Correct Answer - B
Reaction between alcohols and halogen acid follows `S_(N)1` mechanism. In `S_(N)1` mechanism carbocations are formed as intermediates.
Let us consider the formation of carbocations with the given three alcohols.
`CH_(3)-CH_(2)-CH_(2)-OHtoCH_(3)-overset(+)CH+OH^(-)`
In this case, `1^(@)` carbocation is formed. It is least stable. So, here `S_(N)2` mechanism is followed. In this `S_(N)2` mechanism a transitory state is observed in a-carbon is linked with two nucleophiles.
`CH_(3)-CH_(2)-underset(CH_(3))underset(|)"CH"-OHtounderset(2^(@)"carbocation" ("more stable than" 1^(@) "carbocation"))(CH_(3)-underset(CH_(3))underset(|)overset(+)"CH"+OH^(-))`
`CH_(3)-CH_(2)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OHtounderset(3^(@)"carbocation" ("most satble"))(H_(3)C-CH_(2)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C^(+))+OH^(-))`
The reaction proceeded with stable carbocation. Higher the stability of carbocation, higher will be the possibilities of attack of `X^(-)` ion to the carbocation.
As, the tertiary carbocation is most stable so the possibilities of attack of `X^(-)` ion are more prominent in case of tertiary carbocations. Thus, attack of `X^(-)` ion to carbocation is proceeded with tertiary carbocation as follows
`underset(3^(@) "carbocation")(H_(3)C-CH_(2)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C^(oplus))+X)toH_(3)C-CH_(2)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-X`
So, the correct option is (b).
Note Higher the stability of intermediate, higher will be the reactivity of compound and higher will be the yield of the desired product
