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`

Options

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

D

Explanation

`H_(3)C-underset(H)underset(|)overset(CH_(3))overset(|)C-CH_(2)-Br underset(CH_(3)OH)overset(CH_(3)Obar)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 For example
With `CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-O^((-))` it gives mainly elimination The
reaction involves carbocation intermediate
`CH_(3)-underset(H)underset(|)overset(CH_(3))overset(|)(C)-Coverset(o+)(H)_(2)`
but as it is a primary carbocation it will rearrange to give a tertiary carbocation which completes the reaction
`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
Note: Rearrangement can be done in two ways

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