(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)\[{{(C{{H}_{3}})}_{2}}CHOH+HCl\xrightarrow{anh.ZnC{{l}_{2}}}\] |
A) (IV) only
B) (III) and (IV) only
C) (I), and (IV) only
D) (I) and (II) only
Correct Answer: C
Solution :
(I) and (IV) can be used due to presence of anhydrous \[ZnC{{l}_{2}}\](III) gives alkyl halides due to formation of more stable carbocation.You need to login to perform this action.
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