In the reaction sequence `underset("(small)")(C_6H_5)-CH_3overset(Cl_2//hv)rarr[X]overset(HOH//NaOH)rarr[Y]` [X] and [Y] will respectively be :

Options

A. `C_6H_5-CH_2Cl and C_6H_5-CH_2OH`
B. `C_6H_5-CH_2Cl and C_6H_5-CHO`
C. `C_6H_5-"CCl"_3 and C_6H_5COOH`
D. `C_6H_5-"CCl"_3 and C_6H_5COONa`

Answer

To solve the given problem, we need to analyze the reaction sequence step by step.

### Step 1: Identify the starting material
The starting material is toluene, which is represented as \( C_6H_5CH_3 \).

### Step 2: Reaction with \( Cl_2 \) under UV light
When toluene is treated with chlorine gas (\( Cl_2 \)) in the presence of ultraviolet light (hv), a free radical substitution reaction occurs. The chlorine molecule dissociates into two chlorine radicals due to the UV light.

### Step 3: Formation of the first product [X]
1. The chlorine radical attacks the methyl group of toluene, resulting in the formation of a benzyl radical (\( C_6H_5CH_2^{\cdot} \)) and hydrochloric acid (HCl).
2. The benzyl radical can then react with another chlorine radical, leading to the formation of \( C_6H_5CH_2Cl \) (benzyl chloride).
3. Further substitution can occur, resulting in the formation of \( C_6H_5CCl_3 \) (trichlorobenzene) as the final product [X].

### Step 4: Reaction of [X] with \( HOH \) and \( NaOH \)
Now, when the product [X] (trichlorobenzene) is treated with water (HOH) and sodium hydroxide (NaOH), a nucleophilic substitution reaction occurs:
1. The hydroxide ions (\( OH^- \)) will replace the chlorine atoms on the trichlorobenzene.
2. This results in the formation of a compound where all three chlorine atoms are replaced by hydroxyl groups, leading to the formation of a triol.
3. Upon dehydration (loss of water), the product formed is \( C_6H_5C(OH)C(=O)H \), which is benzaldehyde.

### Final Products
- [X] is \( C_6H_5CCl_3 \) (trichlorobenzene).
- [Y] is \( C_6H_5CHO \) (benzaldehyde).

### Summary of the Solution
- [X] = Trichlorobenzene (\( C_6H_5CCl_3 \))
- [Y] = Benzaldehyde (\( C_6H_5CHO \))

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