Identify IUPAC Name of Hydrocarbon with Molar Mass 72

To determine the correct IUPAC name of the hydrocarbon $(x)$, we follow a step-by-step analysis based on its molar mass and structure:

Step 1: Determine the Molecular Formula

A hydrocarbon with a molar mass of $72\text{ g mol}^{-1}$ corresponds to an alkane with the general formula $\text{C}_n\text{H}_{2n+2}$.

The molar mass $M$ of $\text{C}_n\text{H}_{2n+2}$ is calculated as: $M = 12n + 1(2n + 2) = 14n + 2$

Given $M = 72\text{ g mol}^{-1}$: $14n + 2 = 72$ $14n = 70 \implies n = 5$

Thus, the molecular formula of the hydrocarbon is $\text{C}_5\text{H}_{12}$.

---

Step 2: Analyze the Isomers of $\text{C}_5\text{H}_{12}$

A primary ($1^\circ$) carbon atom is defined as a carbon atom attached to at most one other carbon atom (i.e., $-\text{CH}_3$ groups in alkanes).

Let me examine the three structural isomers of $\text{C}_5\text{H}_{12}$:

1. n-pentane: $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3$

   • Primary ($1^\circ$) carbon atoms = $2$
   • Secondary ($2^\circ$) carbon atoms = $3$

2. 2,2-Dimethylpropane (Neopentane): $\text{H}_3\text{C}-\text{C}(\text{CH}_3)_2-\text{CH}_3$

   • Primary ($1^\circ$) carbon atoms = $4$
   • Quaternary ($4^\circ$) carbon atom = $1$

3. 2-Methylbutane (Isopentane): $\overset{1}{\text{C}}\text{H}_3-\overset{2}{\text{C}}\text{H}(\text{CH}_3)-\overset{3}{\text{C}}\text{H}_2-\overset{4}{\text{C}}\text{H}_3$

   • Primary ($1^\circ$) carbon atoms = $3$ (the three terminal $-\text{CH}_3$ groups)
   • Secondary ($2^\circ$) carbon atom = $1$ ($-\text{CH}_2-$)
   • Tertiary ($3^\circ$) carbon atom = $1$ ($-\text{CH}-$)

---

Conclusion

The hydrocarbon containing exactly three primary carbon atoms and a molar mass of $72\text{ g mol}^{-1}$ is 2-Methylbutane.

Correct Option: C