Determine Primary Carbon Atoms in Alkane Y

To determine the total number of primary carbon atoms in alkane $\text{Y}$, we can break down the problem into the following logical steps:

Step 1: Determine the molecular formula of alkane $\text{Y}$

The general chemical equation for the complete combustion of one mole of an alkane with formula $\text{C}_n\text{H}_{2n+2}$ is given by:

$\text{C}_n\text{H}_{2n+2} + \left( \frac{3n + 1}{2} \right) \text{O}_2 \longrightarrow n\text{CO}_2 + (n + 1)\text{H}_2\text{O}$

Given that $1\text{ mole}$ of alkane $\text{Y}$ requires $8\text{ moles}$ of oxygen ($\text{O}_2$) for complete combustion:

$\frac{3n + 1}{2} = 8$

Solving for $n$:

$3n + 1 = 16$ $3n = 15 \implies n = 5$

Therefore, the molecular formula of alkane $\text{Y}$ is $\text{C}_5\text{H}_{12}$ (pentane).

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Step 2: Identify the correct isomer of $\text{C}_5\text{H}_{12}$

The alkane $\text{C}_5\text{H}_{12}$ has three structural isomers:

1. $n$-pentane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$) – yields $3$ monochlorinated isomers.
2. Isopentane ($\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3$) – yields $4$ monochlorinated isomers.
3. Neopentane ($\text{C}(\text{CH}_3)_4$) – yields only $1$ monochlorinated product because all $12$ hydrogen atoms are chemically equivalent.

Since chlorination of $\text{Y}$ yields only one monochlorinated product $(\text{Z})$, alkane $\text{Y}$ must be neopentane (2,2-dimethylpropane).

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Step 3: Count the primary carbon atoms in neopentane

The structural formula of neopentane is:

$$\begin{array}{c} \text{CH}_3 \\ | \\ \text{H}_3\text{C}-\text{C}-\text{CH}_3 \\ | \\ \text{CH}_3 \end{array}$$

• The central carbon atom is attached to four other carbon atoms, so it is a quaternary ($4^\circ$) carbon.
• The four methyl ($-\text{CH}_3$) carbon atoms are each attached to only one other carbon atom, making them primary ($1^\circ$) carbons.

Thus, the total number of primary carbon atoms in alkane $\text{Y}$ is 4.