Diastereoselectivity on attack to carbonyl groups Felkin- Anh method helps us determine the major product diastereomer on nucleophilic attack to a chiral carbonyl compound. Title of the concept Sukumar Honkote Department of Chemistry, IIT Bombay

Master Layout The animation will be broken up into 4 sub animations: ani1, ani2, ani3 and ani4. The colours for the text, background and the various atoms is upto the discretion of the animator. The colour of the same atom or species should be same throughout the animation.

Definitions of the components: 1 1. Electrophile: is a reagent that participates in a chemical reaction by accepting an electron pair in order to bond to a nucleophile. 2. Nucleophile: is a reagent that participates in a chemical reaction by donating an electron pair in order to bond to n electrophile 3. Diastereoselectivity: the preference for the formation of one or more than one diastereomer over the other in an organic reaction. 4. Syn product: two substituents on adjacent carbons oriented in same direction 5. Anti product: The same substituents oriented in opposite direction. 6. Bürgi-Dunitz angle: the angle of attack of a nucleophile at a carbonyl center. It is 107° with respect to the carbonyl bond. 7.Conformers: The different shapes of an organic compound obtained due to its ability to rotate about single bonds 8. 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. 2 3 4 5

Ani1 ACTION DESCRIPTION TEXT & SOUND Text in centre The animations on ‘Conformational Analysis’ and ‘wedge- dash to Newman Projection’ are pre-requisites for this animation 1. text Felkin- Anh method can be used to determine the major product in reactions involving attack by a nucleophile on a carbonyl compound. We shall study the basis and the method to thus utilize Felkin- Anh method. 2. text The major product formed is influenced mainly by the steric affects of the substituents attached to the α carbon 3. Animation ‘example1’ starts 5. Clicks on syn box Word appears in red INCORRECT 6. Clicks on anti box Word appears in green CORRECT 7. Text Let us understand the Felkin- Anh method to get the correct answer always

Master Layout (Ani2) Fig1 Fig1 Fig2 Fig3

In Fig2. and fig3. the angle between M & L , M & S and S & L is always 120° while the angle between R & O is always at 180°. In fig3, the angle between the balls blue & green, blue & red and red & green is always 120° while the angle between black & brown balls is always at 180°. In fig3, the sizes of the balls relative to others have to remain same with respect to others. Red ball will be the smallest and green the largest. In fir2 and fig3 initially the angles between O & L and black ball & green ball is 30°. The violet ball, nu denotes the nucleophile

Fig1 Fig2 Fig3 ACTION DESCRIPTION TEXT & AUDIO 1. Text Let us consider a molecule 2. Fig 1 appears The word Figure 1 should be written below it where L is a large group like Ph S is a small group like H M is a medium group like Me R is any alkyl or aryl group. 3. Fig 2 appears Fig 2 appears beside fig. 1. The word Figure 2 should be written below it. Figure 2 is the Newman projection of the molecule. 4. Fig 3 appears The word Figure 3 should be written below it. Figure 3 is the ball and stick representation of the Newman Projection of the molecule. The balls in this figure denote the sizes of the substituents. Green ball represents L, black ball represents O, Red ball represents S, blue ball represents M and brown ball represents R. 4. The nucleophile will attack the stable and sterically least hindered conformers. Let us find the most stable conformer.

Animation page

Angle between L & O Normalised energy Figures a Figures b -30 0.375 Fig 2 Fig 3 1 Fig 4a Fig 4b 30 0.725 Fig 5a Fig 5b 60 0.823 Fig 6a Fig 6b

90 Fig 7a Fig 7b 120 0.5 Fig 8a Fig 8b 150 0.35 Fig 9a Fig 9b 180 0.9 Fig 10a Fig 10b

210 0.425 Fig 11a Fig 11b 240 0.86 Fig 12a Fig 12b 270 0.2 Fig 13a Fig 13b 300 0.475 Fig 14a Fig 14b

330 Fig 2 Fig 3

ACTION DESCRIPTION AUDIO & TEXT 1. Animation page The animation page appears with fig2 and fig3 shown in their respective boxes. 2. Rotation in fig2 rotate L, S and M anticlockwise while keeping R & O stationary. The angles between L,S and M are constant at 120 °. 3. Rotation in fig3 Similarly rotate green, red and blue balls anticlockwise while keeping brown & black balls stationary. The angles between green, red and blue balls are constant at 120 °. 4. Initially the angle between L & O is -30 °. Initially the angle between green ball & black ball is -30 °. 5. The green, blue and red balls will also rotate according to the rotation of L,M and S resp. Eg. Initially L is at -30° wrt O, M is 90° wrt O and S is 210° wrt O hence simultaneously green ball is -30° wrt Black ball, blue ball is 90° wrt Black ball and green ball is 210° wrt Black ball. 6. The graph2 will get plotted as the rotation occurs. The values of the various points of the graph are given in the MSexcel file book1. So if L has moved 65° wrt O then graph would have been plotted only till 65 ° rotation. 7. When two balls overlap, the brown and black balls will always overlap the red, green or blue balls. 8. When angle between L & O becomes 90°, the Fig 7a obtained is put in ‘most stable conformers’. 9. When the angle between L & O becomes 270°, the Fig 13a is put in ‘most stable conformers’ along side Fig 7a

Master Layout The angle between any 2 adjacent letters is 60°. Eg. Angle between L & R is 60°.

