Multi-Step Aromatic Synthesis & Alkene Stability: 04-BS-12, May 2015, Q4

Question 4: Multi-Step Aromatic Synthesis & Alkene Stability (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) Benzene → meta-substituted target molecules. Both targets need a –CH₃ and a second group (–COOH or –NH₂) in a meta relationship. Neither group can be installed directly in that relationship by simple Friedel–Crafts chemistry on its own (an alkyl group is an ortho/para director, and neither –COOH nor –NH₂ can be built directly by a clean Friedel–Crafts step) — the standard workaround is to install a temporary strong meta-director first (–NO₂), alkylate meta to it, then convert the nitro group into whichever final functional group is needed.

Step 1 — nitration. Benzene is nitrated (HNO₃/H₂SO₄, generating the electrophile NO₂⁺) to install the first, strongly deactivating meta-director.
HNO3
→
H2SO4
Q4a, step 1 — nitration
$$\mathrm{C_6H_6 + HNO_3 \xrightarrow{H_2SO_4} C_6H_5NO_2 + H_2O}$$
Step 2 — Friedel–Crafts methylation, meta to –NO₂. CH₃Cl/AlCl₃ alkylates the ring; because –NO₂ directs incoming electrophiles meta to itself, the new methyl group lands exactly where the final target needs it.
CH3Cl
→
AlCl3
Q4a, step 2 — FC methylation meta to NO2

Check: in practice, Friedel–Crafts alkylation is very sluggish (often impractical) on a ring already carrying a strong deactivator like –NO₂; this nitrate-first/alkylate-second sequence is presented here as the directing-logic route this exam level expects (it correctly places the substituents), not as a claim about realistic laboratory yield.
Step 3 — reduction of the nitro group. Catalytic hydrogenation (H₂/Pd) or Fe/HCl reduces –NO₂ to –NH₂, giving 3-methylaniline directly — this is the complete route for target (ii).
H2/Pd
→
Q4a, step 3 — nitro reduction (target (ii))
$$\mathrm{CH_3C_6H_4NO_2 + 3H_2 \xrightarrow{Pd} CH_3C_6H_4NH_2 + 2H_2O}$$
Steps 4–6 (target (i) only) — diazotisation, Sandmeyer, hydrolysis. Starting from the 3-methylaniline just made, diazotise with NaNO₂/HCl at 0–5°C to form the diazonium salt, then displace N₂ with a Sandmeyer reaction (CuCN) to install a nitrile, and finally hydrolyse the nitrile (H₃O⁺, heat, or NaOH then acidify) all the way to the carboxylic acid:
NaNO2/HCl
→
CuCN
→
H3O+
→
Q4a, steps 4-6 — diazotisation, Sandmeyer, hydrolysis (target (i))
$$\mathrm{ArNH_2 \xrightarrow{NaNO_2,\ HCl,\ 0\text{-}5^\circ C} ArN_2^+ \xrightarrow{CuCN} ArCN \xrightarrow{H_3O^+,\ \Delta} ArCOOH}$$ giving 3-methylbenzoic acid for target (i).

Target	Route
(i) 3-methylbenzoic acid	nitrate → FC methylate (meta) → reduce to amine → diazotise → Sandmeyer CuCN → hydrolyse nitrile
(ii) 3-methylaniline	nitrate → FC methylate (meta) → reduce –NO2 to –NH2

b) Ranking trans-3-hexene, cis-3-hexene and cis-2,5-dimethyl-3-hexene. All three share the identical disubstitution pattern at the double bond (one alkyl group on each sp² carbon), so the substitution-count argument from Question 3 cannot distinguish them — here stability is decided by steric strain between the two alkyl groups, which depends on both geometry (cis vs. trans) and how bulky those two groups are.

Q4b — the three hexene stereo/substitution isomers

Trans-3-hexene — most stable. The two ethyl groups sit on opposite sides of the double bond, as far apart as possible: essentially no steric clash between them.
Cis-3-hexene — middle. Same two ethyl groups, but now on the same side of the double bond, close enough to experience real steric (van der Waals) repulsion — this destabilises the cis isomer relative to its trans counterpart by a well-established, measurable amount (a larger heat of hydrogenation for the cis form).
Cis-2,5-dimethyl-3-hexene — least stable. Same cis geometry as (ii), but each substituent is now a bulkier isopropyl group, (CH₃)₂CH–, instead of a plain ethyl. Bulkier groups crowded onto the same side of the double bond suffer substantially more steric strain than the cis-ethyl case — the extra methyl branch on each substituent has nowhere to go but into the other substituent's space.

Ranking (decreasing stability): trans-3-hexene > cis-3-hexene > cis-2,5-dimethyl-3-hexene.

Compound	Geometry	Substituent bulk	Rank
trans-3-hexene	trans	ethyl/ethyl	1 (most stable)
cis-3-hexene	cis	ethyl/ethyl	2
cis-2,5-dimethyl-3-hexene	cis	isopropyl/isopropyl	3 (least stable)

Concept & Theory

Topic. Meta-director synthetic strategy for 1,3-disubstituted benzenes; steric (cis) strain as a second, independent alkene-stability factor beyond substitution count.

Key relations. –NO₂ is a deactivating meta director; diazonium salts (ArN₂⁺) are versatile synthetic handles (Sandmeyer reactions swap N₂ for Cl, Br, CN, etc.). Alkene stability: for a fixed substitution pattern, trans > cis, and bulkier cis substituents are destabilised further still.

Why this works. Building the meta relationship requires a director that is already in place before the second group is added — you cannot install two groups independently and hope they land meta by chance, since both common directing groups (alkyl, –NH₂) are ortho/para directors. Steric strain in a cis alkene grows with the size of the eclipsing/crowded groups because the van der Waals repulsion between two bulky substituents on the same face is larger than between two small ones.

Common pitfall. Trying to alkylate directly onto aniline or benzoic acid (both give the wrong ring position, and a free –NH₂ also over-activates/poisons a Friedel–Crafts catalyst); assuming all cis alkenes are equally destabilised regardless of substituent size.

Source. McMurry, Organic Chemistry, 9th ed., Ch. 16 (electrophilic aromatic substitution, directing effects, diazonium chemistry), Ch. 6 (cis/trans alkene stability).

Question 4 of 5: Multi-Step Aromatic Synthesis & Alkene Stability

Question 4: Multi-Step Aromatic Synthesis & Alkene Stability (20 marks)