(a)
Interpretation: For a given compound set of compounds, the given
Concept Introduction: Gabriel synthesis plays a very important role for preparing amines. 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
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.
Sodium cyanoborohydride is a strong reducing agent than sodium borohydride. It reduces the carbonyl group into amine group in a rapid way. So, it is called as reductive amination reactions.
Aldehyde or ketone group is reacted with primary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce secondary amines.
Aldehyde or ketone group is reacted with secondary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce tertiary amines.
(b)
Interpretation: For a given compound set of compounds, the given amines are to be prepared via Gabriel synthesis, reductive amination starting from potassium phthalimide
Concept Introduction: Gabriel synthesis plays a very important role for preparing amines. 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.
Sodium cyanoborohydride is a strong reducing agent than sodium borohydride. It reduces the carbonyl group into amine group in a rapid way. So, it is called as reductive amination reactions. Aldehyde or ketone group is reacted with ammonia in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce primary amines.
Aldehyde or ketone group is reacted with primary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce secondary amines.
Aldehyde or ketone group is reacted with secondary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce tertiary amines.
(c)
Interpretation: For a given compound set of compounds, the given amines are to be prepared via Gabriel synthesis, reductive amination starting from potassium phthalimide
Concept Introduction: Gabriel synthesis plays a very important role for preparing amines. 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.
Sodium cyanoborohydride is a strong reducing agent than sodium borohydride. It reduces the carbonyl group into amine group in a rapid way. So, it is called as reductive amination reactions. Aldehyde or ketone group is reacted with ammonia in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce primary amines.
Aldehyde or ketone group is reacted with primary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce secondary amines.
Aldehyde or ketone group is reacted with secondary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce tertiary amines.
(d)
Interpretation: For a given compound set of compounds, the given amines are to be prepared via Gabriel synthesis, reductive amination starting from potassium phthalimide
Concept Introduction: Gabriel synthesis plays a very important role for preparing amines. 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.
Sodium cyanoborohydride is a strong reducing agent than sodium borohydride. It reduces the carbonyl group into amine group in a rapid way. So, it is called as reductive amination reactions. Aldehyde or ketone group is reacted with ammonia in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce primary amines.
Aldehyde or ketone group is reacted with primary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce secondary amines.
Aldehyde or ketone group is reacted with secondary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce tertiary amines.
(e)
Interpretation: For a given compound set of compounds, the given amines are to be prepared via Gabriel synthesis, reductive amination starting from potassium phthalimide
Concept Introduction: Gabriel synthesis plays a very important role for preparing amines. 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.
Sodium cyanoborohydride is a strong reducing agent than sodium borohydride. It reduces the carbonyl group into amine group in a rapid way. So, it is called as reductive amination reactions. Aldehyde or ketone group is reacted with ammonia in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce primary amines.
Aldehyde or ketone group is reacted with primary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce secondary amines.
Aldehyde or ketone group is reacted with secondary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce tertiary amines.
(f)
Interpretation: For a given compound set of compounds, the given amines are to be prepared via Gabriel synthesis, reductive amination starting from potassium phthalimide
Concept Introduction: Gabriel synthesis plays a very important role for preparing amines. 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.
Sodium cyanoborohydride is a strong reducing agent than sodium borohydride. It reduces the carbonyl group into amine group in a rapid way. So, it is called as reductive amination reactions. Aldehyde or ketone group is reacted with ammonia in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce primary amines.
Aldehyde or ketone group is reacted with primary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce secondary amines.
Aldehyde or ketone group is reacted with secondary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce tertiary amines.
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Chapter 22 Solutions
ORGANIC CHEMISTRY 1 TERM ACCESS
- Don't used hand raiting and don't used Ai solutionarrow_forward13.84. Chlorine atoms react with methane, forming HCI and CH3. The rate constant for the reaction is 6.0 × 107 M¹ s¹ at 298 K. When the experiment was run at three other temperatures, the following data were collected: T (K) k (M-1 s-1) 303 6.5 × 107 308 7.0 × 107 313 7.5 x 107 a. Calculate the values of the activation energy and the frequency factor for the reaction. b. What is the value of the rate constant in the lower stratosphere, where T = 218 K?arrow_forwardMy Organic Chemistry textbook says about the formation of cyclic hemiacetals, "Such intramolecular reactions to form five- and six-membered rings are faster than the corresponding intermolecular reactions. The two reacting functional groups, in this case OH and C=O, are held in close proximity, increasing the probability of reaction."According to the book, the formation of cyclic hemiacetals occurs in acidic conditions. So my question is whether the carbonyl group in this reaction reacts first with the end alcohol on the same molecule or with the ethylene glycol. And, given the explanation in the book, if it reacts first with ethylene glycol before its own end alcohol, why would it? I don't need to know the final answer. I need to know WHY it would not undergo an intermolecular reaction prior to reacting with the ethylene glycol if that is the case. Please do not use an AI answer.arrow_forward
- Don't used hand raiting and don't used Ai solutionarrow_forwardHighlight in red each acidic location on the organic molecule at left. Highlight in blue each basic location on the organic molecule at right. Note for advanced students: we mean acidic or basic in the Brønsted-Lowry sense only. Cl N شیخ x Garrow_forwardQ4: Draw the mirror image of the following molecules. Are the molecules chiral? C/ F LL CI CH3 CI CH3 0 CI CH3 CI CH3 CH3arrow_forward
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