Question

For the compound having molecular formula $\mathrm{C_4H_9Br}$, write :
(a) the isomer which is most reactive towards $\mathrm{S_N1}$ displacement.
(b) the isomer which, on reacting with Na metal in the presence of dry ether, gives 2,5-Dimethylhexane.

Hint:

(a) most stable 3° carbocation = tert-butyl bromide; (b) Wurtz couples two isobutyl halves.

Answer 1

(a) answer: The most reactive towards $\mathrm{S_N1}$ is 2-bromo-2-methylpropane, $\mathrm{(CH_3)_3C\text{-}Br}$, which is tert-butyl bromide. An $\mathrm{S_N1}$ reaction begins when the C−Br bond breaks on its own to give a carbocation, so the faster the carbocation forms and the more stable it is, the faster the reaction. Tert-butyl bromide gives a tertiary carbocation $\mathrm{(CH_3)_3C^+}$, which is the most stable kind because the three methyl groups push electron density onto the positive carbon. That is why this isomer reacts fastest by $\mathrm{S_N1}$.

(b) answer: The isomer is 1-bromo-2-methylpropane, $\mathrm{(CH_3)_2CHCH_2Br}$, which is isobutyl bromide. In the Wurtz reaction, sodium metal joins two alkyl halide molecules end to end by removing the two bromine atoms. Here two isobutyl groups couple together to give 2,5-dimethylhexane: \[ 2(CH_3)_2CHCH_2Br + 2Na \rightarrow (CH_3)_2CHCH_2CH_2CH(CH_3)_2 + 2NaBr \]