Propene when reacted with water in the presence of dilute $H_{2} S O_{4}$ gives

Propan-1-ol

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Propan-2-ol

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2-Methylpropan-1-ol

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2-Methylpropan-2-ol

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Solution

The correct option is B Propan-2-ol
Since propene is an unsymmetrical alkene, the given hydration reaction takes place in accordance to Markovnikov’s rule, to form propan-2-ol. The double bond is broken and the OH group attaches at the second carbon.
The reaction is as follow:
$C H_{3} C H = C H_{2} + H_{2} O d i l . H_{2} S O_{4} - ----- \to C H_{3} C H (O H) C H_{3}$

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C C l_{4}

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