Q. A hydrocarbon of molecular formula $C_{6}H_{10}$ reacts with sodamide and the same on ozonolysis followed by hydrogen peroxide oxidation gives two molecules of carboxylic acids, one being optically active. Then, the hydrocarbon may be
A) 1-hexyne
B) 3-hexyne
C) 3-methyl-1-pentyne
D) 3,3-dimethyl-1-butyne
Explanation
(c)
$CH_{3}$
|
$CH_{3} - CH_{2} - CH - C \equiv CH\underset{+ O_{3}}{\overset{+ NaNH_{2}\ \ }{\rightarrow}}$
3-methyl-1-pentyne
[image:13_image258.emf]
$\ \ \ \ \$ $CH_{3}$
|
$CH_{3} - CH - CH - COOH + HCOOH$
Optically active carboxylic acid.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.