it can form bonds with up to four other atoms, allowing for the massive versatility required to form the foundation of life. This versatility is compounded by the fact that carbon, located near the center of the periodic table, can form bonds with many different elements. In addition, because carbon atoms are fairly small, the bonds that they form are strong and stable. Two types- first is ionic, in which electrons are transferred from one atom to another; the second is covalent, in which electrons are shared between atoms.

Answer 8

Within each electron shell, there can be several types of orbitals (s, p, d, f, etc., corresponding to the azimuthal quantum numbers l = 0, 1, 2, 3, and so forth). Each type of atomic orbital has a specific shape. An s-orbital is spherical and symmetrical, centered on the nucleus. A p-orbital is composed of two lobes located symmetrically about the nucleus and contains a node (an area where the probability of finding an electron is zero) in the nucleus. A d-orbital is composed of four symmetrical lobes and contains two nodes. Four of the d-orbitals are clover-shaped, and the fifth looks like a donut wrapped around the center of a p-orbital

Answer 9

•The first three quantum numbers, n, l, and m, describe the size, shape, and number of atomic orbitals an element possesses. The principal quantum number n, which can equal 1, 2, 3, etc., corresponds to the energy levels in an atom and is essentially a measure of size. The smaller the number, the closer the orbital is to the nucleus, and the lower its energy. n, the principal quantum number, describes the size of the orbital and can be any integer value from 1 to ∞. l, the azimuthal quantum number, describes the shape of the orbital and is calculated as n− 1 (l = 0 is an s-orbital, l = 1 is a p-orbital, etc.). m, the magnetic quantum number, denotes the orientation of the orbital in space and can be any integer value from +l to – l. s, the spin quantum number, distinguishes between the two electrons in an orbital by arbitrarily assigning one of the electrons a spin of +1/2 and the other a spin of − ½

Answer 10

two atomic orbitals combine forming a molecular orbital, obtained by adding or subtracting the wave functions of the atomic orbitals. If the signs of the wave functions are the same, a lower-energy (more stable) bonding orbital is produced. If the signs are different, a higher-energy (less stable) antibonding orbital is produced. Molecular orbital formed by head-to-head or tail-to-tail overlap resulting sigma bond, all sigma bonds are sigma bonds accommodating two electrons with the bond holding atoms more closely together and stronger than longer bonds.

Answer 11

consists of both a σ bond and a π bond (and a triple bond consists of a σ bond and two π bonds). Pi bonds are weaker than σ bonds in isolation, but the strength is additive, making double and triple bonds stronger than single bonds. Two orbitals line up in a parallel (side-by-side) fashion, with their electron clouds overlapping and molecular orbital, a pi (π) bond, one pi bond on top of an existing single bond, to form a double bond, single bond and two pi bonds form a triple bond, with double and triple bonds hindering rotation locking the atoms into position. A pi bond cannot exist independently of a sigma bond, formation of a sigma bond will p-orbitals of adjacent carbons be parallel and in position to form pi bond. The more bonds formed between atoms, the shorter the overall bond length. Double bond is shorter than single bond, triple bond shorter than double bond, shorter the bond length, the greater the strength of bond. Individual pi bonds are weaker than sigma bonds; geometric isomers are converted between cis and trans conformations. Breaking a single sigma bond requires far more energy, so it happens much less often.

Answer 12

Carbon has the electron configuration 1s22s22p2 and, therefore, needs four electrons to complete its octet. Although 1 sigma and 3 pi available how are bonds in methane equal? Mixing different types of orbitals forms hybrid orbitals. Just as with molecular orbitals, we can use math to combine three p-orbitals and one s-orbital. The result? Four identical sp3 orbitals with a brand-new, hybrid shape. Hybridization is a way of making all of the bonds to a central atom equivalent to each other. The sp3 orbitals are the reason for the tetrahedral shape that is a hallmark of carbon-containing compounds. All four of these orbitals point toward the vertices of a tetrahedron to minimize repulsion, which explains why carbon prefers a tetrahedral geometry. The hybridization is accomplished by promoting one of the 2s electrons into the 2pz orbital. Produces four valence orbitals, each with one electron, which can be mathematically mixed to provide the hybrids.

Answer 13

we have one s and three p, so the bond therefore has 25% s character and 75% p character. 109.5o apart. Saturated hydrocarbons- they have no double bonds, and thus each carbon is saturated with the maximum number of hydrogens it can hold. This is reflected in the general formula for alkanes, CnH2n + 2.

Answer 14

• one s-orbital is mixed with two p-orbitals (33% s, 66% p), three sp2 hybrid orbitals are formed, alkenes, third p-orbital of each carbon is left un-hybridized— this is the orbital that participates in the pi bond. The three sp2 orbitals are oriented 120° apart, which allows for maximum separation. Hybridized p-orbital is tied up in the pi bond of the double bond. Two of the sp2 hybrids will participate in C– H single bonds, and the other hybrid orbital will line up with the pi bond and form the C=C double bond.

Answer 15

• triple bond two of p-orbitals to form pi bonds and the third p-orbital combine to from two sp-hybrid orbitals 50% s character, 50% p character. These orbitals are oriented 180° apart, which explains the linear structure of molecules such as acetylene.

Answer 16

•a primary carbon atom (written as 1°) is bonded to only one other carbon atom. A secondary (2°) carbon is bonded to two, a tertiary (3°) to three, and a quaternary (4°) to four other carbon atoms. Hydrogen atoms attached to 1°, 2°, or 3° carbon atoms as 1°, 2°, and 3°. Hydrogens attached to carbons will always carry the same name as their parent carbon atom.

Answer 17

* ↑ Chain length: ↑ boiling point, ↑ melting point, ↑ density. ↑ branching: ↓ boiling point, ↓ density. * Physical properties for higher molecular weight as it increases, so do the melting point, boiling point, and density. This is logical; the heavier the molecule is, the harder it should be for the molecule to break away from others and enter the higher-energy gas phase. That's why, at room temperature, straight-chain compounds of up to 4 carbons are found in the gaseous state, chains of 5 to 16 carbons exist as liquids, and the longer-chain compounds are waxes and harder solids.

Answer 18

slightly lower boiling points than their straight-chain isomers. This is because greater branching reduces the surface area of the molecule available for interactions with neighboring molecules. The weakened intermolecular attractive forces (van der Waals forces) result in a decreased boiling point. Melting point- the melting point does follow the same trend— after all, with greater branching, the molecules cannot stack up as close to each other as they would prefer. But for a given number of carbons, what really matters is how symmetrical the molecule is: the more symmetrical a molecule, the higher the melting point.

Answer 19

reaction of alkanes with molecular oxygen to form CO2, water and heat, proceed through radical and is often incomplete with a lot of carbon monoxide present as well Alkane + O2→ CO2 + H2O + heat

Answer 20

R + X2→ RX. Halogenation, in which one or more hydrogen atoms are replaced with a halogen atom (Cl, Br, or I) via a free-radical substitution mechanism •Initiation- diatomic halogens homolytically cleaved (two electrons of sigma bond split equally by heat or ultraviolet light, formation of two free radicals, neutral species with unpaired electrons, extremely reactive and readily attack anything near them, •Propagation- radical produces another radical to continue the reaction, radical reacts with alkane removing hydrogen atom to form HX and create an alkyl radical, reacting with X2 to form alkyl halide generating another radical halogen, begins and ends with a radical product of first step is starting material of second, and product of the starting material for the first step of the chain reaction •Termination- Propagation will continue until two free radicals combine with one another to form a stable molecule, ending the reaction. The more stable the intermediate is, the more likely the reaction is to occur. Larger alkanes have many hydrogens available for the free radical to attack. Bromine radicals react fairly slowly and primarily attack the hydrogens on the carbon atom that can form the most stable free radical. Always think of carbocations and free radicals as electron-deficient because both species have fewer than the eight electrons needed to be complete. In this way, we can remember the trends of stability of both of these species: 3° > 2° > 1° > methyl.

Answer 21

molecule broken down by heat (pyro-), cracking, reduce average molecular weight of heavy oils and increase production of more desirable volatile compounds, C-C bonds cleaved, producing smaller-chain alkyl radicals recombine to form variety of alkanes.

Answer 22

a radical transfers a hydrogen atom to another radical, producing an alkane and an alkene

Answer 23

can take part in nucleophilic substitution reactions, nucleophiles electron-rich species attracted to positively charged/positively polarized atoms, electrophiles. Alkyl halides most common undergoing nucleophilic substitution reactions

Answer 24

nucleophilicty roughly correlated to basicity if same attacking atom, stronger the base more likely to attract positively charged proton bronstead-lowry definition, nucleophilic strength measures how much atom wants to find positive charge with a strong correlation. nucleophilicity strength decreases in the order of RO->HO->RCO2->ROH>H2O

Answer 25

if the attacking atoms differ, or if the conditions differ, nucleophilic ability doesn't necessarily correlate with basicity. For example, in a protic solvent (solvents with protons in solution, such as water or alcohols), large atoms tend to be better nucleophiles because they can shed the solvating protons surrounding them and are more polarizable. When it comes to nucleophilicity, size matters in protic solvents (those capable of hydrogen bonding or donating protons). The solvent, which decreases its ability to act as a nucleophile, can easily surround the smaller atoms and a larger atom thus becomes more nucleophilic in comparison. Big that it can be polarized, meaning the electrons can shift around, making some areas more negative than others, giving it a much better chance to make it to the electrophile while it is still negative enough to attack. Basicity and nucleophilicity are directly related because both imply that the compound wants to donate electrons, but the solvent may stabilize these species and alter that trend. In aprotic solvents, the trend stays intact. With the nucleophilicty strength decreasing in the order CN->I->RO->HO->Br->Cl->F->H2O

Answer 26

without protons, nucleophiles don't have proton coats, not solvated, nucleophilic strength related to opposite to protic order of nucleophilic strength same as base strength. F->CL->Br->I-

Answer 27

depends on how good a leaving group is, leaving group are weak bases, stable anions or neutral species, easily accommodate electron pair Good leaving groups, weak bases, conjugate bases of strong acids, opposite of base strength in cases of halogens. Weak bases make good leaving groups because they are able to spread out electron density, making the species more neutral or more stable. I->Br->Cl->F-

Answer 28

designation for unimolecular nucleophilic substitution reaction, rate of reaction depends only on substrate, original molecule, rate-determining step dissociation of species to form stable, positively charged species of carbocation. k [RX] the dissociation of a molecule into a carbocation and a good leaving group, followed by the combination of the carbocation with a nucleophile, understanding of the intermediate will be essential in determining all of the facts surrounding the reaction, including the rate and the products To get carbon into a carbocation- polar protic solvents with lone electron pairs, because the electron-rich groups can solvate the carbocation and help stabilize it. Carbocations are also stabilized by charge Rate = delocalization. The more highly substituted the cation is, the more stable it will be.

Answer 29

• To get particular product original substituent better leaving group than nucleophile, reverse reaction outcompete forward reaction, carbocation such strong electrophile, pick up anything comes near it with lone electrons, does not require strong nucleophiles, the more reactive the nucleophile, more SN2 reaction will result. Rate of reaction depends upon slowest step, rate limiting or rate=determining step, limiting speed of reaction, depends only on concentration of original molecule, rate increased by anything accelerating formation of carbocation. Structural factors- highly substituted alkyl halides allow for distribution of positive charge over greater number of carbon atoms, forming most stable carbocation. Solvent effect- highly polar solvents better at surrounding and isolating ions than less polar solvents, polar protic solvents water, and solvation stabilizes intermediate state. Nature of leaving group- weak bases dissociate more easily from alkyl chain making better leaving group increasing rate of carbocation formation

Answer 30

• An intermediate is distinct from a transition state. An intermediate is a well-defined species with a finite lifetime and must be at a relative minimum energy for this to occur. On the other hand, a transition state is a theoretical structure used to define a mechanism. The transition state represents a maximum (in energy) between two minima on a reaction coordinate.

Answer 31

• SN2The penta-coordinate (five-coordinate) transition state, where one bond is forming and the other is breaking, is the hallmark of the SN2 reaction. The increased crowding around that central atom will guide the rate and products of the reaction. Rate = k [Nu][RX]. Certain conditions, the formation of a carbocation is unlikely, if not downright impossible. Even under such conditions, substitution reactions can still proceed, but they must occur by a different mechanism that avoids the carbocation altogether. Enter SN2: An SN2 (bimolecular nucleophilic substitution) reaction involves a strong nucleophile pushing its way into a compound while simultaneously displacing the leaving group, in one concerted step. Only one-step, it must be the rate-determining step. The reaction is called bimolecular because the rate-determining step involves two molecules. SN2 reactions, the nucleophile actively displaces the leaving group in a backside attack. For this to occur, the nucleophile must be strong, and the substrate cannot be sterically hindered. Primary substrates are the most likely to undergo SN2 reactions, followed by secondary, whereas tertiary substrates are just too crowded to participate in this mechanism. Nucleophile attacks the reactant from the backside of the leaving group, forming a trigonal bi-pyramidal transition state (sp2). As the reaction progresses, the bond to the nucleophile strengthens, while the bond to the leaving group weakens. The leaving group is displaced as the bond to the nucleophile becomes complete. Single step two reacting species so concentrations of both have a role in determining rate meeting in solution, second order kinetics. The molecule with a leaving group, often an alkyl halide or a tosylate) and the nucleophile.

Answer 32

* Stereochemistry- SN1 mechanism involves a carbocation intermediate in which the carbon only has three groups bound to it. With only three substituents, the molecule takes on a planar shape, with 120° between each of the bonds. That means that the carbons are sp2 hybridized. Because the molecule is planar (and therefore achiral), the nucleophile can attack either the top or the bottom of the compound. This means that as long as the end product has four different groups, we can have two different products, * Depending on whether the nucleophile attacks from the top or the bottom. SN1 leads to loss of stereochemistry; SN2 leads to a relative inversion of stereochemistry owing to backside attack. Be careful, though, because the absolute configuration may remain the same if the leaving group and the nucleophile do not maintain the same priority. If the original compound was optically active, the product will be a racemic mixture and, thus, no longer optically active. * SN2 stereochemistry- nucleophiles must attack backside of molecule, as leaving group leaves from other side, molecule will flip and inversion of configuration. One important thing to be aware of is that this inversion of stereochemistry will lead to an inversion of the absolute configuration only if the leaving group and the nucleophile have the same priority (R will be changed to S, and vice versa). If the nucleophile and the leaving group have different priorities, even though the molecule will still flip, the designation will not be changed.

Answer 33

Physical Properties- melting and boiling points increase with increasing molecular weight and similar to corresponding alkanes, Trans-alkenes generally have higher melting points than cis-alkenes because they are more symmetric, which allows for better packing in the solid state. However, trans-alkenes tend to have lower boiling points than cis-alkenes because they are less polar.

Answer 34

Polarity- result of an asymmetrical distribution of electrons in a molecule. This causes the molecule to have one partially negative region and another that is partially positive. In alkenes, unequal electron distribution creates dipole moments that point from the electropositive alkyl (single-bonded) groups toward the electronegative alkene (double bond). That is, the sp3 carbons donate electrons to the sp2 carbons. Sp3 carbons have less s-character (25%) than sp2 carbons (33%), and s-electrons can be found at the positive nucleus, making them more stable.

Answer 35

elimination reactions alcohols or alkyl halides, elimination carbon backbones kicks off/eliminates hydrogen and halide (dehydrohalogenation), molecule of water (dehydration) forming double bond, • water (dehydration) forming double bond, Elimination- E1- unimolecular rate of reaction depends on concentration of only one species, substrate, departure of leaving group and formation of carbocation intermediate, proton of adjacent carbon (β-carbon) removed by weak base double bond form from electrons from now broken carbon-hydrogen bond. E1 favored by polar protic solvents, ability to form stable carbocation, highly branched carbon chains, good leaving groups, absence of good nucleophile, mechanisms occur under similar conditions, competitive and occur simultaneously with higher temp favoring E1.

Answer 36

• Controlling E2 over SN2- 1. Steric hindrance, highly substituted carbon chains forming most stable alkenes, E2 easily and SN2 rarely, bulk of base larges impact on dominant mechanism hard time getting to backside easier to pluck off hydrogen from neighboring chain. 2. Stronger base favors E2 over SN2, SN2 favored over E2 by weak Lewis bases or strong nucleophiles, cyanide or iodide, strong base pull of hydrogen before reaching carbon atom resulting in E2 or higher temperatures. SN2 depends on properties of substrate and base. Oxidized, it loses electrons, and if it is reduced, it gains electrons, if species is reduced, it will be getting more hydrogen; if it is oxidized, it will be losing hydrogen and, thus, gaining double bonds or oxygen atoms.

Answer 37

Bimolecular elimination- rate of an E2 reaction depends on two species, the substrate and the base. As a strong base (such as the ethoxide ion, C2H5O– removes a proton, and then a halide ion • Anti to the proton leaves, resulting in the formation of a double bond. Two possible products, double bond can form on either side of departing halide, more substituted double bond more stable and constitute a larger percentage of products. Double bond form either side of departing halide, more substituted double bond more stable and constitute larger percentage of products, if molecule can form either geometric isomer, trans isomer predominate, more stable than strained cis.

Answer 38

reducing alkene by adding hydrogen to double bond with aid of metal catalyst, platinum (Pt), palladium (Pd), and nickel (Ni). Occasionally rhodium, iridium, or ruthenium is used, Takes place on surface of metal, face of the π -bond becomes coordinated to the metal surface where molecular hydrogen is bound. That's why the two hydrogen atoms are added to the same face of the double bond, reaction takes place where the two touch. This type of addition is called syn addition.

