`R -Mg-Br(A)` on reaction with `H_(2)O` forms a gas `(B)`, which occupied `1.4 L//g` at `NTP`. What is product when `R-Br` reacts with benzene in presence of `AlCl_(3)`?

Answer

To solve the problem step by step, let's break down the information given and analyze it:

### Step 1: Identify the Reaction of R-Mg-Br with H2O
The compound R-Mg-Br (A) is a Grignard reagent. When it reacts with water (H2O), it undergoes hydrolysis. The general reaction can be represented as:

\[ \text{R-Mg-Br} + \text{H}_2\text{O} \rightarrow \text{R-H} + \text{Mg(OH)Br} \]

In this reaction, R-H is formed, which is typically an alkane or an alcohol depending on the nature of R.

### Step 2: Determine the Gas B
The problem states that the reaction produces a gas (B) that occupies 1.4 L/g at NTP (Normal Temperature and Pressure).

At NTP, 1 mole of gas occupies 22.4 L. To find the molar mass of gas B, we can use the following relationship:

\[ \text{Molar Volume} = \frac{\text{Volume}}{\text{Mass}} \Rightarrow \text{Molar Mass} = \frac{22.4 \, \text{L}}{1.4 \, \text{L/g}} \approx 16 \, \text{g/mol} \]

This suggests that gas B is likely methane (CH4), which has a molar mass of approximately 16 g/mol.

### Step 3: Identify the Reaction of R-Br with Benzene
Now, we need to consider the reaction of R-Br with benzene in the presence of AlCl3. The presence of AlCl3 indicates that this is a Friedel-Crafts alkylation reaction.

The general reaction can be represented as:

\[ \text{R-Br} + \text{C}_6\text{H}_6 \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{R} + \text{HBr} \]

In this reaction, the alkyl group R from R-Br is introduced to the benzene ring, forming an alkylbenzene (C6H5R) and releasing HBr as a byproduct.

### Final Answer
The product when R-Br reacts with benzene in the presence of AlCl3 is an alkylbenzene (C6H5R).

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