(a)
Interpretation: The structures of 3 and 4 with diastereomeric centers have to be drawn
Concept Introduction:
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.
If a carbon is attached with four different groups without any elements of symmetry, then that center has a chirality center. Diastereomers (sometimes called diastereoisomers) are a type of a stereoisomer. Diastereomerism occurs when two or more stereoisomers of a compound have different configurations at one or more stereocenters and are not mirror images of each other. When two diastereoisomers differ from each other at only one stereocenter, they are epimers. Each stereocenter gives rise to two different configurations and thus increases the number of stereoisomers by a factor of two.
To find: Draw the structure of 3 and the reason for its production as a diastereomeric mixture
Do reductive amination in compound 2
(b)
Interpretation: The structures of 3 and 4 with diastereomeric centers have to be drawn
Concept Introduction:
An amide group is formed from an amine compound and the ester group.
If a carbon is attached with four different groups without any elements of symmetry, then that center has a chirality center. Diastereomers (sometimes called diastereoisomers) are a type of a stereoisomer. Diastereomerism occurs when two or more stereoisomers of a compound have different configurations at one or more stereocenters and are not mirror images of each other. When two diastereoisomers differ from each other at only one stereocenter, they are epimers. Each stereocenter gives rise to two different configurations and thus increases the number of stereoisomers by a factor of two.
To find: Draw the structure of 4 and determine the possibility of its production as a diastereomeric mixture
Decide the nature of the
Want to see the full answer?
Check out a sample textbook solution
Chapter 22 Solutions
Organic Chemistry
- MnO2 acts as an oxidant in the chlorine synthesis reaction.arrow_forwardIn Potassium mu-dihydroxydicobaltate (III) tetraoxalate K4[Co2(C2O4)4(OH)2], indicate whether the OH ligand type is bidentate.arrow_forwardImagine an electrochemical cell based on these two half reactions with electrolyte concentrations as given below: Oxidation: Pb(s) → Pb2+(aq, 0.10 M) + 2 e– Reduction: MnO4–(aq, 1.50 M) + 4 H+(aq, 2.0 M) + 3 e– → MnO2(s) + 2 H2O(l) Calculate Ecell (assuming temperature is standard 25 °C).arrow_forward
- : ☐ + Draw the Fischer projection of the most common naturally-occurring form of aspartate, with the acid group at the top and the side chain at the bottom. Important: be sure your structure shows the molecule as it would exist at physiological pH. Click and drag to start drawing a structure. ✓arrow_forwardFor a silver-silver chloride electrode, the following potentials are observed: E°cell = 0.222 V and E(saturated KCl) = 0.197 V Use this information to find the [Cl–] (technically it’s the activity of Cl– that’s relevant here, but we’ll just call it “concentration” for simplicity) in saturated KCl.arrow_forwardA concentration cell consists of two Sn/Sn2+ half-cells. The cell has a potential of 0.10 V at 25 °C. What is the ratio of [Sn2+] (i.e., [Sn2+left-half] / [Sn2+right-half])?arrow_forward
- Electrochemical cell potentials can be used to determine equilibrium constants that would be otherwise difficult to determine because concentrations are small. What is Κ for the following balanced reaction if E˚ = +0.0218 V? 3 Zn(s) + 2 Cr3+(aq) → 3 Zn2+(aq) + Cr(s) E˚ = +0.0218 Varrow_forwardConsider the following half-reactions: Hg2+(aq) + 2e– → Hg(l) E°red = +0.854 V Cu2+(aq) + 2e– → Cu(s)E°red = +0.337 V Ni2+(aq) + 2e– → Ni(s) E°red = -0.250 V Fe2+(aq) + 2e– → Fe(s) E°red = -0.440 V Zn2+(aq) + 2e– → Zn(s) E°red = -0.763 V What is the best oxidizing agent shown above (i.e., the substance that is most likely to be reduced)?arrow_forwardCalculate the equilibrium constant, K, for MnO2(s) + 4 H+(aq) + Zn(s) → Mn2+(aq) + 2 H2O(l) + Zn2+(aq)arrow_forward
- In the drawing area below, draw the condensed structures of formic acid and ethyl formate. You can draw the two molecules in any arrangement you like, so long as they don't touch. Click anywhere to draw the first atom of your structure. A C narrow_forwardWrite the complete common (not IUPAC) name of each molecule below. Note: if a molecule is one of a pair of enantiomers, be sure you start its name with D- or L- so we know which enantiomer it is. molecule Ο C=O common name (not the IUPAC name) H ☐ H3N CH₂OH 0- C=O H NH3 CH₂SH H3N ☐ ☐ X Garrow_forward(Part A) Provide structures of the FGI products and missing reagents (dashed box) 1 eq Na* H* H -H B1 B4 R1 H2 (gas) Lindlar's catalyst A1 Br2 MeOH H2 (gas) Lindlar's catalyst MeO. OMe C6H1402 B2 B3 A1 Product carbons' origins Draw a box around product C's that came from A1. Draw a dashed box around product C's that came from B1.arrow_forward
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
ChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill Education
Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill Education
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Elementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY