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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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).

Similar Questions

Which of the following processes does $NOT$ occur during the preparation of $CHCl_3$ from $C_2H_5OH$ and bleaching powder?

From each of the following pairs,select the compound that will react faster with sodium iodide in acetone:
Pair-$A$: $(1) \, 2-$Chloropropane,$(2) \, 2-$Bromopropane
Pair-$B$: $(3) \, 1-$Bromobutane,$(4) \, 2-$Bromobutane

For the reaction of $3$-bromohexane with alcoholic $KOH$,find $(x)$,where $(x)$ is the total number of $E_2$ elimination products obtained (including stereoisomers).

What is obtained when nitric acid reacts with chloroform?

The major product obtained in the following reaction is $C_2H_5ONa + (CH_3)_3C-Cl \rightarrow$

Vedclass Test Series

Exam Paper Generator

Online Exam Module