Question

An organic compound 'A' with emperical formula $C _6 H _6 O$ gives sooty flame on burning Its reaction with bromine solution in low polarity solvent results in high yield of $B .B$ is

• A.
• B.
• C.
• D.

Hint:

In reactions with bromine in a low polarity solvent like \(\text{CS}_2\), phenols undergo electrophilic substitution, where the bromine replaces a hydrogen atom on the benzene ring.

Answer 1

The Correct Option is C

To solve this problem, let's break down the information given: 

• The compound 'A' has an empirical formula of \(C_6H_6O\), which suggests a compound with six carbon atoms, six hydrogen atoms, and one oxygen atom. Aromatic compounds like phenol often burn with a sooty flame, indicating a higher carbon content relative to hydrogen and oxygen.
• When 'A' reacts with bromine in a low polarity solvent, it produces a compound 'B' in high yield. In such conditions, the reaction is likely an electrophilic aromatic substitution.

Given these points, compound 'A' is most likely phenol \((C_6H_5OH)\), as it is a common aromatic compound with the provided empirical formula and burns with a sooty flame.

When phenol reacts with bromine in low polarity solvents, such as carbon tetrachloride or chloroform, bromine substitutes at the ortho, meta, and para positions on the aromatic ring. However, due to the activating -OH group, the para position predominantly reacts to give para-bromophenol in high yield.

Hence, the correct structure for compound 'B' is para-bromophenol.

Let's consider the incorrect options to understand why they do not fit:

• 1-Bromophenol and 2-bromophenol (ortho and meta products) are less likely due to the positioning and yield considerations already mentioned. The para position is typically more favored.
• The last option, unrelated to bromination products, can be ruled out based on the question's chemical context.

Therefore, compound 'B' is indeed para-bromophenol, which is consistent with the behavior of phenol reacting with bromine in these conditions.