For isomeric hydrocarbons with molecular form | vedclass

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(B) The molecular formula $C_9H_{12}$ corresponds to a degree of unsaturation of $4$,which is consistent with an aromatic ring (benzene ring) plus thr

Vedclass · Accepted Answer

$2$

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(B) The molecular formula $C_9H_{12}$ corresponds to a degree of unsaturation of $4$,which is consistent with an aromatic ring (benzene ring) plus three carbons in the side chain$(s)$. $A$ negative Baeyer's test indicates the absence of aliphatic unsaturation (no double or triple bonds in the side chains). For an aromatic ring to have four different non-aliphatic (aromatic) substitution sites,the ring must have two substituents such that the remaining four positions on the ring are chemically distinct. Considering the isomers of $C_9H_{12}$ (propylbenzenes,ethylmethylbenzenes,and trimethylbenzenes),the two isomers that satisfy this condition are $1$-ethyl-$2$-methylbenzene ($o$-ethyltoluene) and $1$-ethyl-$3$-methylbenzene ($m$-ethyltoluene). In these structures,the four remaining hydrogen-bearing carbons on the benzene ring are non-equivalent,providing four distinct sites for electrophilic aromatic substitution. Thus,the total number of such isomers is $2$.

For isomeric hydrocarbons with molecular formula $C_9H_{12}$ that give a negative Baeyer's test,determine the total number of isomers that possess four different non-aliphatic substitution sites.

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