A) \[C{{H}_{3}}-\overset{C{{H}_{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{\overset{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{CH-COOH\,}}\,}}\,\]
B) \[C{{H}_{2}}=CHC{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\]
C) \[C{{H}_{3}}-\overset{N{{H}_{2}}\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{\overset{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{CH-C{{H}_{3}}\,}}\,}}\,\]
D) \[C{{H}_{3}}-C{{H}_{2}}-\overset{N{{H}_{2}}\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{\overset{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{CH-C{{H}_{3}}}}\,}}\,\]
Correct Answer: D
Solution :
\[C{{H}_{3}}-C{{H}_{2}}-\underset{H\,\,\,\,\,\,\,\,\,}{\overset{N{{H}_{2}}}{\mathop{\underset{|\,\,\,\,\,\,\,\,\,\,\,}{\overset{|\,\,\,\,\,\,\,\,\,\,}{\mathop{{{C}^{*}}\,\,\,-\,}}}\,}}}\,C{{H}_{3}}\] Chiral centre is present. Hence, it exists as optical isomers or enantiomorphs.
You need to login to perform this action.
You will be redirected in
3 sec
