\[A:C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\] |
\[B:C{{H}_{2}}=C{{H}_{2}}\] |
\[C:C{{H}_{3}}-C{{H}_{3}}\] |
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 \[C{{H}_{3}}CH_{2}^{\centerdot }+C{{H}_{3}}CH_{2}^{\centerdot }\xrightarrow{{}}\underset{n-bu\tan e}{\mathop{C{{H}_{3}}C{{H}_{2}}-C{{H}_{2}}C{{H}_{3}}}}\,\]You need to login to perform this action.
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