Question

Compound 'X' with molecular formula \(C_4H_9Br\) reacts with aqueous KOH to give an alcohol. The rate of this reaction depends only on the concentration of the compound 'X'. When an optically active isomer 'Y' of the compound 'X' was treated with aqueous KOH solution, the rate of reaction was found to be dependent on concentration of compound 'Y' and aqueous KOH both.

(a) Write down the structural formula of both 'X' and 'Y'.

(b) Out of 'X' and 'Y', which one will undergo racemisation and why?

(c) Out of 'X' and 'Y', which one will form product with inversion of configuration and why?

Hint:

The order of the reaction tells us the mechanism. A reaction whose rate depends only on the alkyl halide follows \(S_N1\) (first order) and is favoured by tertiary halides.

Answer 1

Concept:
The order of the reaction tells us the mechanism. A reaction whose rate depends only on the alkyl halide follows \(S_N1\) (first order) and is favoured by tertiary halides. A reaction whose rate depends on both the alkyl halide and the hydroxide ion follows \(S_N2\) (second order) and is favoured by primary or simple chiral halides. \(S_N1\) gives a planar carbocation leading to racemisation; \(S_N2\) goes through backside attack leading to inversion.

Step 1 (a):
'X' reacts by \(S_N1\) (rate depends only on X), so X is the tertiary halide: tert-butyl bromide, \((CH_3)_3C-Br\) (2-bromo-2-methylpropane). 'Y' is an optically active (chiral) isomer of \(C_4H_9Br\) that reacts by \(S_N2\) (rate depends on both Y and KOH); the chiral isomer is sec-butyl bromide, \(CH_3-CH_2-CHBr-CH_3\) (2-bromobutane).

Step 2 (b):
'X' (tert-butyl bromide) reacts by \(S_N1\). It forms a planar carbocation intermediate which can be attacked by the nucleophile from either face equally, producing both configurations. So 'X' undergoes racemisation. (Note: tert-butyl bromide itself is not chiral, but the \(S_N1\) mechanism it follows is the one that causes racemisation when a chiral substrate reacts this way.)

Step 3 (c):
'Y' (2-bromobutane) reacts by \(S_N2\). Here the hydroxide ion attacks the carbon from the side opposite to the leaving bromide (backside attack), so the configuration is turned inside out. Therefore 'Y' forms the product with inversion of configuration.

Answer: (a) X = \((CH_3)_3C-Br\) (2-bromo-2-methylpropane), Y = \(CH_3CH_2CHBrCH_3\) (2-bromobutane). (b) X, because it follows the \(S_N1\) mechanism through a planar carbocation, giving racemisation. (c) Y, because it follows the \(S_N2\) mechanism with backside nucleophilic attack, giving inversion of configuration.