1. Wedge- dash to Newman Projection Sub-domain : Stereochemistry Sukumar Honkote Department of Chemistry, IIT Bombay Newman projection is instrumental in predicting the most stable conformer but molecular representation is done by wedge-dash method. Thus the conversion is essential to represent the most stable conformer.‏

2. Master layout or diagram 5 3 2 4 1 Bond coming out of the plane wedge Bond above the plane Bond on the plane Bond going in the plane Bond below the plane dash Wedge – dash representation Carbon center (C) Here bond of C-R 2, C- R 3 are on the plane, while of C-R 1 is above the plane and C-R 4 below the plane, R 1 R 2 R 3 and R 4 are general notation for the various atoms connected. Thus we see here C- R 2 and C-R 3 are the 2 bonds defining the plane While representing a molecule we consider a plane in which 2 or more bonds exist. The other bonds are above or below the plane. For convenience we shall denote the horizontal as the plane

3. Master Layout 5 3 2 4 1 This is a general molecule C1 C2 C1 C2 Newman Projection of the molecule. We view the molecule along C1-C2 bond with C1 in front. As shown earlier represents bonds in the horizontal plane. Thus R1 & R2 are horizontal. B and C are above the plane and hence above the horizontal similarly A and D are below the plane. The angle between any two adjacent lines above is 60°. Thus B-C1- R2 angle is 60°. Thus C-C1-R2 is 120 °. Bond on the plane Note: All diagrams except the wedge- dash diagrams shall be redrawn by the animator

4. Definitions of the components: 5 3 2 4 1 1. Conformers: The different shapes of an organic compound obtained due to its ability to rotate about single bonds 2. Newman Projection: A projection of the molecule about two adjacent atoms having a single bond to determine the most stable conformer. 3. Prochiral Centre: The carbon having a C=X (where X= N, O or S) and the two other substituents attached are both different. Here if a nucleophile attacks this centre a chiral centre is formed.

5. Explain the process 1 5 3 2 4 1.The need of the concept shall be explained. 2. The scientific authenticity behind the concept is explained. 3. The animation will consist of 2 parts: a) with no double bond b) with a double bond. 4. The user will be given the choice initially. 5. After the user selects one choice he is shown how the conversion is done correctly. 6. Then he is given the choice again to select one of the two parts. He can replay the previous part or select the other part. 7. There will be option for the questionnaire. Click shift + F5 from the next slide to view how the animation should be Note: It should be mentioned that the prerequisite for this animation is conformational analysis.

6. C1 C2

7. C1 C2 ≡

8. C1 C2 ≡ C1 C2

9. C1 C2 ≡ C1 C2 ACTIONDESCRIPTIONTEXT & AUDIO TextThis is the conversion of the wedge-dash representation of the molecule into its Newman Projection 2.Such conversion is essential to determine the most stable compound formed in a reaction by analyzing conformationally. 3.Such treatment is essential for reactions such as nucleophilic attack on carbonyl compound. 4. NoteNote that the knowledge of conformational analysis is a prerequisite for this animation. 5. textLet us see how the conversion takes place

10. With no double bondWith double bond Select the molecule system ActionDescriptionAudio & text 1. Selection of the molecule system Allow the user to click on one of two boxes containing the image Select the molecule system by clicking on desired box 2. For now we assume the first box is selected

11. ActionDescriptionAudio & text 1. textAll R’s can be alkyl/ aryl groups or functional groups or combinations of them.

12. ActionDescriptionAudio & text 1. textFor this animation we assume R 5 and R 3 are alkyl /aryl groups such that R 5 -C1-C2-R 3 form the longest chain. C2 C1

13. ActionDescriptionAudio & text 1.We shall view the molecule along C1-C2 bond and draw the Newman Projection with C1 in front. C2 C1

14. ActionDescriptionAudio & text 1. textThe groups denoted by will be in the plane (horizontal). Hence they will be horizontal in the Newman projection. 2.The groups denoted by will be above the plane (horizontal). Hence they will be above the horizontal in the Newman projection. 3.The groups denoted by will be above the plane (horizontal). Hence they will be below the horizontal in the Newman projection. 4.Any plane can be used as reference but for convenience, we use the horizontal plane. C2 C1 ≡

15. ActionDescriptionAudio & text 1. textA molecule can change its shape by rotating about a single bond. Thus we can rotate the molecule about the C1-C2 bond. 2.By convention, the longest chain, R 5 -C1-C2-R 3 will be in the horizontal plane. C2 C1 ≡ C2

16. ActionDescriptionAudio & text 1. Rotate about C1 After arrows denoting C1 and C2 have been removed, we rotate R 5, R 6 and R 2 anticlockwise by 60 °. The angle between R 5, R 6 and R 2 with each other should be constant at 120 °. The rotation can be either clockwise or anticlockwise. The direction of rotation does not make any difference. C2 C1 ≡