Answer 39

π -bond is higher in energy, and thus, more reactive than the σ -bond. Therefore, it can be broken without breaking the σ -bond. As a result, we can add compounds to double bonds while leaving the carbon skeleton intact. Although many different addition reactions are possible, most operate via the same essential mechanism. Because the electrons of the π -bond are reactive, they are attracted to molecules seeking to accept an electron pair (Lewis acids). Recall that these seekers (or “ lovers”) of electrons are called electrophiles. Lewis Acids Accept electron pairs, Lewis bases donate electron pairs.

Answer 40

electrons in the double bond act as a Lewis base and bond with the partially positive hydrogen of HX molecules. This first step yields a carbocation intermediate. In cases where the alkene is asymmetrical, the initial protonation produces the most stable carbocation. Be careful here; this means that the proton will actually add to the least substituted carbon atom (the carbon atom with the most protons). This addition will form the carbocation on the adjacent carbon (that is more substituted), because alkyls substituents help stabilize carbocations.

Answer 41

• and the goal are to produce the most stable carbocation. In the second step, the halide ion combines with the carbocation to yield an alkyl halide; Markovnikov's rule refers to the addition of a group to the more substituted carbon of the double bond. It does so because the more stable carbocation intermediate (the more highly substituted) will form in the slow first step and the nucleophile will then attack that positive charge in the fast second step.

Answer 42

addition of diatomic halogens to a double bond is a rapid process. In fact, it is frequently used as a diagnostic tool to test for the presence of double bonds. The double bond acts as a nucleophile again in this reaction, attacking one half of an X2 molecule and displacing X– from the other side. The second halogen is just acting as the leaving group. The first step results in a cyclic halonium ion intermediate (bromonium and chloronium ions are known to exist). X– then attacks the ion on the opposite face because the bromine is blocking the other side (anti-addition!), resulting in the dihalo compound. If this reaction is carried out in a nucleophilic solvent, the cyclic halonium ion can be attacked by solvent molecules before the second halogen ion gets a chance to do so.

Answer 43

Water can be added to alkenes under acidic conditions (most commonly H2SO4), double bond is protonated according to Markovnikov's rule, forming the most stable carbocation. This carbocation then reacts with water, yielding a protonated alcohol, which then loses a proton to become an alcohol. Perform this reaction at low temperatures because at high temperatures, the reverse reaction, acid-catalyzed dehydration, is heavily favored. Remember, if heat is part of the reaction, look for the formation of a double bond. Hydration of the double bond can also be achieved under slightly milder conditions with oxymercuration-reduction. As with the harsher acid-catalyzed hydration, Markovnikov radiochemistry is observed.

Answer 44

When peroxides or UV light are present, expect free radical reactions that do not follow Markovnikov's rule. HX to alkenes: through a mechanism that uses free-radical intermediates. This reaction occurs in the presence of peroxides, oxygen, or ultraviolet light. Free-radical additions disobey the Markovnikov rule because X• adds first to the double bond, producing the most stable free radical. Halogen will end up on the least substituted carbon, quite similar to standard electrophilic additions, where H+ adds first to produce the most stable carbocation. The important thing to realize here is that both of these mechanisms are in place to create the most stable intermediate. This reaction is useful for HBr, but it is not practical for HCl or HI, because they are energetically unfavorable. In anti-Markovnikov reactions, the most stable radical forms on the most substituted carbon (just as the most stable carbocation formed before), but because the halogen adds first, it ends up on the least substituted carbon. Remember, the most stable intermediate and least energetic transition state will always determine the favored products.

Answer 45

Diborane (B2H6; often written as borane: BH3) adds readily to double bonds. The boron atom (owing to its incomplete octet) is a Lewis acid and attaches to the less sterically hindered carbon atom. At the same time, a hydride is transferred to the adjacent carbon (a concerted mechanism). The second step is an oxidation-hydrolysis with peroxide and aqueous base that directly transfers water to the bond with boron, producing an alcohol with overall anti-Markovnikov, syn orientation

Answer 46

Alkenes can be oxidized with KMnO4 (potassium permanganate), although depending on the reaction conditions, we can end up with drastically different products, If we make our conditions as mild as possible, using cold, dilute KMnO4, the product simply has – OH groups added to each side of the double bond. Such products are called 1,2 diols (vicinal diols), or glycols, and they have syn orientation. • Cold, and dilute conditions should always make you think of a mild or weak reaction (adding alcohols to a double bond). Hot, acidic conditions should make you think of rigorous or strong reactions (breaking the double bond altogether and forming carboxylic acids). • Kick it up a notch and use a hot, basic solution of potassium permanganate, followed by an acid wash, nonterminal alkenes are cleaved to form two molar equivalents of carboxylic acid, and terminal alkenes are cleaved to form a carboxylic acid and carbon dioxide. If the nonterminal double-bonded carbon is disubstituted, a ketone will be formed. Under these intense conditions, we simply chop the double bond in half and make those cleaved carbons as oxidized as possible.

Answer 47

more selective cleaves the double bond in half, it only oxidizes the carbon to an aldehyde (or a ketone if the starting molecule is disubstituted) under reducing conditions (Zn/H+ or (CH3) 2S, Ozonolysis under oxidizing conditions (H2O2) yields also cleaves double bonds but will only oxidize primary carbons to aldehydes or secondary carbons to ketones.

Answer 48

Alkenes can also be oxidized with peroxycarboxylic acids, which are strong oxidizing agents. Peroxyacetic acid (CH3CO3H) and m-chloroperoxybenzoic acid (mCPBA) are commonly used. The unique thing about this reaction is that the products are epoxides (also called oxiranes). This reaction is an example of syn addition

Answer 49

Polymerization is the creation of long, high-molecular-weight chains (polymers) composed of repeating subunits (called monomers). Polymerization usually occurs through a radical mechanism, although anionic and even cationic polymerizations are commonly observed.

Answer 50

- hydrocarbons that possess one or more carbon– carbon triple bonds. All triple bonds form straight lines with 180° between carbons as a result of the SP hybridization. Nomenclature: -yne and specify the position of the triple bond when it is necessary ethyne, which is almost exclusively called acetylene. Frequently, compounds are named as derivatives of acetylene. • Physical properties- similar to those of analogous alkenes and alkanes. In general, similar to alkanes and alkenes, the shorter-chain compounds are gases, but alkynes boil at somewhat higher temperatures than their corresponding alkenes. Internal alkynes, like alkenes, boil at higher temperatures than terminal alkynes. Asymmetrical distribution of electron density causes alkynes to have dipole moments larger than those of alkenes but still small in magnitude. Thus, we can assume that solutions of alkynes will be slightly polar, or at least more polar than a solution of alkenes. The acidity of the hydrogen on a terminal alkyne is the one major difference from all other hydrocarbon molecules. If anything about alkynes. Terminal alkynes can stabilize a negative charge fairly well, something that is uncommon for carbon atoms. As stated earlier, this stabilization stems from the 50 percent s-character. Recall that s-electrons have some probability of being found near the carbon nucleus (negative electrons are happier by the positive nucleus). This property is exploited in some of the reactions of alkynes

Answer 51

triple bonds are through two rounds of elimination of a geminal (remember twins, on the same carbon) or vicinal (neighbors in the vicinity, on neighboring carbons) dihalide. Reaction requires high temperatures and strong base, already existing triple bond into a new carbon skeleton. To do this, a terminal triple bond is converted into a nucleophile by removing its acidic proton with a strong base (NaNH2 or n-BuLi), producing an acetylide ion. Remember that terminal alkynes are fairly acidic, so this is a reasonable process. Once formed, the ion will perform nucleophilic displacements on primary alkyl halides at room temperature,

Answer 52

Reduction- reduction as adding bonds to hydrogen and oxidation as adding bonds to oxygen. Alkynes, just like alkenes, can be hydrogenated (reduced) with a catalyst to produce alkanes. If we want alkenes as our final product, we need to stop the reduction after addition of just one equivalent of H2. First uses Lindlar's catalyst, which is palladium on barium sulfate (BaSO4) with quinoline, a heterocyclic aromatic poison that stops the reaction at the alkene stage. Because the reaction occurs on a metal surface, the alkene product is the cis-isomer. The second method uses sodium in liquid ammonia at temperatures below − 33° C (the boiling point of ammonia) and produces the trans-isomer of the alkene via a free-radical mechanism.

Answer 53

Electrophilic: products form according to Markovnikov's rule. Addition can be stopped at the intermediate alkene stage or carried further. Of course, if we want to go all the way to the alkane stage, we will need two equivalents of reactants

Answer 54

Radicals add to triple bonds just as they do to double bonds, with anti-Markovnikov orientation. Be aware that the reaction product is usually the trans-isomer. This is because the intermediate vinyl radical can isomerize to its more stable form. The most stable electron-deficient species forms. In this reaction, we get an anti-Markovnikov product.

Answer 55

addition of boron to triple bonds occurs by the same method as addition of boron to double bonds. Addition is syn, and the boron atom adds first, boron is bound to three different substituents. If we follow the reaction with an acetic acid wash, the boron atom can be removed, and each substituent will have a proton from acetic acid in its place. This produces three cis-alkenes as shown. Terminal alkyne, a disubstituted borane is used to prevent further boration of the vinylic intermediate to an alkane. The vinylic borane intermediate can be oxidatively cleaved with hydrogen peroxide (H2O2), creating an intermediate vinyl alcohol (an enol), which tautomerizes to the more stable carbonyl compound (via keto-enol tautomerism)

Answer 56

alkynes can be oxidatively cleaved with either hot, basic potassium permanganate, KMnO4 (followed by acidification) or with ozone. Both alkenes and alkynes will give the same product when reacted with hot, basic KMnO4. The difference here is with ozone. Remember that when alkenes react with ozone under reducing conditions (Zn/CH3COOH), they yield • Aldehydes or ketones. However, under oxidizing conditions (H2O2), carboxylic acids are obtained instead of aldehydes. When alkynes are reacted with ozone, they yield carboxylic acids or CO2. Note that triple bonds (with two π -bonds) will add two oxygen’s to each carbon. Terminal alkynes produce CO2. Notice how ozone acts a lot like hot basic KMnO4 when reacting with alkynes.

Answer 57

• A functional group is the part of a molecule where most of its chemical reactions occur. It is the part that effectively determines a compound's chemical properties in addition to many of its physical properties.

Answer 58

Alkyl- non-aromatic hydrocarbon group whereas aryl is an aromatic hydrocarbon group. acyl- Alkanoyl (alkyl + carbonyl). Also referred to as an acyl group has an alkyl group at one end.

Answer 59

* Primary alcohols may be oxidized to aldehydes in the presence of the oxidizing agent PCC (pyridinium chlorochromate). Ketones undergo reduction to secondary alcohols in the presence of NaBH4 (sodium borohydride) * Protonation of an Alcohol- can act as either electrophile or nucleophile, when alcohol protonated becomes positively-charge, good leaving group and participate in electrophile, the alcohol acts as a nucleophile. Because of the polarized C-O bond, the oxygen has a partial negative charge and will acquire a proton from HBr. Once protonated, the molecule bears a good leaving group in H2O. The result is formation of a carbocation, which is electron-deficient. This carbocation will react electrophilically with Br-.

Answer 60

• Alkyl halide acts as electrophile because of polar halogen, alcohol acts as nucleophile because of oxygen lone pairs, alkene nucleophile because pi cloud is electron-rich, alkyne is nucleophile, aromatic is nucleophilic, carbonyl is electrophilic because of polar C=O, amine is nucleophilic because of nitrogen lone pairs.

Answer 61

the boiling points of alcohols are significantly higher than others due to hydrogen bonding, hydrogen attached to highly electronegative atoms. with phenol have the most acidic alcohol with the highest melting and boiling points and is slightly soluble in water,

Answer 62

alkyl groups donate electron density, they help stabilize a positive charge but will destabilize a negative charge. Acidity decreases as more alkyl groups (electron donating) are attached because they destabilize the alkoxide anion. Resonance or electron-withdrawing groups stabilize the alkoxide anion, making the alcohol more acidic. More bonds to oxygen = more oxidized. Remember that a double bond counts as two.

Answer 63

Reduced, or oxidized Reducing agents have a lot of Hs (NaBH4 and LiAlH4), and oxidizing agents have a lot of Os (KMnO4 and CrO3). When you see a transition metal (such as Cr or Mn) with lots of oxygen (Na2Cr2O7, CrO3, or KMnO4), think oxidation. 1) Nucleophilic substitution: SN1, SN2. • 2. Electrophilic addition to a double bond • 3. Nucleophilic addition to a carbonyl Alcohols can participate in SN1/SN2 reactions but only if you turn the – OH into a better leaving group by one of the following methods: • Protonate it • Convert to a tosylate. • Form an inorganic ester. • Phenols good for EAS- the lone pair from the oxygen makes it an electron-donating substituent. Therefore, the – OH will be activating and ortho/para directing.

Answer 64

* Addition Reactions- addition of water to double bonds, an addition reaction that prepared alcohols. Alcohols can also be prepared from the addition of organometallic compounds to carbonyl groups; * Substitution Reactions- Both SN1 and SN2 reactions can be used to produce alcohols under the proper conditions, * Reduction Reactions- Alcohols can be prepared from the reduction of aldehydes, ketones, carboxylic acids, or esters. Lithium aluminum hydride (LiAlH4, or LAH) and sodium borohydride (NaBH4) are the two most common reducing reagents, AH is the powerful one, and it will reduce just about anything (even esters, amides, and carboxylic acids) all the way to an alcohol. NaBH4 is weaker, so although it, too, will reduce aldehydes, ketones, or acyl chlorides, it cannot reduce esters, carboxylic acids, or amides. * Phenol Synthesis- Phenols may be synthesized from arylsulfonic acids with hot NaOH. However, this reaction is useful only for phenol or its alkylated derivatives, as most other functional groups are destroyed by the harsh reaction conditions. A more versatile method of synthesizing phenols is by the hydrolysis of diazonium salts

Answer 65

* Elimination Reactions- Alcohols can be dehydrated in a strongly acidic solution (usually H2SO4) to produce alkenes. The mechanism of this dehydration reaction is E1 for secondary and tertiary alcohols but E2 for primary alcohols. We need an acidic solution so that the – OH group can be protonated and converted to a good leaving group. * milder method employs POCl3 (phosphorus oxychloride), which follows an E2 mechanism for primary and secondary alcohols. Again, it converts the – OH group into a good leaving group. * Substitution Reactions- displacement of hydroxyl groups in substitution reactions is rare because the hydroxide is on is a poor leaving group. If such a transformation is desired, the hydroxyl group must be made into a good leaving group. As we said before, protonating the alcohol makes water a good leaving group for SN1 reactions. Even better, the alcohol can be converted into a tosylate (p-toluenesulfonate) group, which is an excellent leaving group for SN2 reactions. * conversion of alcohols to alkyl halides. A common method involves the formation of inorganic esters, which readily undergo SN2 reactions. Alcohols react with thionyl chloride to produce an intermediate inorganic ester (a chlorosulfite) and pyridine. The chloride ion, through an SN2 mechanism, attacks the backside of the carbon bearing the oxygen and the chlorosulfite group. The reaction generates SO2 and Cl– , forming the desired alkyl chloride with inversion of configuration. An analogous reaction to this, where the alcohol is treated with PBr3 (in pyridine) instead of thionyl chloride, produces alkyl bromides. In both cases, as with tosylates, the poor alcohol-leaving group is converted to a good leaving group. * Phenols readily undergo electrophilic aromatic substitution reactions because the lone pairs on the oxygen donate electron density to the ring. This means the – OH group is strongly activating and, thus, an ortho/para-directing ring substituent * Oxidation Reactions- The oxidation of alcohols generally involves some form of chromium (VI) as the oxidizing agent, which is reduced to chromium (III) during the reaction. Every oxidizing agent we discuss here is a strong oxidizing agent (will convert a primary alcohol into a carboxylic acid) except for PCC. PCC (pyridinium chlorochromate, C5H6NCrO3Cl) is a “ mild” (anhydrous) oxidant, which means it only partially oxidizes primary alcohols. It stops after the primary alcohol has been converted to an aldehyde because PCC lacks the water necessary to hydrate the aldehyde (aldehydes are easily hydrated). When aldehydes are hydrated (geminal diols or 1,1-diols), they can be oxidized to carboxylic acids. PCC will also form ketones from 2° alcohols, so the only difference between PCC and all of the other oxidizing agents is how they react with 1° alcohols. Tertiary alcohols are already as oxidized as they can be and so do not react with any of the oxidizing agents. * Another reagent used to fully oxidize primary and secondary alcohols is an alkali (either sodium or potassium) dichromate salt. This means it will oxidize 1° alcohols to carboxylic acids and secondary alcohols to ketones * An even stronger oxidant is chromium trioxide, CrO3. This is often dissolved with dilute sulfuric acid in acetone, a reaction called Jones's oxidation * Treatment of phenols with oxidizing reagents produces compounds called quinones (2,5-cyclohexadiene-1,4-diones)

Answer 66

An ether is a compound with two alkyl (or aryl) groups bonded to an oxygen atom. The general formula for ethers is ROR. Ethers can be thought of as disubstituted water molecules. important solvents, but ethers are aprotic and unreactive. * An ether is a compound with two alkyl (or aryl) groups bonded to an oxygen atom. The general formula for ethers is ROR. Ethers can be thought of as disubstituted water molecules. important solvents, but ethers are aprotic and unreactive. * Nomenclature- alkoxyalkanes, with the smaller chain as the prefix and the larger chain as the suffix. There is also a common system of nomenclature in which ethers are named as alkyl ethers, with the substituents alphabetized. For example, methoxyethane would be named ethyl methyl ether. Exceptions to these rules occur for cyclic ethers, as there aren't two different alkyl groups attached to them. Remember that smaller rings have more angle strain, making them less stable and more reactive. * Physical Properties- Note that the lack of hydrogen bonding is significant when determining the physical properties of ethers. do not undergo hydrogen bonding. Ethers have no hydrogen atoms bonded to the oxygen atoms, so they can't possibly participate in hydrogen bonding. Ethers therefore boil at relatively low temperatures compared with alcohols; in fact, they boil at approximately the same temperatures as alkanes of comparable molecular weight. Ethers are only slightly polar and, therefore, only slightly soluble in water. They are inert to most organic reagents and so are frequently used as solvents.

