Chapter 7: Problem 152
How many optically active products are obtained on heating lactic acid?
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 7: Problem 152
How many optically active products are obtained on heating lactic acid?
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for freeThe product obtained on reaction of malonic acid with \(\mathrm{P}_{2} \mathrm{O}_{5} / \Delta\) is
The product obtained in the reaction is (A) Benzamide (B) Benzonitrile (C) Ammoniumbenzoate (D) Aniline
For the following reaction sequence, the product \(\mathrm{R}\) is
The decreasing order of rate of esterification of the following acids with MeOH is (I) \(\mathrm{Me}-\mathrm{CH}_{2} \mathrm{COOH}\) (II) Me \(_{2} \mathrm{CHCOOH}\) (III) \(\mathrm{Me}_{3} \mathrm{C}-\mathrm{COOH}\) (A) \(\mathrm{I}>\mathrm{II}>\mathrm{III}\) (B) III \(>\mathrm{II}>\mathrm{I}\) (C) II \(>\mathrm{III}>\mathrm{I}\) (D) II > I > III
Acid amide is not obtained as a product in the reaction: (A) \(\mathrm{Me}_{2} \mathrm{CNOH} \stackrel{\text { Conc. } \mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}\) (B) \(\mathrm{MeCOOH} \stackrel{\mathrm{N}_{3} \mathrm{H}}{\longrightarrow}\) (C) \(\mathrm{Me}_{2} \mathrm{CO} \frac{\mathrm{N}_{3} \mathrm{H}}{\mathrm{H}_{2} \mathrm{SO}_{4}}\) (D) \(\mathrm{MeCN} \frac{95 \%}{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.