Which of the following haloalkanes will give a racemic modification on reaction with aqueous NaOH ?
(A) Benzyl Bromide
(B) Vinyl Bromide
(C) 2-Bromo-2-methylpropane
(D) 2-Bromo-2-methylbutane
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → None of (A), (B), (C) and (D)
Note: None of the given options was correct in the NTA question paper. To provide additional question practice to the students, the question has been slightly modified so that the question has one clear answer.
A racemic modification is a mixture of the two enantiomers in equal amounts. With aqueous NaOH it forms through the SN1 mechanism:
the C–X bond breaks first to give a planar carbocation (sp2 carbon);
\( \ce{OH^{-}} \) can attack this carbocation equally from either face, so an optically active halide gives both configurations in equal amounts.
Two conditions must therefore hold together: the halide must react by SN1 (tertiary, benzylic or allylic halides, which give a stable carbocation), and the carbon that carries the halogen must be a stereocentre (four different groups), so that the halide is optically active and the two faces of the carbocation lead to two different products.
(A) Benzyl Bromide — Incorrect. It reacts by SN1 through the benzyl carbocation, \( \ce{C6H5CH2^{+}} \), which is stabilised by resonance with the ring. But the –CH2Br carbon carries two H atoms, so it is not a stereocentre; the product, benzyl alcohol, is a single achiral compound.
\( \displaystyle \ce{C6H5CH2Br + OH^- -> C6H5CH2OH + Br^-} \)
(B) Vinyl Bromide — Incorrect. The C–Br bond has partial double bond character and the vinyl carbocation is very unstable, so it does not undergo substitution with aqueous NaOH at all (and it has no stereocentre).
(C) 2-Bromo-2-methylpropane — Incorrect. It is a tertiary halide and reacts by SN1 through the tert-butyl carbocation, \( \ce{(CH3)3C^{+}} \). But C-2 carries three identical \( \ce{CH3} \) groups, so it is not a stereocentre; the product, 2-methylpropan-2-ol, is achiral.
\( \displaystyle \ce{(CH3)3C-Br + OH^- -> (CH3)3C-OH + Br^-} \)
(D) 2-Bromo-2-methylbutane — Incorrect. It is also tertiary and reacts by SN1 through a planar carbocation. But C-2 carries two \( \ce{CH3} \) groups, so it is not a stereocentre; attack from either face gives the same achiral alcohol, 2-methylbutan-2-ol.
\( \displaystyle \ce{(CH3)2C(Br)CH2CH3 + OH^- -> (CH3)2C(OH)CH2CH3 + Br^-} \)