Answer 67

SN2 need a strong nucleophile (alkoxides are good) and an unhindered substrate (not bulky or highly substituted). Williamson Ether synthesis- produces ethers from the reaction of metal alkoxides with primary alkyl halides or tosylates. The alkoxides behave as nucleophiles and displace the halide or tosylate via an SN2 reaction, producing an ether. With alkoxides only attacking nohindered halides. the reaction cannot be accomplished with the methoxide ion attacking a bulky alkyl halide substrate. Can be used on phenols with relatively mild reaction conditions sufficient because of their acidity •Cleavage of straight chain ethers is acid-catalyzed. Cleavage of epoxides can be acid-catalyzed (the nucleophile — e.g., H2O, ROH attacks the more substituted carbon of the epoxide) or base-induced (the nucleophile e.g., RMgX, LiAlH4, OH− attacks the least substituted carbon of the epoxide). Base-induced cleavage has mostly SN2 character, whereas acid-catalyzed cleavage seems to have some SN1 character.

Answer 68

* Cyclic ethers can be prepared in a number of ways, but you are likely to see it via internal SN2 displacement. Intramolecular reactions are favored because, as we know, the rate and equilibrium of the reaction are affected by the reagent concentrations. With intramolecular reactions, the reagents encounter fairly high concentrations of each other; they are basically tied together * Another way to make cyclic ethers is by the oxidation of an alkene with a peroxy acid (general formula RCOOOH) such as mCPBA (m-chloroperoxybenzoic acid). This reaction will also produce an epoxide or oxirane,

Answer 69

Peroxide Formation- Ethers react with the oxygen in air to form highly explosive compounds called peroxides (general formula ROOR), don't need to know mechanism Cleavage- Cleavage of straight-chain ethers will take place only under vigorous conditions, usually at high temperatures in the presence of HBr or HI. Cleavage is initiated by protonation of the ether oxygen. The reaction then proceeds by an SN1 or SN2 mechanism, depending on the conditions and the structure of the ether. alcohol products usually react with a second molecule of hydrogen halide to produce an alkyl halide. Because epoxides are highly strained cyclic ethers, they are ready to react and, thus, susceptible to SN2 reactions. Unlike straight-chain ethers, these reactions can be catalyzed by acid or reacted with base (nucleophiles), symmetrical epoxides, either carbon can be nucleophilically attacked. However, in asymmetrical epoxides, the most substituted carbon is nucleophilically attacked when catalyzed with acid, and the least substituted carbon is attacked with a nucleophile under basic conditions. Base (nucleophile) induced cleavage has mostly SN2 character, so it occurs at the least hindered (least substituted) carbon. Because the environment is basic, it provides a better nucleophile than an acidic environment. In contrast, acid-catalyzed cleavage is thought to have some SN1 character and some SN2 character. The epoxide's oxygen is protonated, converting it to a better leaving group. As a result, the carbons share a bit of the positive charge. Because substitution stabilizes this charge (remember, 3° carbons make the best carbocations), the more substituted carbon becomes a good target for nucleophilic attack.

Answer 70

* The only difference between the two is what is attached to the carbonyl. A ketone has two alkyl (or aryl) groups bonded to the carbonyl, whereas an aldehyde has one alkyl (or aryl) group and one hydrogen. This hydrogen tells us that the aldehyde must be at an end of the chain, or at the terminal position. An aldehyde is a terminal functional group; it defines the C– 1 position. A ketone, on the other hand, will always be mid-chain and can never be a terminal functional group. * Nomenclature: aldehyde-al, if the aldehyde is attached to a ring, we use the suffix – carbaldehyde. Second, if the aldehyde does not hold priority in the molecule, we name it as a substituent with the prefix formyl– . * Ketones are quite logically named with the suffix – one. As opposed to aldehydes, the location of the carbonyl group on a ketone must be specified with a number, with the exception of propanone (acetone), butanone, and cyclic ketones (where it is assumed that the carbonyl occupies the number one position). Naming ketones with the common system of nomenclature is similar to naming ethers. We simply list the two-alkyl groups alphabetically, followed by the word ketone. When it is necessary to name the ketone as a substituent, we use the prefix Oxo– * form– will be seen also in formic acid (a one-carbon carboxylic acid), and acet– is seen in many two-carbon compounds (acetylene, acetic acid, and acetyl CoA).

Answer 71

carbonyl has the unique ability to behave as either a nucleophile (as in condensation reactions) or an electrophile (as in nucleophilic addition reactions), essentially taking on opposite characteristics. This gives the carbonyl more opportunities to react with other molecules. dipole of a carbonyl is greater than the dipole of an alcohol because the carbonyl lacks a hydrogen, which would cancel some of the C– O dipole. In solution, the dipole moments associated with these polar carbonyl groups line up, causing an elevation in boiling point relative to their associated alkanes. However, even though the dipoles are more polar than alcohols, the elevation in boiling point is less than that in alcohols because no hydrogen bonding is involved. In general, aldehydes are more reactive towards nucleophiles than ketones. The carbonyl group (in aldehydes and ketones) has a dipole moment. Oxygen is more electronegative— it is an “ electron hog,” pulling the electrons away from the carbon, making the carbon electrophilic. While the dipole moments in the carbonyl increase their intermolecular forces and boiling points relative to alkanes, it is not as significant as the effect of hydrogen bonding seen in alcohols.

Answer 72

* an aldehyde can be obtained from the partial oxidation of a primary alcohol, and a ketone can be obtained from the oxidation of a secondary alcohol. Primary alcohols can be carefully oxidized to aldehydes. Secondary alcohols can be oxidized to ketones. he only reagent you will see that can oxidize a primary alcohol to an aldehyde (and not all the way to a carboxylic acid) is PCC (a dry, nonhydrating oxidizing reagent). We have many more options to oxidize a secondary alcohol into a ketone, because there is no risk of oxidizing too far. Sodium or potassium dichromate, chromium trioxide (Jones's reagent), or even PCC will perform this oxidation quite well. * Alternatively, double bonds can be oxidatively cleaved to form aldehydes and/or ketones. Whether you get an aldehyde or ketone depends on whether you start with a mono- or disubstituted double bond. This is another one of those reactions where the name tells us almost everything we need to know: Ozonolysis breaks double bonds using ozone. it produces aromatic ketones (aldehydes if R = H) in the form of R-CO-Ar.

Answer 73

Enolization and reactions of enols- Aldehydes and ketones exist in the traditional keto form (C=O) and as the less common enol tautomer (enol = ene + ol). The enol form can act as a nucleophile. Alpha protons (protons attached to the carbons adjacent to carbonyls) are relatively acidic (pKa ≈ 20) owing to resonance stabilization of the conjugate base. push the leftover electrons up toward the oxygen and see how the extra negative charge can be spread among several atoms. The hydrogen atom that detaches from the α -carbon has a good probability of reattaching to the partially negative oxygen instead of the carbon. Therefore, aldehydes and ketones exist in solution as a mixture of two isomers, the familiar keto form, and the enol form, representing the unsaturated alcohol (ene = the double bond, ol = the alcohol, so ene + ol = enol). The two isomers, which differ only in the placement of a proton (and the double bond), are called tautomers. The equilibrium between the tautomers lies far to the keto side, so there will be many more keto-isomers in solution. enolization or, less specifically, tautomerization. Not resonance structures. • Enols are important intermediates in many reactions of aldehydes and ketones. The enolate carbanion, which acts as a nucleophile, can be created with a strong base, such as lithium diisopropyl amide (LDA) or potassium hydride (KH). A 1,3-dicarbonyl is extra acidic because there are two carbonyls to delocalize negative charge and, as such, is often used to make the carbanion. Once formed, the nucleophilic carbanion reacts via an SN2 mechanism with alkyl halides (a favorite of most nucleophiles) or α , β -unsaturated carbonyl compounds in reactions called Michael additions. In this reaction, the carbanion attaches to the unsaturated carbonyl at the β -position owing to its resonance forms.

Answer 74

In acid-catalyzed hydration reactions, H2O is the nucleophile, and the carbonyl carbon is the electrophile. formation of acetals and ketals, alcohol is the nucleophile and the carbonyl carbon is the electrophile. • Addition reactions- nucleophilic addition to a carbonyl. When a nucleophile attacks, it forms a covalent bond to the carbon, breaking the π bond in the C=O. The electrons from the π bond are pushed up onto the oxygen atom, which, being more electronegative, is relatively happy to hold the negative charge for a while, generating a tetrahedral intermediate. If no good leaving group is present, the carbonyl will not re-form, and the final product will be nearly identical to the intermediate, except that the – O– will usually accept a proton to become a hydroxyl (– OH) group. However, if a good leaving group is present, the carbonyl double bond can re-form and push off the leaving group • Hydration- In the presence of water, aldehydes and ketones react to form geminal diols (1,1-diols), the nucleophilic oxygen in water attacks the electrophilic carbonyl carbon. This hydration reaction normally proceeds slowly, but we can increase the rate by adding a small amount of acid or base.

Answer 75

* Acetal and Ketal Formation- When one equivalent of alcohol (the nucleophile in this reaction) is added to an aldehyde or ketone, the product is a hemiacetal or a hemiketal. hemiacetal or hemiketal because it will still contain one hydroxyl group. Because the reactions here have only proceeded halfway, the products are termed hemi. In basic conditions, the * reaction would stop here. When two equivalents of alcohol are added, the reaction proceeds all the way and the product is an acetal or a ketal. Because aldehydes contain a hydrogen on the carbonyl, both hemiacetals and acetals will still contain a hydrogen as a distinguishing characteristic. The reaction proceeds via the same mechanism as hydration, and it is catalyzed by anhydrous acid. Acetals and ketals, which are comparatively inert, are frequently used as protecting groups for carbonyl functionalities. Ethylene glycol is a popular protecting group, as it is a diol and both alcohol groups come from the same molecule. Molecules with protecting groups can easily be converted back to carbonyls with aqueous acid and heat. • Reactions with Hydrogen Cyanide (HCN)- Hydrogen cyanide is a classic nucleophile. alkynes are fairly acidic because of their triple bonds. Well, HCN has both a triple bond and an electronegative nitrogen atom, so it is even more acidic (pKa ≈ 9.2). After the hydrogen dissociates, the nucleophilic cyanide anion can attack the carbonyl carbon atom. Reactions with aldehydes and ketones produce stable compounds called cyanohydrins (once the oxygen has been reprotonated). The cyanohydrin gains its stability from the newly formed C– C bond. (In contrast, when a carbonyl reacts with HCl, a weak C– Cl bond is formed, and the resulting chlorohydrin isn't very stable.)The ylide can act as a nucleophile and attack the carbonyl.

Answer 76

• Reactions with Hydrogen Cyanide (HCN)- Hydrogen cyanide is a classic nucleophile. alkynes are fairly acidic because of their triple bonds. Well, HCN has both a triple bond and an electronegative nitrogen atom, so it is even more acidic (pKa ≈ 9.2). After the hydrogen dissociates, the nucleophilic cyanide anion can attack the carbonyl carbon atom. Reactions with aldehydes and ketones produce stable compounds called cyanohydrins (once the oxygen has been reprotonated). The cyanohydrin gains its stability from the newly formed C– C bond. (In contrast, when a carbonyl reacts with HCl, a weak C– Cl bond is formed, and the resulting chlorohydrin isn't very stable.)The ylide can act as a nucleophile and attack the carbonyl.

Answer 77

* an aldehyde acts both as an electrophile (in its keto form) and a nucleophile (in its enol or enolate form). For example, when acetaldehyde (ethanal) is treated with a catalytic amount of base, an enolate ion is produced. The enolate is more nucleophilic than the enol because it is negatively charged. This nucleophilic enolate ion can react with the carbonyl group (an electrophile) of another acetaldehyde molecule. The key is that you have both species in the same flask; this is why you do not convert all of your aldehyde into an enolate. The product is 3-hydroxybutanal, which contains both alcohol and aldehyde * functional groups. This type of compound is called an aldol, from aldehyde and alcohol. With a strong base and higher temperatures, dehydration occurs: We kick off a water molecule and form a double bond, producing an α , β -unsaturated aldehyde one type of aldehyde or ketone. If there are multiple aldehydes or ketones, we can't easily control which will act as the nucleophile and which will act as the electrophile, and a mixture of products will result unless one of the molecules is missing an α -hydrogen (like benzaldehyde).

Answer 78

* The Wittig reaction ultimately converts C=O to C=C (aldehydes/ketones to alkenes). the Wittig reaction forms carbon– carbon double bonds by converting aldehydes and ketones into alkenes. formation of a phosphonium salt from the SN2 reaction of an alkyl halide with the nucleophile triphenylphosphine, (C6H5)3P. This compound is simply a phosphorus atom that has three aromatic phenyl groups attached to it. With its lone pairs and the added electron density from the phenyl groups, the phosphorus makes a great nucleophile and readily attacks the partially positive carbon on the alkyl halide. This phosphonium salt is then deprotonated (losing the proton α to the phosphorus) with a strong base, yielding a neutral compound called an ylide (pronounced “ ill-id” ) or phosphorane. The ylide form is a zwitterion (a molecule with both positive and negative charges), and the phosphorane form has a double bond between carbon and phosphorus. (The phosphorus atom may be drawn as pentavalent because it can use its low-lying 3d atomic orbitals.) * Notice that an ylide is a type of carbanion and, thus, has nucleophilic properties. When combined with an aldehyde or ketone, an ylide attacks the carbonyl carbon, giving an intermediate called a betaine (a specific kind of zwitterion), which forms a four-membered ring with an ionic bond between the oxygen and the phosphorus. This ringed intermediate is called an oxaphosphetane, and it decomposes to yield an alkene and triphenylphosphine oxide

Answer 79

* The Wittig reaction ultimately converts C=O to C=C (aldehydes/ketones to alkenes). the Wittig reaction forms carbon– carbon double bonds by converting aldehydes and ketones into alkenes. formation of a phosphonium salt from the SN2 reaction of an alkyl halide with the nucleophile triphenylphosphine, (C6H5)3P. This compound is simply a phosphorus atom that has three aromatic phenyl groups attached to it. With its lone pairs and the added electron density from the phenyl groups, the phosphorus makes a great nucleophile and readily attacks the partially positive carbon on the alkyl halide. This phosphonium salt is then deprotonated (losing the proton α to the phosphorus) with a strong base, yielding a neutral compound called an ylide (pronounced “ ill-id” ) or phosphorane. The ylide form is a zwitterion (a molecule with both positive and negative charges), and the phosphorane form has a double bond between carbon and phosphorus. (The phosphorus atom may be drawn as pentavalent because it can use its low-lying 3d atomic orbitals.) * Notice that an ylide is a type of carbanion and, thus, has nucleophilic properties. When combined with an aldehyde or ketone, an ylide attacks the carbonyl carbon, giving an intermediate called a betaine (a specific kind of zwitterion), which forms a four-membered ring with an ionic bond between the oxygen and the phosphorus. This ringed intermediate is called an oxaphosphetane, and it decomposes to yield an alkene and triphenylphosphine oxide

Answer 80

* carboxyl group (a carbonyl attached to a hydroxyl group) they're acids, so they like to give away protons. Second, when a carboxylic acid does give away a proton, the leftover electrons are resonated between two oxygen atoms, making the molecule more likely to give off the proton in the first place (pKa = 3– 6). Third, a carboxylic acid has a hydrogen bond donor and acceptor in the same functional group, leading to large intermolecular forces and high boiling points. Fourth, the carbonyl carbon is electropositive, so it makes a great electrophile. Fifth, carboxylic acids, a terminal functional group, have the highest priority in nomenclature, so we're talking about the top dog here. Sixth and most important, carboxylic acids occur widely in nature and are synthesized by all living organisms, including you. * carboxylic acids have the highest priority, so they are always named by adding the suffix – oic acid to the alkyl root. This also means that the chain is always numbered so that the carboxyl group receives the lowest possible number. Additional substituents are named in the usual fashion form– for one carbon and acet– for two. Cyclic carboxylic acids are usually named as cycloalkane carboxylic acids. The carbon atom to which the carboxyl group is attached is numbered 1, and all other groups are given the lowest possible numbers. Salts of carboxylic acids are named beginning with the cation, followed by the name of the acid with the ending – ate replacing – ic acid. Two carboxylic acid groups- first six straight-chain terminal dicarboxylic acids are oxalic (2C), malonic (3C), succinic (4C), glutaric (5C), adipic (6C), and pimelic (7C) acids. Their IUPAC names are ethanedioic acid, propanedioic acid, butanedioic acid, pentanedioic acid, hexanedioic acid, and heptanedioic acid

Answer 81

hydrogen bonding- Carboxylic acids are polar and form hydrogen bonds with each other. Not only can carboxylic acids hydrogen bond, they can hydrogen bond really well, because there are two different sites that can participate in hydrogen bonding. As a result, carboxylic acids form dimers: pairs of molecules connected by two hydrogen bonds. Multiple hydrogen bonds elevate the boiling and melting points of carboxylic acids even higher than those of the corresponding alcohols. As usual, the boiling points also increase with increasing molecular weight. hydrogen bonding- Carboxylic acids are polar and form hydrogen bonds with each other. Not only can carboxylic acids hydrogen bond, they can hydrogen bond really well, because there are two different sites that can participate in hydrogen bonding. As a result, carboxylic acids form dimers: pairs of molecules connected by two hydrogen bonds. Multiple hydrogen bonds elevate the boiling and melting points of carboxylic acids even higher than those of the corresponding alcohols. As usual, the boiling points also increase with increasing molecular weight.

