Question

1 mol of alkane (X) for complete combustion required 8 mol of oxygen. When it is brominated in presence of sunlight it gives only one product. How many primary carbons are present in alkane (X)?

Hint:

For alkanes like octane, all carbons are primary in a straight-chain structure. The number of primary carbons can be determined by counting the carbons in such structures.

Answer 1

Correct Answer: 8

Step 1: General combustion reaction of alkane.
For the combustion of an alkane, the general reaction is: \[ \text{C}_n\text{H}_{2n+2} + \left( n + \frac{1}{2} \right) \text{O}_2 \rightarrow n \text{CO}_2 + (n+1) \text{H}_2\text{O} \]
Step 2: Relating oxygen requirement with alkane.
Given that 1 mol of alkane (X) requires 8 mol of oxygen, we can compare the stoichiometry of oxygen. For an alkane with the formula \( \text{C}_n\text{H}_{2n+2} \), the oxygen required for complete combustion is: \[ \left( n + \frac{1}{2} \right) \text{O}_2 \] Given that oxygen required is 8 moles, we equate: \[ n + \frac{1}{2} = 8 \] Solving for \( n \), we get: \[ n = 7.5 \]
Step 3: Conclusion about the number of carbons.
Since \( n \) must be an integer, the value of \( n = 8 \) satisfies the equation. Thus, alkane (X) has 8 carbon atoms.
Step 4: Bromination in presence of sunlight.
The fact that only one product is formed during bromination under sunlight indicates that all hydrogen atoms are attached to primary carbons. This suggests that alkane (X) is octane (\( \text{C}_8\text{H}_{18} \)), where all carbons are primary.
Step 5: Counting the primary carbons.
In octane, all 8 carbons are primary because each carbon is attached to only one other carbon. Therefore, the number of primary carbons is 8.