Organic Chemistry HL only

O C OR 20.1 Introduction Esters More functional groups Contain functional group COOR, where R = an alkyl group such as CH3 (methyl group).

name of this alkyl group forms first part of name. Naming esters C O R name of this alkyl group forms first part of name. second part of name is derived from the name of the carboxylic acid salt.

Name the following ester ethyl propanoate C O CH3CH2 CH2CH3 this is an ethyl group. this is a propanoate group.

Name the following ester methyl methanoate C O H CH3 this is a methyl group. this is a methanoate group.

Draw out the structure of ethyl benzenecarboxylate C O C6H5 CH2CH3

Draw out the structure of ethyl butanoate C O CH3CH2CH2 CH2CH3

C NH2 O Amide suffix –amide For example CH3CONH2 ethanamide

We will also come across amides where one of the hydrogen atoms attached to the N atom is replaced by an alkyl group these are referred to as N-substituted amides. The structure shown is N-methylethanamide. CH3C O NHCH3

C N Nitriles For example CH3CH2CN propanenitrile suffix -nitrile For example CH3CH2CN propanenitrile Remember to include the nitrile group as part of the C chain.

C NH2 Amines For example CH3NH2 methylamine H2NCH2COOH suffix –amine or prefix amino- For example CH3NH2 methylamine H2NCH2COOH aminoethanoic acid

-amine suffix tends to be only used for short chain amines e. g -amine suffix tends to be only used for short chain amines e.g. propylamine. Amino- prefix is used for longer names so CH3CH2CH2CH2CH2CH2NH2 is 1-aminohexane What would the structure of 2-aminohexane look like? For secondary and tertiary amines, the longest alkyl chain attached to the N is identified and we then name in a similar fashion to N-substituted amides. So for example:

N H CH3 C2H5 N-methylethanamine N H C2H5 C3H7 N-ethylpropanamine N C2H5 CH3 N,N-dimethylethanamine

20.2 Nucleophilic Substitution Reactions The examples of nucleophilic substitution looked at earlier are not the only examples. Other nucleophiles which will react with halogenoalkanes include H2O, NH3 and CN-. Using water as the nucleophile would produce an alcohol but the reaction is much slower than with hydroxide ions as the hydroxide ions have a negative charge so are attracted more strongly to the d+ on the C atom.

tertiary > secondary > primary The order of reactivity with these nucleophiles is: C – I > C – Br > C – Cl As explained earlier, this is due to increasing bond strengths. From experimentation we also found: tertiary > secondary > primary This is harder to explain but it is thought that the activation energy to form the tertiary carbocation intermediate in SN1 is less than that required to form the transition state in SN2.

1. With cyanide ions CH3CH2I (ethanol) + CN-(aq) CH3CH2CN + I- propanenitrile This is an important reaction for organic chemists as it increases the length of the carbon chain. The cyanide group can then be converted to other functional groups:

CH3CH2COOH CH3CH2CN CH3CH2CH2NH2 H2O propanoic acid CH3CH2CN H2 Ni CH3CH2CH2NH2 1-aminopropane or propylamine The mechanism for the nucleophilic substitution reaction is:

H CH3 NC Br - Br C CH3 H Transition state C - CN NC H CH3 C Br - + SN2

2. With ammonia CH3CH2Br + NH3 CH3CH2NH2 + HBr aminoethane

- + SN2 C H CH3 Br H2N H Br C CH3 H Transition state NH3 H2N H CH3 C

C2H5NH2 + C2H5Br  (C2H5)2NH + HBr There is still a lone pair of electrons on the N atom in ethylamine so this can react with a further molecule of bromoethane to form a secondary amine. C2H5NH2 + C2H5Br  (C2H5)2NH + HBr N-ethylethanamine The secondary amine still contains a lone pair so further reaction occurs forming a tertiary amine.

(C2H5)2NH + C2H5Br  (C2H5)3N + HBr N,N-diethylethanamine The tertiary amine still contains a lone pair so further reaction occurs forming a quaternary ammonium salt. (C2H5)3N + C2H5Br  (C2H5)4N+Br- tetraethylammonium bromide

A better method for making a primary amine involves catalytic (Ni catalyst) hydrogenation of a nitrile. The yield is larger and there is no further reaction possible. CH3CH2CN + 2H2  CH3CH2CH2NH2 propanenitrile propylamine

20.3 Elimination Reactions When warm aqueous sodium hydroxide reacts with a halogenoalkane, an alcohol is formed via a nucleophilic substitution reaction. The hydroxide ions behave as a nucleophile so: C2H5Br + OH-(aq)  C2H5OH + Br- However, if the sodium hydroxide is dissolved in hot ethanol and the mixture heated under reflux a different product is formed, an alkene.

