${\left(C{H}_3\right)}_{3}COH+C{H}_{3}MgBr\underset{ }{\to }$Hydrocarbon (A). A is____

$RMgX\frac{Active\left(H\right)}{or{H}^{+}}RH$, this is the common reaction, hence the reaction is $C{H}_{3}MgBr\underset{ }{\overset{{\left(C{H}_3\right)}_{3}CO-H}{\to }}C{H}_4$