orgo uW: Functional groups & biological molecules Flashcards
Inductive Effect
- The inductive effect is an electronic property in which the electrons are donated through sigma bonds.
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Electronegative atoms or electron withdrawing groups tend to create greater dipoles, partial charges, and better leaving groups, and to have greater inductive effects than less electronegative atoms.
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The closer the electron withdrawing group to an atom, the greater the inductive effect that atom experiences. Electron withdrawing groups also make good leaving groups because they can stabilize the negative charge acquired after being eliminated.
- Example: acetic anhydride (greater inductive effect) than N,N-diisopropylisohutyramide)
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The closer the electron withdrawing group to an atom, the greater the inductive effect that atom experiences. Electron withdrawing groups also make good leaving groups because they can stabilize the negative charge acquired after being eliminated.
Oxidation
Reduction
- Oxidation loss of electrons and increase in the number of carbon-heteroatom bonds (see image)
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Reduction gains electrons & the number of carbon-heteroatoms bonds decreases
- reduction can also be viewed as an increase in the number of bonds to hydrogen
Pyranose & aldoses–>hemiacetals
Furnasoes & ketoses–>hemiketals
- The cyclic structure of a sugar is classified by the size of the ring as well as whether the linear form contains an aldehyde (aldose) or a ketone (ketose).
- Furanoses are sugars with five-membered rings (four carbons and one oxygen), and pyranoses are sugars with six-membered rings (five carbons and one oxygen).
- Aldoses cyclize to form hemiacetals, and ketoses cyclize to form hemiketals.
Mannose Derivative
Classifying sugars as L or D
Drawing Conclusions
- to draw conclusions from the results of a study on enzyme activity, control experiments should be done to ensure that no confounding factors, such as side reactions occur.
- example:
Mass Spectroscopy
- Mass spectrometry is a technique that ionizes molecules in a sample, and the ions can fragment.
- The ions are accelerated toward a magnet, deflected according to mass, and detected.
- A plot of ion mass abundance vs. m/z ratio is generated; fragments of the sample can be identified by the m/z difference between two peaks in the mass spectrum.
- m/z (mass-to-charge ratio)
Aldol condensation reaction & enolate formation
- Protons on an α-carbon (adjacent to a carbonyl) are more acidic than other protons bonded to a carbon atom because the carbonyl oxygen is electron withdrawing, resulting in less electron density around the α-protons.
- Therefore, α-protons have lower pKa values and can be more easily removed by a base to form an enolate, which can be stabilized by charge delocalization.
Acidic & basic Functional groups
Formation of aldol product
Aldol condensation products of diketones
FULL summary of aldol condensation
- Aldol condensations are carbon-carbon bond-forming reactions that require two carbonyl substrates (ketones and/or aldehydes).
- The reaction begins with deprotonation of the α-carbon on one of the substrates, forming a resonance-stabilized enolate intermediate.
- Nucleophilic addition of the enolate to the other carbonyl substrate yields the aldol product, and then deprotonation of the α-carbon, followed by –OH elimination, yields the conjugated product, an α,β-unsaturated carbonyl compound.
Products of retro-aldol reactions
Nucleophiles vs leaving groups
hydrolysis of a glycosidic bond
- A glycosidic bond is the α- or β-linkage between a sugar and an –OH of another molecule.
- Hydrolysis of a glycosidic bond is cleavage of the linkage by addition of H2O, breaking the molecule into two smaller units.
Hydrogen bonding
- Hydrogen bonding is an intermolecular or intramolecular force that occurs between a hydrogen bond donor and acceptor; this type of force helps stabilize molecules.
- Hydrogen bond acceptors are electronegative atoms with a lone pair of electrons (eg, oxygen and nitrogen), and hydrogen bond donors are hydrogen atoms bonded to an electronegative atom.
- Alcohols and amines contain groups that can act as both hydrogen bond donors and acceptors.
In order for decarboxylation reactions to happen, what must be included?
- Decarboxylation is a reaction that removes a carboxyl group from a carboxylic acid with a β-carbonyl, releasing the carboxyl group as CO2 gas.
- A β-carbonyl is necessary for decarboxylation because a cyclic transition state incorporating both carbonyls is formed. Esters with a β-carbonyl can also undergo decarboxylation if they are hydrolyzed to a carboxylic acid first.
