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Concept explainers
(a)
Interpretation:
The chemical conditions to remove the acetyl group from N-Acetylglucosamide have to be determined.
Concept introduction:
Amide: One
Amide Formation: Amide is formed when a carboxylic acid reacts with an amine or ammonia.
- Primary amide is produce when a carboxylic acid reacts with ammonia.
- Secondary and tertiary amide is produce when a carboxylic acid reacts with primary and secondary amine respectively.
- Ammonium salt is formed when tertiary amine and a carboxylic acid reacts forming an ionic compound with a carboxylate acid anion and a trialkyl ammonium cation since there is no hydrogen atom in trialkyl amine
Amide Hydrolysis: In presence of base or acid with continuous heating, amide reacts with water to form the corresponding amine and carboxylic acid.
- Carboxylic acid and amine salt are obtained when amide hydrolysis takes place in presence of acid.
- Carboxylate anion and amine are obtained when amide hydrolysis takes place in presence of base.
(b)
Interpretation:
The structure of products formed during the acid hydrolysis of N-Acetylglucosamide has to be drawn.
Concept introduction:
Carboxylic acid: One
Amide: One
Amide Formation: Amide is formed when a carboxylic acid reacts with an amine or ammonia.
- Primary amide is produce when a carboxylic acid reacts with ammonia.
- Secondary and tertiary amide is produce when a carboxylic acid reacts with primary and secondary amine respectively.
- Ammonium salt is formed when tertiary amine and a carboxylic acid reacts forming an ionic compound with a carboxylate acid anion and a trialkyl ammonium cation since there is no hydrogen atom in trialkyl amine
Amide Hydrolysis: In presence of base or acid with continuous heating, amide reacts with water to form the corresponding amine and carboxylic acid.
- Carboxylic acid and amine salt are obtained when amide hydrolysis takes place in presence of acid.
- Carboxylate anion and amine are obtained when amide hydrolysis takes place in presence of base.
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Chapter 17 Solutions
Fundamentals of General, Organic, and Biological Chemistry (8th Edition)
- Problem 15 of 15 Submit Using the following reaction data points, construct Lineweaver-Burk plots for an enzyme with and without an inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Using the information from this plot, determine the type of inhibitor present. 1 mM-1 1 s mM -1 [S]' V' with 10 μg per 20 54 10 36 20 5 27 2.5 23 1.25 20 Answer: |||arrow_forward12:33 CO Problem 4 of 15 4G 54% Done On the following Lineweaver-Burk -1 plot, identify the by dragging the Km point to the appropriate value. 1/V 40 35- 30- 25 20 15 10- T Км -15 10 -5 0 5 ||| 10 15 №20 25 25 30 1/[S] Г powered by desmosarrow_forward1:30 5G 47% Problem 10 of 15 Submit Using the following reaction data points, construct a Lineweaver-Burk plot for an enzyme with and without a competitive inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. 1 -1 1 mM [S]' s mM¹ with 10 mg pe 20 V' 54 10 36 > ст 5 27 2.5 23 1.25 20 Answer: |||arrow_forward
- Problem 14 of 15 Submit Using the following reaction data points, construct Lineweaver-Burk plots for an enzyme with and without an inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Using the information from this plot, determine the type of inhibitor present. 1 mM-1 1 s mM -1 [S]' V' with 10 μg per 20 54 10 36 20 5 27 2.5 23 1.25 20 Answer: |||arrow_forward12:36 CO Problem 9 of 15 4G. 53% Submit Using the following reaction data points, construct a Lineweaver-Burk plot by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Based on the plot, determine the value of the catalytic efficiency (specificity constant) given that the enzyme concentration in this experiment is 5.0 μ.Μ. 1 [S] ¨‚ μM-1 1 V sμM-1 100.0 0.100 75.0 0.080 50.0 0.060 15.0 0.030 10.0 0.025 5.0 0.020 Answer: ||| O Гarrow_forwardProblem 11 of 15 Submit Using the following reaction data points, construct a Lineweaver-Burk plot for an enzyme with and without a noncompetitive inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. 1 -1 1 mM [S]' 20 V' s mM¹ with 10 μg per 54 10 36 > ст 5 27 2.5 23 1.25 20 Answer: |||arrow_forward
- Problem 13 of 15 Submit Using the following reaction data points, construct Lineweaver-Burk plots for an enzyme with and without an inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Using the information from this plot, determine the type of inhibitor present. 1 mM-1 1 s mM -1 [S]' V' with 10 μg per 20 54 10 36 20 5 27 2.5 23 1.25 20 Answer: |||arrow_forward12:33 CO Problem 8 of 15 4G. 53% Submit Using the following reaction data points, construct a Lineweaver-Burk plot by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Based on the plot, determine the value of kcat given that the enzyme concentration in this experiment is 5.0 μM. 1 [S] , мм -1 1 V₁ s μM 1 100.0 0.100 75.0 0.080 50.0 0.060 15.0 0.030 10.0 0.025 5.0 0.020 Answer: ||| Гarrow_forward1:33 5G. 46% Problem 12 of 15 Submit Using the following reaction data points, construct a Lineweaver-Burk plot for an enzyme with and without an uncompetitive inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. 1 -1 1 mM [S]' 20 V' s mM¹ with 10 μg per 54 10 36 > ст 5 27 2.5 23 1.25 20 Answer: |||arrow_forward
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