Question

Give reasons for the following :
(i) Ethyl iodide undergoes $S_{N}2$ reaction faster than ethylbromide.
(ii) $(\pm )$2 - Butanol is optically inactive.
(iii) C - X bond length in halobenzene is smaller than C - X bond length in $C{H}_3-X.$

Answer 1

(i) Ethyl iodide undergoes $S_{N}2$ reaction faster than ethyl bromide because $I^{-}$ is a better leaving group than $B{r}^{-}.$

(ii) $(\pm )$ 2- Butanol represents a racemic mixture of two enantiomers. The optical rotation due to one enantiomer in one direction and hence it is optically inactive.

(iii) C atom bearing halogen in halobenzene is $s{p}^2$ hybridised while in $C{H}_3-X$ it is $s{p}^3$ hybridised. As $s{p}^2$ hybrid orbital is smaller than $s{p}^3$ hybrid orbital due to greater s - character, the C - X bond length in halobenzene is smaller than in $C{H}_3-X.$