Alcohols Phenols and Ethers - Result Question 4
####4. Which of the following reaction (s) can be used for the preparation of alkyl halides? [2015 RS]
(I) $CH_3 CH_2 OH+HCl$
$\xrightarrow{\text{ anh. }} \xrightarrow{ZnCl_2}$
(II) $CH_3 CH_2 OH+HCl \longrightarrow$
(III) $(CH_3)_3 COH+HCl \longrightarrow$
(IV) $(CH_3)_2 CHOH+HCl \xrightarrow{\text{ anh. } ZnCl_2}$
(a) (I), (III) and (IV) only
(b) (I) and (II) only
(c) (IV) only
(d) (III) and (IV) only
Show Answer
Solution:
- (a)
$ \begin{matrix} CH_3-CH_2 OH+ZnCl_2 \to R & \stackrel{\oplus}{O}-ZnCl_2 \\ (R=CH_3-CH_2-) & \stackrel{H}{H} \tag{I}\\ \text{ (I) } \end{matrix} $
$ R-\underset{H}{\stackrel{\oplus}{O}-ZnCl_2} \to R^{\oplus}+[HOZnCl_2]^{\ominus} $
(I)
Carbocation is formed as intermediate in the $S _{N_1}$ mechanism which these reaction undergoes. (III) Tertiary carbocation easily formed due to the stability.
$ CH_3-\underset{ _CH^{3}}{\stackrel{\oplus}{C}}-CH_3+H_2 O $
(IV) In the presence of $ZnCl_2, 2^{\circ}$ carbocation is formed from $(CH_3)_2-\underset{H}{C}-OH$ i.e., $CH_3-CH-CH_3$
$ZnCl_2$ is a lewis acid and interact with alcohol. In the absence of $ZnCl_2$ formation of primary carbocation is difficult.
