Organic Chemistry Flashcards

Question

How are primary alcohols oxidised

Answer 1

All primary alcohols are oxidized by acidied potassium dichromate(VI), rst to aldehydes then to carboxylic acids.

Answer 2

Secondary alcohols are oxidized to ketones, which cannot undergo urther oxidation.

Answer 3

Tertiary alcohols cannot be oxidized by acidied dichromate(VI) ions as they have no hydrogen atoms attached directly to the carbon atom containing the -OH group. It is not true to say that tertiary alcohols can never be oxidized, as they burn readily, but when this happens the carbon chain is destroyed.

Answer 4

Because of the greater electronegativity of the halogen atom compared with the carbon atom halogenoalkanes have a polar bond. Reagents that have a non-bonding pair o electrons are attracted to the electron-decient carbon atom in halogenoalkanes and a substitution reaction occurs. Such reagents are called nucleophile The halogen breaks off as a negatve ion

Answer 5

The extra stability provided by the delocalization o the electrons in the benzene ring means that benzene, unlike simple alkenes, does not readily undergo addition reactions

Answer 6

Benzene has a high electron density so reacts with electrophiles Benzene + chlorine in the presence o aluminium chloride (the electrophile is Cl+ ) to orm chlorobenzene Benzene + nitric acid in the presence o suluric acid and heat (the electrophile is NO + ) to orm nitrobenzene and water.

Answer 7

Alcohols react with carboxylic acids in the presence o a small amount o concentrated suluric acid, which acts as a catalyst, to orm an ester and water

Answer 8

A condensation reaction involves the reaction between two molecules to produce a larger molecule with the elimination o a small molecule such as water or hydrogen chloride.

Answer 9

In a process known as transesterication fats can react with alcohols in the presence o sodium hydroxide (which acts as a catalyst) to orm alkyl esters, which can be used as biouel and glycerol

Answer 10

Protic solvents which are polar, such as water or ethanol, avour the SN1 mechanism as they support the breakdown of halogenoalkanes into carbocations and halide ions

Answer 11

Aprotic solvents which are less polar such as ethoxyethane, favour an S 2 mechanism involving a transition state

Answer 12

The nature ofthe nucleophile The eectiveness o a nucleophile depends on its electron density. Anions tend to be more reactive than the corresponding neutral species. The nature of the halogen Iodoalkanes react aster than bromoalkanes, which in turn react aster than chloroalkanes. This is due to the relative bond enthalpies as the CI bond is much weaker than the CCl bond and thereore breaks more readily. The nature ofthe halogenoalakane Tertiary halogenoalkanes react aster than secondary halogenoalkanes which in turn react aster than primary halogenoalkanes. The SN1 route which involves the ormation o an intermediate carbocation is aster than the S 2 route which involves a transition N state with a relatively high activation energy.

Answer 13

The reaction can occur in the dark which suggests that a ree radical mechanism is not involved. The double bond in the ethene molecule has a region o high electron density above and below the plane o the molecule. Hydrogen bromide is a polar molecule due to the greater electronegativity o bromine compared with hydrogen. The hydrogen atom (which contains a charge o +) rom the HBr is attracted to the double bond and the HBr bond breaks, orming a bromide ion. At the same time the hydrogen atom adds to one o the ethene carbon atoms leaving the other carbon atom with a positive charge. A carbon atom with a positive charge is known as a carbocation. The carbocation then combines with the bromide ion to orm bromoethane. Because the hydrogen bromide molecule is attracted to a region o electron density it is described as an electrophile and the mechanism is described as electrophilic addition.

Answer 14

Electrophilic addition also takes place when bromine adds to ethene in a non-polar solvent to give 1,2-dibromoethane. Bromine itsel is non-polar but as it approaches the double bond o the ethene an induced dipole is ormed by the electron cloud.

Answer 15

Asymmetric alkenes contain different groups attached to the carbon atoms o the C=C bond.

Answer 16

MARkOVNIkOVfS RULE allows us to predict the. major products of. the reaction between an asymmetrical alkene and a hydrogen halide. The rule states that the Hydrogen from the hydrogen halide will bind with the carbon atom ( from the carbon atoms with double bonds) with most hydrogens bonded. The halogen will bind to the other carbon atom which was oriiginally in the carbon bond.