Answer 82

acidity is due to resonance stabilization of the carboxylate anion, conjugate base and can be enhanced by adding electronegative groups or other potential resonance structures. The more stable the conjugate base is, the more likely the proton is to leave, and thus, the stronger the acid. Other ways to stabilize the negative charge (and thus increase acidity) are: Electron-withdrawing groups (e.g., halides), Groups that allow more resonance stabilization (e.g., benzyl or allyl substituents), Substituents on carbon atoms near a carboxyl group will influence its acidity. Electron-withdrawing groups, such as – Cl or – NO2, further absorb the negative charge and increase acidity. Electron-donating groups, such as – NH2 or – OCH3, donate additional electron density and destabilize the negative charge, making the compound less acidic. The more of these groups that exist and the closer they are to the acid, the stronger the acid is. When the hydroxyl proton dissociates from the acid, the negative charge left on the carboxylate group is delocalized between the two oxygen atoms. In dicarboxylic acids, one – COOH group (which is electron withdrawing owing to the partial positive charge on carbon) influences the other, making the compound more acidic than the analogous monocarboxylic acid. The catch here is that once the proton leaves and the carboxylate anion has formed, it will make the second carboxyl group less acidic. Think about it: If the second group were deprotonated, it would create a doubly charged species in which the two negative charges repel each other. Because this is unfavorable, the second proton is even less acidic than the alpha proton of a monocarboxylic acid. Beta-dicarboxylic acids are notable for the high acidity of the α -hydrogens located on the carbon between the two-carboxyl groups (pKa∼ 10). Loss of this acidic hydrogen atom produces a carbanion, which is stabilized by the electron-withdrawing effect of two carboxyl groups. This acidity also applies to the α -hydrogens of β -ketoacids and other molecules that share this 1,3-dicarbonyl structure.

Answer 83

oxidation reactions- Carboxylic acids (most oxidized) can be prepared via oxidation of aldehydes, primary alcohols, and certain alkyl benzenes. The oxidant is usually potassium permanganate, KMnO4. secondary and tertiary alcohols cannot be oxidized to carboxylic acids because of valence limitations. the nucleophile is essentially a carbanion that is coordinated with a positively charged magnesium, and the electrophile is the carbon of the CO2 (which is similar to any other carbonyl or carboxylic carbon).

Answer 84

Organometallic reagents, such as Grignard reagents, react with carbon dioxide (CO2) to form carboxylic acids. This reaction is useful for the conversion of tertiary alkyl halides into carboxylic acids, which, as we just mentioned, cannot be accomplished through other methods. Note that this reaction adds one carbon atom to the chain (because we are adding the CO2 and not just oxidizing a carbon already on the molecule).

Answer 85

• Nitriles, also called cyanides, are compounds containing the functional group – C≡ N. The cyanide anion (– C≡ N) carries the negative charge on the carbon atom, making it a great nucleophile but not a great base. It will displace primary and secondary halides in typical SN2 fashion. Nitriles can then be hydrolyzed under either acidic or basic conditions, producing carboxylic acids and ammonia (or ammonium salts) conversion of alkyl halides into carboxylic acids, additional carbon introduced

Answer 86

When long-chain carboxylic acids react with sodium or potassium hydroxide, they form salts. This can be done in practice by mixing fat (triglycerides: three carboxylic acids connected by a glycerol) with lye (sodium hydroxide). These salts (which we call soaps) are useful because they can solvate nonpolar organic compounds in aqueous solutions since they possess both a nonpolar tail and a polar carboxylate head. placed in aqueous solution, soap molecules arrange themselves into spherical structures called micelles. polar heads face outward, where they can be solubilized by water, and the nonpolar hydrocarbon chains are oriented toward the inside of the sphere, protected from the solvent. Nonpolar molecules, such as grease, can dissolve in the hydrocarbon interior of the spherical micelle, whereas the micelle as a whole is hydrophilic owing to its polar shell. Thus, when you wash your hands, the soap molecules arrange themselves in a micelle around nonpolar dirt and grease, and the whole micelle dissolves in water and rinses off.

Answer 87

many of the reactions in which carboxylic acids (and their derivatives) participate proceed via a single mechanism: nucleophilic acyl substitution. This mechanism is similar to nucleophilic addition to a carbonyl Nucleophilic substitution concludes with re-formation of the C=O double bond and elimination of a leaving group

Answer 88

LiAlH4 and not the less reactive NaBH4 can only reduce reduction- Carboxylic acids and esters. Protonating the C=O makes the C even more ripe for nucleophilic attack. Acid chlorides are among the highest-energy (least stable and most reactive) members of the carbonyl family. carboxylic acids occupy the most oxidized end of the oxidation– reduction continuum. That means that they can't be f further oxidized. However, carboxylic acids can definitely be reduced. Lithium aluminum hydride (LAH, LiAlH4) reduces carboxylic acids to their corresponding alcohols. Aldehyde intermediates may be formed in the course of the reaction, but they, too, will be reduced to the alcohol. The reaction occurs by nucleophilic addition of hydride to the carbonyl group.

Answer 89

we react carboxylic acids with alcohols under acidic conditions. This is a condensation reaction, so water is a side product. n an acidic solution, the O on the C=O can be protonated. This enhances the polarity of the bond, putting more positive charge on the C and making it even more susceptible to nucleophilic attack. This condensation reaction occurs most rapidly with primary alcohols.

Answer 90

acid halides, are compounds with carbonyl groups bonded to halides. Several reagents can convert carboxylic acids into acyl halides, but thionyl chloride, SOCl2. acid chlorides are reactive. The greater electron withdrawing power of the – Cl− makes the carbonyl carbon even more susceptible to nucleophilic attack than the carbonyl carbon of a carboxylic acid. As such, acid chlorides are frequently used as intermediates in the conversion of carboxylic acids to esters and amides

Answer 91

losing a carbon. 1,3-Dicarboxylic acids and other β -keto acids may spontaneously decarboxylate when heated. Under these conditions, the carboxyl group is lost and replaced with hydrogen. Because both the electrophile and nucleophile are in the same molecule, the reaction proceeds through a six-membered ring transition state the enol that is initially formed from the destruction of the ring tautomerizes to the more stable keto form.

Answer 92

acyl halides, anhydrides, amides, and esters. Each of these molecules replaces the – OH on the carboxyl group with – X, – OCOR, – NH2, or – OR, respectively. Order of carboxylic acid derivative reactivity: acyl halides > anhydrides > esters = carboxylic acids > amides. The carbonyl carbon acts as an electrophile and is attacked by a nucleophile. In the second step, the leaving group takes the extra electrons. Note: Aldehydes and ketones do not have a leaving group. That is why they undergo nucleophilic additions.

Answer 93

Nomenclature- acid or alkanoyl halides, acyl group is RCO–. halogen leaving group, acyl halides are the most reactive of the carboxylic acid derivatives. oic acid ending of the carboxylic acid to – oyl halide. • Syntehesis: most common acyl halides are acid chlorides, although you may occasionally encounter acid bromides and iodides. Acid chlorides are prepared by reacting a carboxylic acid with thionyl chloride, SOCl2. SO2 and HCl are the other products, and the evolution of SO2 drives this reaction. Alternatively, PCl3 or PCl5 (or PBr3, to make an acid bromide) will accomplish the same transformation. Since acyl halides are the most reactive carboxylic acid derivative • reactions: nucleophilic acyl substitution: Hydrolysis- eact rapidly with water to form their carboxylic acid and HCl. The electrophile, of course, is the carbonyl carbon and the nucleophile is the benzene ring. So, we have either a nucleophilic acyl substitution or an electrophilic aromatic substitution, depending on your perspective. Conversion into Esters- alcohol as the nucleophile. The leaving group is still chlorine, which can pick up a hydrogen in solution, making HCl as a side product. Conversion into Amides- Acyl halides can be converted into amides (compounds of the general formula RCONR2) by an analogous reaction with amines. The lone pairs on nucleophilic amines, such as ammonia, attack the carbonyl group, displacing chloride. The side product is ammonium chloride (a salt— remember, this is an ionic compound), formed from excess ammonia and HCl. Whereas acyl halides react with amines to form amides, if we were to react ketones with amines, the product would be an imine • Friedel-Crafts Acylation- aromatic rings acylated by electrophilic aromatic substitution, with nucleophile the aromatic ring, and electrophile is carbonyl. when the acyl chloride attacks, its bond to chlorine is almost completely broken by the reaction with the Lewis acid AlCl3. The pi electrons of the aromatic system act as a nucleophile, attacking the electrophilic acyl cation, or acylium ion (RCO+) with the product a alkyl aryl ketone. • Reduction- can be reduce to alcohols or selectively reduce to intermediate aldehydes using a bulky hydride reagent with only one hydride to transfer: LiAlH(OC(CH3)3)3. This will prevent the reducing agent from reducing the compound all the way to an alcohol.

Answer 94

* Anhydride means “ without water.” Anhydrides are formed from the condensation of two acid molecules (with loss of water). Nomenclature- acid anhydrides, condensation dimers of carboxylic acids with the general formula RCOOCOR, symmetrical anhydrides have anhydride substituted for acid, anhydrides are asymmetrical, you simply name the two chains alphabetically, followed by “ anhydride”. succinic, maleic, and phthalic anhydrides are cyclic anhydrides arising from intramolecular condensation or dehydration of diacids. * Synthesis: The carbonyl compound is the electrophile, the nucleophile is water, and the leaving group is a carboxylic acid (which is not as good a leaving group as the halogens in the acyl halide reactions). anhydrides are the product of a condensation reaction between two carboxylic acids. The hydroxide group of one acid acts as the nucleophile, and (as always) the carbonyl is the electrophile. One molecule of water is lost in the condensation. Anhydrides can also be synthesized by the reaction of an acid chloride and a carboxylate anion * convert compounds to amides are the same as all the others we've seen; they only differ in the attacking nucleophile. Simply heating carboxylic acids can form certain cyclic anhydrides. reaction is driven forward by the increased stability of the newly formed ring; as such, only five- and six-membered ring anhydrides are easily made. Just as with all anhydride formations, the hydroxyl group of one – COOH acts as the nucleophile, attacking the carbonyl on the other – COOH. * Acid chlorides can be converted into any of the other derivatives, anhydrides can make any of the less reactive derivatives (esters and amides), and esters can be converted into amides. * Reactions: Anhydrides react under the same conditions as acid chlorides, but because they are more stable, they are less reactive. The reactions are slower, and no matter what the nucleophile, they will produce a carboxylic acid side product instead of the HCl produced by acid halides. Cyclic anhydrides are also subject to the following reactions, which cause ring opening at the anhydride group along with formation of the new functional groups hydrolysis- we can break up anhydrides into two equivalents of carboxylic acids by exposing them to water. For these reactions to be useful, the anhydride must be symmetric, otherwise we would get a mixture of products.

Answer 95

* Conversion into Amides: Anhydrides can also be cleaved by ammonia, producing amides and carboxylic acid. One of our products is a carboxylic acid, and we're carrying out the reaction in an environment filled with ammonia. That means we've now got an acid in a basic environment. The two will react, forming a salt, specifically ammonium carboxylate. final products will be an amide and the ammonium salt of a carboxylate anion. * Conversion into Esters and Carboxylic Acids-reaction will form esters and carboxylic acids * Acylation- when we add AlCl3 or another Lewis acid catalyst the reaction will occur readily. our Lewis acid catalyst binds with the entire RCOO group of the anhydride instead of with just the chloride of an acyl chloride

Answer 96

RCONR2, -amid, alkyl substituents listed as prefixes, specified by N. peptide bond is an amide linkage that possesses double-bond character from resonance and is the most stable carboxylic acid derivative. Synthesized by acid chlorides with amines, or acid anhydride and ammonia. Loss of hydrogen required which only primary and secondary amines undergo. • Reactions- amides, most stable of carboxylic acid derivatives, bound to carbonyl takes extreme acid or base to break amide linkage. Hydrolysis- use nucleophilic substitution under acidic conditions. Allow carbonyl oxygen to be protonated more susceptible to nucleophilic attack by water molecule with a product of carboxylic and ammonia. Or strong not protonated and nucleophile is hydroxide product deprotonated and is carboxylate ion. • Hofmann Rearrangement=amide to primary amine with the loss of a carbon as CO2. Initial reactants bromine and sodium hydroxide reacting to form sodium hypobromite involving formation of nitrene, nitrogen analog of carbine, nitrene attached to carbonyl only six electrons, electrophile looking for more electrons, electron deficiency resolved by rearrangement, R group bound to carbonyl carbon shifted directly to electron-deficient nitrene. Stabilizing nitrogen resulting in another double bond to carbonyl carbon, forming isocyanate, double bond on either side of the carbon one to oxygen and one to nitrogen, isocyanate molecule is hydrolyzed to form amine, with CO2 as leaving group. • Reduction- amides reduced with lithium aluminum hydride (LAH) to corresponding amine, no loss of carbon,

Answer 97

• acid +alcohol, alkyl or aryl alkanoates. Synthesis- acidic conditions mixtures of carboxylic acids and alcohols condensing, losing water into esters- Fischer esterification or obtained from reaction of acid chlorides or anhydrides with alcohols phenolic aromatic esters produced, although aromatic acid chlorides less reactive than aliphatic nonaromatic acid chlorides, adding base catalyst. Tryacylglycerols esters, fats, long chains of carboxylic acids, fatty acids with glycerol as alcohol and free fatty acids RCOOH, Saponification- process of fats hydrolyzed under basic conditions to produce soaps, acidification of soap regenerates fatty acids. • Reactions: hydrolysis- esters, can be hydrolyzed, hydrolysis of esters produce carboxylic acids and alcohols, esters and carboxylic acids are equally reactive, catalyzing the reaction by using either acidic or basic conditions. Basic- oxygen on C=O not protonated and nucleophiles Oh instead of water. • Grignard reagents, RMgX, are essentially equivalent to R– nucleophiles. • different alcohol attacks an ester, the ester is transformed, and transesterification results. • Claisen condensation- instead of an aldehyde acting as an electrophile and nucleophile in its enol form, ester acting as an electrophile and nucleophile (enolate), keto-ester. Long hydrocarbon chains of lipids synthesized in systems. Acetyl coenzyme A performs function of ethyl acetate, long chain compounds with even numbers more common than built up form units of two carbon atoms. • Conversion into amides- nitrogen bases will attack the electron-deficient carbonyl carbon atom, displacing alkoxide to yield an amide and an alcohol side product, • Transesterification- alcohols act as nucleophiles and displace alkoxy groups on esters, process transforms one ester into another, transesterification. more powerful LAH is necessary to reduce the ester. Remember that NaBH4 is weaker and more selective (aldehydes and ketones) than LiAlH4 but both are essentially equivalent to an H– nucleophile. • Grignard addition- negatively charged carbon Grignard reagent adds to carbonyl group esters, resulting in ketone, carbonyl re-formed and alkoxy kicked off, two equivalents of Grignard reagent used to produce tertiary alcohols with good yield. (The intermediate ketone can be isolated only if the alkyl groups are sufficiently bulky to prevent further attack.) proceeding like nucleophilic substitution second round is nucleophilic addition, carbonyl is turned into an alcohol instead of being re-formed. Phosphodiester bonds should look familiar to you from your studies of molecular biology. • Condensation Reactions- two moles of ethyl acetate react under basic conditions to produce a β -keto-ester, specifically, ethyl 3-oxobutanoate, or acetoacetic ester. reaction proceeds by addition of an enolate anion (created by the basic conditions deprotonating the α carbon) to the carbonyl group of another ester, followed by displacement of an ethoxide ion • reduction- Esters can be reduced to primary alcohols with LAH but not with the weaker NaBH4. allows for selective reduction in molecules with multiple functional groups. • Phosphate Esters- Although phosphoric acid derivatives are not carboxylic acid derivatives, they do form esters. Phosphoric acid and the mono- and diesters are acidic (more so than carboxylic acids), so they usually exist as anions. Like all esters, they can be cleaved under acidic conditions into the parent acid and alcohols. Phospholipids are the main component of cell membranes, and phospholipid/carbohydrate polymers form the backbone of nucleic acids, the hereditary material of life. The nucleic acid derivative adenosine triphosphate (ATP), the fuel that drives our cellular engines, can give up or regain one or more phosphate groups. ATP facilitates many biological reactions by releasing phosphate groups (via hydrolysis) to other compounds, thereby increasing their reactivity. This reaction is downhill in free energy, so it is thermodynamically favorable and drives many biological reactions.

Answer 98

Spectroscopy science of characterizing and identifying compounds based on the degree to which they absorb electromagnetic radiation. measures the energy differences between the possible states of a molecular system by determining the frequencies of electromagnetic radiation (light) absorbed by the molecules. These possible states are quantized energy levels associated with different types of molecular motion, such as molecular rotation, vibration of bonds, nuclear spin transitions, and electron absorption. Different types of spectroscopy measure different types of molecular properties, allowing us to identify the presence of specific functional groups and even to determine how they are connected.

