An alkane of molecular weight 86 g/mol on monochlorination gives two products. The alkane is

Options

A. 2 - Methylbutane
B. n - butane
C. 2, 2- Dimethyl propane
D. 2, 3- Dimethyl butane

Answer

To solve the problem of identifying the alkane with a molecular weight of 86 g/mol that gives two products upon monochlorination, we can follow these steps:

### Step 1: Determine the General Formula of Alkanes
The general formula for alkanes is given by \( C_nH_{2n+2} \).

### Step 2: Set Up the Equation for Molecular Weight
The molecular weight of the alkane can be expressed as:
\[ 12n + (2n + 2) = 86 \]
Where:
- \( 12n \) is the contribution from carbon (C),
- \( 2n + 2 \) is the contribution from hydrogen (H).

### Step 3: Simplify the Equation
Combine the terms:
\[ 12n + 2n + 2 = 86 \]
This simplifies to:
\[ 14n + 2 = 86 \]

### Step 4: Solve for \( n \)
Subtract 2 from both sides:
\[ 14n = 84 \]
Now, divide by 14:
\[ n = 6 \]

### Step 5: Write the Molecular Formula
Using \( n = 6 \), we can write the molecular formula of the alkane:
\[ C_6H_{2(6) + 2} = C_6H_{14} \]

### Step 6: Identify the Alkane
The alkane \( C_6H_{14} \) is hexane.

### Step 7: Analyze Monochlorination Products
Hexane can undergo monochlorination to produce two different products. This is because there are two distinct types of hydrogen atoms in hexane:
1. Primary hydrogens (attached to the terminal carbons).
2. Secondary hydrogens (attached to the internal carbons).

### Step 8: Identify the Products
When monochlorination occurs, the two products formed are:
1. 1-chlorohexane (from a primary hydrogen).
2. 2-chlorohexane (from a secondary hydrogen).

### Conclusion
Thus, the alkane of molecular weight 86 g/mol that gives two products upon monochlorination is **hexane (C6H14)**.

---