Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Q20P (page 1000)

Why is excess ammonia used in the preceding reaction?

Short Answer

Expert verified

Reductive amination involves the formation of amine from carbonyl compounds particularly aldehyde or ketone via from the imine intermediate.

Step by step solution

01

Reductive amination

Reductive amination involves the formation of amine from carbonyl compounds particularly aldehyde or ketone via from the imine intermediate.

02

Explaination

In this reaction excess ammonia is used because one equivalent of ammonia first reacts with acidic proton and forms NH+4ion and loses its nucleophilicity. Then the remaining ammonia substitutes the Br-. So, excess of ammonia is used in this reaction.

The reaction is represented as follows:

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the configuration about each of the asymmetric centres in aspartame?

After the polypeptide shown below was treated with maleic anhydride, it was hydrolyzed by trypsin. (After a polypeptide is treated with maleic anhydride,

trypsin will cleave the polypeptide only on the C-side of arginine.)

Gly-Ala-Asp-Ala-Leu-Pro-Gly-Ile-Leu-Val-Arg-Asp-Val-Gly-Lys-Val-Glu-Val-Phe-Glu-Ala-Gly-

Arg-Ala-Glu-Phe-Lys-Glu-Pro-Arg-Leu-Val-Met-Lys-Val-Glu-Gly-Arg-Pro-Val-Gly-Ala-Gly-Leu-Trp

a. After a polypeptide is treated with maleic anhydride, why does trypsin no longer cleave it on the C-side of lysine?

b. How many fragments are obtained from the polypeptide?

c. In what order will the fragments be eluted from an anion-exchange column using a buffer of pH = 5?

a. What percentage of the α-amino group of lysine will be protonated at its pI?<25%50% >75%

b. Answer the same question for theε-amino group of lysine.

Treatment of a polypeptide with 2-mercaptoethanol yields two polypeptides with the following primary structures:

Val-Met-Tyr-Ala-Cys-Ser-Phe-Ala-Glu-Ser

Ser-Cys-Phe-Lys-Cys-Trp-Lys-Tyr-Cys-Phe-Arg-Cys-Ser

Treatment of the original intact polypeptide with chymotrypsin yields the following peptides:

1. Ala, Glu, Ser 3. Tyr, Val, Met 5. Ser, Phe, 2 Cys, Lys, Ala, Trp

2. 2 Phe, 2 Cys, Ser 4. Arg, Ser, Cys 6. Tyr, Lys

Determine the positions of the disulfide bridges in the original polypeptide.

Show the steps in the synthesis of the tetrapeptide Leu- Phe- Ala- Val
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free