Answer 17

For tertiary halogenoalkanes the predominant mechanism is SN1 and for primary halogenoalkanes it is SN2. Both mechanisms occur for secondary halogenoalkanes.

Answer 18

The rate determining step (slow step) in an SN1 reaction depends only on the concentration of the halogenoalkane, rate = k[halogenoalkane]. For SN2, rate = k[halogenoalkane][nucleophile].

Answer 19

Benzene is the simplest aromatic hydrocarbon compound (or arene) and has a delocalized structure of π bonds around its ring. Each carbon to carbon bond has a bond order of 1.5. Benzene is susceptible to attack by electrophiles.

Answer 20

When hydrogen ions react with propene two dierent carbocation intermediates can be ormed. The first one has the general formula RCH+ and is a primary carbocation The second one has two R- groups attached to the positive carbon ion R2CH+ and is known as a secondary carbocation A tertiary carbocation has the general ormula R3CH+ . The R-groups (alkyl groups) tend to push electrons towards the carbon atom they are attached to which tends to stabilize the positive charge on the carbocation. This is known as a positive inductive effect. This effect will be greatest with tertiary carbocations and smallest with primary

Answer 21

Consider the reaction o iodine chloride ICl with but-1-ene. Since iodine is less electronegative than chlorine the iodine atom will act as the electrophile and add frst to the alkene. The major product will thus be 2-chloro-1-iodobutane (page 91 of revision book)

Answer 22

Benzene reacts with a mixture o concentrated nitric acid and concentrated suluric acid when warmed at 50 C to give nitrobenzene and water. Note that the temperature should not be raised above 50 C otherwise urther nitration to dinitrobenzene will occur. The electrophile is the nitryl cation NO2 + nitronium ion). The concentrated suluric acid acts as a catalyst. Its unction is to protonate the nitric acid which then loses water to orm the electrophile. In this reaction nitric acid is acting as a base in the presence o the more acidic suluric acid. The NO2+ is attracted to the delocalized pi bond and attaches to one o the carbon atoms. This requires considerable activation energy as the delocalized pi bond is partially broken. The positive charge is distributed over the remains o the pi bond in the intermediate. The intermediate then loses a proton and energy is evolved as the delocalized pi bond is reormed. The proton can recombine with the hydrogensulate ion to regenerate the catalyst

Answer 23

Sodium tetrahydridoborate NaBH4 can be used in the presence o protic solvents such as water or ethanol but is ineffectual at reducing carboxylic acids. The stronger reducing agent lithium aluminium hydride LiAlH4 must initially be used in aprotic solvents such as ether as it reacts with water, then the reaction is acidied to obtain the product. Another reducing agent that can be used is hydrogen itsel in the presence o a nickel, platinum or palladium catalyst.

Answer 24

Aldehydes are reduced to primary alcohols, with NaBH4 or H+ as a catalyst

Answer 25

Ketones are reduced to secondary alcohols, with NaBH4 or H+ as a catalyst

Answer 26

Carboxylic acids are reduced to primary alcohols, with LiAlH4 in ethers or H+ as catalyst

Answer 27

The reduction of nitrobenzene to phenylamine is usually carried out in two stages. Stage 1. Nitrobenzene is refuxed with a mixture o tin and concentrated hydrochloric acid. The tin provides the electrons by acting as the reducing agent and the product is the phenylammonium ion. Stage 2. The addition o sodium hydroxide solution releases the free amine

Answer 28

The more steps there are in an organic synthesis, then the lower the nal yield is likely to be, because some material will be lost during each step.

Answer 29

Isomers are compounds that are composed o the same elements in the same proportions but dier in properties because o dierences in the arrangement o atoms

Answer 30

Structural isomers share the same molecular ormula but have dierent structural ormulas.

Answer 31

Stereoisomers have the same structural ormula but dier in their spatial arrangement

Answer 32

A cis-isomer is one in which the substituents are on the same side o the double bond. In a trans-isomer the substituents are on opposite sides o the double bond.