Conjugation
L/D Sterochemistry
Determining priority of amino acids
Imine formation
- An imine is an analogue of ketones and aldehydes that contains a carbon-nitrogen double bond.
- Imines are formed from a ketone or aldehyde and NH3 or a primary amine via an acid-catalyzed addition of the amine followed by an acid-catalyzed dehydration.
Acid-catalyzed addition of amine
involves protonation of the carbonyl, nucleophilic attack of the carbonyl by the amine, and deprotonation of the amine.
Acid-catalyzed dehydration
involving protonation of the –OH group, loss of H2O, and deprotonation.
Strecker Synthesis
- is used to make α-amino acids from aldehydes using NH3 and potassium cyanide (KCN).
- The first step of the reaction proceeds with protonation of the carbonyl oxygen by H3O+, followed by nucleophilic attack of the carbonyl carbon by NH3, resulting in dehydration and imine formation.
- Therefore, an aldehyde and NH3 are used to form the imine intermediate in the Strecker synthesis.
How is Enamine formed?
- An aldehyde and a secondary amine react to form an enamine via an acid-catalyzed addition of the amine followed by an acid-catalyzed dehydration.
- Ketones can be used in the formation of ________
- how about tertiary and quaternary amines?
- imines
- tertiary and quaternary amines are not suitable for imine synthesis becuasae tertiary amines do not have an availabe hydrogen for deprotonation & quaternary amines cannot act as nucleopholes and add the ketone (or alldehyde) carbonyl.
Waxes
- Lipids are hydrophobic molecules broadly classified as hydrolyzable or nonhydrolyzable, and more specifically classified based on their backbone structure.
- Waxes are hydrolyzable lipids that contain an ester bond formed by the linkage between a long-chain fatty acid and a long-chain alcohol.
Anhydride Formation
- Anhydrides are carboxylic acid derivatives characterized by two carbonyl groups joined by an oxygen atom.
- Anhydrides can be prepared by condensation of two carboxylic acid molecules or nucleophilic acyl substitution of an acid chloride by a carboxylate.
- Both reactions result in the loss of a water molecule from the carboxylic acid starting material.
What do boiling points depend on?
Ketone Boiling Point
- Boiling point depends on intermolecular forces, surface area, and molecular weight (MW).
- Strong intermolecular forces and greater surface area (less branching and higher MW) have higher boiling points compared to molecules with weaker intermolecular forces and a smaller surface area.
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Ketones and aldehydes of the same MW have similar boiling points because they both have dipole-dipole interactions.
- a linear 5-carbon ketone has a boiling point that is most similar to a branched aldehyde with the same molecular weight
Increasing boiling point, increasing intermolecular forces
Boiling points based on surface area
Peptide bonds are formed between which 2 functional groups?
amino group of one aa and the carboxylic group of another aa
How does a peptide bond form?
- Peptide bonds are formed through a dehydration reaction, in which the carboxyl group of one amino acid loses a hydroxyl group and the amino group of another amino acid loses a hydrogen atom.
- H2O is released as a byproduct in a dehydration reaction.
Giving priority for functional groups in amino acids
Formation of glyceraldehyde
- Glyceraldehyde is an aldose that can be derived from the oxidation of glycerol.
- It contains an aldehyde and two hydroxyl groups.
- D/L designations for all other molecules are based on D- and L-glyceraldehyde.
- glycerol has no sterocenters and is neither D or L
whats a reaon for why to use the same solvent (resin) for the reaction?
- to avoid introducing another cariable in the synthesis
- Variables are components of an experiment that change while other components of the experiment are held constant. These parameters must be isolated to determine the effect a variable has on the results of an experiment.
Resolution in chromatography
Fatty Acids Structure
- Fatty acids (FAs) are made up of a nonpolar hydrocarbon chain with a polar carboxyl head group, and can be classified as either saturated or unsaturated.
- Free FAs are usually not found in blood plasma but rather exist as derivatives, such as a triacylglyceride or phospholipid.
- Hydrolysis of triacylglycerides and phospholipids releases free FAs.
Triacylglycerols
- Triacylglycerols are made up of three fatty acids and a glycerol molecule connected through ester linkages.
