(a)
Interpretation:
Draw the structural formula the compound formed by the reaction of given compound with sodium borohydride.
Concept Introduction:
Answer to Problem 57P
Explanation of Solution
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Therefore, the product formed will be as follows:
(b)
Interpretation:
Draw the structural formula the compound formed by the reaction of given compound with sodium borohydride.
Concept Introduction:
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Answer to Problem 57P
Explanation of Solution
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Therefore, the product formed will be as follows:
(c)
Interpretation:
Draw the structural formula the compound formed by the reaction of given compound with sodium borohydride.
Concept Introduction:
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Answer to Problem 57P
Explanation of Solution
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Therefore, the product formed will be as follows:
(d)
Interpretation:
Draw the structural formula the compound formed by the reaction of given compound with sodium borohydride.
Concept Introduction:
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Answer to Problem 57P
Explanation of Solution
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Therefore, the product formed will be as follows:
(e)
Interpretation:
Draw the structural formula the compound formed by the reaction of given compound with sodium borohydride.
Concept Introduction:
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Answer to Problem 57P
Explanation of Solution
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Therefore, the product formed will be as follows:
(f)
Interpretation:
Draw the structural formula the compound formed by the reaction of given compound with sodium borohydride.
Concept Introduction:
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Answer to Problem 57P
Explanation of Solution
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Therefore, the product formed will be as follows:
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Chapter 16 Solutions
Introduction To General, Organic, And Biochemistry
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- need help please and thanks dont understand only need help with C-F Learning Goal: As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT. The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7). Part A - Difference in binding free eenergies Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol. The margin of error is 2%. Part B - Compare difference in free energy to the thermal…arrow_forwardneed help please and thanks dont understand only need help with C-F Learning Goal: As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT. The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7). Part A - Difference in binding free eenergies Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol. The margin of error is 2%. Part B - Compare difference in free energy to the thermal…arrow_forwardPlease correct answer and don't used hand raitingarrow_forward
- need help please and thanks dont understand a-b Learning Goal: As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT. The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7). Part A - Difference in binding free eenergies Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol. The margin of error is 2%. Part B - Compare difference in free energy to the thermal energy Divide the…arrow_forwardPlease correct answer and don't used hand raitingarrow_forwardPlease correct answer and don't used hand raitingarrow_forward
- Can you tell me if my answers are correctarrow_forwardBunsenite (NiO) crystallizes like common salt (NaCl), with a lattice parameter a = 4.177 Å. A sample of this mineral that has Schottky defects that are not supposed to decrease the volume of the material has a density of 6.67 g/cm3. What percentage of NiO molecules is missing? (Data: atomic weight of Ni: 58.7; atomic weight of O: 16).arrow_forwardA sample of aluminum (face-centered cubic - FCC) has a density of 2.695 mg/m3 and a lattice parameter of 4.04958 Å. Calculate the fraction of vacancies in the structure. (Atomic weight of aluminum: 26.981).arrow_forward
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