Answer 33

Both cistrans and E/Z isomerism occur when rotation about a bond is restricted or prevented. This is because the double bond in an alkene is made up of a sigma and a pi bond. The pi bond is formed from the combination of two p orbitals, one from each of the carbon atoms. These two p orbitals must be in the same plane to combine. Rotating the bond would cause the pi bond to break so no rotation is possible. Cistrans isomerism can also occur in disubstituted cycloalkanes. The rotation is restricted because the CC single bond is part o a ring system.

Answer 34

E/Z terminology is quite easy to apply and depends on what are known as the Cahn-Ingold-Prelog (CIP) rules for determining the priority of the atoms or groups attached to the two carbon atoms of the double bond. In simple terms the higher the atomic number of the attached atoms to each carbon atom the higher the priority. If. both entities of highest priotiry lie on the same side we haze Z-isomerism, if both of highest priority are on on oppisite sides. there is E isomerism. (Page 95 in revision book)

Answer 35

Optical isomerism is shown by all compounds that contain at least one asymmetric or chiral carbon atom within the molecule, that is, one that contains our dierent atoms or groups bonded to it, also known as a stereocentre. The two isomers are known as enantiomers and are mirror images o each other

Answer 36

The two dierent isomers are optically active with plane-polarized light. Normal light consists o electromagnetic radiation which vibrates in all planes. When it is passed through a polarizing flter the waves only vibrate in one plane and the light is said to be plane-polarized. The two enantiomers both rotate the plane o plane-polarized light. One o the enantiomers rotates it to the let and the other rotates it by the same amount to the right.

Answer 37

I a molecule contains two or more stereocentres then several dierent stereoisomers are possible. They are known as enantiomers i they are mirror images and as diastereomers i they are not mirror images o each other. Diastereomerism occurs when two or more stereoisomers o a compound have dierent confgurations at one or more (but not all) o the equivalent stereocentres. This is particularly important with many sugars and some amino acids. Diastereomers have dierent physical properties and dierent chemical reactivity.

Answer 38

The optical activity o enantiomers can be detected and measured by an instrument called a polarimeter. It consists o a light source, two polarizing lenses, and between the lenses a tube to hold the sample o the enantiomer dissolved in a suitable solvent. When light passes through the frst polarizing lens (polarizer) it becomes plane-polarized. That is, it is vibrating in a single plane. When the sample is placed between the lenses the analyser must be rotated by X degrees, either clockwise (dextrorotatory) or anticlockwise (laevorotatory) to give light o maximum intensity. The two enantiomers rotate the plane o plane-polarized light by the same amount but in opposite directions. If both enantiomers are present in equal amounts the two rotations cancel each other out and the mixture appears to be optically inactive. Such a mixture is known as a racemic mixture or racemate.

Answer 39

Boiling point increases with increasing carbon number. This is because of the increased instantaneous induced dipoles causing stronger London (dispersion) forces between the molecules as their molecular size increases

Answer 40

A primary carbon atom is attached to the functional group and also to at least two hydrogen atoms. Molecules with this arrangement are known as primary molecules. For example, ethanol, C2H5OH, is a primary alcohol and chloroethane, C2H5Cl, is a primary halogenoalkane.

Answer 41

A secondary carbon atom is attached to the functional group and also to one hydrogen atom and two alkyl groups. These molecules are known as secondary molecules. For example, propan-2-ol, CH3CH(OH)CH3, is a secondary alcohol, and 2-chloroethane, CH3CHClCH3, is a secondary halogenoalkane.

Answer 42

A tertiary carbon atom is attached to the functional group and is also bonded to three alkyl groups and so has no hydrogen atoms. These molecules are known as tertiary molecules. For example, 2-methylpropan-2-ol

Answer 43

If the N is attached to one Alkyl group then it is primary, two secondary and three tertiary

Answer 44

All carbon–carbon bond lengths in benzene are equal and intermediate in length between single and double bonds as each bond contains a share of three electrons between the bonded atoms.