Answer 99

•Infrared (IR) spectroscopy measures molecular vibrations, which can be seen as bond stretching, bending, or combinations of different vibrational modes. To get an IR spectrum, simply pass IR light (2,500 to 25,000 nm) through a sample and record the absorbance pattern. The useful absorptions of IR light occur at wavelengths of 3,000 to 30,000 nm. When light of these frequencies/wavenumbers is absorbed, the molecules enter excited vibrational states Wavenumbers (cm− 1) and are an analog of frequency. In an IR spectrum, percent transmittance is plotted versus frequency, where percent transmittance equals absorbance minus one (%T = A − 1); this means that maximum absorptions appear as the bottom of valleys on the spectrum. Symmetric stretches do not show up in IR spectra because they involve no net change in dipole moment. * More complex vibration patterns, caused by the motion of the molecule as a whole, can be seen in the 1,500 to 400 cm− 1 region. fingerprint region, as it is characteristic of each individual molecule. the vibration must result in a change in the bond dipole moment. This means that molecules that do not experience a changing dipole moment, such as those composed of atoms with the same electronegativity or molecules that are symmetrical, do not exhibit absorption. For example, we cannot get an absorption from O2 or Br2, but we can from HCl or CO. Symmetric bonds will also be silent (the triple bond in acetylene). Infrared spectroscopy is best used for identification of functional groups. The most important peaks to know are those for the – OH (broad peak above 2,900 cm− 1) and carbonyl groups (sharp peak near 1,700 cm− 1). * Characteristic absorptions- alcohol (or anything else with an – OH group), which absorbs around 3,300 cm− 1 with a broad peak, and the second is the carbonyl, which absorbs around 1,700 cm− 1 with a sharp peak (compared with the – OH stretch). bond between any atom and hydrogen always has a relatively high frequency and how, as we add more bonds between carbon atoms, the frequency at which they will absorb increases. N– H bonds are in the same region as O– H bonds, except they have a sharp peak instead of a broad one IR can only tell you what types of bonds are present, not how many. The bonds in organic compounds absorb strongly in the infrared region of the electromagnetic spectrum. Since each type of bond absorbs light at a different, predictable frequency, IR spectroscopy can be used to identify which functional groups are present in a molecule. The x-axis is frequency, measured in cm-1, and ranges from 4000 to 400. The y-axis is transmittance, so valleys on our graph correspond to frequencies that are strongly absorbed. The CH peak is always sharp, unlike the OH peak, which is broad. The NH peak will generally be to the left of the CH peak.

Answer 100

(MRI) is a noninvasive medical diagnostic tool that uses proton NMR. Multiple cross-sectional scans of the patient's body are taken, and the various chemical shifts of absorbing protons are translated into specific colors. This produces a picture that shows the relative density of specific types of protons; for instance, a dark blue area on the MRI output screen indicates a high concentration of the types of protons giving rise to the range of resonances that correlate to dark blue. Comparison with normal MRI then allows the diagnostician to detect abnormalities in the scanned region. NMR spectroscopy is based on the fact that certain nuclei have magnetic moments that are oriented at random. When such nuclei are placed in a magnetic field, their magnetic moments tend to align either with or against the direction of this applied field. Nuclei whose magnetic moments are aligned with the field are said to be in the α -state (lower energy), whereas those whose moments are aligned against the field are said to be in the β -state (higher energy). The nuclei can then be irradiated with radio frequency pulses that match the energy gap between the two states, which will excite some lower-energy nuclei into the α -state. The absorption of this radiation leads to excitation at different frequencies, depending on an atom's magnetic environment. In addition, the nuclear magnetic moments of each atom are affected by nearby atoms that also possess magnetic moments. A typical NMR spectrum is a plot of frequency versus absorption of energy during resonance. Because different NMR spectrometers operate at different magnetic field strengths, a standardized method of plotting the NMR spectrum has been adopted. Uses an arbitrary variable, called chemical shift (represented by the symbol δ ), with units of parts per million(ppm) of spectrometer frequency. The chemical shift is plotted on the x-axis, and it increases toward the left (referred to as downfield). To make sure that we know just how far downfield compounds are, we use TMS (tetramethylsilane) as the calibration standard to mark 0 ppm. Nuclear magnetic resonance is most commonly used to study 1H nuclei (protons) and 13C nuclei, although any atom possessing a nuclear spin (any nucleus with an odd atomic number or odd mass number) can be studied, such as 19F, 17O, 5N, and 31P. Nuclei with odd mass or odd atomic numbers, or both, will have a magnetic moment when placed in a magnetic field. Not all nuclei have magnetic moments (12C, for example).

Answer 101

•Certain nuclei, most notably hydrogen, possess a magnetic moment, which causes the nuclei to act like tiny bar magnets. Normally, all the nuclei are oriented randomly. However, if an external magnetic field is applied, the nuclei will tend to align themselves with it. • A nucleus aligned with the external field represents a low energy state, while flipping a nucleus so it is aligned against the field represents a high-energy state. A photon of just the right energy could flip a nucleus from low to high energy. The energy required to affect this flipping depends on the chemical environment the nucleus is exposed to. For example, if the nucleus is in close proximity to oxygen, the oxygen will strip away some of the electron density surrounding the nucleus, exposing it somewhat. Such a nucleus will therefore flip at a higher frequency than one that is fully shielded. Nuclear Magnetic Resonance (NMR)- 1H-NMR- most 1H nuclei come into resonance 0 to 10 ppm downfield from TMS. Each distinct set of nuclei gives rise to a separate peak. This means that if multiple nuclei are in relatively identical locations, they will give the same peak. he single proton attached to the dichloromethyl group (Ha) is in a different magnetic environment from the three protons on the methyl group (Hb), so the two classes will resonate at different frequencies. The three protons on the methyl group are magnetically equivalent and resonate at the same frequency because this group rotates freely and, on average, each proton sees an identical environment. Think of a proton as being surrounded by a shield of electrons. As we add electronegative atoms or have resonance structures that pull electrons away from the proton, we Deshield and move Downfield. These atoms pull electron density away from the surrounding atoms, thus deshielding the proton from its electron cloud. The more the proton's electron density is pulled away, the less it can shield itself from the applied magnetic field, resulting in a reading further downfield. With this same reasoning, we know that if we had an electron-donating group, such as the silica atom in TMS, it would help shield the 1H nucleus and give it a position further upfield. That's why we use TMS as the reference peak, assigned to a value of zero; everything else will be more deshielded than TMS. The splitting of the peak represents the number of adjacent hydrogens. A peak will be split into n + 1 peaks, where n is the number of adjacent hydrogens. if the nucleus has a dense electron cloud around it, it will take a stronger electromagnetic field to excite it. Similarly, a nucleus whose electron cloud is partially stripped away by nearby electronegative atoms, also referred to as a deshielded nucleus, will be easier to excite.

Answer 102

* the magnetic environment of Ha can be affected by Hb, and vice versa. Thus, at any given time, Ha can experience two different magnetic environments, because Hb can be in either the α - or the β -state. The different states of Hb influence the nucleus of Ha (because the two H atoms are within three bonds of each other), causing slight upfield and downfield shifts. There is approximately a 50 percent chance that Hb will be in one of the two states, so the resulting absorption is a doublet, two peaks of identical intensity, equally spaced around the true chemical shift of Ha. Ha and Hb will both appear as doublets, because each one is coupled with one other hydrogen. To determine the number of peaks present (doublet, triplet, etc.), we use the n + 1 rule, where n is the number of protons that are three bonds away from our proton of interest. The magnitude of this splitting, measured in hertz, is called the coupling constant. * the peaks for alkyl groups are upfield (1 to 3 ppm), peaks for alkenes are further downfield (4 to 7 ppm), peaks for aldehydes are even further downfield (9 to 10 ppm), and carboxylic acids are the furthest downfield (10 to 12 ppm). Although there are technically four different states, α β has the same effect as β α (just as 3 × 4 is equal to 4 × 3), so both of these resonances occur at the same frequency. This means we will have three unique frequencies, α α , β β , and α β /β α . Ha will appear as three peaks (a triplet) centered on the true chemical shift, with an area ratio of 1:2:1. both hydrogens are attached to the same carbon, they will be magnetically identical (this doesn't apply to alkenes, since they can't freely rotate around the double bond). These hydrogens are within three bonds of one other hydrogen, Ha. This means that they will appear as a doublet, but because there are two of them, the peak for Hb will be taller than the peak for Ha. * add up the number of coupled hydrogens and add one (for our proton of interest itself) to determine the number of peaks. In addition, peaks that have more than four shifts will sometimes be referred to as a multiplet

Answer 103

* determining the relative number of protons and their relative chemical environments * showing how many adjacent protons there are by splitting patterns * showing certain functional groups * Each resonance, which includes the group of peaks that are part of a multiplet, represents a single proton or a group of equivalent protons. * The relative area of each peak reflects the ratio of the protons producing each peak. * The position of the peak (upfield or downfield) due to shielding or deshielding effects reflects the chemical environment of the protons.

Answer 104

• 13C-NMR- difference is that 13C-NMR signals occur 0 to 210 δ downfield from the carbon peak of TMS, quite a bit further than the 0 to 12 δ downfield that we saw with 1H-NMR. We should also note that 13C atoms are rare (although not as rare as the 14C atoms used for carbon dating). In fact, 13C atoms account for only 1.1 percent of all carbon atoms. This has two effects: First, a much larger sample is needed to run a 13C spectrum (about 50 mg compared with 1 mg for 1H-NMR), and second, coupling between carbon atoms is generally not observed (the probability of two 13C atoms being adjacent is 0.011 × 0.011, or roughly 1 in 1,000). f a carbon atom is attached to two protons, it can experience three different states of those protons (α α , α β /β α , and β β ), and the carbon signal is split into a triplet with the area ratio 1:2:1. ability to record a spectrum without the coupling of adjacent protons. This is called spin decoupling, and it produces a spectrum of singlet’s, each corresponding to a separate, and magnetically equivalent carbon atom.

Answer 105

1. The number of nonequivalent nuclei, determined from the number of peaks 2. The magnetic environment of a nucleus, determined by the chemical shift 3. The relative numbers of nuclei, determined by integrating the peak areas in 1H-NMR 4. The number of neighboring nuclei, determined by the splitting pattern observed (except for 13C in the spin-decoupled mode)

Answer 106

•UV-vis utilizes light of shorter wavelength, higher frequency, and higher energy than IR spectroscopy. You must remember that while aromatic molecules are always conjugated, conjugated molecules are not always aromatic. Ultraviolet Spectroscopy (UV)- Ultraviolet spectra are obtained by passing ultraviolet light through a sample that is usually dissolved in an inert, nonabsorbing solvent, and the absorbance is plotted against wavelength. The absorbance is caused by electronic transitions between orbitals. The biggest piece of information we get from this technique is the wavelength of maximum absorbance, which tells us the extent of conjugation within conjugated systems; the more conjugated the compound, the lower the energy of the transition and the greater the wavelength of maximum absorbance.

Answer 107

* Mass Spectroscopy- no absorption of electromagnetic radiation is involved) and because it destroys the compound, so we cannot reuse the sample once the analysis is complete. Most mass spectrometers use a high-speed beam of electrons to ionize the sample (eject an electron), a particle accelerator to put the charged particles in flight, a magnetic field to deflect the accelerated cationic fragments, and a detector that records the number of particles of each mass that exit the deflector area. M+ (parent ion peak) tells us the molecular weight can be useful. The initially formed ion is the molecular radical-cation (M+), which results from a single electron being removed from a molecule of the sample. This unstable species usually decomposes rapidly into a cationic fragment and a radical fragment. Because there are many molecules in the sample and usually more than one way for the initially formed radical-cation to decompose into fragments, a typical mass spectrum is composed of many lines, each corresponding to a specific mass/charge ratio (m/z). The spectrum itself plots mass/charge on the horizontal axis and relative abundance of the various cationic fragments on the vertical axis. * Characteristics- tallest peak (highest intensity) belongs to the most common ion, called the base peak, and is assigned the relative abundance value of 100 percent. The peak with the highest m/z ratio (the peak furthest to the right. the molecular ion peak(parent ion peak) or simply M+. Because this is the original compound with one electron missing, the charge value is usually 1; hence, the m/z ratio can usually be read as the mass of the fragment itself. * Application- fragmentation patterns provide information that helps us identify distinguish certain compounds because of molecular mass. a mass spectrum would give us unambiguous data distinguishing the two. * Mass spectrometry works by ionizing a sample, giving the resulting cations a velocity using a particle accelerator, and then using a magnetic field to deflect the cation stream into a detector. The detector does not detect the mass of the incoming particles. It detects the mass/charge ratio. the sample loses a single electron, and the resulting molecular cation (M+) has a charge of +1 and mass equivalent to the molecule’ s MW. However, such molecular cations are generally unstable, and rapidly break down in characteristic ways. A large molecule may have dozens of peaks on a mass spectrum due to various fragments and radicals breaking off from the parent cation. Base peak - the tallest peak on the graph is called the base peak. It is the most abundant ion, and is assigned a relative abundance of 100. * Molecular ion peak, or parent ion peak (M+) - though it goes by several different names, by far the most important peak is the one with the highest mass to charge ratio, or the peak furthest to the right. This peak usually represents the initial molecular cation. In general, the value for the molecular ion peak is equivalent to the molecular weight of the original molecule.

Answer 108

The π bond is in some ways weaker than the σ bond; therefore the π bond can be broken without breaking the σ bond. A further advantage of the π bond is that the electrons are exposed and are easily attacked by electrophiles (molecules that are seeking an electron pair * The Markovnikov rule states that the addition of a protic acid (HX) to an alkene occurs such that the proton attaches to the carbon with the smallest number of alkyl substituents, and thus producing the most stable carbocation. The proton will add to the carbon with the greatest number of hydrogens. adding an alkyl halide to an alkene we add the halide to the most substituted carbon atom in the double bond and the proton to the least substituted carbon atom of the double bond. * Free radical additions occur when peroxides, oxygen, UV light or other radical causing conditions are present. Free radical additions disobey Markovnikov's rule because the X· adds first to the double bond, and the hydrogen radical then adds second. * The double bond acts as a nucleophile and attacks an X2 molecule, displacing X. Note that this addition is anti, because the X- attacks the cyclic halonium ion in a standard SN2 displacement. The anti-addition of X2 proceeds via an ionic mechanism as in the case of Markovnikov addition. But the intermediate is a cyclic ion. * Syn additions are additions to the same side of a double bond. catalytic hydrogenation, which is the reductive process of adding molecular hydrogen to a double bond with an acid or metal catalyst. Because the face of the double bond is coordinated to the metal surface, the two hydrogen atoms will add to the same side of the double bond.

Answer 109

* Elimination Reactions- introduce a double bond into a molecule * E1 mechanisms are a two-step mechanism proceeding through a carbocation intermediate. The rate is dependent on the concentration of only one species, the substrate. Therefore, increasing the concentration of the nucleophile will not increase the reaction rate. In the first step of an E1 mechanism, the leaving group departs producing a carbocation. In the second step, the base removes a proton. Since the rate of the reaction is determined by the formation of the carbocation, and polar protic solvents stabilize the positive charge on carbocations, E1 reactions are favored by highly polar protic solvents as well as more branched carbon chains, good leaving groups, and weak nucleophiles. E2, a bimolecular elimination reaction, proceeds through a concerted (one step) mechanism. The rate of the reaction is determined by the two participating species, the substrate and the nucleophile. * E2 Reactions-favored by the presence of a strong base. When the substrate is a primary halide and the base is an ethoxide ion, elimination is not favored. Elimination would not be favored by a primary halide substrate because the base can easily approach the carbon bearing the leaving group. The reaction would then proceed through an SN2 mechanism, with the substituted product being the major one. An E2 elimination reaction is favored by secondary and tertiary halides because of the steric hindrance of the substrate. E2 reactions are favored by a more hindered substrate. Elimination reactions are favored at higher temperatures because eliminations have a higher energy of activation than substitution reactions. The reason for the higher energy of activation is that more bonds are broken and formed during an elimination reaction compared to a substitution one. By giving more molecules enough energy to surmount the energy barriers, increasing the temperature increases the rates of both substitution and elimination reactions. However, because the activation energy for elimination reactions is higher, the proportion of molecules able to overcome this barrier is significantly higher than that occurring at lower temperatures. E2 reaction mechanisms are favored by the use of a strong bulky base. teric hindrance of the bulky methyl groups of the tert-butoxide ion favors the elimination mechanism over the substitution one. The larger the pKa of the conjugate acid, the stronger the base. The stronger the base, the more likely that the reaction will proceed through an E2 mechanism. We see that the hydroxide ion is the strongest base among the compounds described above because its conjugate acid (water) has a high pKa and is thus very weak. (We know that water is the weakest acid because it has the largest pKa.) with e2 favoring highly substituted double bonds

Answer 110

many alkyl halides are subjected to dehydrohalogenation, more than one product can be formed. Used a more stable base, major product more highly substituted with the most stable product. Carrying out dehydrohalogenations with a bulky base such as potassium tert-butoxide in tert- butyl alcohol favors the formation of the less substituted alkene. Because of steric bulk of the base associated with solvent made larger, large ion difficulty removing internal 2o hydrogen atoms because of crowding of the transition site. With the elimination of the less substituted alkene following Hoffman rule.

Answer 111

Forming Alkenes- dehydragention of alcohols. Heating most alcohols with a strong acid causes them to lose a molecule of water and form an alkene. Elimination reaction favored at high temperatures using bronstead Lowry acids for proton donors. • 1. The experimental conditions (temperature and concentration of acid) that are required for alcohol dehydration are closely related to the structure of the individual alcohol. Ease of alcohols undergoing dehydration. • 2. Some primary and secondary alcohols also undergo rearrangements during dehydration to the most stable carbocation stability to be more highly substituted.