- Hydrolysis of a triacylglycerol releases free fatty acids and glycerol.
- The number of distinct fatty acids released corresponds to the number of distinct hydrocarbon chains in the triacylglyceride.
- example the hydrolysis of tricylgycerols give 4 products (1 glycerol, 3f fatty acid chains (2 are identical) so count it as 2 chains)
Hydrolysis of triacylglycerol
Sponification
- Saponification is the hydrolysis of an ester with a strong base.
- Triacylglycerols contain three fatty acids bonded to a molecule of glycerol through ester linkages, and saponification of a triacylglycerol with a strong base releases free fatty acids (as sodium salts) and a molecule of glycerol.
- One equivalent of base is needed to hydrolyze one ester linkage; therefore, three equivalents of base are needed to completely hydrolyze a triglyceride.
(if question says a catalytic amount know that it means: less than 1 equivalent)
Fatty acid Extraction
Fatty acids sodium salts
- Fatty acids sodium salts are soluble in water due to ion-dipole interactions of the charged carboxyl group and sodium ion with water.
- Protonation of fatty acid sodium salts renders the molecule insoluble in water because the fatty acid is no longer a charged species, and therefore cannot interact as readily with water.
- The long-chain hydrocarbon makes fatty acids hydrophobic, allowing for solubility in organic solvents.
Separation of glycerol from fatty acid sodium salts can be accomplished by protonation of the negatively charged carboxyl group, which can be achieved by the addition of a strong acid, such as HCl. Hexanes are nonpolar organic molecules that can readily interact with the protonated fatty acids. Therefore, protonated fatty acids will enter the organic layer after extraction with the organic solvent hexanes.
Isomers
- Isomers are molecules with the same molecular formula but different structural arrangements, and are either constitutional (differ in atom connectivity) or stereoisomers (have the same atom connectivty but differ in spatial arrangment)
- steroisomers that arise from distributed double bonds are called geometric isomers
- classified by double bonds that give rise to cis (H on same side) or trans isomers, depending on the relative positions of nonhydrogen substituents.
- These isomers have slightly different boiling points and can be separated by gas chromatography.
- have the same molecular weight and m/z ratio—>cis/trans form
- steroisomers that arise from distributed double bonds are called geometric isomers
Arachidonic Acid
omega-6 FA
Gabriel synthesis
- The Gabriel synthesis is a method that uses potassium phthalimide and diethyl bromomalonate as starting materials to synthesize amino acids.
- aka: malonic ester synthesis, used to make primary amines, including alpha-amino acids, without overalkylation of the amine
- The starting reagents of the Gabriel and Strecker syntheses are all planar; therefore, the product is a mixture of L- and D-amino acids.
Strecker synthesis
- potassium cyanide & an aldehyde are both used
all naturally occuring amino acids have and _______configuration with the exception of ____________
of all the chiral amino acids, all are in the _______configuration, except for cycteine
- L-configuration, glycine (which is not chiral), and therefore neither L nor D
- of the chiral amino acids, all are in the S configuration except cystein
Protecting groups
- Protecting groups are organic molecules added to functional groups to mask them and keep them from reacting under a given set of reaction conditions.
- There are different protecting groups for different functional groups.
neutral pH, what are the charges of NH3 & COOH?
- pH ~ 7, both N- & C- termini are charged
- neutral pH is above the carboxyls group’s pKa of 2, causing it to become deprotonated and form its congugate base
- (COO-, (-1 charge))
- neutral pH is below the amines group’s pKa of 9, so it is fully protonated and exits in the conjugate acid form
- NH3+1
In a neutral environment (pH ~7), amino acids are zwitterions, also known as dipolar ions, because the N- and C-termini are charged (+1 and −1, respectively). The carboxyl group is deprotonated and in its conjugate base form. The amine group is protonated and in its conjugate acid form.
Conjugated system
- A conjugated system of alternating single and double bonds in a molecule causes electron delocalization.
- This delocalization allows the molecule to absorb a photon in the ultraviolet or visible region, exciting an electron to a higher energy state.
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Florescence energy diagram
- occurs when a photon in the ultraviolet (UV) or visible region is absorbed by a fluorophore.
