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Protonation of 1,2-butadine gives carbocation(A) and protonation of 2-butene gives carbocation(B). In both case the most stable crbocation possible is produced but no carbocation rearrangements take place. Which of the two carbocation is more stable and why?

$\begin{array}{1 1}a)\;A\; is\; more\; stable\; because\; it is\; a\; terminal \;carbocation,according\; to\; markonikov's\; rule.\\b)\; A \;is\; more\; stable\; because\; it\; is\; stabilized\; by\; resonance \;and\;hyper conjugation.\\c)\; A\; is\; more\; stable\; because \;cationic\; carbon\; is\; sphybridized.\\d)\; B\; is\; more \;stable\; because\; it\; has \;2\; alkyl\; groups \;that\; can\;provide \;extra \;stabilization\; by\; hyperconjugation \;and \;inductive\;effect.\end{array}$

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answered Dec 3, 2013 by sreemathi.v

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