question_answer

\[C{{H}_{3}}C{{H}_{2}}Br\] undergoes Wurtz reaction. We may expect some of the following product

\[A:C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\]

\[B:C{{H}_{2}}=C{{H}_{2}}\]

\[C:C{{H}_{3}}-C{{H}_{3}}\]

Select correct product.

A)  Only A        

B)  A and B

C)  A, B and C     

D)  A and C

Correct Answer: C

Solution :

 \[{{C}_{2}}{{H}_{5}}Br+N{{a}^{+}}\xrightarrow{{}}C{{H}_{3}}CH_{2}^{\bullet }+NaBr\] Intermediate \[C{{H}_{3}}CH_{2}^{\bullet }\] combines to form \[C{{H}_{3}}C{{H}_{2}}-C{{H}_{2}}C{{H}_{3}}\]   (main)   and  also \[C{{H}_{2}}=C{{H}_{2}}\] and \[C{{H}_{3}}C{{H}_{3}}\] by disproportion \[C{{H}_{3}}CH_{2}^{\centerdot }+C{{H}_{3}}CH_{2}^{\centerdot }\xrightarrow{{}}\underset{n-bu\tan e}{\mathop{C{{H}_{3}}C{{H}_{2}}-C{{H}_{2}}C{{H}_{3}}}}\,\]