A) Ethyl alcohol
B) Propyl alcohol
C) Isobutyl alcohol
D) Methyl alcohol
Correct Answer: A
Solution :
\[\underset{C{{H}_{2}}}{\overset{C{{H}_{2}}}{\mathop{||}}}\,+{{H}_{2}}S{{O}_{4}}\to \underset{C{{H}_{2}}HS{{O}_{4}}}{\overset{C{{H}_{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}\,\xrightarrow{{{H}_{2}}O}\]\[\underset{C{{H}_{2}}OH}{\overset{C{{H}_{3}}\,\,\,\,\,\,\,\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}\,+{{H}_{2}}S{{O}_{4}}\] Except ethyl alcohol, no other primary alcohol can be prepared by this method as the addition of \[{{H}_{2}}S{{O}_{4}}\] follows Markownikoff?s rule. Generally secondary and tertiary alcohols are obtained.
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