C2H5Br + OH-(alc)  C2H4 + H2O + Br- This is an elimination reaction and the sodium hydroxide is behaving as a base (proton acceptor) rather than a nucleophile. C2H5Br + OH-(alc)  C2H4 + H2O + Br- Overall HBr is eliminated (removed) from the halogenoalkane. There are two possible mechanisms known as E1 and E2.

E1 - - - E2 CH3 H C Br H H CH3 C C H + H acting as a base OH CH3 H C propene Br - OH - H OH E2 acting as a base

RCH2CH2X alcohol nucleophilic substitution RCH3CH2OH + X- + OH- (aqueous) hydroxide acts as a nucleophile RCH2CH2X + OH- (ethanol) hydroxide acts as a base elimination RCH=CH2 + H2O + X- alkene

20.4 Condensation Reactions A condensation reaction involves two molecules reacting together to form a larger molecule with the elimination of a small molecule such as water or hydrogen chloride. Esterification is an example of a condensation reaction. Esterification involves the formation of an ester from a carboxylic acid and an alcohol in the presence of concentrated sulphuric acid.

For example, ethyl ethanoate is formed when ethanoic acid and ethanol are mixed in the presence of a few drops of concentrated sulphuric acid. CH3COOH + C2H5OH Ý CH3COOC2H5 + H2O The conc sulphuric acid acts as a catalyst and also shifts the equilibrium to the right hand side by removing water. The water is formed from the –OH of the carboxylic acid combining with the H from the alcohol.

Uses of Esters Solvents – esters are volatile and polar. Polarity means that they act as solvents for many polar organic compounds. Low b.p. means that they evaporate from less volatile solutes. e.g. ethyl ethanoate is used as the solvent in glues such as polystyrene cement. Plasticisers – plastics are often not flexible as chains cannot move over each other easily. The addition of plasticisers allows chain movement. Over time these additives escape so the plastic becomes brittle and stiff.

pentyl ethanoate – pear 2,2,dimethylpropyl ethanoate – banana Food flavourings – many esters have a sweet, often fruity smell and are used as artificial food flavouring. For example – pentyl ethanoate – pear 2,2,dimethylpropyl ethanoate – banana octyl ethanoate – orange ethyl butanoate – pineapple pentyl pentanoate – apple

RCOOH + R’NH2  RCONHR’ + H2O Amide Formation Carboxylic acids also undergo a condensation reaction with amines (or ammonia) to form an amide. The –OH group from the carboxylic acid reacts with one of the hydrogen atoms attached to the N atom to form water. RCOOH + R’NH2  RCONHR’ + H2O R’ = an H atom  amide = an alkyl group  N-substituted amide

CH3C O OH CH3C O NH2 + NH3  + H2O CH3C O OH CH3C O NHCH3 + CH3NH2  Why is this reaction of biological importance?

C COOH R H NH2 carboxylic acid group primary amino group Two amino acid molecules can react together to form an amide. What is the correct IUPAC name of the two amino acids on the next slide? Deduce the structure of two possible condensation products.

C COOH H NH2 glycine C COOH CH3 H NH2 alanine

The two amino acid units are held together by what is known as the peptide bond (peptide link). The two condensation products are known as dipeptides.

H2NCHCOOH R1 + H2NCHCOOH R2 H2NCH – C – N – R1 CHCOOH R2 H O + H2O peptide link H2NCH – C – N – R1 CHCOOH R2 H O + H2O

On one end of the dipeptide is an amine group and on the other is a carboxylic acid group so further reaction at each end is possible. The result is a CONDENSATION POLYMER. In this instance a polyamide.

Condensation Polymers Condensation polymers are formed by the reaction between molecules having two functional groups, involving the loss of small molecules such as H2O, CH3OH or HCl. There are two types of condensation polymer: polyamides polyesters

Polyamides The reaction between a dicarboxylic acid and a diamine leads to the formation of a polyamide. For example Nylon-6,6 is formed from hexanedioic acid and hexane-1,6-diamine. Nylon-6,6 is so called as it is made from a 6 C diacid and a 6 C diamine.

nHOC(CH2)4COH O + nH2N(CH2)6NH2 HN(CH2)6NHC(CH2)4C O - C(CH2)4C repeating unit + 2n-1H2O

As with naturally occurring polyamides (polypeptides and proteins), synthetic polyamides are susceptible to hydrolysis and can be broken down into component monomer units. Consequently, polyamides are biodegradable.

