Hydrogenation, Multi-Step Synthesis & Carbon Classification: 04-BS-12, December 2015, Q5

Question 5: Hydrogenation, Multi-Step Synthesis & Carbon Classification (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.

Check: Part (a) of this problem is set up identically to Question 4(a)(i) on this same paper (the same reagent, methylenecyclohexane + H₂/Pd, is asked twice, worth 5 points each time). This is answered faithfully as printed rather than silently assumed to be a typo for a different substrate; the mechanism and product are identical to Q4(a)(i) below.

a) Methylenecyclohexane + H₂/Pd (identical to Q4a(i)). As in Question 4, Pd hydrogenates the exocyclic C=C only:

→

Fig. — hydrogenation of the exocyclic alkene only

\mathrm{C_7H_{12} + H_2 \xrightarrow{Pd} \boxed{C_7H_{14}}}

Product: methylcyclohexane.

b) 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.

Fig. Q5b — the two meta-substituted targets

(i) 3-Methylaniline.

Nitrate benzene. $$\mathrm{C_6H_6 \xrightarrow{HNO_3/H_2SO_4} C_6H_5NO_2}$$
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.
Reduce the nitro group to the amine. $$\mathrm{1\text{-methyl-3-nitrobenzene} \xrightarrow{Fe/HCl} \boxed{\text{3-methylaniline}}}$$ 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:

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

c) CH₃CH₂C(CH₃)₂CH₂CH₂CH₃ is 3,3-dimethylhexane. A carbon's class is read directly off how many other carbons it is bonded to: 1→primary, 2→secondary, 3→tertiary — and this structure has a carbon bonded to four, which the p/s/t scheme the question offers does not label.

Fig. Q5c — C3 is bonded to four other carbons (quaternary) — not tertiary

Main chain, C1–C6. C1 (CH₃, bonded only to C2) and C6 (CH₃, bonded only to C5) are each bonded to one carbon → primary. C2 (CH₂, bonded to C1 and C3), C4 (CH₂, bonded to C3 and C5), and C5 (CH₂, bonded to C4 and C6) are each bonded to two carbons → secondary.
C3 — the trap. C3 is bonded to C2, C4, and both methyl branches — four carbon neighbours in total, with zero hydrogens left on C3 itself. This is a quaternary carbon, a distinct category the p/s/t scheme does not cover; it is not tertiary, even though it superficially "has three things attached" if the branches are miscounted.
The two methyl branches on C3. Each branch CH₃ is bonded only to C3 — one carbon neighbour each → both primary.

Part	Result
a	methylcyclohexane
b(i)	3-methylaniline via nitration → meta-methylation → nitro reduction
b(ii)	3-methylbenzoic acid via (i) → diazotization → Sandmeyer (CuCN) → hydrolysis
c	4 primary (C1, C6, both methyls), 3 secondary (C2, C4, C5), 0 tertiary, 1 quaternary (C3)

Concept & Theory

Topic. Chemoselective hydrogenation; directing-group strategy for meta-disubstituted benzenes; diazonium/Sandmeyer chemistry; primary/secondary/tertiary/quaternary carbon classification.

Key relations. Nitro = meta-director (deactivating); alkyl = ortho/para-director (activating). Carbon class = number of carbon neighbours (1/2/3/4 → p/s/t/quaternary).

Why this works. Choosing which group to install first on the ring controls where the second group can land, since every existing substituent's directing effect governs the next electrophilic attack; a diazonium salt is uniquely versatile because it can be converted into many different final groups while never disturbing the ring position it occupies. A quaternary carbon has no C–H bonds left at all, so it cannot participate in reactions that require abstracting a C–H at that position — the classification has real chemical consequences, not just bookkeeping value.

Common pitfall. Trying to alkylate the ring first and then nitrate meta to the alkyl group — alkyl groups are ortho/para-directors, so this ordering can never deliver a clean meta relationship. Mislabelling a quaternary carbon as tertiary by only counting "three other groups drawn nearby" instead of literally counting carbon-to-carbon bonds.

Source. McMurry, Organic Chemistry, 9th ed., Ch. 7 (hydrogenation), Ch. 16 (EAS directing effects, diazonium chemistry), Ch. 3 (carbon classification).

Question 5 of 5: Hydrogenation, Multi-Step Synthesis & Carbon Classification

Question 5: Hydrogenation, Multi-Step Synthesis & Carbon Classification (20 marks)