A chloro compound (A) showed the following properties. It decolourised bromide water. It reacts with `H_(2)` in presence of catalyst. It gave a brick red precipitate with an ammonical `Cu_(2)Cl_(2)`. When vaporized, 1.49g of (A) gave 448 ml of vapours at STP. On reaction with excess of hydrogen and nickel gave 'B'. Calculate the molecular weight of 'B'.

Answer

To solve the problem step by step, we will analyze the given information about the chloro compound (A) and derive the molecular weight of compound (B).

### Step 1: Analyze the properties of compound (A)
- **Decolorizes bromine water**: This indicates that compound (A) has unsaturation, likely a carbon-carbon double or triple bond.
- **Reacts with H₂ in presence of a catalyst**: This further confirms the presence of unsaturation, as hydrogenation occurs across double or triple bonds.
- **Gives a brick-red precipitate with ammonical Cu₂Cl₂**: This property is characteristic of terminal alkynes (compounds with a triple bond at the end of the carbon chain).

### Step 2: Determine the molar mass of compound (A)
- **Mass of (A)**: 1.49 g
- **Volume of vapors at STP**: 448 mL = 448 × 10⁻³ L
- At STP, 1 mole of gas occupies 22.4 L.

Using the ideal gas law, we can find the number of moles of (A):
\[ \text{Number of moles} = \frac{\text{Volume at STP}}{\text{Volume of 1 mole at STP}} = \frac{448 \times 10^{-3}}{22.4} = 0.02 \text{ moles} \]

Now, we can calculate the molar mass (M) of compound (A):
\[ M = \frac{\text{Mass}}{\text{Number of moles}} = \frac{1.49 \text{ g}}{0.02 \text{ moles}} = 74.5 \text{ g/mol} \]

### Step 3: Determine the structure of compound (A)
From the previous analysis:
- The molecular weight of (A) is 74.5 g/mol.
- The presence of chlorine (Cl) and the terminal alkyne structure suggests the formula could be C₄H₇Cl.

Let’s set up the equation based on atomic masses:
- Chlorine (Cl) = 35.5 g/mol
- Carbon (C) = 12 g/mol
- Hydrogen (H) = 1 g/mol

Assuming the structure is Cl-CH₂-C≡C-H (terminal alkyne):
\[ \text{Molecular weight} = 35.5 + (4 \times 12) + (7 \times 1) = 35.5 + 48 + 7 = 90.5 \text{ g/mol} \]

However, this does not match 74.5 g/mol, so we adjust our assumption.

### Step 4: Calculate the molecular weight of compound (B)
When compound (A) reacts with excess hydrogen in the presence of nickel, it will fully saturate:
\[ \text{A} \rightarrow \text{B} \quad \text{(Cl-CH₂-CH₂-CH₃)} \]
The structure of compound (B) is C₃H₇Cl (C₄H₉Cl after saturation).

Now, we calculate the molecular weight of compound (B):
\[ \text{Molecular weight of B} = 35.5 + (4 \times 12) + (9 \times 1) = 35.5 + 48 + 9 = 92.5 \text{ g/mol} \]

### Conclusion
The molecular weight of compound (B) is **78.5 g/mol**.