Question

An isomer of C$_8$H$_{18}$ is X. This has five primary, one tertiary and one quaternary carbon. What is X?

• A. 3-Ethylpentane
• B. 3,3-Dimethylpentane
• C. 2,2,3-Trimethylbutane
• D. 2,4-Dimethylpentane

Hint:

Count carbon types: primary (CH$_3$), secondary (CH$_2$), tertiary (CH), quaternary (C). Verify molecular formula matches C$_n$H$_{2n+2}$. Draw structures to confirm carbon connections.

Answer 1

The Correct Option is C

1. For C$_8$H$_{18}$, a branched alkane, we need 5 primary (CH$_3$), 1 tertiary (CH), 1 quaternary (C) carbons. Total carbons = 8.
2. 3-Ethylpentane (C$_7$H$_{16}$): Not C$_8$H$_{18}$, incorrect.
3. 3,3-Dimethylpentane: 5 C total (2 CH$_3$, 2 CH$_2$, 1 C), incorrect.
4. 2,2,3-Trimethylbutane: Structure (CH$_3$)$_3$C-CH(CH$_3$)-CH$_3$; 5 CH$_3$, 1 CH, 1 C, total 7 carbons (C$_7$H$_{16}$), but fits pattern, likely intended.
5. 2,4-Dimethylpentane: 4 CH$_3$, 2 CH$_2$, 1 CH, no quaternary, incorrect.
6. Thus, (3) 2,2,3-Trimethylbutane is correct assuming C$_7$H$_{16}$ typo.