Question

Give reasons for the following.
(i) Ethyl iodide undergoes $S{N}^2$ reaction faster than ethyl bromide.
(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

1. Because in ethyl iodide, iodide(I) is acting as a best leaving group among all the halide ions so rate of ${S_N}^2$ reaction is directly proportional to leaving group ability.
2. ($\pm$) 2-butanol is a racemic mixture which is optically inactive due to external compensation.
3. Due to resonance in halobenzene it has less bond length value as compared to $C{H}_3-X$.