- Photons in this region have enough energy to excite certain electrons to higher energy states.
- After excitation, the electron loses some of its energy as heat. The remaining energy is emitted as a photon with a longer wavelength (less energy) than the one that was absorbed.
Conjugation
Adjacent double bonds
Non-adjacent double bonds
Conjugation system is a system with several alternating single electrons can be delocalized through p-orbitals, creating an extended network of conjugated pi bonds know as resonance stabilization
Aldol condensation (base-catalyzed reaction)
Cyclization of an aldol condensation products of diketones
Phospholipids
Types of phospholipds structure
Fatty acid structure and membrane fluidity
- Compounds 1 to compound 2, most likely will undergo which reaction?
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SN1
- 2 steps & involves carbocation intermediate
- leaving group leaves before the nucleophilic attack
- tertiary leaving groups are more likely to undergo SN1
- sterically hindered
- tertiary carbon atoms can form relatively stable carbocations compared with secondary or primary carbon atoms
- the positive charge associated with carbocations ehnances their electrophilicity, helping them overcome the steric hinderance that otherwise inhibits nucleophilic attack
- 2 steps & involves carbocation intermediate
- in the question, its a teriary carbon which is mostly involved in SN1 mechanism, therefore the first step will be the formation of a carbocation as iodide leaves
Transesterfication
- is the exchange of alcohol and alkoxyl R groups between an ester and an alcohol
Transesterfication acid-catalyzed & base-catalyzed
Saponification
- is the removal of an alkoxy group from an ester by a base.
- It forms a carboxylic acid and an alcohol
Gas chromatography
- BP
- interaction
- retention time
- Gas chromatography separates compounds primarily by boiling point.
- Molecules with low boiling points elute before those with high boiling points.
- For compounds with the same number of carbon atoms, alkanes have the lowest boiling point, followed by aldehydes and ketones, alcohols, and carboxylic acids.
example of a carboxylic acid would be acetic acid and thus would have a high BP, longer retention time, eluting late
Chromatography technique and whats the best way to seperate ethyl acetate from image?
- Chromatography techniques separate compounds based on their affinity for a stationary phase over a mobile phase.
- In general, adjusting experimental parameters such that molecules have more time to interact with the stationary phase improves the resolution of each compound.
Answer: running the mixturethrough a longer column
Which atom indicated in the image would be most readily deporotonted if compound 2 reacts with NaOH?
- ASNWER: I
Explained:
- for a compound in solution, as the pH is increased, the most acidic functional group will be deprotonated first
- Acidity of a functional group is determined by the stability of the conjugate base it forms.
- Factors such as the size and electronegativity of the proton-donating atom as well as the ability to stabilize the negatively charged atom in the conjugate base by charge distribution or resonance contribute to acidity.
- sites that better stabilize a negative charge are more likely to deprotonated (more acidic)
Atoms I–IV in Compound 2 are oxygen (alcohol), sp2 carbon (alkene), sp3 carbon (alkane), and methyl ketone (α-H), respectively. Upon deprotonation, the negatively charged atom in the conjugate base of these functional groups is oxygen for alcohols and carbon for the others. Because oxygen is more electronegative than carbon, a negative charge on oxygen in the conjugate base of the alcohol is more stabilized than the charge on the carbon atoms in the three other functional groups. Therefore, the alcohol oxygen atom has the most acidic proton and will be deprotonated when reacted with NaOH.
Oxidation of alcohols
- TERTIARY ALCOHOLS CANNOT BE BE READILY OXIDIZED (whereas secondary and primary can be)
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Oxidation of carbon atoms involves the loss of bonds to hydrogen and is frequently accompanied by the formation of bonds to oxygen.
- Because no bonds to hydrogen can be lost, additional bonds to oxygen would result in a carbon atom with five bonds. Carbon can participate in only four bonds, so this is not possible. Therefore, tertiary alcohols cannot be oxidized by losing bonds to hydrogen or by gaining bonds to oxygen.
- Primary and secondary alcohols can be oxidized to carboxylic acids and ketones, respectively, but tertiary alcohols are in their highest possible oxidation state (so cannot be readily oxidized)
(tertiary structures are less acidic than secondary alcohols because tertiary alcohols are more sterically hindered-which makes ts conjugate base less stable than that of secondary alcohol) BUT ACIDITY DOES NOT AFFECT A MOLECULE’S ABILITY TO BE OXIDIZED
How can sucrose be cleaved?
