Question

Alkyl halides and alcohols easily undergo nucleophilic substitution either through $S_{N}1$ or $S_{N}2$ mechanism. The relative case of these two processes depends upon the nature of the substrate (alkyl group) as well as leaving group, nature of nucleophile and also upon the nature of the solvent.

$S_{N}1$ mechanism involves the formation of carbocation as intermediate while $S_{N}2$ mechanism involves the formation of a transition pentavalent state. $S_{N}1$ is the main mechanism in $3^0$ alkyl halides and alcohols, while $S_{N}2$ mechanism is the path adopted by most of the $1^o$ alkly halides; $2^o$ alkyl halides may follow $S_{N}1$ as well as $S_{N}2$.

What happens when $1^{o}ROH$ react with $HX$?

• A. The $1^o$ carbocations are unstable and hence are not formed.
• B. The carbocations are unstable and hence undergo rearrangement.
• C. The $1^o$ alcohol forms protonated alcohol and then follow $S_{N}2$ path.
• D. $X^{-}$ makes an $S_{N}2$ attack at $\alpha -C$ in $1^{o}ROH$.

Answer 1

Answer: (A), (C)

The $1^o$ carbocations are unstable and hence are not formed.
The $1^o$ alcohol forms protonated alcohol and then follow $S_{N}2$ path.

$1^o$ carbocations are not formed as they are unstable. $1^o$ alcohols first form protonated alcohols and then halide ions attack it forcing $H_{2}O$ to leave in an $S_{N}2$ path.