Question

An alkane (Y) requires 8 moles of oxygen for complete combustion and on chlorination with \(Cl_2/h\nu\), (Y) gives only one monochlorinated product (Z). The total number of primary carbon atoms in (Y) is _____.

Answer 1

Correct Answer: 4

Step 1: Understanding the Question:
1. Use the combustion data to find the molecular formula of the alkane.
2. Use the chlorination data to determine its specific isomer structure.
3. Count the primary (\(1^{\circ}\)) carbons in that structure.
Step 2: Key Formula or Approach:
General combustion equation for an alkane (\(C_{n}H_{2n+2}\)):
\[ C_{n}H_{2n+2} + \left( \frac{3n+1}{2} \right) O_{2} \rightarrow nCO_{2} + (n+1)H_{2}O \]
Step 3: Detailed Explanation:
1. Finding 'n':
Moles of \(O_{2}\) per mole of alkane \(= \frac{3n+1}{2} = 8\).
\[ 3n + 1 = 16 \Rightarrow 3n = 15 \Rightarrow n = 5 \]
The alkane is Pentane (\(C_{5}H_{12}\)).

2. Finding Isomer structure:
There are 3 isomers of pentane: n-pentane, isopentane, and neopentane.
The question states it gives only one monochlorinated product. This means all the hydrogen atoms in the molecule must be chemically equivalent.
- n-pentane: Gives 3 products (1-chloro, 2-chloro, 3-chloro).
- Isopentane: Gives 4 products.
- Neopentane (\(2,2-\text{dimethylpropane}\)): All 12 hydrogens are equivalent (attached to primary carbons). It gives only one product.

3. Counting Primary Carbons:
The structure of neopentane is a central carbon surrounded by four methyl groups:
\[ C(CH_{3})_{4} \]
- The central carbon is quaternary (\(4^{\circ}\)).
- The four carbons in the methyl groups are each attached to only one carbon, so they are all primary carbons.
Step 4: Final Answer:
The total number of primary carbon atoms is 4.