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  4. (CH3)3COH + CH3MgBr arrow Hydrocarbon (A); (A) is

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Q. $(CH_3)_3COH + CH_3MgBr \rightarrow$ Hydrocarbon $(A); (A)$ is

Hydrocarbons

Solution:

$ROH + CH_3MgBr \rightarrow CH_4 + Mg <{}^{Br}_{OR}$