(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
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Chapter 22 Solutions
EBK ORGANIC CHEMISTRY AS A SECOND LANGU
- Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. :0: :0 H. 0:0 :0: :6: S: :0: Select to Edit Arrows ::0 Select to Edit Arrows H :0: H :CI: Rotation Select to Edit Arrows H. < :0: :0: :0: S:arrow_forward3:48 PM Fri Apr 4 K Problem 4 of 10 Submit Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. Mg. :0: Select to Add Arrows :0: :Br: Mg :0: :0: Select to Add Arrows Mg. Br: :0: 0:0- Br -190 H 0:0 Select to Add Arrows Select to Add Arrows neutralizing workup H CH3arrow_forwardIarrow_forward
- Draw the Markovnikov product of the hydrobromination of this alkene. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for this problem. + Explanation Check 1 X E 4 1 1 1 1 1 HBr Click and drag to start drawing a structure. 80 LE #3 @ 2 $4 0 I அ2 % 85 F * K M ? BH 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center & 6 27 FG F10 8 9 R T Y U D F G H P J K L Z X C V B N M Q W A S H option command H command optiarrow_forwardBe sure to use wedge and dash bonds to show the stereochemistry of the products when it's important, for example to distinguish between two different major products. Predict the major products of the following reaction. Explanation Q F1 A Check F2 @ 2 # 3 + X 80 F3 W E S D $ 4 I O H. H₂ 2 R Pt % 05 LL ee F6 F5 T <6 G Click and drag to start drawing a structure. 27 & A 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center Acce Y U H DII 8 9 F10 4 J K L Z X C V B N M T H option command F11 P H commandarrow_forwardCurved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the arrows to draw the intermediate and product in this reaction or mechanistic step(s). Include all lone pairs and charges as appropriate. Ignore stereochemistry. Ignore inorganic byproducts. H :0: CH3 O: OH Q CH3OH2+ Draw Intermediate protonation CH3OH CH3OH nucleophilic addition H Draw Intermediate deprotonation :0: H3C CH3OH2* protonation H 0: H CH3 H.arrow_forward
- Predicting the reactants or products of hemiacetal and acetal formation uentify the missing organic reactants in the following reaction: H+ X+Y OH H+ за Note: This chemical equation only focuses on the important organic molecules in the reaction. Additional inorganic or small-molecule reactants or products (like H2O) are not shown. In the drawing area below, draw the skeletal ("line") structures of the missing organic reactants X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Explanation Check Click and drag to start drawing a structure. ? olo 18 Ar © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibilityarrow_forwardcan someone please answer thisarrow_forwardPlease, please help me figure out the the moles, molarity and Ksp column. Step by step details because I've came up with about three different number and have no idea what I'm doing wrong.arrow_forward
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