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(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-STD.WILEY PLUS CARD
- Could you please solve the first problem in this way and present it similarly but color-coded or step by step so I can understand it better? Thank you!arrow_forwardCould you please solve the first problem in this way and present it similarly but (color-coded) and step by step so I can understand it better? Thank you! I want to see what they are doingarrow_forwardCan you please help mne with this problem. Im a visual person, so can you redraw it, potentislly color code and then as well explain it. I know im given CO2 use that to explain to me, as well as maybe give me a second example just to clarify even more with drawings (visuals) and explanations.arrow_forward
- Part 1. Aqueous 0.010M AgNO 3 is slowly added to a 50-ml solution containing both carbonate [co32-] = 0.105 M and sulfate [soy] = 0.164 M anions. Given the ksp of Ag2CO3 and Ag₂ soy below. Answer the ff: Ag₂ CO3 = 2 Ag+ caq) + co} (aq) ksp = 8.10 × 10-12 Ag₂SO4 = 2Ag+(aq) + soy² (aq) ksp = 1.20 × 10-5 a) which salt will precipitate first? (b) What % of the first anion precipitated will remain in the solution. by the time the second anion starts to precipitate? (c) What is the effect of low pH (more acidic) condition on the separate of the carbonate and sulfate anions via silver precipitation? What is the effect of high pH (more basic)? Provide appropriate explanation per answerarrow_forwardPart 4. Butanoic acid (ka= 1.52× 10-5) has a partition coefficient of 3.0 (favors benzene) when distributed bet. water and benzene. What is the formal concentration of butanoic acid in each phase when 0.10M aqueous butanoic acid is extracted w❘ 25 mL of benzene 100 mL of a) at pit 5.00 b) at pH 9.00arrow_forwardCalculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 Group of answer choices 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/molearrow_forward
- Calculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 choices: 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/molearrow_forwardCalculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0arrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forward
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