Question

KOH (alcoholic) + CH3 C(CH3)2 CH2Br (1), The reactants in the condition given will under
go:

A. Nucleophilic substitution reaction. "B. Elimination reaction. C. Nucleophilic addition,
D. None of the above.
2nd​

Answer 1

Answer:

Nucleophilic Substitution reaction

Answer 2

Alcoholic KOH normally does give you an elimination reaction, but here there's no chance for an E-2 mechanism and the same goes for E-1, since a 1° carbocation can't be stable enough for Bromine to leave by itself.

So I think it will be a nucleophilic substitution reaction.