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In the orange box, I think that the blue-carbon should be a carbocation instead. So, option A should be correct.

But, the answer is B. Why is it so?

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    $\begingroup$ migratory aptitude in cationic reactions is secondary > primary > methyl because the migrating carbon has to bear some of the positive charge in the transition state, see e.g. chemistry.stackexchange.com/a/16470/16683 $\endgroup$ – orthocresol Aug 3 at 15:27
  • $\begingroup$ Thanks @orthocresol, actually I was starting the migrating-$\ce{C}$ from that quaternary carbon and so I was confused :P $\endgroup$ – rv7 Aug 4 at 2:19
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As orthocresol commented (vide supra), migratory aptitude in cationic reactions has been given as: $\ce{H}=\mathrm{Ph} \gt \ce{(CH3)3C \gt (CH3)2CH \gt CH3CH2 \gt CH3}$ (Wikipedia and Migratory aptitude in pinacol-pinacolone rearrangement). In your question, the secondary carbocation formed when protonated has three groups, which can migrate to give more stable tertiary carbocation. They, with their migratory aptitude, are: $\ce{-R^1CH2(CH3)CH\!- \gt R^2CH2CH2\!- \gt H3C\!-}$ (where $\ce{R^1}$ and $\ce{R^2}$ are continuation of rest of cyclopentane ring).

Now, if $\ce{H3C\!-}$ group migrated, it'd give option (D); if $\ce{R^2CH2CH2\!-}$ group migrated, it'd give option (A); and if $\ce{R^1CH2(CH3)CH\!-}$ group migrated, it'd give option (B). If nothing migrated you have option (C). Although all 4 products are possible in the reaction mixture ((C) would be very trace amount), asking is the major product, and hence is is the product given but migration of group with highest migratory aptitude. Therefore, it should be the product given in option (B).

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