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Roni posted an Question

November 10, 2020 • 20:57 pm

30 points

IIT JAM
Chemistry (CY)

25 goc please ans my following questions as soon as possible

2 Answer(s)

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Suman Kumar

Both rings become anti aromatic. Such kind of compound show Quasi aromatic. They become aromatic by breaking the double bond between two rings.
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Hence option C is correct
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Priyanshu kumar Best Answer

Both rings become anti aromatic. Such kind of compound show Quari aromaticity. They become aromatic by breaking the double bond between two rings. As (III) become antiaromatic after breaking of bond. Thus the bond will not break easily and hence C−C rotation is less. Stability order of ring is 6−member>5−member>7−member>4−member>3−member Thus In (I) and (II) both ring become aromatic. C=C bond will break in (II) more easily as both the ring are more stable. Thus the order of C−C bond rotation is II>I>III
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option C is correct
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but sir aromacity of cyclopropenyl cation >cycloheptatrienyl cation
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check stability of rings...
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yes sir ..
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Dinesh khalmaniya 1

option C Here check the stability of compounds. I and II become Aromatic in ionic form so here only single bond exist. in I there is small rings so we know the small rings are more aromatic hene more stabilized in ionic form so rotation barrier will be least. III is non aromatic so purely double bond so highest rotation barrier
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sir I thinking same concept but ans is c ... why ?
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let me check
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yes , i made some mistake
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Stability order of ring is 6−member>5−member>7−member>4−member>3−member
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just remember this order
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ok sir thanks
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