Amines are basic in nature and show many types of reactions such as substitution, Hoffmann elimination, carbylamine, acylation, diazotisation etc.
Certain nitrogeneous compound with molecular mass 180 shows an increase in its molecular mass to 348 after treatment with acetyl chloride. The number of possible -`NH_(2)` groups in the molecule is...................

Options

A. 5
B. 4
C. 3
D. 6

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
We know that when an amine (RNH₂) reacts with acetyl chloride (CH₃COCl), one hydrogen atom from the amine is replaced by an acetyl group (CH₃CO). This reaction will increase the molecular mass of the original amine.

### Step 2: Calculate the Increase in Molecular Mass
The initial molecular mass of the nitrogenous compound is given as 180 g/mol. After treatment with acetyl chloride, the molecular mass increases to 348 g/mol.

To find the increase in mass:
\[ \text{Increase in mass} = \text{Final mass} - \text{Initial mass} = 348 - 180 = 168 \text{ g/mol} \]

### Step 3: Determine the Mass Contribution of Acetyl Group
When one hydrogen atom is replaced by an acetyl group, we need to calculate the mass of the acetyl group (CH₃CO):
- Carbon (C) = 12 g/mol (2 Carbons)
- Hydrogen (H) = 1 g/mol (3 Hydrogens)
- Oxygen (O) = 16 g/mol (1 Oxygen)

The mass of the acetyl group (CH₃CO) is:
\[ \text{Mass of CH₃CO} = 12 + 3 + 16 = 31 \text{ g/mol} \]
However, we need to subtract the mass of the hydrogen that is replaced:
\[ \text{Mass of replacement for one hydrogen} = 31 - 1 = 30 \text{ g/mol} \]

### Step 4: Calculate the Number of Amino Groups
Now, we need to find out how many hydrogen atoms (or -NH₂ groups) were replaced to account for the total increase in mass of 168 g/mol.

Let \( x \) be the number of -NH₂ groups in the compound. The total mass increase due to the replacement of \( x \) hydrogen atoms is:
\[ \text{Total increase} = x \times 30 \]
Setting this equal to the total increase in mass:
\[ x \times 30 = 168 \]
Solving for \( x \):
\[ x = \frac{168}{30} = 5.6 \]
Since \( x \) must be a whole number, we round down to the nearest whole number, which is 5.

### Conclusion
The number of possible -NH₂ groups in the molecule is **5**.