Interpretation: For a given reaction, the expected product has to be found via reductive amination
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.
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.
To find: Plan the expected product via reductive amination using the given starting material
Draw the structures of the starting material and decide the reaction conditions
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Chapter 22 Solutions
Organic Chemistry
- Nonearrow_forwardRedraw the molecule below as a skeletal ("line") structure. Be sure to use wedge and dash bonds if necessary to accurately represent the direction of the bonds to ring substituents. Cl. Br Click and drag to start drawing a structure. : ☐ ☑ Parrow_forwardK m Choose the best reagents to complete the following reaction. L ZI 0 Problem 4 of 11 A 1. NaOH 2. CH3CH2CH2NH2 1. HCI B OH 2. CH3CH2CH2NH2 DII F1 F2 F3 F4 F5 A F6 C CH3CH2CH2NH2 1. SOCl2 D 2. CH3CH2CH2NH2 1. CH3CH2CH2NH2 E 2. SOCl2 Done PrtScn Home End FA FQ 510 * PgUp M Submit PgDn F11arrow_forward
- given cler asnwerarrow_forwardAdd curved arrows to the reactants in this reaction. A double-barbed curved arrow is used to represent the movement of a pair of electrons. Draw curved arrows. : 0: si H : OH :: H―0: Harrow_forwardConsider this step in a radical reaction: Br N O hv What type of step is this? Check all that apply. Draw the products of the step on the right-hand side of the drawing area below. If more than one set of products is possible, draw any set. Also, draw the mechanism arrows on the left-hand side of the drawing area to show how this happens. O primary Otermination O initialization O electrophilic O none of the above × ☑arrow_forward
- Nonearrow_forwardCan I get a drawing of what is happening with the orbitals (particularly the p orbital) on the O in the OH group? Is the p orbital on the O involved in the ring resonance? Why or why not?arrow_forward1) How many monochlorination products-including stereochemistry- are there for the molecule below:arrow_forward
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