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In the following reaction
${CH}_{3} - {CH}_{2} - CH (O H) - {CH}_{3} \to_{475 K}^{H_{2} {SO}_{4}}$

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${CH}_{3} CH = {CHCH}_{3}$ predominates

${CH}_{2} = {CHCH}_{2} {CH}_{3}$ predominates

Both are formed in equal amounts

The amount of production depends on the nature of catalyst

answer is A.

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Detailed Solution

${CH}_{3} - {CH}_{2} - {CH}_{2} - {CH}_{3} \to_{475 K}^{H_{2} {SO}_{4}}$

$\underset{More summetrical (major product)}{{CH}_{3} - CH = CH - {CH}_{3}}$

${CH}_{3} - {CH}_{2} - {CH}_{2} - {CH}_{3} \to_{475 K}^{H_{2} {SO}_{4}}$

$\begin{matrix} {CH}_{2} = CH - {CH}_{2} - {CH}_{3} \\ Less symmetrical or \\ unsymmetrical \\ (minor product) \end{matrix}$

It is based on Saytzeff's rule. According to this more symmetrical or more alkylated alkene predominates.

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