Answer 112

* The first thing to do is to determine the hybridization of the atom the hydrogen is bonded to. If the hydrogen is bonded to an sp3-hybridized carbon, the shift will be between 0 and 5. If the hydrogen is bonded to an sp2-hybridized carbon, the shift will between 5 and 10. * In effect, we simply divide the spectrum in half, and put sp3 on the right, sp2 on the left. * For example, the hydrogens in benzene are all bonded to sp2 carbons. Therefore we expect them to have shifts greater than 5 ppm. However, there are no electronegative atoms in benzene, so we don’ t expect it to be much greater than five. If we look up the relevant values, we see that aromatic hydrogens tend to show up around 6-7 ppm, which fits in nicely with our prediction. * On the other hand, what about the aldehyde hydrogen? Not only is it bonded to an sp2 carbon, that sp2 carbon is in turn bonded to a highly electronegative oxygen. Therefore, aldehyde protons show up at a 9-10 ppm. Even carboxylic acid protons are only 10-12 ppm! * Basically, each signal will be split into n+1 peaks, where n is the number of neighboring protons that signals possess. Therefore, if a proton had two neighbors, it would become a triplet. If it had three neighbors, it would be a quartet, and so on. * Neighboring protons are defined as any protons that are within three bonds, but which are chemically distinct from the protons being split. * protons are the most common target, any nucleus that contains an odd number of protons or which possesses an odd mass number will do\\

Answer 113

e- Amines have the general formula NR3, so they normally have a lone pair of electrons. They are classified according to the number of alkyl (or aryl) groups to which they are bound. A primary(1° ) amine is attached to one alkyl group, a secondary(2° ) amine to two, and a tertiary(3° ) amine to three. A nitrogen atom attached to four alkyl groups will carry a positive charge, as it loses its lone pair forming the fourth bond. This ion is called a quaternary ammonium compound, and it exists as a salt owing to its positive charge. named as alkylamines. name all of the substituents bound to the nitrogen alphabetically and then add “ amine.” – amine for the final e of the longest alkane to which the nitrogen is attached. N is used to label substituents attached to the nitrogen in secondary or tertiary amines, and we have to list a separate N for each different substituent attached to the nitrogen. another functional group on the molecule with higher priority (more oxidized), the prefix amino–. az in its name, it contains a nitrogen

Answer 114

* Amides are the condensation products of carboxylic acids and amines; * Carbamates are compounds with the general formula RNHC(O)OR'. Because they have nitrogen attached to a carbonyl, carbamates also fall into the category of amides a special type of amide because they have an oxygen on the other side of the carbonyl with an alkyl or aryl group attached to it, which means carbamates are a hybrid of an amide and an ester. Carbamates are derived from compounds called isocyanates (general formula RNCO) isocyanates have a carbon double-bonded to both an oxygen and a nitrogen, the carbon is positively polar and ripe for nucleophilic attack. When the isocyante is attacked by alcohol, a carbamate is formed. Carbamates are also called urethanes, and they can form polyurethanes;

Answer 115

•Enamines are the nitrogen analogs of enols: Instead of a hydroxide group, an amine group is attached to the carbon– carbon double bond. •Imines contain nitrogen– carbon double bonds; •all amino acids are amines. Peptide bonds between amino acids in proteins are amide bonds. Urea, a molecule used to store ammonia for removal from the body, is also an amide. •Nitriles, or cyanides, are compounds with a triple bond between a carbon atom and a nitrogen atom. They are named with either the prefix cyano– or the suffix – nitrile. Nitro compounds contain the nitro group, NO2 contain an N2 functionality, with two nitrogens at the end of a chain resonating between a double and a triple bond. They tend to lose the N2 as nitrogen gas and form carbenes, highly reactive carbons with only six valence electrons. This is usually seen as a carbon with two R groups and a lone pair of electrons. Azides are compounds with a linear N3 functionality (double bonds between three nitrogens). When azides lose nitrogen gas (N2), they form nitrenes, the nitrogen analogs of carbenes. Nitrenes tend to have their six valence electrons distributed in one bond to an R group and two lone pairs of electrons.

Answer 116

• Properties- the boiling points of amines lie between those of alkanes and alcohols. primary and secondary amines, which have hydrogens available, will have elevated boiling points, although not as high as corresponding alcohols because the bonds will be a bit weaker. Tertiary amines are not able to hydrogen bond. as molecular weight increases, so do boiling points. Primary and secondary amines can form hydrogen bonds, but because nitrogen is not as electronegative as oxygen, the hydrogen bonds of amines are not as strong as those of alcohols. Tertiary amines, on the other hand, cannot hydrogen bond at all (they have no hydrogen!) and, thus, have lower boiling points than their other amine counterparts.

Answer 117

* nitrogen atom in an amine is approximately sp3 hybridized. Nitrogen must bond to only three substituents to complete its octet; a lone pair occupies the last sp3 orbital. The lone pair of electrons on nitrogen is the determining characteristic of nitrogen chemistry. It endows nitrogen-containing compounds with their basic and nucleophilic properties. In addition, nitrogen is more electronegative than carbon but less than oxygen, which will indicate the distribution of electron density on a molecule. * Nitrogen atoms bonded to three different substituents are technically chiral because of the geometry of the orbitals. However, these enantiomers cannot be isolated because they interconvert rapidly in a process called nitrogen inversion: an inversion of the sp3 orbital occupied by the lone pair * activation energy for this process is only 6 kcal/mol, so the nitrogen will not be optically active. However, at very low temperatures or if the structure prevents the inversion of the molecule, it will be optically active. * amines are bases, so they readily accept protons to form ammonium ions (NH4+). Due to the electron-donating nature of R groups, the pKb value of alkyl amines is around 4, making them slightly more basic than ammonia (pKb = 4.76) but less basic than hydroxide (pKb = − 1.7). Aromatic amines such as aniline (pKb = 9.42) are far less basic than aliphatic (nonaromatic) amines, because the electron-delocalizing effect of the ring reduces the basicity of the amino group. The presence of other substituents on the ring also alters the basicity of anilines: Electron-donating groups (such as − OH, − CH3, and − NH2) increase basicity, whereas electron-withdrawing groups (such as NO2) reduce basicity. * Amines also function as very weak acids. The pKas of amines are around 35; thus, a very strong base is required for deprotonation.

Answer 118

• Synthesis- alkylation of ammonia- direct- the product will actually be a better nucleophile than the original nucleophile, because the electron-donating properties of the alkyl group cause further substitution. Alkyl halides react with ammonia to produce alkyl ammonium halide salts, Ammonia functions as a nucleophile and displaces the halide atom. When the salt is treated with base, the alkyl amine product is formed. This reaction can often lead to side products. The alkyl amine formed is itself nucleophilic because of the lone pair on the nitrogen, and it can react with other alkyl halide reactants to form more complex products.

Answer 119

Gabriel Synthesis- The Gabriel synthesis converts a primary alkyl halide to a primary amine, deprotonated phthalimide only able to go through reaction once so no side products. Phthalimide, acts as a good nucleophile when deprotonated. It displaces the halide ion of an alkyl halide, forming N-alkylphthalimide. Since the nitrogen atom is tertiary, it will not react with other alkyl halides. When the reaction is complete, the N-alkylphthalimide can be hydrolyzed with aqueous base to produce our product, the alkylamine •Reduction- Amines can be obtained from other nitrogen-containing compounds via reduction reactions. Amines can be formed by the following:SN2 reactions: Ammonia reacting with alkyl halides, Gabriel synthesis, Reduction of: Amides, Aniline and its derivatives, Nitriles, Imines

Answer 120

• Nitro compounds- we can easily reduce nitro compounds to primary amines. The most common reducing agent is iron or zinc used with dilute hydrochloric acid can be used on aromatic compounds converts deactivating to an activating group • nitriles- Nitriles can be reduced with hydrogen and a metal catalyst, or with lithium aluminum hydride (LAH) yielding primary amines • imines- reductive amination. start with an aldehyde or ketone and react it with ammonia, a primary amine, or a secondary amine. This reaction yields a primary, secondary, or tertiary amine, respectively, and the carbonyl becomes an − OH group (a carbinolamine). After the nitrogen transfers a proton to the OH group, the carbinolamine loses water to form an imine. If the unstable imine is then exposed to hydrogen and a metal catalyst, it will undergo reduction in much the same way that a carbonyl does, producing the amine •amides- can be reduced with LAH to form amines exhaustive methylation- Hoffman elimination. amine is converted to quaternary ammonium iodide by treatment with excess methyl iodide. he nitrogen now has methyl groups in all the positions where it used to have hydrogens or lone pairs. Treatment with silver oxide and water displaces the iodide ion and converts the molecule to ammonium hydroxide, which, when heated, undergoes elimination to form an alkene and an amine. he predominant alkene formed is the least substituted, in contrast with normal elimination reactions, where the predominant alkene product is the most substituted. The least substituted alkene (least stable) is formed because of the bulk of the quaternary ammonium salt leaving group.

Answer 121

Extraction- the transfer of a dissolved compound (the desired product) from a starting solvent into a solvent in which the product is more soluble. Extraction is based on the fundamental concept that like dissolves like. polar substance will dissolve best in polar solvents and a nonpolar substance will stick with the nonpolar solvents. If we selectively take advantage of this characteristic, we can extract our desired product, leaving most of the impurities behind in the first solvent. two solvents are immiscible (form two layers that do not mix. The two layers are temporarily mixed (when shaken) so that solute can pass from one solvent to the other. Extraction depends on the rules of solubility and polarity— “ like dissolves like.” Remember the three intermolecular forces that affect solubility: Hydrogen bonding: Compounds that can do this, such as alcohols or acids, will move most easily into the aqueous layer. Dipole– dipole interactions: These compounds are less likely to move into the aqueous layer. Van der Waals (London) forces: With only these interactions, compounds are least likely to move into the aqueous layer. • The water (aqueous) and ether (organic) phases will separate on their own if we give them enough time to settle in a specialized piece of glassware called a separatory funnel. Gravitational forces cause the heavier (more dense) layer to sink to the bottom of the funnel, and the bottom layer can be removed. the organic layer will be on the top and the aqueous layer will be on the bottom, although the opposite can also occur because their relative densities determine the order of the layers. Once we drain out the aqueous layer from the bottom of the funnel, we will have removed most of the more dense from the mixture, small amount will remain dissolved in order to retrieve go through extraction several times. Another way to take advantage of solubility properties is to perform the reverse of the extraction we just described and remove unwanted impurities. This process is called a wash because it is washing the product of unwanted impurities. • the desired product should have solubility that depends on temperature— it should be more soluble at high temperature and less so at low temperatures. In contrast, impurities should be equally soluble at various temperatures. The recrystallization proceeds with the solute molecules fitting into the defined crystal lattice, which excludes the impurities.

Answer 122

* Filtration- isolates a solid from a liquid. we pour our liquid-solid mixture onto a paper filter that allows only the solvent to pass through. At the end of a filtration, we have a solid (often called the residue) on the filter paper and a flask full of the liquid that passed through the filter, known as the filtrate. * gravity filtration. In gravity filtration, the solvents own weight pulls it through the filter. wever, the pores of the filter become clogged with solid, slowing the rate of filtration and possibly resulting in a big mess. For this reason, when we use gravity filtration we want the substance of interest to be in solution (dissolved in the solvent), and for any impurities to remain undissolved. because we want to ensure that the product remains dissolved, gravity filtration is usually carried out with hot solvent * vacuum filtration- separate solid-liquid mixture. solvent is forced through the filter by a vacuum on the other side. To do this, you need a specific flask that has a valve on the side to attach the vacuum. This method works much faster, so vacuum filtration is used when you need to isolate relatively large quantities of solid, particularly when the solid is the desired product

Answer 123

•recrystallization- when we use vacuum filtration to isolate a solid product, significant impurities may still be present. At this point, we can use recrystallization to purify our solid product even further. In this process, we dissolve our product in a minimum amount of hot solvent, and let it recrystallize as it cools— but we need to be really picky about our choice of solvent. We want to make sure that our solid product is soluble in the solvent at high temperatures only. In other words, we want our product to be insoluble at cold (or even room) temperature. This way, when we heat up the solvent, the entire solid product will dissolve (our desired product and all the impurities). When it cools, however, only our desired product will recrystallize out of solution, because the solute's defined crystal lattice tends to exclude the impurities. We must also consider the polarity of the solvent because polar solvents dissolve polar compounds and nonpolar solvents dissolve nonpolar compounds. a solvent with intermediate polarity is generally desirable for recrystallization. In addition, our solvent should have a low enough freezing point so that the solution may be sufficiently cooled without a chance of freezing. mixed solvent system may be used for recrystallization. First, we dissolve the crude compound in a solvent in which it is highly soluble. Then we slowly add another solvent in which the compound is less soluble. This second solvent is added in drops, just until the solid begins to precipitate. We then heat the solution again, enough to redissolve the precipitate. After it dissolves, we slowly cool the mixture to induce crystal formation, which can then be isolated with vacuum filtration mixed solvent system may be used for recrystallization. First, we dissolve the crude compound in a solvent in which it is highly soluble. Then we slowly add another solvent in which the compound is less soluble. This second solvent is added in drops, just until the solid begins to precipitate. We then heat the solution again, enough to redissolve the precipitate. After it dissolves, we slowly cool the mixture to induce crystal formation, which can then be isolated with vacuum filtration

Answer 124

* Sublimation- sublimation occurs when a heated solid turns directly into a gas without passing through an intermediate liquid stage. Sublimation can be used as a method of purification because the impurities found in most reaction mixtures will not sublime easily. perform sublimation, we produce vapors that condense on a chilled glass tube called a cold finger piece of glassware chilled by packing it with dry ice or running cold water through it. Most sublimations are performed under vacuum because at lower pressures, compounds will be less likely to pass through a liquid phase and will sublime instead. Another benefit of using a vacuum is that the low pressure reduces the temperature required for sublimation (just as low-pressure conditions reduce the temperature for evaporation) and, thus, there is less danger that the compound will decompose. The optimal conditions depend on the compound we are purifying (because each compound has a different phase diagram), although most compounds (other than water) * To make a solid sublime, you must either * raise the temperature at a low enough pressure; or * lower the pressure at a very cold temperature. * Ever wonder why the ice cubes in your freezer seem to be shrinking if they stay in there long enough?

Answer 125

* Centrigution- Inside a centrifuge, the solution (in test tubes) is subjected to centrifugal forces, causing the compounds of greater mass and density to settle toward the bottom of the test tubes, while lighter compounds remain near the top (the same as what would happen if you just let the test tube sit, only much faster). We use this method of separation frequently in biochemistry, when we separate large particles such as cells, organelles, and biological macromolecules. * Centrifugation is generally used to separate large things from each other. For example, you can centrifuge blood to separate cells (red blood cells, white blood cells, and platelets) from plasma; centrifuge cell debris to separate out organelles of interest, such as mitochondria; centrifuge (at extremely high speeds— called ultracentrifugation) to separate big DNA molecules, such as bacterial chromosomes, from smaller ones, such as plasmids. Large quantities of pure DNA are obtained using this method.

Answer 126

* Distillation- Distillation takes advantage of differences in boiling point to separate two liquids by vaporization and condensation and separates liquids that are soluble in each other. The liquid with the lower boiling point will vaporize first, and the vapors rise up the distillation column and condense in a water-cooled condenser, which then drip down the condenser into a vessel that catches the distillate. The temperature is kept low enough so the liquid with the higher boiling point will not boil and, thus, will remain liquid in the initial container. * Simple distillation- because there are no special factors involved in this distillation, it should only be used to separate liquids that boil below 150° C and that have at least a 25° C difference in boiling point. This way, the temperature is low enough that the compounds won't degrade, and there is a large enough difference in boiling points that we won't accidentally cause the second compound to boil off into the distillate. The apparatus itself consists of a distilling flask containing the two liquids, a distillation column consisting of a thermometer and a condenser, and a receiving flask to collect the distillate * vacuum distillation- t liquids boil when their vapor pressure equals atmospheric pressure. In vacuum distillation, we lower the atmospheric pressure so that the liquid can boil at lower temperatures. We use vacuum distillation whenever we want to distill a liquid that has a boiling point over 150° C. By using a vacuum, we are lowering the pressure over the surface of the liquid. This decreases the temperature that the liquid must reach to boil. don't have to worry about degrading the compound with excessively high heat, and we get the liquid to boil faster, a win-win situation * Fractional distillation- is like a million simple distillations in one, which is why it is much more precise. When we want to separate two liquids with similar boiling points (less than 25° C apart), we use fractional distillation. In this distillation, we use a fractioning column to connect the distilling flask to the condenser. A fractioning column is basically any column filled with inert objects, such as glass beads or steel wool, which act to increase the surface area of the column. As the vapors rise up the column they will condense on the available surfaces, and then as more heat rises, the condensation will re-evaporate and rise up further, thus recondensing even higher on the column. Each time the condensations evaporate, the vapors will contain a greater proportion of the lower-boiling point component. By the time the vapors make it to the top of the fractioning column, the vapor is composed of only our desired substance, which then condenses in the condenser and drips down to the receiving flask.

Answer 127

•Chromatography- tool that uses physical and chemical properties to separate and identify compounds from a complex mixture The more similar the compound is to its surroundings (whether by polarity, charge, etc.), the more it will stick and move slowly through its surroundings. Chromatography separates compounds based on how strongly they adhere to the solid, or stationary, phase (or, in other words, how easily they come off into the mobile phase). placing our sample onto a solid medium called the stationary phase, or adsorbent. We then run the mobile phase, usually a liquid (or gas in gas chromatography) through the stationary phase. This will displace (elute) the sample and carry it through the stationary phase. Depending on the substance and the polarity of the mobile phase, it will adhere to the stationary phase with different strengths, causing the different substances to migrate at different speeds. This is called partitioning, and it represents an equilibrium between the two phases. Different compounds will have different equilibrium constants and elute at different rates. This results in each compound separating within the stationary phase, allowing us to isolate them individually. speed at which substances move through media to measure how far each substance travels in a given amount of time (as in TLC), or we time how long it takes the substance to elute off the column (as in column or gas chromatography) so it can be collected. All chromatography is about how “ like” the substance is to the mobile and stationary phases except for size-exclusion chromatography.

