The major product [B] in the following sequen | vedclass

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(D) Step $1$: Hydroboration-oxidation of $CH_3-C(CH(CH_3)_2)=CH-CH_2CH_3$ follows anti-Markovnikov addition of water to form the alcohol $[A]$,$CH_3-C

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$CH_3-C(CH_3)=C(CH_3)-CH_2CH_3$

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(D) Step $1$: Hydroboration-oxidation of $CH_3-C(CH(CH_3)_2)=CH-CH_2CH_3$ follows anti-Markovnikov addition of water to form the alcohol $[A]$,$CH_3-CH(CH(CH_3)_2)-CH(OH)-CH_2CH_3$.Step $2$: Acid-catalyzed dehydration of $[A]$ using $dil. H_2SO_4/\Delta$ proceeds via a carbocation intermediate.Step $3$: The secondary carbocation undergoes a $1,2-H^-$ shift to form a more stable tertiary carbocation.Step $4$: Elimination of a proton $(H^+)$ from the most stable carbocation leads to the most substituted alkene (Saytzeff product).The final major product $[B]$ is $CH_3-C(CH_3)=C(CH_3)-CH_2CH_3$ (which is $2,3-dimethylpent-2-ene$).

The major product $[B]$ in the following sequence of reactions is
$CH_3-C(CH(CH_3)_2)=CH-CH_2CH_3$ $\xrightarrow[(i) B_2H_6]{(ii) H_2O_2, OH^{-}} [A]$ $\xrightarrow[\Delta]{dil. H_2SO_4} [B]$

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The major product $[B]$ in the following sequence of reactions is$CH_3-C(CH(CH_3)_2)=CH-CH_2CH_3$ $\xrightarrow[(i) B_2H_6]{(ii) H_2O_2, OH^{-}} [A]$ $\xrightarrow[\Delta]{dil. H_2SO_4} [B]$