The major product formed in the following reation is
`CH_(3)-underset(H)underset(|)overset(CH_(3))overset(|)C-CH_(2)-Br underset(CH_(3)OH)overset(CH_(3)ONa)rarr` .

Options

A. `CH_(3)-overset(CH_(3))overset(|)underset(H)underset(|)C-CH_(2)OCH_(3)`
B. `CH_(3)-underset(OCH_(3))underset(|)(CH)-CH_(2)CH_(3)`
C. `CH_(2)-overset(CH_(2))overset(|)(C)=CH_(2)`
D. `CH_(3)-underset(OCH_(2))underset(|)overset(CH_(3))overset(|)(C)-CH_(3)`

Answer

D

Explanation

`H_(3)C-overset(CH_(3))overset(|)underset(H)underset(|)C-CH_(2)-Broverset(CH_(3)O^(-))underset(CH_(3)OH)rarrA?`
Alkyl halide is `1^(@)`. Keep in mind `1^(@)` halide give product by `S_(N)2//E-2` mechanism and `1^(@)` halide always gives substitution reaction except when strongly hindered base is used.
ex. With `CH_(3)-overset(CH_(3))overset(|)underset(CH_(3))underset(|)C-O^(-)` it gives mainly elimination. The reaction involves carbocation intermediate.
i.e. `underset(("tertiary carbocation"))(CH_(3)-overset(CH_(3))overset(|)underset(H)underset(|)C-overset(oplus)CH_(2))` ltbr Stability of carbocation : `3^(@)gt2^(@)gt1^(@)gtoverset(oplus)CH_(3)` It is because the stability of a charged system is increased by dispersal of the charge. The more stable the carbocation, the faster it is formed. N.B.-Rearrangement can be done in two ways.