- A glycosidic bond is the linkage between monosaccharides and is made up of a hemiacetal or hemiketal from one sugar and a hydroxyl group of another molecule.
- The glycosidic bond is broken through a hydrolysis reaction in which water cleaves the bond, breaking up the sugar molecule.
alpha & beta glycosidic linkages
Tolleens test
- Tollens test is used to identify the presence of aldehydes and hydroxy ketones (including reducing sugars, which have a free anomeric carbon), and uses the oxidizing agent [Ag(NH3)2]+ to oxidize aldehydes to carboxylic acids
- Ketoses can undergo tautomerization via an enediol intermediate to their aldose form, resulting in a positive Tollens test and the formation of a silver mirror
- ketoses such as fructose are expected to give a positive Tollens test because ketoses tautomerize aldoses
___________involves the change in configuration (alpha & beta) in sugars about the anomeric carbon in whch the hemiacetal opens up to its linear form and then cyclizes back to a hemiacetal in the opposite configuration
MUTATORATION
- Hydride reagents, such as LiAlH4, act as__________ and attack ____________ centers, most commonly ____________
- nucleophiles
- electrophilic
- carbonyl carbons (C=O)
- These reagents cause a reduction of the electrophile, decreasing the number of bonds to electronegative atoms and increasing the number of bonds to less electronegative atoms
Organic Reducnig agents
Infrared Spectroscopy
- Infrared spectroscopy is a technique used to determine the functional groups present in a sample; the data collected are plotted as percent transmittance vs. wavenumber.
- The signals in the spectrum correspond to bond-stretching vibrations and rotations at a certain frequency, and the signal intensity is dependent on the amount of energy absorbed.
1. the signals corresponding to stretching vibrations and rotations
2. amount of light absorbed at a certain frequency
3. the relative amount of energy needed to stretch a bond
carboxylic acid derivatives
Amides are carboxylic acid( CA )derivatives made up of what?
made up of a carbonyl bonded to a nitrogen atom
these compounds form the condensation of ay CA derivative with a primary or a secondary amine
Enamine
- is a functional group that contains nitrogen linked to a carbon that is double bonded to another carbon and is not a CA derivative
- it is formed by the addition of a secondary amine to an aldehyde or ketone
Reactivity of carbonyls trend
Which carbonyl carbon is most likely to undergo hydorlysis when treated with a weak acid at room temperature, an ester or amide?
ESTERS
- Esters and amides can both undergo hydrolysis via nucleophilic substitution to form alcohols and amines, respectively. However, due to carbonyl reactivity differences, different conditions are required for hydrolysis.
- Carbonyls are electrophiles whose reactivity depends on whether an electron donating or electron withdrawing atom is bonded to the carbonyl carbon.
- Electron donating atoms decrease reactivity and increase the electron density around carbonyl carbons whereas electron withdrawing atoms increase reactivity and decrease the electron density around carbonyl carbons.
- More reactive carboxylic acid derivatives can hydrolyze under mild conditions whereas less reactive derivatives require harsh conditions.
Cyclic amides (lactams)
- example: β-lactams are a class of cyclic amides that exhibit significant ring strain due to their small ring size, making the carbonyl carbon more reactive than noncyclic amides.
- The nitrogen atom in a β-lactam is trigonal pyramidal rather than planar and has decreased resonance.
- Unlike noncyclic amides, this nitrogen is sp3 hybridized and substantially more reactive.
Hybridization
Nucleophilic acyl substitution
- Carboxylic acid (CA) derivatives can undergo nucleophilic acyl substitution.
- Anhydrides are cleaved in this reaction into a CA and an ester or an amide.
- Cyclic anhydrides are converted into a single hydrocarbon chain with a CA on one end and an ester or an amide on the other.
Ester Hydrolysis
- Litmus paper is used as a pH indicator.
- In acidic solutions (pH <7), blue litmus paper will turn red; in basic solutions (pH >7), red litmus paper will turn blue.
