Question 5 of 5: Multi-Step Synthesis, Enantiomers & Double Benzylic Oxidation

Question 5: Multi-Step Synthesis, Enantiomers & Double Benzylic Oxidation (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

a) Both targets need the amine or acid group placed meta to a fixed methyl group. A methyl group installed by Friedel–Crafts alkylation is itself an ortho/para-director, so it cannot be used to place a second substituent meta to itself. The standard workaround is to install the target position first as a nitro group (a meta-director), alkylate meta to it, and only then convert the nitro group into whatever the target needs.

(i) 3-Methylaniline.

1. Nitrate benzene. $$\mathrm{C_6H_6 \xrightarrow{HNO_3/H_2SO_4} C_6H_5NO_2}$$
2. Friedel–Crafts methylate, directed meta by the nitro group. $$\mathrm{C_6H_5NO_2 \xrightarrow{CH_3Cl,\ AlCl_3} 1\text{-methyl-3-nitrobenzene}}$$
   Check: Friedel–Crafts alkylation is, in practice, very sluggish on a ring as strongly deactivated as nitrobenzene (nitrobenzene is even used as an FC solvent for this reason). This step is presented at the directing-group-logic level this "propose a synthesis" question is testing; a laboratory chemist would more likely reach the same meta relationship via a diazonium-salt blocking/replacement strategy.
3. Reduce the nitro group to the amine. $$\mathrm{1\text{-methyl-3-nitrobenzene} \xrightarrow{Fe/HCl} \text{3-methylaniline}}$$ The methyl and the new NH₂ stay meta to each other throughout, since reduction does not move ring substituents.

(ii) 3-Methylbenzoic acid — continues from (i). Diazotizing the aniline and displacing the diazonium group with cyanide (Sandmeyer) installs a nitrile at the same ring position (still meta to the methyl); acid hydrolysis of the nitrile then gives the acid:

1. Diazotize the amine from (i). $$\mathrm{\text{3-methylaniline} \xrightarrow{NaNO_2,\ HCl,\ 0\text{-}5^\circ C} \text{diazonium salt}}$$
2. Sandmeyer reaction with CuCN. $$\mathrm{\text{diazonium salt} \xrightarrow{CuCN} \text{3-methylbenzonitrile}}$$
3. Acid hydrolysis of the nitrile. $$\mathrm{\text{3-methylbenzonitrile} \xrightarrow{H_3O^+,\ \Delta} \text{3-methylbenzoic acid}}$$

b) Enantiomers. An enantiomer is one of a pair of stereoisomers that are non-superimposable mirror images of each other — they arise whenever a molecule has a stereocentre (commonly a carbon bonded to four different groups) and have identical connectivity, but differ in the 3-D spatial arrangement at that centre, the way a left hand differs from a right hand. In 2-bromobutane, C2 is bonded to four different groups (CH₃, Br, H, CH₂CH₃), making it a stereocentre with exactly two possible spatial arrangements:

The two arrangements are non-superimposable (no rotation makes one lie exactly on the other) and are related exactly as mirror images — the defining property of an enantiomeric pair.

c) m-Xylene + H₂CrO₄. Both methyl groups sit directly on the ring (both benzylic), so hot chromic acid oxidises both side chains fully to –COOH:

$$\mathrm{CH_3\text{-}C_6H_4\text{-}CH_3 \xrightarrow{H_2CrO_4,\ \Delta} HOOC\text{-}C_6H_4\text{-}COOH}$$

Product: isophthalic acid (benzene-1,3-dicarboxylic acid).