A) propane
B) 2, 3-dimethylbutane
C) hexane
D) allyl bromide
Correct Answer: B
Solution :
\[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-Cl\xrightarrow[KOH]{alc.}\underset{(B)}{\mathop{C{{H}_{3}}CH=C{{H}_{2}}}}\,\]\[\xrightarrow{HBr}C{{H}_{3}}-\underset{(C)}{\mathop{\overset{Br}{\mathop{\overset{|}{\mathop{C}}\,}}\,H}}\,-C{{H}_{3}}\xrightarrow[ether]{Na}\] \[\underset{(D)}{\mathop{C{{H}_{3}}\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{C}}\,}}\,H-\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{C}}\,}}\,HC{{H}_{3}}}}\,\]You need to login to perform this action.
You will be redirected in
3 sec