I. \[C{{H}_{3}}C{{H}_{2}}OH+HCl\xrightarrow{anh.ZnC{{l}_{2}}}\] |
II. \[C{{H}_{3}}C{{H}_{2}}OH+HCl\xrightarrow{\,}\] |
III. \[{{(C{{H}_{3}})}_{3}}COH+HCl\xrightarrow{\,}\] |
IV. \[{{(C{{H}_{3}})}_{2}}CHOH+HCl\xrightarrow{\text{anh}\text{.}\,\text{ZnC}{{\text{l}}_{\text{2}}}}\] |
A) I, III and IV
B) I and II
C) Only IV
D) III and IV
Correct Answer: A
Solution :
In (I) and (IV) in the presence of Lucas reagent \[(HCl+anh.ZnC{{l}_{2}})\] alcohols give alkyl halides while in (III) alkyl halide is formed due to \[{{S}_{N}}1\] reaction.You need to login to perform this action.
You will be redirected in
3 sec