A) \[C{{H}_{3}}.C{{H}_{2}}.C{{H}_{2}}.\underset{\text{Cl}}{\overset{\text{I}}{\mathop{\underset{\text{ }\!\!|\!\!\text{ }}{\overset{\text{ }\!\!|\!\!\text{ }}{\mathop{\text{C}}}}\,}}}\,-H\]
B) \[C{{H}_{3}}C{{H}_{2}}.\underset{\text{I}}{\mathop{\underset{|}{\mathop{CH}}\,}}\,.\underset{Cl}{\mathop{\underset{|}{\mathop{C{{H}_{2}}}}\,}}\,\]
C) \[C{{H}_{3}}C{{H}_{2}}-\underset{Cl}{\overset{I}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-C{{H}_{3}}\]
D) \[C{{H}_{3}}.\underset{Cl}{\mathop{\underset{|}{\mathop{CH}}\,}}\,C{{H}_{2}}.C{{H}_{2}}I\]
Correct Answer: C
Solution :
This reaction occurs according to Markovnikov's rule which states that when an unsymmetrical alkene undergo hydrohalogenation, the negative part goes to that C-atom which contain lesser no. of H-atom. \[C{{H}_{3}}-C{{H}_{2}}-C\equiv CH+HCl\xrightarrow{{}}\] \[C{{H}_{3}}-C{{H}_{2}}-\overset{Cl}{\mathop{\overset{|}{\mathop{C}}\,}}\,=C{{H}_{2}}\xrightarrow{HI}\] \[C{{H}_{3}}-C{{H}_{2}}-\underset{I}{\overset{Cl}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}}\,-C{{H}_{3}}\]
You need to login to perform this action.
You will be redirected in
3 sec
