A) chain isomerism
B) position isomerism
C) chain as well as position isomerism
D) functional isomerism
Correct Answer: C
Solution :
\[C{{H}_{3}}-\underset{Br}{\overset{C{{H}_{3}}}{\mathop{\underset{\mathbf{|}}{\overset{\mathbf{|}}{\mathop{C}}}\,}}}\,-C{{H}_{3}}\] \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br\] Here, the chain of carbon atoms and the position of functional group both are different, they exhibit chain as well as position isomerism.
You need to login to perform this action.
You will be redirected in
3 sec