Ani 3 Action Description Text & audio 1. Screen becomes blank 2. Fig 7a and Fig 13a appear at the top The most stable conformers of the molecule 3. Fig 7b and Fig 13b appear in the centre The ball and stick image of the Newman projections 4. Nucleophile shown the violet ball nu appears beside fig 7b Let us see how the nucleophile attacks. 5. Title The word ‘nucleophile’ appears below the violet ball for two seconds only 6. The animation then looks like above

Action Description Text & audio 7. Consider the attack of nu to fig 7b only 8. Angle shown The Fig 7b becomes as shown in fig 15 for 2 seconds only to show the angle. nu will still be shown any nucleophile attacking the carbonyl group will do so from the Bürgi–Dunitz angle—about 107° from the C=O bond 9. The attack by the nucleophile nu approaches the intersecting point of the lines connecting the balls of fig. 7b. nu approaches the centre at an angle of 107° wrt to the line connecting black and brown ball 10. Repulsion as shown in Fig 16 In its approach nu will collide with the green ball (as shown by fig 16). Then it gets repelled back (to be fig 16a). Show this 4 times with slight change in the angle of approach. Approach is hindered by the large group. No product is formed from this attack. 11. Return to start Now the fig changes that to Fig 7b.

Action Description Text & audio 12. Consider the attack of nu to fig 7b only Let us see the nucleophilic attack from the other side. 13. Angle shown The Fig 7b becomes as shown in fig 17 for 2 seconds only to show the angle. any nucleophile attacking the carbonyl group will do so from the Bürgi–Dunitz angle—about 107° from the C=O bond 13. nu appears Fig 7b becomes as fig 18a by the appearance of nu. 14. Repulsion as shown in Fig 18 In its approach nu will collide with the blue ball (as shown by fig 18). Then it gets repelled back(as shown by fig 18a). Show this 2 times. 15. 3rd time Now the approach angle changes slightly such that it just grazes the blue ball Approach is less hindered by the smaller group. This attack gives the minor product 16. The bond formation When it is past the blue ball a black line forms connecting nu to the intersecting point 17. The bond movement As soon as the black line between intersecting point and nu starts to form the black ball and the brown ball move 30° to the left. The double bond between C and O breaks to form C-O single bond. Now the hybridization changes from sp2 to sp3. 18. Stable conformer The final image looks like Fig 21. The angle between all adjacent balls in 60°. The fig 7a changes to that of fig 22a. The bonds C-O and C-R move to give the most stable, staggered conformation. This is the minor product.

Action Description Text & audio 19. Consider the attack of nu to fig 13b only Let us see the nucleophilic attack on the other conformer. 20. We shall not consider the attack from the side of the large group. It was determined earlier that the reaction does not happen due to high steric hindrance. 21. Angle shown The Fig 13b becomes as shown in fig 19 for 2 seconds only to show the angle. any nucleophile attacking the carbonyl group will do so from the Bürgi–Dunitz angle—about 107° from the C=O bond 22. Nu appears The fig 13b changes to that of fig. 19a 23. The attack Now nu approaches the intersecting point at an angle of 107° wrt to the line connecting black and brown ball from the left. It does not collide with any ball. Approach is least hindered by the small group. This attack gives the major product 24. Bond formation When nu is past the red ball a black line forms connecting nu to the intersecting point 25. The bond movement As soon as the black line between intersecting point and nu starts to form the black ball and the brown ball move 30° to the right. The double bond between C and O breaks to form C-O single bond. Now the hybridization changes from sp2 to sp3. 26. Stable conformer The final image looks like Fig 20. The angle between all adjacent balls in 60°. The fig 13a changes to that of fig 21a. The bonds C-O and C-R move to give the most stable staggered conformation. This is the major product.

ACTION DESCRIPTION AUDIO & TEXT 1. Blank screen Fig 20 and Fig 21 appear beside each other with words “major” and “minor” written above them respectively 2. Fig 21a and Fig 22a appear exactly below them respectively These are the Newman projections of the major and minor products. The nucleophile is denoted by nu. 3. Fig 21c and fig 22c appear in place of fig 21a and 22a with words “major” and “minor” written above them respectively These are the wedge-dash representation of the products. (Note: see the Newman Projection to Wedge – dash animation to learn the correct method of conversion) 4. The above diagram appears below This is the final equation 5. Animation ends

The following is the Summary : Using the Felkin–Anh model Draw Newman projections of the conformations of the starting material that place a large group perpendicular to C=O Allow the nucleophile to attack along the least hindered trajectory, taking into account the Bürgi–Dunitz angle Draw a Newman projection of the product that arises from attack in this way. Rotate about the carbon (anticlockwise or clockwise) such that the longest chain is horizontal. This is considered the plane. The bonds above the plane of longest chain (the horizontal plane) are denoted by a wedge bond and the bonds below the plane of longest chain (the horizontal plane) are denoted by a dash bond.

QUESTIONS 1.For the reaction of the following molecule with which nucleophile the percentage major product will be most compared with minor Product? A: Et- B: Me- C: nBu- D: H- answer is C as a bulky nucleophile is more stereoselective compared with other nucleophiles of similar reactivity. 2. The major product of the reaction of 1-phenyl, 1-methyl acetone with KCN in presence of hydrocyanic acid would be? A: B: answer is a 3. The relationship between the products formed in the previous question is that they are: A: enantiomers b: diastereomers c: anomers d: mesomers Answer is b

4. The propensity to be attacked by a nucleophile is maximum for which of the following molecules: A: acetone b: 1 phenyl acetone c: 1,1 diphenyl acetone d: 1,1,1 triphenyl acetone

Links for further reading Books: Organic Chemistry by Clayden, Greevs, Warren and Wothers Thank You