(a)
Interpretation: Using a Gabriel synthesis, a give set of primary amine compounds have to be synthesized along with their corresponding starting materials.
Concept Introduction: The general formula for primary amine is –NH2. There are several methods available to prepare primary
Step-1: 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.
Step-2: Formation of R−N bond by SN2 nucleophilic substitution
The negative charged nitrogen atom in phthalimide can easily attract the positive side of R−X. In primary
Step-3: 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.
The starting material of Gabriel synthesis is alkyl halide. It can be prepared in a number of ways such as bromination-elimination-addition reactions, reduction-displacement of
(b)
Interpretation: Using a Gabriel synthesis, a give set of primary amine compounds have to be synthesized along with their corresponding starting materials.
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.
Step-1: 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.
Step-2: Formation of R−N bond by SN2 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 SN2 nucleophilic substitution reaction. Halogen atom is going away as halide anion.
Step-3: 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.
The starting material of Gabriel synthesis is alkyl halide. It can be prepared in a number of ways such as bromination-elimination-addition reactions, reduction-displacement of function group reactions, direct bromination reactions, etc. Some of the reagents such as HBr, bromine in photolytic condition, phosphorous tribromide are used to achieve the halide group preparation.

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Chapter 23 Solutions
ORG.CHEM EBOOK W/BBWILEY PLUS>CUSTOM<
- The statements in the tables below are about two different chemical equilibria. The symbols have their usual meaning, for example AG stands for the standard Gibbs free energy of reaction and K stands for the equilibrium constant. In each table, there may be one statement that is faise because it contradicts the other three statements. If you find a false statement, check the box next to t Otherwise, check the "no false statements" box under the table. statement false? AG"1 no false statements: statement false? AG-0 0 InK-0 0 K-1 0 AH-TAS no false statements 2arrow_forwardComplete the following esterification reactions by drawing the line formulas of the carboxylic acid and alcohol required to form the ester shown. catalyst catalyst catalyst apricot fragrancearrow_forwardShow the saponification products of the following ester: You don't need to draw in the Na+ cation. catalyst, A catalyst, A catalyst, Aarrow_forward
- What would happen if the carboxylic acid and alcohol groups were on the same molecule? In essence, the molecule reacts with itself. Draw the structure of the products formed in this manner using the reactants below. If two functional groups interact with one another on the same molecule, this is called an “intramolecular" (within one) rather than "intermolecular" (between two or more) attack. OH OH catalyst OH HO catalyst catalyst HO OHarrow_forwardQ3: Write in the starting alkyl bromide used to form the following products. Include any reactants, reagents, and solvents over the reaction arrow. If more than one step is required, denote separate steps by using 1), 2), 3), etc. H OH racemic OH OH 5 racemicarrow_forwardDraw the Lewis structure of the SO3-O(CH3)2 complex shown in the bottom right of slide 2in lecture 3-3 (“Me” means a CH3 group) – include all valence electron pairs and formal charges.From this structure, should the complex be a stable molecule? Explain.arrow_forward
- please add appropriate arrows, and tell me clearly where to add arrows, or draw itarrow_forwardWhat I Have Learned Directions: Given the following reaction and the stress applied in each reaction, answer the question below. A. H2(g) + Cl2(g) 2 HCl(g) Stress applied: Decreasing the pressure 1. What is the Keq expression? 2. What will be the effect in the number of moles of HCl(g)? 3. What will be the Equilibrium Shift or the reaction? B. Fe3O4(s) + 4 H2(g) + heat 53 Fe(s) + 4 H₂O(g) Stress applied: Increasing the temperature 1. What is the Keq expression?. 2. What will be the effect in the volume of water vapor collected? 3. What will be the Equilibrium Shift or the reaction? C. 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) + heat Stress applied: Increasing the volume of the container 1. What is the Keq expression?. 2. What will be the effect in the amount of H₂O? 3. What will be the Equilibrium Shift or the reaction?arrow_forwardConsider the solubility products (Ksp values) for the following compounds:SrSO4 (Ksp = 7.6 x 10−7), BaSO4 (Ksp = 1.5 x 10−9), SrCO3 (Ksp = 7.0 x 10−10), BaCO3 (Ksp = 1.6 x 10−9)Which anion is the harder base, CO32− or SO42−? Justify your answer.arrow_forward
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