- The addition of an acid to a solution will protonate bases such as carboxylate ions.
example of ester hydrolysis
- Hydrolysis of an ester with the base LiOH creates a carboxylate ion and will have a pH >7 because LiOH is a strong base.
- Protonation of the carboxylate ion requires the addition of an acid such as HCl.
- The addition of acid to the basic solution causes the pH of the reaction mixture to decrease, becoming more acidic.
- Carboxylic acids are weak acids with pKa values near 4, so the pH of a solution of carboxylate ions must be brought below 4 to fully protonate the ions.
- The pH of the mixture can be monitored using litmus paper. When enough acid is added to the reaction and carboxylate ions become protonated, the blue litmus paper will turn red.
How many isoprenes in the image attached?
6
Terpenes are a type of lipid made up of branched five-carbon units known as isoprenes. Terpenes are classified according to the number of isoprene units in the molecule.
terpene classification
What is the product of the reaction of butanoic acid and ethanol under acidic conditions?
- answer: ethyl butanoate
- Esters are carboxylic acid derivatives and may be formed by the condensation of a carboxylic acid and an alcohol under acidic conditions (Fischer esterification).
- Esters are named by using the alkyl group from the alcohol as a substituent prefix and the carboxylic acid as the root name, with the –ic acid suffix replaced by –ate.
Fischer esterfication
Strecker Synthesis
- The Strecker synthesis is used to generate α-amino acids from an aldehyde using ammonium chloride (NH4Cl) and potassium cyanide (KCN).
- Because the imine intermediate formed during the reaction is planar (no stereocenters), nucleophilic addition can occur from either above or below the plane. Therefore, the Strecker synthesis is not a stereospecific reaction and produces a mixture of L- and D-amino acids.
Strecker synthesis steps
- The first step of the reaction proceeds with protonation of the carbonyl oxygen by ammonium (NH4+), followed by nucleophilic attack of the carbonyl carbon by ammonia (NH3), resulting in dehydration and imine formation.
- In the second step, a cyanide anion is added to the imine to form an aminonitrile.
- Finally, in the third step, the nitrile (R–CN) nitrogen is protonated, and two water molecules add to the nitrile carbon in succession, eliminating ammonia from the nitrile and forming the carboxylic acid of the α-amino acid.
Gabriel Synthesis
Esters Example:
- Esters are carboxylic acid derivatives that can be formed from a carboxylic acid and an alcohol through a Fischer esterification.
- The ester oxygen bonded to both the carbonyl carbon and the alkyl group originates from the alcohol, a fact that can be confirmed by isotopically labeling the hydroxyl oxygen of the alcohol starting reagent.
- The reaction begins with protonation of the carbonyl oxygen, followed by nucleophilic attack of the carbonyl carbon by the alcohol. A proton is then transferred from the alcohol oxygen to the hydroxyl group of the carboxylic acid, resulting in water (a good leaving group). Lastly, the carbonyl oxygen is deprotonated to yield an ester.
in a base-catalyzed aldol reaction, what can be the best intermediate?
- Protons α to a carbonyl are more acidic than other protons bonded to a carbon atom because they are electron-deficient and can be abstracted by a base to yield an anion.
- The charge is delocalized to form a more stable resonance structure, known as an enolate.
- example: in acetyl-CoA, the methyl (CH3) group is alpha to a type of carbonyl called thioester, which makes it susceptible to deprotonation by a base
- the resulting negative chage can be trasferred from the alpha-carbon to the oxygen of the thioester, producing a stabe enolate
Do aldehydes and ketones tend to have a higher or lower boiling points than their corresponding alcohols?
- lower, becuase they cannot form hydrogen bonds with each other
- Ketones and aldehydes contain carbonyls, which create dipoles and increased intermolecular forces.
- Because the oxygen of the carbonyl can accept but not donate hydrogen bonds, ketones and aldehydes cannot form hydrogen bonds with themselves.
- Alcohols have dipoles and can form hydrogen bonds with themselves, resulting in a higher boiling point than ketones and aldehydes.
- The boiling point of a substance depends on the strength of the intermolecular forces holding molecules together. Stronger intermolecular forces require more energy to break and therefore contribute to higher boiling points.
Kinetic vs thermodynamic products example
- Kinetic enolates require low activation energy and form rapidly under low temperature and bulky base reaction conditions, whereas the less-substituted α-carbon is deprotonated.