17. ActionDescriptionAudio & text 1. Rotate about C2 we rotate R 4, R 1 and R 3 anticlockwise by 60 °. The angle between R 4, R 1 and R 3 with each other should be constant at 120 °. The rotation can be either clockwise or anticlockwise. C2 C1 ≡

18. ActionDescriptionAudio & text 1. Remove the arrows denoting C1 and C2 2. textThis is the appropriate Newman projection of the given molecule. ≡

19. With no double bondWith double bond Select the molecule system ActionDescriptionAudio & text 1. Selection page appears Allow the user to click on one of two boxes containing the image Select the molecule system by clicking on desired box 2. For now we assume the second box is selected

20. ActionDescriptionAudio & text 1. textAll R’s can be alkyl/ aryl groups or functional groups or combinations of them.

21. ActionDescriptionAudio & text 1. textFor this animation we assume R 1 and R 3 are alkyl /aryl groups such that R 1 -C1-C2-R 3 form the longest chain. 2.Let the group R 2 be larger than R 4. C2 C1 C2

22. ActionDescriptionAudio & text 1.We shall view the molecule along C1-C2 bond and draw the Newman Projection with C2 in front. C2 C1 C2

23. ActionDescriptionAudio & text 1. textFor this animation we assume R 1 and R 3 are alkyl /aryl groups such that R 1 -C1-C2-R 3 form the longest chain. 2.We shall view the molecule along C1-C2 bond and draw the Newman Projection with C1 in front. C2 C1 C2 ActionDescriptionAudio & text 1. textRemove the red arrow The groups denoted by will be in the plane (horizontal). Hence they will be horizontal in the Newman projection. 2.The groups denoted by will be above the plane (horizontal). Hence they will be above the horizontal in the Newman projection. 3.The groups denoted by will be above the plane (horizontal). Hence they will be below the horizontal in the Newman projection. 4.The double bond will always remain in the plane. 5.Any plane can be used as reference but for convenience, we use the horizontal plane. ≡

24. C2 C1 C2 ≡ ActionDescriptionAudio & text 1. textA molecule can change its shape by rotating about a single bond. Thus we can rotate the molecule about the C1-C2 bond. 2.By convention, the longest chain, R 3 -C1-C2-R 1 will be in the horizontal plane. C1 C2

25. C1 C2 ≡ ActionDescriptionAudio & text 1. Rotate about C1 After arrows denoting C1 and C2 have been removed, we rotate R 2, R 3 and R 4 clockwise by 120 °. The angle between R 2, R 3 and R 4 with each other should be constant at 120 °. The rotation can be either clockwise or anticlockwise. The direction of rotation does not make any difference.

26. C2 C1 C2 ≡ ActionDescriptionAudio & text 1. textO and R 3 will repel each other due to steric factors. They cannot be in the same plane. 2.R 3 being the largest will rotate such that it is perpendicular to the carbonyl bond. 1. Rotate about C1 we rotate R 2, R 3 and R 4 clockwise by 90 °. The angle between R 2, R 3 and R 4 with each other should be constant at 120 °.

27. C2 C1 C2 ≡ ActionDescriptionAudio & text 1. textR2 is a large group. Hence there will be considerable repulsion between O and R2. 2.R4 is smaller than R2 as was decided. Hence it will be the group near O. 3.Hence we rotate about C1 to get the most stable conformer. 4. Rotate about C1 we rotate R 2, R 3 and R 4 clockwise by 180 °. The angle between R 2, R 3 and R 4 with each other should be constant at 120 °.

28. C2 C1 ≡ ActionDescriptionAudio & text 1. Remove the arrows denoting C1 and C2 2. textThis is the appropriate Newman projection of the given molecule.

29. References books1) Organic Chemistry Jonathan Clayden, Nick Greeves, Stuart Warren, and Peter Wothers 2) Solomons, Fryhle: Organic Chemistry, 8th Edition

30. QUESTIONNAIRE 1. What is the Newman Projection of this trans-decalin? a: B: c: D: answer is a (Note: the bonds appearing axially upwards or downwards will appear in the same way in the Newman Projection for atom forming the ring.)

31. 2. What is the Newman Projection of this cis-decalin? A: B: c: D: answer is b (Note: the bonds appearing axially upwards or downwards will appear in the same way in the Newman Projection for the atoms forming the ring)

32. 3. The group perpendicular to the carbonyl group in the Newman projection of the most stable conformation is: A: H b: R1 c: R2 Answer is R2 (Note: R2 is bulkier than R1) 4. The stereochemistry of a chiral compound cannot be determined by its Newman projection involving the chiral atom. True or False? Answer is False 5. Draw the Newman Projection of the following Molecule Answer is or