Which one of the following reactions is consistent with the Markovnikov's rule?

(C H_{3})_{2} C H - C H = C H_{2} + H_{2} O \to (C H_{3})_{2} C H - C H_{2} - C H_{2} - O H

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C H_{3} - C H_{2} - C H = C H_{2} + H_{2} O \to C H_{3} - C H_{2} - C H_{2} - C H_{2} - O H

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C H_{3} C H = C H C H_{3} + H_{2} O \to C H_{3} C H_{2} - C H (C H_{3}) - O H

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H_{2} C = C H C H_{3} + H_{2} O \to H_{3} C - C H (O H) - C H_{3}

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Solution

The correct option is D $H_{2} C = C H C H_{3} + H_{2} O \to H_{3} C - C H (O H) - C H_{3}$

Markownikoff's rule can be stated that with the addition of a protic acid HX to an unsymmetrical alkene the hydrogen atom is added to the carbon with the greater number of hydrogen atoms while the X component is added to the carbon with the fewer number of hydrogen atoms.
Anti-Markownifkoff's reaction is exactly the opposite of Markovnikov reaction

In the above question, all reactions are anti-Markovnikov reactions except option-4. this reaction is consistent with Markovnikov' rule.

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