1-Bromopropane is converted to a Grignard reagent with $Mg$ in dry ether. It is then treated with propanone, followed by $H X_{3} O X^{+}$ , to give $A$ . Heating $A$ with conc. $H X_{2} SO X_{4}$ gives the major alkene $B$ . $B$ is

Answer: C

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

adds to propanone and hydrolysis gives

A = (CH X_{3}) X_{2} C (OH) CH X_{2} CH X_{2} CH X_{3}

, 2-methylpentan-2-ol. Acid-catalysed dehydration proceeds through the tertiary cation and follows Zaitsev's rule, losing a hydrogen from the

CH X_{2}

side to give the more substituted alkene

(CH X_{3}) X_{2} C = CH - CH X_{2} CH X_{3}

. This is 2-methylpent-2-ene.