1. 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

2. 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.

3. 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 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

4. 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

5. Master Layout (Ani2)
Fig1
Fig1
Fig2
Fig3

6. 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

7. 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.

8. Animation page

9. 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

10. 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

11. 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

12. 330
Fig 2
Fig 3

14. 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

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

16. 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

17. 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.

18. 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.

19. 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.

20. 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

21. 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.

22. 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

23. 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

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