question_answer

\[C{{H}_{3}}\underset{O}{\mathop{\underset{||}{\mathop{C}}\,}}\,\cdot C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Cl\xrightarrow{C{{H}_{3}}MgBr}A,\]Here A is:

A)  \[C{{H}_{3}}--\underset{OH}{\overset{C{{H}_{3}}}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,\cdot C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Cl\]

B)  \[\underset{O}{\mathop{\underset{||}{\mathop{C{{H}_{3}}CC{{H}_{2}}\cdot C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,}}\,\]

C)  

D)  

Correct Answer: C

Solution :

Ketones, with Grignard reagent, give tertiary alcohols. Here \[3{}^\circ \text{-}\]alcohol, obtained, cyclises to give product (C). The reaction follows as: