Q.In presence of aqueous alkali which of the following will take part easily in SN1 reaction?

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. …

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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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. …