Question 7 of 13: Mechanisms (S N 1 Solvolysis; Exhaustive Methylation) & a Slow-Reacting Hexachlorocyclohexane

Question 7: Mechanisms (S_N1 Solvolysis; Exhaustive Methylation) & a Slow-Reacting Hexachlorocyclohexane

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.

7a.(i) Solvolysis of 1-bromo-1,2-dimethylcyclohexane in ethanol — S_N1

Ethanol is both a weak nucleophile and the solvent (not a strong base) — classic S_N1 solvolysis conditions, and the mixture of two diastereomeric ether products (rather than one clean inversion product) is the diagnostic clue that a planar carbocation intermediate is involved.

1. Ionisation (rate-determining). The C–Br bond heterolyses: a curved arrow runs from the C–Br bonding pair to Br, expelling bromide and generating a planar (sp²), tertiary carbocation at C1 (stabilised by its three alkyl substituents — the C1 methyl and two ring carbons — which is why a tertiary bromide ionises readily in a polar protic solvent). The adjacent C2 stereocentre (bearing the methyl group) is untouched and keeps its original configuration throughout.
2. Nucleophilic capture from either face. Because the cation is planar, an ethanol oxygen lone pair can attack C1 from either the face Br originally occupied (curved arrow: O lone pair → new C–O bond) or the opposite face — there is nothing in a planar cation to prefer one face over the other except modest steric bias from the adjacent methyl group. Attack from each face gives a different diastereomer at C1 (the C2-methyl stereocentre is never disturbed), which is exactly the "mixture of two products" the question shows.
3. Deprotonation. A second ethanol molecule removes the proton from the resulting protonated ether (curved arrow: O–H bonding pair → the base), giving the neutral ether product and a new protonated-ethanol/ethyloxonium species that ultimately equilibrates with bromide to release HBr, the by-product shown.

7a.(ii) Exhaustive methylation of butylamine (Hofmann/Menshutkin methylation)

Excess methyl iodide with K₂CO₃ present drives three successive S_N2 methylations, with the base mopping up each acidic ammonium intermediate in between so the next methylation can proceed on a neutral (nucleophilic) amine:

1. First methylation. The primary amine's nitrogen lone pair attacks CH₃I from the backside (curved arrows: N lone pair → new N–C bond; C–I bonding pair → I⁻), giving a secondary ammonium salt, R–NH₂⁺–CH₃. K₂CO₃ deprotonates this ammonium (curved arrow: N–H bonding pair → carbonate), regenerating a neutral, nucleophilic secondary amine.
2. Second methylation. The secondary amine repeats the same S_N2 attack on a fresh CH₃I, giving a tertiary ammonium salt; K₂CO₃ again deprotonates it to the neutral tertiary amine.
3. Third methylation. The tertiary amine attacks a third CH₃I to give the quaternary ammonium salt. This cation has no remaining N–H to lose, so it cannot be deprotonated further — it is the final, isolable product, paired with I⁻ as the counter-ion (butyltrimethylammonium iodide).

7b. The slow-reacting hexachlorocyclohexane isomer

The isomer that reacts ~7000× slower is the one with chlorines on strictly alternating faces all the way around the ring (drawn above) — this is the only one of the nine stereoisomers in which all six C–Cl bonds can be equatorial simultaneously in a single chair conformation, and the only one in which no chlorine can ever sit anti-periplanar to a hydrogen on a neighbouring carbon.

1. E2 needs a trans-diaxial H–C–C–Cl pair. Anti-periplanar E2 elimination on a cyclohexane requires the leaving group (Cl) to be axial and the hydrogen on an adjacent carbon to be axial on the opposite face (trans-diaxial).
2. Preferred chair: every Cl equatorial. With the chlorines on strictly alternating faces, one chair puts all six C–Cl bonds equatorial (so all six H are axial). No Cl is axial, so nothing can be eliminated from this conformation.
3. Ring-flipped chair: every Cl axial — but still no axial H next to it. Flipping the ring makes all six Cl axial, but it simultaneously makes all six H equatorial. The group that is anti-periplanar to each axial Cl is therefore the axial Cl on the neighbouring carbon (adjacent chlorines are trans, so they are trans-diaxial), never an H. So even the high-energy all-axial chair (six sets of 1,3-diaxial Cl···Cl repulsions) offers no anti-periplanar H–Cl pair.
4. Why every other isomer is fast. Each of the other eight stereoisomers has at least one pair of cis neighbouring chlorines. For a cis-1,2 pair, one Cl is axial and the other equatorial in either chair, so the equatorial one's carbon carries an axial H that is anti-periplanar to the axial Cl — a normal trans-diaxial H/Cl arrangement is available in an ordinary chair. The all-trans (β) isomer has no cis neighbours and can only eliminate through a strained non-chair geometry or a much slower syn pathway, hence ~7000× slower.