In the following reaction,'B' is | vedclass

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(B) The reaction proceeds via the following steps:$1$. Protonation of the alcohol group: The $-OH$ group is protonated by $H_3O^+$ to form a good leav

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(B) The reaction proceeds via the following steps:$1$. Protonation of the alcohol group: The $-OH$ group is protonated by $H_3O^+$ to form a good leaving group $(-OH_2^+)$.$2$. Formation of carbocation: Loss of water molecule generates a secondary carbocation.$3$. Rearrangement: $A$ $1,2-CH_3^-$ shift occurs to form a more stable tertiary carbocation.$4$. Cyclization: The double bond attacks the carbocation to form a new ring.$5$. Deprotonation: Loss of a proton $(H^+)$ from the adjacent carbon leads to the formation of the final stable alkene product,which corresponds to the structure in option $B$.

In the following reaction,'$B$' is

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The product $(A)$ is:

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