question_answer

A bromoalkane \[X\] reacts with magnesium in dry ether to form compound\[Y\]. The reaction of \[Y\] with methanal followed by hydrolysis yield an alcohol having molecular formula\[{{C}_{4}}{{H}_{10}}O\]. The compound \[X\] is:

A)  bromo ethane                 

B)  bromo methane

C)  1-bromo propane          

D)  2-bromo propane

Correct Answer: C

Solution :

\[\underset{1-bromopropane}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-Br+Mg}}\,\]\[\xrightarrow[ether]{dry}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}MgBr\]\[\xrightarrow{HCHO}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}-O-MgBr\]