Question 3 of 5: Multi-Step Synthesis from Benzene & Cycloalkane Combustion

Question 3: Multi-Step Synthesis from Benzene & Cycloalkane Combustion (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 an ortho/para-director, so it cannot place a second substituent meta to itself directly. The standard workaround is to install the ring's eventual amine/acid position first, as a nitro group (NO₂) — nitro is a meta-director — then alkylate meta to it, and finally convert the nitro group into whatever is needed at that position.

(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, extremely sluggish on a ring as strongly deactivated as nitrobenzene (nitrobenzene is even used as an FC solvent precisely because it resists these conditions). This step is presented at the level this exam is testing — directing-group logic for a "propose a synthesis" question — and is the standard textbook answer for reaching a meta-alkyl/amine relationship; a laboratory chemist would more likely use a diazonium-salt blocking/replacement strategy for the same outcome.
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 remain meta to each other throughout, since reduction does not move substituents around the ring.

(ii) 3-Methylbenzoic acid — reuses the (i) synthesis, then two more steps. Diazotizing the aniline from part (i) and displacing the diazonium group with cyanide (Sandmeyer reaction) installs a nitrile at the ipso carbon (with retention of its ring position, still meta to the methyl); acid hydrolysis of the nitrile then gives the carboxylic acid:

1. Diazotize the amine from part (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-hydrolyze the nitrile. $$\mathrm{\text{3-methylbenzonitrile} \xrightarrow{H_3O^+,\ \Delta} \text{3-methylbenzoic acid}}$$ Carbon count check: the nitrile carbon becomes the carboxyl carbon one-for-one, so no ring rearrangement or carbon loss occurs.

b) Combustion of 1-ethyl-3-methylcyclohexane. The molecule is C₉H₁₈ (a cyclohexane ring, C₆H₁₂, with two H's replaced by ethyl and methyl: C₆H₁₀ + C₂H₅ + CH₃ = C₉H₁₈). Complete combustion (excess O₂) converts every carbon to CO₂ and every hydrogen to H₂O; since n = 9 gives a half-integer O₂ coefficient (3n/2 = 13.5), the whole equation is doubled to clear the fraction:

$$\mathrm{2\,C_9H_{18} + 27\,O_2 \longrightarrow 18\,CO_2 + 18\,H_2O}$$

Balance check: C: 18 = 18; H: 36 = 36; O: 27×2=54 both sides.