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

Question

TestHub · Accepted Answer

To identify the hydrocarbon with a molar mass of $72 g m o l^{- 1}$ and three primary carbon atoms, follow these steps:

Determine the Formula: For an alkane ( $C_{n} H_{2 n + 2}$ ), the molar mass is $12 n + 1 (2 n + 2) = 72$ . Solving $14 n = 70$ gives $n = 5$ , so the formula is $C_{5} H_{12}$ .

Analyze Isomers: n-pentane has 2 primary carbons, while 2,2-dimethylpropane has 4 primary carbons.

Identify the Structure: 2-methylbutane ( $C H_{3} - C H (C H_{3}) - C H_{2} - C H_{3}$ ) contains exactly three primary carbon atoms (the three methyl groups).

Conclusion: Therefore, the correct IUPAC name is 2-Methylbutane (Option C).

Chemistry - Nomenclature Question with Solution | TestHub

Options:

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