Propene is treated with $HBr$ in the presence of peroxide to give X. X is converted to its Grignard reagent with Mg in dry ether, which is treated with ethanal and then $H X_{3} O X^{+}$ . The final alcohol is

Answer: A

Peroxide effect gives the anti-Markovnikov bromide,

CH X_{3} CH X_{2} CH X_{2} Br

. Its Grignard reagent

CH X_{3} CH X_{2} CH X_{2} MgBr

adds to ethanal (

CH X_{3} CHO

) to give a secondary alkoxide. Hydrolysis gives

CH X_{3} CH (OH) CH X_{2} CH X_{2} CH X_{3}

, pentan-2-ol.