Polyesters The reaction between a dicarboxylic acid and a diol leads to the formation of a polyester. The most important example is terylene which can be formed from benzene-1,4-dicarboxylic acid and ethane-1,2-diol.

benzene-1,4-dicarboxylic acid COH HOC O + HOCH2CH2OH benzene-1,4-dicarboxylic acid ethane-1,2-diol -H2O C OCH2CH2O OC O n + 2n-1H2O

Polyesters (like esters) are susceptible to hydrolysis and can be broken down into their component monomers. So polyesters (like polyamides) are biodegradeable.

20.6 Stereoisomerism Stereoisomers are compounds that have the same structural formula but their atoms are arranged differently in space. There are two types: Geometrical isomerism and, Optical isomerism.

Geometrical Isomerism Arises due to lack of rotation at C = C in alkenes. C = C H R C = C R H cis isomer trans isomer It is not possible to have geometrical isomerism when there are two identical groups attached to the same C atom in the double bond. So but-1-ene does not exist as geometric isomers whereas but-2-ene does.

The C atoms in the double bond are sp2 hybridised. One of the bonds is a  bond formed by overlap of two sp2 hybrid orbitals. The other is a  bond formed by sideways overlap of p orbitals. These must be in the same plane to overlap. Any attempt to rotate will mean that these will no longer be in the same plane which would mean the bond would be broken.

Consequences of Geometric Isomerism Melting points are influenced by how closely molecules pack together. cis but-2-ene melts at -139 °C, trans but-2-ene melts at -106 °C. cis 1,2-dichloroethene melts at 60 °C, trans 1,2-dichloroethene melts at 48 °C. Draw out the cis and trans isomers of but-2-ene-1,4-dioic acid.

Which of the two isomers will have the highest melting point? Why? trans isomer has an m.p. of 286 °C, cis isomer has m.p. of 131 °C. trans isomer has strong hydrogen bonding between molecules but cis isomer has strong hydrogen bonding within molecules. Which of the two isomers will dehydrate more easily? Why? What will the product be?

Geometric isomerism is also found in cyclic compounds where the rigid structure of the ring prevents free rotation. For example: 1,2-dichlorocyclopropane exists as cis and trans isomers. Draw them! How many isomers are there of dichlorocyclobutane? Draw them and name them.

Optical Isomerism Arises when there are four different groups attached to a carbon atom. This means that the molecule has no centre, plane or axis of symmetry. The molecule is said to be CHIRAL and possesses an asymmetric carbon atom. Two tetrahedral arrangements in space are possible so that one is the mirror image of the other.

For example 2-hydroxypropanenitrile C CN H3C H OH C NC CH3 H HO mirror Stereoisomers of this type are referred to as enantiomers.

Enantiomers have exactly the same physical properties except for their effect on the plane of plane-polarised light. As a result they are referred to as optically active. Plane polarised light is made of waves vibrating in one plane only. When it is passed through a solution of a chiral molecule, the light emerges with its direction of polarisation changed. One enantiomer rotates the light in a clockwise direction and is referred to as the (+) isomer or dextrorotatory (d). Its mirror image rotates the light by exactly the same angle but in an anti-clockwise direction. It is referred to as the (-) isomer or laevorotatory (l).

A mixture of equal amounts of both enantiomers is optically inactive because the effects of each enantiomer is cancelled out. Such a mixture is called a racemic mixture or racemate.

Draw out the structures of the following molecules and state whether they are optically active or not. If they are optically active, highlight any chiral carbon atoms. Butan-2-ol 3-methylhexane 3-methylpentan-3-ol 4. 2-aminopropanoic acid 5. 1-aminopropan-2-ol 6. 2-methylpropan-2-ol

C H OH butan-2-ol C H CH3 3-methylhexane

C H CH3 OH 3-methylpentan-3-ol C NH2 H OH O 2-aminopropanoic acid

C H OH NH2 1-aminopropan-2-ol C H OH CH3 2-methylpropan-2-ol

Many naturally occurring molecules exist as single enantiomers, for example most amino acids, such as 2-aminopropanoic acid (alanine). The chemical properties of enantiomers are identical except in reactions with other optically active substances. Enzymes are stereospecific. This means they will catalyse the reactions of only one of a pair of isomers. (+)-carvone is found in spearmint oil, (-)-carvone is the main constituent of caraway seed oil. Acts much faster when administered as the (+)-enantiomer. thalidomide