Answer 128

thin-layer chromatography(TLC) uses silica gel, a highly polar substance, as its stationary phase. This means that any polar compound will adhere to the gel quite well and thus move (elute) slowly. In addition, when using column chromatography, size and charge both have a role in how quickly a compound moves through the stationary phase. Even strong interactions, such as antibody-ligand binding, are used in chromatography. We can use a plethora of different media as our stationary phase, each one exploiting different properties that allow us to separate out our compound. The adsorbent we use in TLC is either a piece of paper or a thin layer of silica gel or alumina adhered to an inert carrier sheet (glass or plastic). We then place the mixture that we want to separate onto the adsorbent itself; this is called spotting because we apply a small, well-defined spot of our mixture onto the plate. The TLC plate is then developed, which involves placing the adsorbent upright in a developing chamber (usually a beaker with a lid or a wide-mouthed jar) containing a shallow pool of eluant (solvent) at the bottom. We have to make sure that the initial spots on the plate are above the level of the solvent. If not, they'll simply elute off the plate and into the solvent, rather than moving up the plate. If everything's set up correctly, the solvent will creep up the plate via capillary action, carrying the different compounds with it at varying rates. When the solvent front nears the top of the plate, the plate is removed from the chamber and allowed to dry. The mobile phase on the other hand, is usually an organic solvent (often a mixture) of weak to moderate polarity, so it doesn't bind well to the gel. Because of this, nonpolar compounds hang out with the organic solvent and move quickly as the solvent moves up the plate, whereas the more polar molecules are stuck to the gel. Thus the more nonpolar the sample is, the further up the plate it will move. The spots of individual compounds are usually white, which makes them difficult or impossible to see on the white TLC plate. To get around this problem, we can place the developed TLC plate under ultraviolet light, which will show any compounds that are ultraviolet sensitive. . Alternatively, we can use iodine, phosphomolybdic acid, or vanillin (yes, the kind that tastes good) to stain the spots. The problem with this is that the stain will destroy the compound (usually by oxidation), so we can't recover it. This ratio is called the Rf value. . We take the distance that the compound travels and divide it by the, TLC is frequently used only for qualitative identification (determining the identity of a compound). distance that the solvent travels (which will always be a larger number). Reverse-phase chromatography- Preparative or prep TLC- if we really wanted to, we could use TLC on a larger scale as a means of purification. uses a large TLC plate that has a big streak of a mixture on it. As the plate develops, the streak splits into bands of individual compounds, just as it did in the small-scale version. Because the streak is so large, we can scrape the bands off and rinse them with a polar solvent, recovering the pure compounds from the silica.

Answer 129

Reverse-phase chromatography is the exact opposite. Here, the stationary phase is very nonpolar, so polar molecules move up the plate very quickly, whereas nonpolar molecules stick more tightly to the stationary phase.

Answer 130

uses a whole column filled with silica or alumina beads as an adsorbent, allowing for much greater separation. TLC uses capillary action to move the solvent and compounds up the plate, whereas in column chromatography, gravity moves the solvent and compounds down the column. To speed up the process, we can force the solvent through the column with nitrogen gas, a technique called flash column chromatography. In column chromatography, the solvent polarity can easily be changed to help elute our compound. Eventually, the solvent drips out of the end of the column, and we can collect the different fractions that leave the column at varying times. Each fraction contains bands that correspond to different compounds. After collection, we can evaporate the solvent and isolate the compounds we want to keep. Column chromatography is particularly useful in biochemistry because it can be used to separate and collect macromolecules such as proteins or nucleic acids

Answer 131

• in this method, the beads in the column are coated with charged substances, so they attract or bind compounds that have an opposite charge. For instance, a positively charged column will attract and hold the negatively charged backbone of DNA as it passes though the column, either increasing its retention time or retaining it completely. After all other compounds have moved through the column, a salt gradient is used to elute the charged molecules that have stuck to the column.

Answer 132

in this method, the beads used in the column contain tiny pores of varying sizes. These tiny pores allow small compounds to enter the beads, thus slowing them down. Large compounds can't fit into the pores, so they will move around them and travel through the column faster. It is important to remember that in this type of chromatography, the small compounds are slowed down and retained longer. The size of the pores may be varied so that different molecular weight molecules may be fractionated. A common approach in protein purification is to use an ion exchange column followed by a size-exclusion column

Answer 133

• Affinity chromatography- we can also customize columns to bind any substance of interest. For example, if we wanted to purify substance A, we could use a column of beads coated with something that binds A very tightly (hence the name affinity chromatography), such as a receptor for A, A's biological target, or even a specific antibody. This means that A will bind to the column very tightly, and it will likely stay inside the column. Later, we can elute A by washing the column with a free receptor (or target or antibody), which will compete with the bead-bound receptor and ultimately free substance A from the column. The only drawback of the elution is that we now have our inhibitor or receptor bound to our biological target. This inhibitor can be difficult to remove if it binds tightly.

Answer 134

•Gas chromatography (GC) is another method we have for qualitative separation. GC, also called vapor-phase chromatography(VPC). eluant is a gas (usually helium or nitrogen) instead of a liquid. The adsorbent is inside a 30-foot column that is coiled and kept inside an oven to control its temperature. The mixture is then injected into the column and vaporized. The gaseous compounds travel through the column at different rates because they adhere to the adsorbent in the column to different degrees and will separate by the time they reach the end of the column. The requirement for the compounds that we inject is that they be volatile: low melting point, sublimable solids or vaporizable liquids with analysis by computers The compounds are registered by a detector, which records the presence of a compound as a peak on a chart. It is common to separate molecules using GC and then to inject the pure molecules into a mass spectrometer for molecular weight determination (GC-mass spec).

Answer 135

stands for high-performance liquid chromatography, the eluant is a liquid, and it travels through a column of a defined composition. There are a variety of columns whose stationary phase is chosen depending on the target molecule and whose size is chosen depending on the quantity of material that needs to be purified. This is fairly similar to column chromatography. In the past, very high pressures were used, but recent advances allow for much lower pressures, which is why the name changed from high-pressure to high-performance. In HPLC, a small sample is injected into the column, and separation occurs as it flows through. The compounds pass through a detector and are collected as the solvent flows out the end of the apparatus. It functions similarly to GC because computers do all the work for us, but we use liquid under pressure instead of gas. As the whole process is under computer control, sophisticated solvent gradients can be applied to the column to help resolve the various components in our mixture.

Answer 136

* separate mixture of compounds that carry charge subjecting our compounds, usually macromolecules such as proteins or DNA, to an electric field, which moves them according to their net charge and size. Negatively charged compounds (such as DNA) will migrate toward the positively charged anode, and positively charged compounds will migrate toward the negatively charged cathode. The velocity of this migration, known as the migration velocity, v, of a molecule, is directly proportional to the electric field strength, E, and to the net charge on the molecule, z, and is inversely proportional to a frictional coefficient, f, which depends on the mass and shape of the migrating molecules. V=Ez/f. v = migration velocity, E = electric field (units = N/C), z = net charge on the molecule (units = C), f = frictional coefficient. To make your gel run faster: Increase the voltage, ↑ E; Use a lower percentage of agarose or acrylamide, ↓ f. Generally speaking, the more charged the molecule or the stronger the electric field, the faster the molecule will migrate through the medium. Conversely, the bigger and more convoluted the molecule is, the slower it will migrate. * Whereas spontaneous galvanic cells have a positive cathode and a negative anode, nonspontaneous electrolytic cells (which are used for electrophoresis) have a negative cathode and a positive anode. In most forms of electrophoresis, the size of a macromolecule is usually the most important factor— small molecules move faster, whereas large ones move more slowly and may in fact take hours to leave the well. In electrophoresis Anions are attracted to the Anode, whereas Cations are attracted to the Cathode. SDS-PAGE and agarose gel electrophoresis separate molecules based on size.

Answer 137

* separate mixture of compounds that carry charge subjecting our compounds, usually macromolecules such as proteins or DNA, to an electric field, which moves them according to their net charge and size. Negatively charged compounds (such as DNA) will migrate toward the positively charged anode, and positively charged compounds will migrate toward the negatively charged cathode. The velocity of this migration, known as the migration velocity, v, of a molecule, is directly proportional to the electric field strength, E, and to the net charge on the molecule, z, and is inversely proportional to a frictional coefficient, f, which depends on the mass and shape of the migrating molecules. V=Ez/f. v = migration velocity, E = electric field (units = N/C), z = net charge on the molecule (units = C), f = frictional coefficient. To make your gel run faster: Increase the voltage, ↑ E; Use a lower percentage of agarose or acrylamide, ↓ f. Generally speaking, the more charged the molecule or the stronger the electric field, the faster the molecule will migrate through the medium. Conversely, the bigger and more convoluted the molecule is, the slower it will migrate. * Whereas spontaneous galvanic cells have a positive cathode and a negative anode, nonspontaneous electrolytic cells (which are used for electrophoresis) have a negative cathode and a positive anode. In most forms of electrophoresis, the size of a macromolecule is usually the most important factor— small molecules move faster, whereas large ones move more slowly and may in fact take hours to leave the well. In electrophoresis Anions are attracted to the Anode, whereas Cations are attracted to the Cathode. SDS-PAGE and agarose gel electrophoresis separate molecules based on size.

Answer 138

• used to separate pieces of nucleic acid (DNA &RNA). Medium is agarose plant gel, nontoxic and easy to manipulate (unlike sodium dodecyl sulfate-polyacrylamide [SDS-PAGE]). Because every piece of nucleic acid is negatively charged because of its phosphate-sugar backbone, nucleic acids can be separated on the basis of size and shape alone (even without the charge-masking qualities of SDS,) useful to stain agarose gels with a compound called ethidium bromide, which binds to nucleic acids and allows us to visualize our results by fluorescence under ultraviolet light. The staining process, however, is toxic. Agarose gel electrophoresis can also be used to obtain our compound (preparatively) by cutting the desired band out of the gel and eluting out the nucleic acid.

Answer 139

• SDS Page- Sodium dodecyl sulfate-polyacrylamide gel electrophoresis is a useful tool because it separates proteins on the basis of mass alone. Polyacrylamide gel is the standard medium for electrophoresis and functions much in the same way as agarose gel. What makes it interesting is that SDS disrupts all noncovalent interactions. It binds to proteins and creates large chains with net negative charges, thereby neutralizing the protein's original charge and denaturing the protein. As the proteins move through the gel, the only variable affecting their velocity is f, the frictional coefficient, which depends on mass. After separation, we can stain the gel so the protein bands can be visualized and our results recorded. If you do not want to have your protein denatured, so-called “ native” gels may be run without the denaturing effects of SDS.

Answer 140

• Isoelectric focusing- exploits the acidic and basic properties of amino acids: isoelectric focusing. Its isoelectric point, pI, may characterize each protein which is the pH at which its net charge (the sum of all the charges on all of its amino acids) is zero. take a mixture of proteins and place them in a electric field that exists across a gel with a pH gradient (acidic at the positive anode, basic at the negative cathode, and neutral in the middle), the proteins will move until they reach the point that has a pH equal to their pI. At this pH, the protein's net charge is zero, so it will stop moving. Protein with a pI of 9. As we know, this means that when the protein is in an environment with a pH of 9, it will carry no charge; if it is at a pH that is higher or lower than 9, it will carry a charge. If we place this protein onto the gel at a pH of 7, there will be more protons around the protein; a pH of 7 is more acidic than 9: thus, there are more protons in solution. These protons will attach to the available basic sites on the protein, creating a net positive charge on the molecule. This charge will then carry the protein toward the negatively charged cathode, which rests on the basic side of the gradient. As the protein moves closer to the cathode, there are fewer protons in the gel (the pH increases). Eventually, as the concentration of free protons drops and we near a pH of 9, the protons creating the positive charge will dissociate, and the protein will become a neutral zwitterion. A quick way to remember the charge of each end of the gel is to recall that we associate acids with protons, which carry a positive charge, and thus the anode is positively charged. We associate bases with the negatively charged hydroxide ion, which gives us the negatively charged cathode. Because amino acids and proteins are organic molecules, the fundamental principle of acid-base chemistry apply to them as well L at a low pH, [H+] is relatively high. Thus, at a pH < pI, proteins will tend to be protonated and positively charged. As a result, they will migrate towards the negative cathode; at a relatively high (basic) pH, [H+] is lower and proteins will tend to be deprotonated and negatively charged. As a result, they will migrate towards the positive anode.

Answer 141

- polyhydroxylated aldehydes or ketones with the general formula Cn(H2O)n (hence carbo-hydrate). A single carbohydrate unit is called a monosaccharide (simple sugar) and, logically, a molecule with two sugars is called a disaccharide. Oligosaccharides are short carbohydrate chains (oligos is Greek for “ a few” ), whereas polysaccharides are long carbohydrate chains.

Answer 142

• Monosaccharides- Monosaccharides are the simplest units of carbohydrates and are classified by the number of carbons. r, to name monosaccharides, we use the numerical prefix followed by the suffix – ose (think glucose). For example, trioses, tetroses, pentoses, and hexoses have three, four, five, and six carbons, respectively. The basic structure of monosaccharides is exemplified by the simplest of them all, glyceraldehyde. Monosaccharides are the simplest units of carbohydrates and are classified by the number of carbons. Glyceraldehyde is a polyhydroxylated aldehyde, also known as an aldose (aldehyde sugar). The numbering of the carbon atoms in monosaccharides begins with the carbon closest to the carbonyl group. Thus, with aldoses, the aldehyde will always have the number C– 1. D and L are based on the stereochemistry of glyceraldehyde. These are not related to the + and − designations denoting optical rotation. Ketones- The simplest ketone sugar (ketoses) is dihydroxyacetone (see figure 14.2). As we just mentioned, the ketone will receive the lowest possible number; will have the ketone group on C– 2. Notice that on every monosaccharide, every carbon other than the carbonyl carbon will carry a hydroxyl group.

Answer 143

• stereochemistry- In a Fischer projection, if the Lowest – OH is on the Left, the molecule is L. If the – OH is on the Right, it's D (from the Latin root dextro, meaning “ right” designation of D and L configurations. D-Glyceraldehyde was later determined to exhibit a positive rotation (designated as D-(+)-glyceraldehyde), and L-glyceraldehyde a negative rotation (designated as L-(− )-glyceraldehyde) all other based upon relationship to glyceraldehyde. Thus, all D sugars will have the hydroxide of their highest numbered chiral center on the right, and all L sugars will have that hydroxide on the left. Make sure that you are familiar with these three types of stereoisomers. 1. The same sugars, in different optical families, are enantiomers (such as D-glucose and L-glucose). 2. All nonidentical (nonmirror image) sugars within the same family (as long as both are ketoses/aldoses, and have the same number of carbons) are diastereomers.3. Diastereomers that only differ at only one chiral center are known as epimers

Answer 144

• carbonyl carbon is (as always) a good electrophile and the many – OH groups can act as nucleophiles two groups together in same molecule an intramolecular nucleophilic acyl substitution, Anomers differ in configuration only at the newly formed chiral center, which is created by the attack of the alcohol on two different sides of the planar carbonyl carbon. α = trans to the – CH2OH (down in glucose). β = cis to the – CH2OH (up in glucose). fructose is a ketose, whereas glucose, galactose, and mannose are all aldoses. When glucose is reacted with methanol under acid catalysis, the hemiacetal is converted to an acetal via replacement of the anomeric hydroxyl group with an alkoxy group, creating acetal glycoside.

Answer 145

• Because monosaccharides contain both a hydroxyl group (a nucleophile) and a carbonyl group (an electrophile), they can undergo intramolecular reactions to form cyclic hemiacetals (from aldoses) and hemiketals (from ketoses). Because of ring strain, the only cyclic molecules that are stable in solution are six-membered pyranose rings or five-membered furanose rings. Note that the hydroxide group is the nucleophile in the ring formation, so oxygen becomes a member of the ring structure. Like cyclohexane, the pyranose rings adopt a chairlike configuration, and the substituents assume axial or equatorial positions to minimize steric hindrance. When we convert the monosaccharide from its straight-chain Fischer projection to the Haworth projection. any group on the right of the Fischer projection will point down, and any group on the left side of the Fischer projection will point up. Because the oxygen of the hydroxide on the highest-numbered chiral group (the same one that determines whether it is D or L) functions as the nucleophile in ring formation, six-membered rings are formed from six carbon aldoses or seven carbon ketoses. Alternatively, five-membered rings are formed from five carbon aldoses or six carbon ketoses. When we convert a straight-chain monosaccharide into its cyclic form, the carbonyl carbon (C– 1 for glucose) becomes chiral. Cyclic stereoisomers that differ about the new chiral carbon are known as anomers. In fact, the carbon that becomes chiral is labeled the anomeric carbon. When a sugar is drawn in ring form, it is easy to identify the anomeric carbon: Simply find the carbon that is attached to both the oxygen in the ring and a hydroxide group. In glucose, the alpha (α ) anomer has the – OH group of C– 1 trans to the CH2OH substituent (pointing down), whereas the beta (β ) anomer has the – OH group of C– 1 cis to the CH2OH substituent (pointing up). When exposed to water they will spontaneously open up and re-form. Because the substituents on the single bond between C– 1 and C– 2 can rotate freely, either the α or β anomer can be formed. This spontaneous change of configuration about C– 1 is known as mutarotation, and it occurs more rapidly when we catalyze it with an acid or base. Mutarotation results in a mixture that contains both anomers at their equilibrium concentrations (for glucose: 36% α , 64% β ). The α configuration is less favored because the hydroxyl group of the anomeric carbon is axial, adding to the steric strain of the molecule.

Answer 146

Reactions: ester formation- react same way as smaller molecule contain hydroxyl groups, and undergo many of the same reactions as simple alcohols can convert monosaccharide to esters, usng acid anhydride and base with all of hydroxyl groups esterified. Oxidation of monosaccharides- monosaccharides switch between anomeric configurations, hemiacetal rings spen short period of time in open-chain aldehyde form. Can be oxidized to carboxylic acids; oxidized aldoeses are aldonic acids. Because aldoes can be oxidized and considered reducing agents. Any monosaccharide with hemiacetal ring, is a reducing sugar ex. Glucose, fructose, lactose, maltose. Tollen’s reagenet and Benedict’s reagent can be used to detect the presence of reducing sugar. Positive tollen’s test involves reduction of ag+ to form metallic silver. When Benedict’s reagent used red precipitate of Cu2O indicating the presence of reducing agent. Ketose sugars reducing sugars and give positive tollen’s and benedict’s tests. Although ketones cannot be oxidized to carboxylic acids, can isomerize to aldose via keto-enol shifts, while in aldose can react with tollen’s or benedict’s reagents to form carboxylic acid a more powerful oxidizing agent, or dilute nitric acid will oxidize both aldehyde and primary alcohol (C-6) to carboxylic acids. Glycosidic Reactions- hemiacetals will react with alcohol to form acetals. True to form, hemiacetal monosaccharides will react with alcohol under acidic conditions. The anomeric hydroxyl group is transformed into an alkoxy group, yielding a mixture of the α - and β -acetals (with water as a leaving group). The resulting C– O bond is called a glycosidic linkage, and the acetal is known as a glycoside.

