Question

The major product formed in the reaction is:
`CH_(3)-underset(H)underset(|)overset(CH_(3))overset(|)(C)-CH_(2)Brunderset(CH_(3)OH)overset(CH_(3)O^(Ө))to`
A. `CH_(3)-underset(H)underset(|)overset(CH_(3))overset(|)(C)-CH_(3)OCH_(3)`
B. `CH_(3)-underset(OCH_(3))underset(|)(C)H-CH_(2)CH_(3)`
C. `CH_(3)-overset(CH_(3))overset(|)(C)=CH_(2)`
D. `CH_(3)-underset(OCH_(3))underset(|)overset(CH_(3))overset(|)(C)-CH_(3)`

Answer 1

Correct Answer - D
`H_(3)C-underset(H)underset(|)overset(CH_(3))overset(|)(C)-CH_(2)-Brunderset(CH_(3)OH)overset(CH_(3)O^(-))toA,` Alkyl halide is `1^(@)`.
Keep in mind `1^(@)`
Keep in mind `1^(@)` halide give by `S_(N^(2))//E2` mechanism and `1^(@)` halide always gives substitution reaction except when strongly hindered base is base.
ex: With `CH_(3)-underset(CH_(3))underset(|)(|)overset(CH_(3))overset(|)(C)-O^((-))` it gives mainly elimination.
The reaction involves carbocation intermediate.
i.e., `underset(("Primary carbocation"))(CH_(3)-underset(H)underset(|)overset(CH_(3))overset(|)(C)-overset(o+)(CH_(2))`
but as it is a primary carbocation it will rearrange to give a tertiary carbocation, which completes the reaction
`underset("tertiary carbocation")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C^(o+))`
Stability of carbocation: `3^(@) gt 2^(@) gt 1^(@)gt overset(o+)(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.-Rearrangemnt can be done in two ways.