Question

A compound ‘A’ (C₇H₈O) is insoluble in NaHCO₃ solution but dissolve in NaOH and give a characteristic colour with neutral FeCl₃ solution. When treated with Bromine water compound ‘A’ forms the compound B with the formula C₇H₅OBr₃. ‘A’ is

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is B

We can deduce the structure of compound 'A' from the given reactions and properties:

1. Compound 'A' is insoluble in NaHCO₃ solution, which suggests that it is not an acid, so it is not a carboxylic acid.
2. Compound 'A' dissolves in NaOH, which indicates that it is an alcohol (phenol or an alcohol). Since phenols are generally soluble in NaOH and alcohols are also soluble, we must further investigate.
3. Compound 'A' gives a characteristic colour with neutral FeCl₃ solution, which is a test for phenols. This reaction forms a coloured complex with phenolic compounds, indicating that 'A' is likely a phenol or an alcohol that can react in this way.
4. Compound 'A' reacts with Bromine water to form a compound B with the formula C₇H₅OBr, which suggests that 'A' undergoes bromination, typical of alcohols (especially phenols) and the resulting product is a bromo derivative. 

Based on these reactions, the correct compound 'A' is phenol (C₆H₅OH), and it undergoes bromination in the presence of bromine water to form a bromo derivative.

The correct answer is (B) : .

Answer 2

To solve this problem, we must analyze the properties of compound 'A' based on the given reactions and information. The compound 'A' with formula $C_7H_8O$ is insoluble in $NaHCO_3$ but dissolves in $NaOH$, giving a characteristic colour with neutral $FeCl_3$ solution. When treated with bromine water, it forms a compound 'B' with the formula $C_7H_5OBr_3$.

1. Solubility in $NaOH$:
The fact that compound 'A' dissolves in $NaOH$ suggests it is a phenol or a carboxylic acid. Since it's insoluble in $NaHCO_3$, it's less likely to be a carboxylic acid, indicating it is likely a phenol.

2. Reaction with $FeCl_3$:
The characteristic colour with neutral $FeCl_3$ solution confirms the presence of a phenol group.

3. Reaction with Bromine Water:
The reaction with bromine water to form $C_7H_5OBr_3$ suggests that three hydrogen atoms on the benzene ring have been substituted by bromine atoms. This is characteristic of phenols with activating substituents, which direct the bromination to ortho and para positions.

4. Analyzing the Options:
(A) Benzyl alcohol ($C_6H_5CH_2OH$): This is an alcohol, not a phenol, and would not dissolve in $NaOH$ or react with $FeCl_3$ in the way described. Also, it does not directly react with bromine water to form a tribrominated product.
(B) o-Cresol (2-Methylphenol): This is a phenol. It dissolves in $NaOH$ and reacts with $FeCl_3$. Upon treatment with bromine water, the two ortho and one para positions relative to the -OH group will be brominated, leading to the formation of $C_7H_5OBr_3$. This looks like a plausible answer.
(C) m-Cresol (3-Methylphenol): This is a phenol. It dissolves in $NaOH$ and reacts with $FeCl_3$. Upon treatment with bromine water, bromination will occur at two ortho and one para positions, again giving $C_7H_5OBr_3$.
(D) p-Cresol (4-Methylphenol): This is a phenol. It dissolves in $NaOH$ and reacts with $FeCl_3$. Upon treatment with bromine water, bromination will occur at two ortho positions, leading to the formation of $C_7H_5OBr_3$.

5. Differentiating Between C, B and D:
B, C, and D all give similar products on bromination. The key is that the question asks *the* compound A. All the reactions apply to a phenol and give the appropriate brominated product, but since only one answer can be selected, the location of the methyl group needs to be considered. The fact that three bromines are added to the benzene ring means that all the other positions must have been free, relative to the -OH group. Considering all options again:
B. o-Cresol: Br atoms would add to positions 4, 5, and 6. The methyl would be at the 2 position, hydroxyl at 1.
C. m-Cresol: Br atoms would add to positions 2, 4, and 6. The methyl would be at the 3 position, hydroxyl at 1.
D. p-Cresol: Br atoms would add to positions 2, 3, and 5. The methyl would be at the 4 position, hydroxyl at 1.

Final Answer:
The compound 'A' is o-Cresol (Option B).