(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 23 Solutions
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- Curved arrows are used to illustrate the flow of electrons. Follow the curved arrows and draw the structure of the missing reactants, intermediates, or products in the following mechanism. Include all lone pairs. Ignore stereochemistry. Ignore inorganic byproducts. H Br2 (1 equiv) H- Select to Draw Starting Alkene Draw Major Product I I H2O 四: ⑦.. Q Draw Major Charged Intermediate Iarrow_forwardNH (aq)+CNO (aq) → CO(NH2)2(s) Experiment [NH4] (M) [CNO] (M) Initial rate (M/s) 1 0.014 0.02 0.002 23 0.028 0.02 0.008 0.014 0.01 0.001 Calculate the rate contant for this reaction using the data provided in the table.arrow_forward2CIO2 + 20H-1 CIO31 + CIO2 + H2O Experiment [CIO2], M [OH-1], M 1 0.0500 0.100 23 2 0.100 0.100 3 0.100 0.0500 Initial Rate, M/s 0.0575 0.230 0.115 ... Given this date, calculate the overall order of this reaction.arrow_forward
- 2 3 .(be)_[Ɔ+(be)_OI ← (b²)_IƆO+ (be)_I Experiment [1-] M 0.005 [OCI-] 0.005 Initial Rate M/min 0.000275 0.0025 0.005 0.000138 0.0025 0.0025 0.000069 4 0.0025 0.0025 0.000140 Calculate the rate constant of this reaction using the table data.arrow_forward1 2 3 4 I(aq) +OCl(aq) → IO¯¯(aq) + Cl¯(aq) Experiment [I-] M 0.005 [OCI-] 0.005 Initial Rate M/min 0.000275 0.0025 0.005 0.000138 0.0025 0.0025 Calculate the overall order of this reaction using the table data. 0.0025 0.000069 0.0025 0.000140arrow_forwardH2O2(aq) +3 I¯(aq) +2 H+(aq) → 13(aq) +2 H₂O(l)· ••• Experiment [H2 O2]o (M) [I]o (M) [H+]。 (M) Initial rate (M/s) 1 0.15 0.15 0.05 0.00012 234 0.15 0.3 0.05 0.00024 0.3 0.15 0.05 0.00024 0.15 0.15 0.1 0.00048 Calculate the overall order of this reaction using the table data.arrow_forward
- The U. S. Environmental Protection Agency (EPA) sets limits on healthful levels of air pollutants. The maximum level that the EPA considers safe for lead air pollution is 1.5 μg/m³ Part A If your lungs were filled with air containing this level of lead, how many lead atoms would be in your lungs? (Assume a total lung volume of 5.40 L.) ΜΕ ΑΣΦ = 2.35 1013 ? atoms ! Check your rounding. Your final answer should be rounded to 2 significant figures in the last step. No credit lost. Try again.arrow_forwardY= - 0.039 (14.01) + 0.7949arrow_forwardSuppose 1.76 g of magnesium acetate (Mg (CH3CO2)2) are dissolved in 140. mL of water. Find the composition of the resulting electrolyte solution. In particular, list the chemical symbols (including any charge) of each dissolved ion in the table below. List only one ion per row. mEq Then, calculate the concentration of each ion in dwrite the concentration in the second column of each row. Be sure you round your answers to the L correct number of significant digits. ion Add Row mEq L x 5arrow_forward
- A pdf file of your hand drawn, stepwise mechanisms for the reactions. For each reaction in the assignment, you must write each mechanism three times (there are 10 reactions, so 30 mechanisms). (A) do the work on a tablet and save as a pdf., it is expected to write each mechanism out and NOT copy and paste the mechanism after writing it just once. Everything should be drawn out stepwise and every bond that is formed and broken in the process of the reaction, and is expected to see all relevant lone pair electrons and curved arrows.arrow_forwardNonearrow_forwardNonearrow_forward
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