Compare the carbon-carbon bond rotation across the compounds A, B and C:

A > B > C

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B > A > C

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A > C > B

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B > C > A

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Solution

The correct option is B $B > A > C$
Driving force for breaking of double bond is formation of aromatic species.
Order of stability of ring: 6-membered > 5-membered > 7 membered > 4-membered > 3-membered
Hence, (B) will break most easily due to formation of stable rings. (C) will not break easily due to formation of antiaromatic ring.

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∣ ∣ ∣ ∣ ∣ \begin{matrix} (b + c)^{2} & b a & a c b a & (c + a)^{2} & c b c a & c b & (a + b)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

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