In the above reaction,the left-hand side and | vedclass

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(A) $1$. The reaction starts with the protonation of the alcohol group,followed by the loss of water to form a carbocation at the central carbon.$2$.

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Finally both rings will become six-membered each.

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(A) $1$. The reaction starts with the protonation of the alcohol group,followed by the loss of water to form a carbocation at the central carbon.$2$. The carbocation undergoes a ring expansion (or rearrangement) involving the adjacent rings to form a more stable system.$3$. Specifically,the $4$-membered ring $(A)$ expands to a $6$-membered ring,and the $5$-membered ring $(B)$ also expands to a $6$-membered ring through a series of rearrangements and shifts.$4$. The final product is a bicyclic system where both rings are $6$-membered (decalin derivative).$5$. Therefore,both rings expand to become $6$-membered.

In the above reaction,the left-hand side and right-hand side rings are named as '$A$' and '$B$' respectively. They undergo ring expansion. The correct statement for this process is:

Similar Questions

Classify the following species into carbocation,free radical,and carbanion:
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$4. \, CH_3 - \overset{+}{C} H_2$
$5. \, \overset{+}{C} H_3$

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