Which reaction will give CH_3-CH(CH_3)-CH(OH) | vedclass

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(C) The target product is $CH_3-CH(CH_3)-CH(OH)-CH_3$,which is $3$-methylbutan-$2$-ol.Option $A$: Acid-catalyzed hydration $(H_3O^+)$ of $3$-methylbut

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The reaction of $3$-methylbut-$1$-ene with $(i) Hg(OAc)_2, H_2O, (ii) NaBH_4, OH^-$

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(C) The target product is $CH_3-CH(CH_3)-CH(OH)-CH_3$,which is $3$-methylbutan-$2$-ol.Option $A$: Acid-catalyzed hydration $(H_3O^+)$ of $3$-methylbut-$1$-ene $(CH_3-CH(CH_3)-CH=CH_2)$ involves the formation of a carbocation intermediate. The initial protonation gives a primary carbocation,which rearranges to a more stable tertiary carbocation $(CH_3-C^+(CH_3)-CH_2-CH_3)$. Attack by water leads to $2$-methylbutan-$2$-ol.Option $B$: Hydroboration-oxidation follows anti-Markovnikov addition of water,yielding $3$-methylbutan-$1$-ol.Option $C$: Oxymercuration-demercuration involves the Markovnikov addition of water without carbocation rearrangement. The $OH$ group adds to the more substituted carbon of the double bond. For $3$-methylbut-$1$-ene,the double bond is at the terminal position. However,the structure $CH_3-CH(CH_3)-CH(OH)-CH_3$ corresponds to the addition of $OH$ to the $C_2$ position. This reaction proceeds via a cyclic mercurinium ion,and water attacks the more substituted carbon $(C_2)$,resulting in $3$-methylbutan-$2$-ol as the major product.Therefore,option $C$ is the correct reaction.

Which reaction will give $CH_3-CH(CH_3)-CH(OH)-CH_3$ as a major product?

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