$(R)$-3-bromo-3-methylhexane is solvolysed in water. Predict the optical outcome (in terms of $R$/$S$ composition) of the alcohol formed, and explain why using the geometry of the reaction intermediate.

Question

Accepted Answer

3-Bromo-3-methylhexane is a tertiary haloalkane, and water is a weak nucleophile and polar protic solvent -- exactly the combination that favours $S_N1$ . The rate-determining step is ionisation of the substrate to give a planar, $sp^2$ tertiary carbocation, in which the original stereocentre no longer exists as a three-dimensional arrangement at all -- the empty p orbital is equally exposed on both faces. Water then attacks this flat intermediate from either face with roughly equal probability, so the alcohol formed is obtained as a near 1:1 mixture of both configurations, i.e. largely racemic, and the original optical purity of the $(R)$ starting material is mostly lost (with, at best, a slight statistical excess of t …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set SEM31 markMCQ
$Q. Which of the following compounds will give racemic mixture on SN1 reaction with aqueous KOH? (I) CH3-CH(C2H5)-Br, (II) CH3-C(CH3)(C2H5)-Br, (III) CH3-CH(CH3)-CH2Br, (IV) CH3-CH2-CH2-CH2-Br$
(a) I
(b) II
(c) III
(d) IV
›Reveal solutionSolution
A racemic product in SN1 requires the reacting carbon to be a chiral centre. Only (I) 2-bromobutane has four different groups on the C-Br carbon, so its SN1 gives a racemic mixture. Correct option (a).

SN1 proceeds via a planar (sp2) carbocation, which is attacked by the nucleophile from both faces. This gives a racemic (50:50) mixture only if the reacting carbon was a stereocentre (four different groups) so that the two faces lead to enantiomers.

Checking each:

(I) CH3-CH(C2H5)-Br = 2-bromobutane: the C-Br carbon bears CH3, C2H5, H, Br - four different groups (chiral). SN1 -> racemic mixture. Correct. …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following halogen compounds undergoes faster SN1 reaction?$
(a) (structure a) tert-butyl chloride type, Cl on tertiary carbon
(b) (structure b) sec-butyl chloride type, Cl on secondary carbon of shorter chain
(c) (structure c) Cl on secondary carbon of a longer branched chain
(d) (structure d) straight-chain primary chloroalkane, Cl on terminal carbon
›Reveal solutionSolution

S $_N$ 1 reaction rate tracks carbocation stability: 3° > 2° > 1°, so the tertiary halide reacts fastest.

An S $_N$ 1 reaction proceeds through a carbocation intermediate formed in the rate-determining (ionisation) step:

$R\text{-}Cl \xrightarrow{\text{slow}} R^+ + Cl^-$

The rate of this step depends entirely on how stable the carbocation is — more stable carbocations form faster because they need less activation energy. Carbocation stability order is 3° > 2° > 1° due to increasing hyperconjugation and the +I (electron-donating) effect of more alkyl groups around the positive carbon.

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. In presence of aqueous alkali which of the following will take part easily in SN1 reaction?$
(a) CH3CH2CH2Cl
(b) (CH3)2CHCH2Cl
(c) CH3-CH2CH(CH3)-Cl
(d) (CH3)3C-Cl
›Reveal solutionSolution
SN1 reactivity depends on carbocation stability; a tertiary halide forms the most stable (tertiary) carbocation, so it reacts fastest by the SN1 pathway.

Reasoning: In an SN1 reaction, the rate-determining step is the formation of a carbocation intermediate after the leaving group (Cl-) departs. The more stable this carbocation, the faster the SN1 reaction proceeds. Carbocation stability order: 3° > 2° > 1° (due to increasing hyperconjugation and +I inductive electron donation from more alkyl groups).

Among the given options:

(a) CH3CH2CH2Cl — 1° halide, forms an unstable 1° carbocation, reacts very slowly by SN1 (favours SN2 instead).

(b) (CH3)2CHCH2Cl — 1° halide (neopentyl-like/isobutyl), also a poor SN1 substrate. …
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Q. $(R)$ -3-bromo-3-methylhexane is solvolysed in water. Predict the optical outcome (in terms of $R$ / $S$ composition) of the alcohol formed, and explain why using the geometry of the reaction intermediate.

Concept understanding — SN1 Mechanism and Racemization

Q.(R)(R)(R)-3-bromo-3-methylhexane is solvolysed in water. Predict the optical outcome (in terms of RRR/SSS composition) of the alcohol formed, and explain why using the geometry of the reaction intermediate.

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Q. $(R)$ -3-bromo-3-methylhexane is solvolysed in water. Predict the optical outcome (in terms of $R$ / $S$ composition) of the alcohol formed, and explain why using the geometry of the reaction intermediate.

Unlock everything free for 14 days