Answer 147

Disaccharides- a monosaccharide may react with alcohols to give acetals. Notice that monosaccharides also have hydroxide groups, so they, too, can function as the alcohol in reactions with other monosaccharides. When two monosaccharides react in this way, the product is called a disaccharide. most common glycosidic linkage occurs between C– 1 of the first sugar and C– 4 of the second, designated as a 1,4′ linkage. 1,6′ and 1,2′ bonds are also observed. The glycosidic bonds may be either α or β , depending on the orientation of the hydroxyl group on the anomeric carbon. he product has a 1,4′ -α linkage (maltose). Two glucose monosaccharides joined by a 1,4′ -β linkage yield cellobiose. Glycosidic linkages are often cleaved in the presence of aqueous acid. For example, we can cleave the glycosidic linkage of the disaccharide maltose into two molecules of glucose.

Answer 148

•Polysaccharides- Polysaccharides are large chains of monosaccharides linked together by glycosidic bonds. The three most important biological polysaccharides are cellulose, starch, and glycogen. Although these three polysaccharides have different functions, they are all composed of the same monosaccharide, D-glucose. In cellulose, the chain of glucose molecules is linked by 1,4′ -β -glycosidic bonds. Cellulose is the structural component of plants and is not digestible (think fiber), at least by humans. Starch is a polysaccharide that is more digestible by humans. Plants store energy as starch molecules by linking glucose molecules primarily in 1,4′ -α -glycosidic bonds, although occasional 1,6′ -α -glycosidic bonds form branches off the chain. Animals, on the other hand, store their excess glucose as glycogen. Glycogen is similar to starch, except that it has more 1,6′ -α -glycosidic bonds (approximately 1 for every 12 glucose molecules), which makes it a highly branched compound. All three polysaccharides are composed of glucose subunits, but they differ in their configuration about the anomeric carbon and the position of glycosidic bonds, resulting in notable biological differences Key biological polysaccharides are cellulose (1,4′ -β ), starch, and glycogen (mostly 1,4′ -α , although some are 1,6′ -α ).

Answer 149

• contain an amine group Amino acids contain an amine group and a carboxyl group attached to a single carbon atom (the α -carbon). The other two substituents of the α -carbon are a hydrogen atom and a variable side chain referred to as the R-group. It is helpful to think of the α -carbon as the central atom of the amino acid, because it is the atom that has all of the different functional groups attached to it. he α -carbon, with its four different groups, is a chiral (stereogenic) center (except in glycine, the simplest amino acid, where R = H and it only has three different groups attached to it), so all amino acids (except for glycine) are optically active. Naturally occurring amino acids (of which there are 20) are all L-isomers. With the exception of glycine, all amino acids are chiral. Therefore, by convention, the Fischer projection for an amino acid is drawn with the amino group on the left (L = left). L-Amino acids have S configurations, except for cysteine, which is R because of the change in priority caused by the sulfur.

Answer 150

Species that can act as both acids and bases are described as amphoteric (water is also an amphoteric species); their function depends on the pH of their environment. This means that if there are lots of protons in solution (acidic, low pH), the amino acid will pick up a proton, thus functioning as a base. On the other hand, if there are few protons in solution (basic, high pH), the amino acid will donate a proton, thus functioning as an acid. At its isoelectric point, an amino acid is uncharged. amino groups take on positive charges when protonated and carboxyl groups take on negative charges when deprotonated and when put into solution will form zwitterion (from German zwitter, or “ hybrid” ). The two oppositely charged halves of the molecule neutralize each other, so at a neutral pH, amino acids exist in the form of internal salts. Because there are two different locations that can either be protonated or deprotonated, amino acids have at least two different dissociation constants, Ka1 and Ka2 relative to the pH, or Kb1 and Kb2 relative to the pOH. Neutral amino acid in acidic solution will be protonated, amino group is protonated easily because it is protonated even at neutral but need very acidic to protonate carboxyl group. Titration with base: First the carboxyl group is deprotonated, then the amino group. Amino acids pass through at least two buffering stages, one at each pKa. if drop into basic opposite, amino acid fully protonated, carboxyl group is easy to deprotonate because deprotonated at neutral pH takes more basic environment to deprotonate amino group. This means that at a low pH, the amino acid will carry an excess positive charge and at a high pH, it will carry an excess negative charge. The intermediate pH, at which the amino acid exists as a zwitterion, is known as the isoelectric point (pI), or isoelectric pH, of the amino acid. This isoelectric pH must lie between pKa1 and pKa2.

Answer 151

Titration of amino acids- because amino acids have acidic and basic properties, they are great candidates for titration. The titration of each proton occurs as a distinct step, resembling that of a simple monoprotic acid. Thus, the titration curve ends up looking like a combination of two or three monoprotic acids (three if the amino acid has an acidic or basic R-group) all tied together Henderson-Hasselbalch Equation: Used to determine pH given the pKa of an acid and the concentrations of acid and base present. A 1 M glycine solution is acidic, which means glycine exists predominantly as +NH3CH2COOH: fully protonated and with a positive charge. As the solution is titrated with NaOH, the carboxyl group, because it is the most acidic, will be the first to lose a proton. During this stage, the amino acid acts as a buffer, and the pH changes very slowly. When 0.5 moles of base have been added to the amino acid solution, the concentrations of the initial glycine +NH3CH2COOH and +NH3CH2COO− (its zwitterion) are equimolar. At this point, the pH is equal to pKa1, and the solution is buffered against pH changes. This is an important point to remember: When pH = pKa, the solution is in a buffer zone, represented by a flat horizontal line on the graph (because there is no change in pH). The best buffering occurs within one pH unit of the pKa of the titrating group.

Answer 152

A 1 M glycine solution is acidic, which means glycine exists predominantly as +NH3CH2COOH: fully protonated and with a positive charge. As the solution is titrated with NaOH, the carboxyl group, because it is the most acidic, will be the first to lose a proton. During this stage, the amino acid acts as a buffer, and the pH changes very slowly. When 0.5 moles of base have been added to the amino acid solution, the concentrations of the initial glycine +NH3CH2COOH and +NH3CH2COO− (its zwitterion) are equimolar. At this point, the pH is equal to pKa1, and the solution is buffered against pH changes. This is an important point to remember: When pH = pKa, the solution is in a buffer zone, represented by a flat horizontal line on the graph (because there is no change in pH). As we add more base, more and more of the carboxyl groups will become deprotonated. The amino acid starts to lose its buffering capacity, and the pH will start to rise quickly during this phase. By the time 1 mole of base has been added (remember that we started with 1 mole of glycine), glycine exists exclusively as +NH3CH2COO− . Because we have added equal amounts of glycine and base, each molecule of glycine has been deprotonated at the carboxyl group. This means that every amino acid is now electrically neutral; thus, the pH at this point is equal to the isoelectric point (pI) of glycine. This is our second point to remember: When we've added equal amounts of amino acid and base, pH = pI, and we have a vertical line on our graph. As we continue adding base, glycine passes through a second buffering stage, during which the pH change is held steady again. But now, because we've already deprotonated all of the carboxyl groups, the less acidic amino groups start to deprotonate. When 1.5 moles of base have been added, the concentrations of +NH3CH2COO− and NH2CH2COO− are equimolar, and the pH is equal to pKa2. This is our second buffering zone, which we need to remember occurs when pH = pKa2, and once again it appears as a roughly horizontal line on the graph. Continuing with our theme, as we add another 0.5 moles of base (2 moles total), the remaining amino groups are deprotonated, leaving only NH2CH2COO− in solution.

Answer 153

* 1. When adding base, the carboxyl groups lose their protons first; after all of the carboxyl groups are fully deprotonated, the amino groups start to lose their acidic protons. 2. Two moles of base must be added to deprotonate one mole of most amino acids. The first mole deprotonates the carboxyl group, whereas the second deprotonates the amino group. 3. The buffering capacity of the amino acid is greatest at or near the pH of the two dissociation constants, pKa1 and pKa2. At the isoelectric point (which can be found by determining the average of pKa1 and pKa2), its buffering capacity is minimal, and the graph appears as a vertical line. 4. Some amino acids contain acidic or basic side chains. To find the pI of these amino acids, simply average the two acidic pKas if the side chain is acidic. If the side chain is basic, take the average of the two basic pKas. 5. It is possible to perform the titration in reverse, from alkaline pH to acidic pH, with the addition of acid; in that case, the sequence of events is reversed. * Nonpolar amino acids are often found at the core of globular proteins or in transmembrane regions of proteins that are in contact with the hydrophobic portion of the phospholipid membrane. The other types of amino acids (polar, acidic, and basic) are found in regions of proteins that are exposed to the aqueous, polar environment (that is, the surface of the protein). At physiological pH, basic amino acids have a net positive charge. They also have three dissociation constants. You can use pI to predict an amino acid's charge at a given pH. If pH < pI, think positive charge. If pH > , think negative charge.

Answer 154

* Henderson-Hasselbach equation- the ratio of protonated to deprotonated amino acids that exist in solution depends on the pH of the solution. The Henderson-Hasselbalch equation defines this relationship by relating the pH to the ratio of conjugate acid to conjugate base. It also provides a mathematical expression for the dissociation constants of amino acids. When the pKa1 of glycine is known, the ratio of acid to its conjugate base for a particular pH can be determined. * ten times as many zwitterions as there are fully protonated amino acids. We can also do this with the pKa2 of glycine (if it is given); we would simply change the conjugate base to NH2CH2COO− and the conjugate acid to +NH3CH2COO− . We can use the Henderson-Hasselbalch equation experimentally to prepare effective buffer solutions of amino acids. The best buffering regions of amino acids occur within one pH unit of the pKa or pKb.’

Answer 155

The 20 naturally occurring amino acids are grouped into four categories: nonpolar, polar (but uncharged), acidic, and basic

Answer 156

• have R-groups that are saturated hydrocarbons. This means that these R-groups are hydrophobic and, thus, decrease the solubility of the amino acid in water. For this reason, these amino acids prefer to be buried inside proteins, away from the aqueous cellular environment. affects hemoglobin by replacing the hydrophilic amino acid glutamic acid with hydrophobic valine. Because the glutamic acid is normally on the aqueous exterior of the molecule, replacing it with hydrophobic valine causes the molecule to contort into a sickle shape in an attempt to bury the valine in the molecule's interior. Few other nonpolar amino acid side chains, tryptophan has a nitrogen atom with a lone pair, the electrons are resonated through the aromatic ring, and so it does not exhibit basic properties. The tryptophan ring, which is large and hydrophobic, is often a nucleating residue when proteins fold.

Answer 157

have uncharged polar R-groups that are hydrophilic. This polarity increases their solubility in water, so they are often found on the surface of proteins

Answer 158

have an R-group that contains a carboxyl group. Have negative charge at physiological pH 7.4, exist in salt form in the body. Substrate-binding sites of enzymes and reactions that require proton transfer. With proteases enzymes use acid-base properties of acidic side chains. Aspartic acid and glutamic acid each have three groups that must be neutralized during titration (two – COOH and one – NH3+). Therefore, their titration curves are different from the standard curve that we saw for glycine. This also means that these molecules have three distinct dissociation constants— pKa1, pKa2, and pKa3— although the neutralization curves of the two-carboxyl groups overlap to a certain extent. Because of the additional carboxyl group, the isoelectric point is shifted toward an acidic pH and can be found by averaging both of the acidic pKas together. In addition, three groups require three moles of base to deprotonate each mole of the acidic amino acid.

Answer 159

amino acids side chains contain an amino group called basic, carrying a positive charge at physiological pH 7.4. Similar to acidic amino acids, the titration curve of basic amino acids is modified by the additional amino group that must be neutralized. These amino acids also have three dissociation constants, and the neutralization curves for the two amino groups overlap somewhat. The isoelectric point is shifted toward an alkaline pH and can be found by averaging the two basic pKas together. Once again, three moles of acid are needed to neutralize one mole of a basic amino acid.

Answer 160

* Peptides- composed of amino acid subunits called residues. Have carboxyl group on one end and an amino group on the other, two groups combine peptide bond or amide bond forming between them. Peptides, small proteins distinction between peptide and protein vague accepted that peptides contain fewer than 50 residues with two amino acid joined together to form a dipeptide, with three forming tripeptide, and many forming a polypeptide. * Reactions- joined by peptide (amide bonds) between carboxyl group of one amino acid and amino group of another. Condensation reaction occurs with water lost, reverse reaction, hydrolysis cleavage by adding water to the peptide bond catalyzed by an acid or base. With certain enzymes digesting the chain at specific peptide linkages. * Properties- The terminal amino acid with a free α -amino group is known as the amino-terminal or N-terminal residue, whereas the terminal residue with a free carboxyl group is called the carboxy-terminal or C-terminal residue. By convention, peptides are drawn with the N-terminal end on the left and the C-terminal end on the right. amides have two resonance structures, so the true structure is a hybrid with partial double-bond character between the nitrogen and the carbonyl carbon. Rotation about the C– N bond is restricted. This restriction adds to the rigidity and stability of the backbone of proteins. The bonds on either side of the peptide unit, on the other hand, can rotate however they like, because they are only single bonds. Rotation about the C– N bond is restricted. This restriction adds to the rigidity and stability of the backbone of proteins. The bonds on either side of the peptide unit, on the other hand, can rotate however they like, because they are only single bonds.

Answer 161

• Proteins- polypeptides that range from only a few to more than thousand amino acids in length. Act as acting as enzymes, hormones, membrane pores, receptors, and elements of cell structure. Proteins are the main actors of biological systems; after all, our genetic code is simply a grocery list of different protein codes. There are four levels of protein structure— primary, secondary, tertiary, and quaternary. Primary structure consists of the amino acid sequence and covalent bonds. Secondary structure refers to the local structure of a protein as determined by hydrogen bond interactions. Tertiary structure is the three-dimensional shape of the protein. Quaternary structure is the arrangement of polypeptide subunits. Conjugated proteins have prosthetic groups. primary, secondary, tertiary, and quaternary

Answer 162

coded in DNA of organism, sequence of amino acids listed from N-terminus to C-terminus, linked by peptide bond. sequence that determines all higher levels of protein structure. In other words, a protein will assume whatever secondary, tertiary, and quaternary structures are the most energetically favorable for the given primary structure and the environment. This primary structure can be determined in a laboratory using a procedure called sequencing. This is most easily done from the DNA (the gene) that produced the protein.

Answer 163

• local structure of neighboring amino acids. Result of hydrogen bonding between nearby amino acids, with the α -helix and the β -pleated sheet. • α –helix- a rod like structure in which the peptide chain coils clockwise about a central axis. This helix is stabilized by the intramolecular hydrogen bonds between carbonyl oxygen atoms and amide hydrogen atoms four residues away from each other (n + 4 hydrogen bond). The side chains of these amino acids point away from the helix's core, interacting with the cellular environment. A typical protein with this structure is keratin, a fibrous structural protein that is found in our hair and fingernails. β -Pleated Sheet- β -pleated sheets (which may be parallel or anti-parallel), the peptide chains lie alongside each other, forming rows. These chains are held together by intramolecular hydrogen bonds between the carbonyl oxygen atoms on one peptide chain and the amine hydrogen atoms on another. To accommodate the greatest possible number of hydrogen bonds, the β -pleated sheet assumes a rippled, or pleated, shape. R-groups of amino residues point above and below the plane of the β -pleated sheet. Silk fibers are composed of β -pleated sheets.

Answer 164

three-dimensional shape of the protein. It is mostly determined by hydrophilic and hydrophobic interactions between the R-groups of amino acids. This three-dimensional structure is also determined by the distribution of disulfide bonds. A disulfide bond results when two cysteine molecules become oxidized to form cysteine. Disulfide bonds create loops in the protein chain. Disulfide bonds determine how wavy or curly hair is. proline, because of its ring shape, cannot fit into every location in an α -helix, so it causes a kink in the chain. Amino acids with hydrophilic (polar or charged) R-groups tend to arrange themselves toward the outside of the protein, where they interact with the aqueous cellular environment. Amino acids with hydrophobic R-groups tend to be found close together, and they protect themselves from the aqueous environment by burying themselves in the middle of the protein. Proteins are divided into two major classifications on the basis of their tertiary structure. Fibrous proteins, such as collagen, are found as sheets or long strands, whereas globular proteins (think globe), such as myoglobin, are spherical.

Answer 165

Conjugated proteins derive part of their function from covalently attached mole-cules called prosthetic groups. These prosthetic groups can be organic molecules, such as vitamins, or even metal ions, such as iron. Proteins with lipid, carbohydrate, and nucleic acid prosthetic groups are referred to as lipoproteins, glycoproteins, and nucleoproteins, respectively. hemegroup. This heme group is composed of an organic porphyrin ring with an iron atom bound in the center. The heme group itself binds to and carries oxygen; as such, hemoglobin would be inactive without the heme group.

Answer 166

- Denaturation, or melting, is the process by which proteins lose their three-dimensional structure and revert to a random-coil state. Because this process destroys the protein's tertiary structure, it renders it completely functionless. There are several methods we can use to denature a protein— with a detergent, change in pH, temperature, or even solute concentration. When a protein denatures, the damage is usually permanent. However, certain gentle denaturing agents (such as urea) do not permanently disrupt the protein. Removing the reagent might allow the protein to renature (regain its structure and function); that is, the denaturation is reversible.