Using a Gabriel synthesis, a given primary amine compound has to be synthesized. Concept Introduction : The general formula for primary amine is –NH 2 . There are several methods available to prepare primary amines . Among them, Gabriel synthesis plays a very important role for preparing it. In this method, secondary and tertiary amines are not formed as side products. It involves in three steps. Formation of potassium phthalimide (deprotonation) Potassium phthalimide in alkaline KOH acts as the reagent which has negatively charged phthalimide. It is formed by the reaction between phthalimide and potassium hydroxide. Formation of R−N bond by S N 2 nucleophilic substitution The negative charged nitrogen atom in phthalimide can easily attract the positive side of R−X. In primary alkyl halides (R−X), R and X get positive and negative charges, respectively when they ionize. As a result, a bond between nitrogen of phthalimide and carbon of R is formed. This is S N 2 nucleophilic substitution reaction. Halogen atom is going away as halide anion. Formation of primary amine by hydrolysis The resultant product further goes for hydrolysis using hydrazine as the reagent. This reaction also follows nucleophilic substitution reaction. Finally, primary amine is formed with a side product of hydrazine derivative. To find: Using a Gabriel synthesis, prepare the given primary amine compound Identify the correct alkyl halide involved in the formation of the given compound

Question

Want to see the full answer?

Answer 1

Question

Chapter 23.5, Problem 2LTS

Interpretation Introduction

Interpretation: Using a Gabriel synthesis, a given primary amine compound has to be synthesized.

Concept Introduction:

The general formula for primary amine is –NH₂. There are several methods available to prepare primary amines. Among them, Gabriel synthesis plays a very important role for preparing it. In this method, secondary and tertiary amines are not formed as side products. It involves in three steps.

Formation of potassium phthalimide (deprotonation)

Potassium phthalimide in alkaline KOH acts as the reagent which has negatively charged phthalimide. It is formed by the reaction between phthalimide and potassium hydroxide.

Formation of R−N bond by S_N2 nucleophilic substitution

The negative charged nitrogen atom in phthalimide can easily attract the positive side of R−X. In primary alkyl halides (R−X), R and X get positive and negative charges, respectively when they ionize. As a result, a bond between nitrogen of phthalimide and carbon of R is formed. This is S_N2 nucleophilic substitution reaction. Halogen atom is going away as halide anion.

Formation of primary amine by hydrolysis

The resultant product further goes for hydrolysis using hydrazine as the reagent. This reaction also follows nucleophilic substitution reaction. Finally, primary amine is formed with a side product of hydrazine derivative.

To find: Using a Gabriel synthesis, prepare the given primary amine compound

Identify the correct alkyl halide involved in the formation of the given compound

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

1. For each of the reaction "railroads" below, you are either asked to give the structure(s) of the starting material(s) or product(s), or provide reagents/conditions to accomplish the transformation, as indicated by the boxes. a. NaOMe H+ .CO,H HO₂C MeOH (excess) MeOH H3C Br يع CH3 1. LiAlH4 2. H3O+ 3. PBг3 H3C 1. Et-Li 2. H3O+ -CO₂Me -CO₂Me OH CH3 CH3 ল CH3

Predict the intermediate 1 and final product 2 of this organic reaction: NaOMe ག1, ད།་, - + H You can draw 1 and 2 in any arrangement you like. 2 work up Note: if either 1 or 2 consists of a pair of enantiomers, just draw one structure using line bonds instead of 3D (dash and wedge) bonds at the chiral center. Explanation Check Click and drag to start drawing a structure. Х © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | P

What is the total energy cost associated with the compound below adopting the shown conformation? CH3 HH DH CH3