Answer 45

Addition reactions are energetically not favoured as they would involve disrupting the entire cloud of delocalized electrons. The resonance energy would have to be supplied and the product, without the delocalized ring of electrons, would be less stable. Benzene can instead undergo substitution reactions that preserve the stable ring structure

Answer 46

Addition reactions are energetically not favoured as they would involve disrupting the entire cloud of delocalized electrons. The resonance energy would have to be supplied and the product, without the delocalized ring of electrons, would be less stable. Benzene can instead undergo substitution reactions that preserve the stable ring structure

Answer 47

The amount of energy that would have to be supplied in order to overcome the special stability of the delocalised ring

Answer 48

The intermolecular forces decrease when branching occurs as the surface area is reduced

Answer 49

Alkanes contain only C –– C and C –– H bonds, which are both strong bonds: C –– C = 348 kJ mol–1 and C –– H = 412 kJ mol–1. So these molecules will only react in the presence of a strong source of energy, strong enough to break these bonds. As a result, alkanes are stable under most conditions. The C –– C and C –– H bonds are also characteristically non-polar, so these molecules are not susceptible to attack by most common reactants. These two factors taken together mean that alkanes are generally of very low reactivity.

Answer 50

As the C : H ratio increases with unsaturation, there is an increase in the smokiness of the flame, due to unburned carbon.

Answer 51

The reaction does not take place in the dark as the energy of UV light is necessary to break the covalent bond in the chlorine molecule. This splits it into chlorine atoms, which each have an unpaired electron and are known as free radicals. It is known as photochemical homolytic fission. Once formed, these radicals will start a chain reaction in which a mixture of products including the halogenoalkane is formed.

Answer 52

First initiation occurs in which the diatomic Chlorine molecule is broken into two chlorine free radicals in the presence of UV light. This is known as photochemical homolytic fission. Next propagation occurs. Propagation reactions both use and produce free radicals. For example: Cl*+CH4 →CH3*+HCl CH3* + Cl2 → CH3Cl + Cl* There are many possible propagation steps, which all allow the reaction to continue. This is why this type of reaction is often called a chain reaction. Last step is termination. Termination reactions remove free radicals from the mixture by causing them to react together and pair up their electrons. For example: Cl*+Cl*→Cl2 CH3* + Cl*→ CH3Cl CH3* + CH3* → C2H6

Answer 53

Hydrogen reacts with alkenes to form alkanes in the presence of a nickel catalyst at about 150 °C.

Answer 54

Halogens react with alkenes to produce dihalogeno compounds. These reactions happen quickly at room temperature, and are accompanied by the loss of colour of the reacting halogen.

Answer 55

Hydrogen halides, such as HCl and HBr, react with alkenes to produce halogenoalkanes. These reactions take place rapidly in solution at room temperature All the hydrogen halides are able to react in this way, but the reactivity is in the order HI > HBr > HCl due to the decreasing strength of the hydrogen halide bond down Group 17. So HI, with the weakest bond, reacts the most readily.

Answer 56

The –– OH group is polar, and so increases the solubility in water of the molecules, relative to alkanes of comparable molar mass?

Answer 57

Unlike their parent acid and alcohol, esters have no free –– OH groups so they cannot form hydrogen bonds and so are mostly quite insoluble and form a layer on the surface.

Answer 58

Addition reactions, which would lead to loss of the stable arene ring, are generally not favoured as the products would be of higher energy than the reactants. Instead, substitution reactions, in which one or more of the hydrogen atoms is replaced by an incoming group, occur more readily as these lead to products in which the arene ring is conserved.

Answer 59

Because H2O lacks a negative charge it is a weaker nucleophile than OH–.

Answer 60

rate = k [halogenoalkane] [nucleophile]

Answer 61

The nucleophile attacks the electrophilic carbon atom on the opposite side from the leaving group, which causes an inversion of the arrangement of the atoms around the carbon atom

Answer 62

The SN2 mechanism is described as stereospecific because the three-dimensional arrangement of the reactants determines the three-dimensional configuration of the products. This happens because bond formation comes before bond cleavage in the transition state, so the stereochemistry of the carbon attacked is not lost.