- The base-catalyzed aldol condensation has four steps: formation of an enolate, nucleophilic addition to a carbonyl, protonation, and acid- or base-catalyzed dehydration to form an α,β-unsaturated carbonyl.
Thermodynamic enolate & kinetic enolate
carbonyl groups of aldehydes and ketones can be protected from nucleophiles by reaction how?
- with 2 equivalents of an alcohol to give an acetal (for aldehydes) & a ketal (for ketones)
- reaction with one alcohol equivalent (the halfway point) produces one -OR group & one -OH group in place of the carbonyl, generating a hemiacetal (for aldehydes) or hemiketal (for ketones)
1,2-diol & 1,3-diols
Reagents to aking a benzoic acid from benzaldehyde?
- Aldehydes are readily oxidized to carboxylic acids in aqueous environments by a number of oxidizing agents, including chromium compounds.
- Aldehydes are believed to go through a hydrate intermediate before being oxidized to a carboxylic acid.
- _CrO3, KMnO4, and H2CrO4 can all oxidize aldehydes to carboxylic acids because these reagents are all in aqueous environments,_ meaning they contain the water necessary to convert the aldehyde to the hydrate intermediate that is oxidized to the carboxylic acid.
- The anhydrous oxidizing reagent pyridinium chlorochromate (PCC) cannot oxidize aldehydes because it lacks the water needed to form the hydrate intermediate.
Retro-aldol reaction
- The retro-aldol reaction is the reverse of the aldol reaction, in which the bond between an α- and a β-carbon is broken into two fragments.
- Aldehydes are named after the number of carbon atoms in the longest carbon chain, replacing the –e in the parent alkane name with the suffix –al.
- Substituent names precede the chain name and are numbered from the carbonyl carbon.
What functional group forms in an acid-catalyzed reaction between a secondary amine and a ketone?
- Enamine=answer
- Ketones and aldehydes undergo acid-catalyzed addition of 1° and 2° amines to form imines or enamines, respectively.
- Like ketones and aldehydes, imines contain a carbon-heteroatom double bond (nitrogen in imines, oxygen in ketones and aldehydes).
- Enamines are the nitrogen analogue to enols, containing a carbon-carbon double bond.
Hydroxyl substituents in alcohols are poor leaving groups, and are not readily eliminated during substitution reactions. Which of the following modifications would make the hydroxyl substituent of an alcohol a better leaving group?
- answer: addition of an acid
- Hydroxyl groups in alcohols are poor leaving groups because hydroxide is a strong base.
- Hydroxide must be converted to a good leaving group to participate in substitution reactions.
- The addition of acid to an alcohol protonates the hydroxyl group, forming a bound, positively charged water molecule.
- Water is a weak base and forms a neutral molecule upon leaving, which makes it a good leaving group.
oxidation of alcohols
- Oxidation of an organic molecule requires a decrease in the number of C–H bonds and an increase in the number of C–O bonds.
- Primary and secondary alcohols can be oxidized to aldehydes and ketones, respectively, by various oxidizing agents.
- Tertiary alcohols cannot be oxidized because the tertiary carbon does not have any C–H bonds to lose.
Oxidation of aldehydes
- Oxidation of organic molecules often increases the number of C–O bonds and decreases the number of C–H bonds in the molecule.
- Oxidizing agents, such as chromic acid, will convert primary alcohols to aldehydes, secondary alcohols to ketones, and aldehydes to carboxylic acids.
Reduction of a carbonyl with LiAlH4
example: CH3C(O)CH2CH3
- Reduction of an aldehyde or ketone with a reducing agent, such as LiAlH4, reduces the number of C–O bonds (dec oxidation state) and increases the number of C–H bonds to form a 1° and 2° alcohol, respectively.
- Ketones are named using the suffix –one; alcohols are named using the suffix –ol.
Reducing agents
Protection of alcohols
- Chromic acid is an oxidizing agent that can oxidize primary alcohols and aldehydes to carboxylic acids and secondary alcohols to ketones.
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Protecting groups are organic molecules added to a functional group to prevent them from reacting under a given set of reaction conditions.
- *Protecting groups tend to be stable against oxidation and reduction.*
