The reaction proceeds with 98% racemisation. The reaction may follow:

S_{N} 1

mechanism

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S_{N} 2

mechanism

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E1 mechanism

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E2 mechanism

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Solution

The correct option is A $S_{N} 1$ mechanism
The mixture containing two enantiomers in equal proportions will have zero optical rotation, as the rotation due to one isomer will be cancelled by the rotation due to the other isomer. Such mixture is racemic modification. The process of conversion of enantiomer into a racemic mixture is called as racemisation. In case of optically active alkyl halides $S_{N} 1$ reactions are accompanied by racemisation. This is because the carbocation formed in slow step is planar.

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C_{6} H_{5} - C H_{3} | C | H - B r + H_{2} O ⟶ H O - C H_{3} | C | H - C_{6} H_{5} + H B r

S_{N} 1

Consider the reactions

(i) $(C H_{3})_{2} C H - C H_{2} B r C_{2} H_{5} O H - ---- \to (C H_{3})_{2} C H - C H_{2} O C_{2} H_{5} + H B r$

(ii) $(C H_{3})_{2} C H - C H_{2} B r C_{2} H_{S} O^{-} - ---- \to (C H_{3})_{2} C H C H_{2} O C_{2} H_{5} + B r^{-}$

The mechanism of reactions (i) & (ii) are respectively

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