Answer 63

Aprotic solvents are those which are not able to form hydrogen bonds as they do not contain –OH or –NH bonds, although they may have strong dipoles. This means they solvate the metal cations (Na+ ) rather than the nucleophile (OH ). The unsolvated, bare nucleophile has a higher energy state and this increases the reaction rate.

Answer 64

The alkyl groups around the carbon of the carbon–halogen bond causes what is called steric hindrance, meaning that these bulky groups make it difficult for an incoming group to attack this carbon atom. Instead, the first step of the reaction involves the halogenoalkane ionizing by breaking its carbon–halogen bond heterolytically. As the halide ion is detached, this leaves the carbon atom with a temporary positive charge, which is known as a carbocation intermediate. This is then attacked by the nucleophile in the second step of the reaction, leading to the formation of a new bond. Another reason which favours this mechanism in tertiary halogenoalkanes is that the carbocation is stabilized by the presence of the three alkyl groups, as each of these has an electron-donating or positive inductive effect. This stabilizing effect helps the carbocation to persist for long enough for the second step to occur.

Answer 65

rate = k [halogenoalkane]

Answer 66

The carbocation intermediate has a planar shape, which means the nucleophile can attack from any position in the second step. This means that the S stereospecific.

Answer 67

Polar, protic solvents contain – OH or – NH and so are able to form hydrogen bonds. They are effective in stabilizing the positively charged intermediate by solvation involving ion–dipole interactions

Answer 68

As the electronegativity of the halogens decreases in going down the group from fluorine to iodine, the carbon of the carbon–halogen bond becomes progressively less electron deficient and so less vulnerable to nucleophilic attack. So from this we would expect: fluoroalkane > chloroalkane > bromoalkane > iodoalkane Bond energy data show that the carbon– halogen bond decreases in strength from fluorine to iodine. As the substitution reaction involves breaking this bond, we would expect the ease of breaking bonds to be: C –– I > C –– Br > C –– Cl > C –– F Reaction rate data indicate that (b) above, the strength of the carbon–halogen bond, dominates the outcome here. The relative rate of reaction of the different halogens in halogenoalkanes when all other variables are kept constant is therefore: iodoalkanes > bromoalkanes > chloroalkanes > fluoroalkanes

Answer 69

-The carbon atoms of the double bond are sp2 hybridized, forming a planar triangular shape, with bond angle 120°. This is a fairly open structure that makes it relatively easy for incoming groups to attack. * The pi (π) bond represents an area of electron density above and below the plane of the bond axis. Because electrons in the π bond are less closely associated with the nuclei, it is a weaker bond than the σ bond and so breaks more easily during the addition reactions. * Because it is an area of electron density, the π bond is attractive to electrophiles, species that either are electron deficient or that become electron deficient in the presence of the π bond. When this bond breaks, reactants attach at each carbon atom:

Answer 70

Bromine is a non-polar molecule, but as it approaches the electron rich region of the alkene, it becomes polarized by electron repulsion. The bromine molecule is polarized by an alkene The bromine atom nearest the alkene’s double bond gains a δ+ charge and acts as the electrophile. The bromine molecule splits heterolytically, forming Br+ and Br−, and the initial attack on the ethene in which the π bond breaks is carried out by the positive ion, Br+ This step is slow, resulting in an unstable carbocation intermediate in which the carbon atom has a share in only six outer electrons and carries an overall positive charge. This unstable species then reacts rapidly with the negative bromide ion, Br−, forming the product 1,2-dibromoethane. For hydrogen halides the same process occurs with the H+ making the first attack after heterolytic fission!

Answer 71

The reaction has high activation energy and so proceeds rather slowly. This is because the first step in the mechanism, in which an electron pair from benzene is attracted to the electrophile, leads to a disruption of the symmetry of the delocalized π system. The unstable carbocation intermediate that forms has both the entering atom or group and the leaving hydrogen temporarily bonded to the ring. Loss of the hydrogen ion, H+, from this intermediate leads to the electrically neutral substitution product as two electrons from the C–H bond move to regenerate the aromatic ring.

Answer 72

The oxidation of thiols gives compounds called disulfides. Two thiol molecules are required for the oxidation to form the disulfide.