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(B) The reaction of $1$-butene $(CH_3-CH_2-CH=CH_2)$ with water in the presence of an acid catalyst $(H^+)$ follows Markovnikov's rule.According to Ma

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Secondary alcohol

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(B) The reaction of $1$-butene $(CH_3-CH_2-CH=CH_2)$ with water in the presence of an acid catalyst $(H^+)$ follows Markovnikov's rule.According to Markovnikov's rule,the negative part of the addendum $(OH^-)$ attaches to the carbon atom with fewer hydrogen atoms.In $1$-butene,the double bond is between $C_1$ and $C_2$. $C_2$ has one hydrogen atom,while $C_1$ has two.Therefore,the $OH$ group attaches to $C_2$,resulting in the formation of $2$-butanol $(CH_3-CH_2-CH(OH)-CH_3)$.Since the $OH$ group is attached to a secondary carbon atom,the product is a secondary alcohol.

According to Markovnikov's rule,what is the product formed when $1$-butene reacts with water?

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