MCQ
$(CH_3)_2CHBr$ $\xrightarrow[{(2)\,CuI}]{{(1)\,Li}}\left( A \right)\xrightarrow{{{{\left( {C{H_3}} \right)}_2}CHC{H_2}Br}}B$
$'B'$ is
- A$(CH_3)_2CHCH_2CH_3$
- B$CH_3)_2CHCH(CH_3)_2$
- ✓$(CH_3)_2CHCH_2CH(CH_3)_2$
- D$(CH_3)_2CHCH_2CH_2CH(CH_3)_2$
✓
Answer
Correct option: C.
$(CH_3)_2CHCH_2CH(CH_3)_2$
c
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