A) 3
B) 4
C) 1
D) 2
Correct Answer: D
Solution :
2-methylbutane has 4 types of hydrogen atoms. |
\[{{\operatorname{CH}}_{3}}-\underset{{}}{\overset{\overset{{}}{\mathop \underset{|}{\mathop C{{H}_{3}}}\,}\,}{\mathop{CH}}}\,-C{{H}_{2}}C{{H}_{3}}\] |
Hence, on monochlorination, it gives 4 product. |
(i) |
It will exist in from of enantiomeric pair. |
(ii) |
It has no chiral carbon, so no enaniomerism possible. |
(iii) |
It has no chiral carbon atom, so no enantiomerism. |
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