In the reaction:
`underset("(Major)")(A) overset(C_(2)H_(5)OH)larr (CH_(3))_(3)CBr underset(C_(2)H_(5)OH)underset(C_(2)H_(5)O^(-)Na^(+))to underset("(Major)")(B)`
A and B are respectively:

Options

A. A is `(CH_(3))_(2)C=CH_(2)` and B is `(CH_(3))_(3)COC_(2)H_(5)`
B. A is `(CH_(3))_(3)COC_(2)H_(5)` and B is `(CH_(3))_(2)C=CH_(2)`
C. Both A and B are `(CH_(3))_(2)C=CH_(2)`
D. Both A and B are `(CH_(3))_(3)COC_(2)H_(5)`.

Answer

B

Explanation

With a weak nucleophile such as `C_(2)H_(5)OH` substitution takes place to form tert-butyl ethyl ether `(CH_(3))_(3)C–O–C_(2)H_(5)(A)`. With a strong base like `C_(2)H_(5)O^(-)`, tert-butyl bromide undergoes elimination to form `(CH_(3))_(2)CH=CH_(2)Br`.