Two isomers (A) and (B) with molar mass 184 | vedclass

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(C) $1$. Empirical formula calculation: Moles of $C = 52.2/12 = 4.35$,$H = 4.9/1 = 4.9$,$Br = 42.9/80 = 0.536$. Ratio: $C:H:Br = 8:9:1$. Empirical for

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$H_3C-CHBr-C_6H_5$ and $p$-bromobenzyl bromide

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(C) $1$. Empirical formula calculation: Moles of $C = 52.2/12 = 4.35$,$H = 4.9/1 = 4.9$,$Br = 42.9/80 = 0.536$. Ratio: $C:H:Br = 8:9:1$. Empirical formula is $C_8H_9Br$. Molar mass is $184 \ g/mol$,which matches $C_8H_9Br$. $2$. Oxidation with $KMnO_4$: Isomer $(A)$ gives benzoic acid,implying it has an alkyl group attached to the benzene ring at the alpha position (e.g.,$C_6H_5-CH(Br)-CH_3$). Isomer $(B)$ gives $p$-bromobenzoic acid,implying it has an ethyl group at the para position relative to the bromine atom (e.g.,$p-Br-C_6H_4-CH_2CH_3$). $3$. Isomer $(A)$ is optically active: $C_6H_5-CH(Br)-CH_3$ has a chiral center. $4$. Isomer $(A)$ gives a precipitate with $AgNO_3$: The benzylic bromide is reactive. $5$. Therefore,$(A)$ is $1$-bromo-ethylbenzene $(H_3C-CHBr-C_6H_5)$ and $(B)$ is $1$-bromo$-4-$ethylbenzene ($p$-ethylbromobenzene).

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