Question

Identify the correct IUPAC name of hydrocarbon (x) containing three primary carbon atoms and with molar mass 72 g mol$^{-1}$.

• A. 1, 1 - Dimethylcyclopropane
• B. 2, 2 - Dimethylpropane
• C. 2 - Methylbutane
• D. n-pentane

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
First, determine the molecular formula from the molar mass. Then, analyze the structure of the isomers to count the number of primary ($1^\circ$) carbon atoms.
Step 2: Detailed Explanation:
Molar mass of alkane $C_nH_{2n+2}$ is $14n + 2 = 72 \implies 14n = 70 \implies n = 5$. The formula is $C_5H_{12}$.
Isomers of $C_5H_{12}$:
1. n-pentane: $CH_3-CH_2-CH_2-CH_2-CH_3$. Primary carbons = 2 (the ends).
2. 2-methylbutane (Isopentane): $CH_3-CH(CH_3)-CH_2-CH_3$. The three $CH_3$ groups at the ends are primary carbons. Total = 3.
3. 2,2-dimethylpropane (Neopentane): $C(CH_3)_4$. All four methyl groups are primary carbons. Total = 4.
4. 1,1-dimethylcyclopropane has formula $C_5H_{10}$, so its molar mass is 70, which is incorrect.
Step 3: Final Answer:
2-Methylbutane is the correct hydrocarbon with 3 primary carbons and molar mass 72.