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Chapter 26: Problem 646

A compound \(\mathrm{A}\) has the molecular formula \(\mathrm{C}_{7} \mathrm{H}_{4} \mathrm{BrClO}\). Treatment of A With diazomethane gave B, which upon heating with silver oxide and water was converted to another compound, C, \(\mathrm{C}_{8} \mathrm{H}_{7} \mathrm{BrO}_{2}\). The nmr spectrum of \(\mathrm{C}\) is as follows: \(\delta 2.7\) singlet, \(\mathrm{A}=2 ; \delta 7.0 \mathrm{multiplet}, \mathrm{A}=4 ; \delta 12.0\) singlet, \(\mathrm{A}=1 .\) What are the structures of compounds \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) ?

Short Answer

Expert verified
In conclusion, the structures of the compounds are as follows: Compound A: \(2-Bromo-4-chlorobenzoic\, acid\) Compound B: \(2-Bromo-4-chlorobenzoate\, methyl\, ester\) Compound C: \(2-Bromo-4-hydroxybenzoic\, acid\)

Step by step solution

01

Identify Key Reactions

1. Treatment of compound A with diazomethane gave compound B. This gives us a clue that compound A might be a carboxylic acid, as diazomethane can convert carboxylic acids into methyl esters. 2. Heating compound B with silver oxide and water led to compound C. This suggests that compound B might be a methyl ester, as Ag2O and water can hydrolyze esters to form carboxylic acids.
02

Determine the Functional Groups in A

Based on the reactions, we have reasons to believe that compound A is a carboxylic acid, and it has a Cl and a Br atom in its structure. The molecular formula of A is C7H4BrClO, which means that there are two double bonds or one triple bond. The presence of four hydrogen atoms in the formula supports the presence of an aromatic ring in the structure.
03

Analyze the NMR Spectrum of C

1. δ 2.7 singlet, A=2: This peak suggests the presence of two identical protons that are not split by neighboring protons. This may correspond to a methylene group. 2. δ 7.0 multiplet, A=4: This peak indicates four aromatic protons on the benzene ring. 3. δ 12.0 singlet, A=1: This peak is characteristic of a hydrogen bonding with an oxygen atom, and it can be assigned to the acidic hydrogen of a carboxylic acid group.
04

Identify the Structure of Compound A

Now, let's consider the possible structures of compound A. Based on the above analysis, we know that compound A contains an aromatic ring (benzene) and a carboxylic acid group: 1. Compound A should have a Br atom and Cl atom attached to the benzene ring. 2. The positions of Br, Cl, and the carboxylic acid group are crucial. We will have ortho, meta, and para isomers for both halogens when placing them on the benzene ring. Since there are only four hydrogens, the halogens and the carboxylic acid cannot be ortho/substituted. So, the carboxylic acid must be in a meta position. Considering these points, the structure of compound A should be \(2-Bromo-4-chlorobenzoic\, acid\).
05

Identify the Structure of Compound B

As we determined that compound A is a carboxylic acid and diazomethane converts carboxylic acids into methyl esters, compound B will be a methyl ester. Therefore, the structure of compound B should be \(2-Bromo-4-chlorobenzoate\, methyl\, ester\).
06

Identify the Structure of Compound C

We know that the process of hydrolyzing a methyl ester (compound B) with silver oxide and water yields a carboxylic acid and a hydroxyl group. Based on the NMR spectrum, we can compare this to the structure of compound C, C8H7BrO2. The structure of compound C should be \(2-Bromo-4-hydroxybenzoic\, acid\). In conclusion, the structures of the compounds are as follows: Compound A: 2-Bromo-4-chlorobenzoic acid Compound B: 2-Bromo-4-chlorobenzoate methyl ester Compound C: 2-Bromo-4-hydroxybenzoic acid

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