Question

Which of the following compound has all the four types ($1^o,2^o,3^o,4^o$) of carbon atoms?

• A. 2,3,4-Trimethylpentane
• B. Neo-Pentane
• C. 2,2,4-Trimethylpentane
• D. None of the above

Answer 1

The correct option is C 2,2,4-Trimethylpentane

2,2,4-Trimethylpentane has all the four types ($1^o,2^o,3^o,4^o$) of carbon atoms.

$H_3\underset{C{H}_3}{\underset{|}{\stackrel{1}{C}-\stackrel{C{H}_3}{\stackrel{2|}{C}}}}-\stackrel{3}{C}{H}_2-\underset{C{H}_3}{\underset{|}{\stackrel{4}{C}}}H-\stackrel{5}{C}{H}_3$

$C_1$ and $C_5$ are $1^o$, $C_3$ is $2^o$, $C_4$ is $3^o$ and $C_2$ is $4^o$
2,3,4-Trimethylpentane has $1^o,3^o$ C atoms.

Neo-Pentane has $1^o,4^o$ C atoms.

Option C is correct.