Question

Which of the following is optically active?

• A. Butane
• B. 4-methyl heptane
• C. 3-methyl heptane
• D. 2-methyl heptane

Answer 1

Answer: (C)

Optically active compound must have at least one chiral carbon atom. (Carbon attached to 4 different groups/atoms)
$\underset{\begin{array}{c}\\ Bu\tan e\end{array}}{H-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-H}$ $\underset{\begin{array}{c}\\ 4-methylheptane\end{array}}{H-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-H}$ $\underset{\begin{array}{c}\\ 3-methylheptane\end{array}}{H-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ C{H}_3\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C^{\ast }}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-H}$ $\underset{\begin{array}{c}\\ 2-methylheptane\end{array}}{H-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}H\\ |\end{array}}{C}}-\underset{\begin{array}{c}|\\ H\end{array}}{\stackrel{\begin{array}{c}C{H}_3\\ |\end{array}}{C}}-C{H}_3}$ $\because$ choice has one asymmetric carbon atom $\therefore$ it it optically active