2. SN2 REACTIONS/ WALDEN INVERSION Author: Dhruv Joshi, Dept. of Chemistry, IITB Mentor: E Siva Subramaniam Iyer & Prof. A.Dutta This animation explains the mechanism of Nucleophilic Substitution (bimolecular) reactions

3. Learning objectives: 5 3 2 4 1 After interacting with this Learning Object, the learner will be able to:  identify the factors involved in a SN2 reaction

4. Definitions of the components: 5 3 2 4 1 1. Nucleophile – The atom (or group of atoms) which is usually negatively charged, and attacks the substrate (which may /may not have a positive charge) ‏ 2. Substrate – The group of atoms which is attacked by the nucleophile (usually has a “leaving group” attached at the point of attack) ‏ 3. Leaving Group – The atom (or group of atoms) which is capable of detaching itself from the molecule and existing in the solution with a negative charge.

5. Concept: 5 3 2 4 1 The SN2 reaction is a single-step displacement of a leaving group by a nucleophile. During the transition state, the bond to the nucleophile forms at the same time that the bond to the leaving group breaks. Therefore the nucleophile is required to approach from the back, and configuration at carbon is inverted.

6. Explain the process 1 5 3 2 4 In this step, use an example to explain the concept. It can be an analogy, a scenario, or an action which explains this concept/process/topic Try to use examples from day-to-day life to make it more clear You have to describe what steps the animator should take to make your concept come alive as a series of moving images. Keep the examples simple to understand, and also to illustrate/animate.

7. Analogy 1 5 3 2 4 The substrate is shaped like an umbrella (with the carbon atom at the centre and the hydrogen atoms at the edges) When the nucleophile attacks, the “umbrella” turns inside- out (like an umbrella does when there is excessive wind, for example)‏ Simultaneously, the leaving group leaves the atom and it is left in the “inside out” version of what it was initially.

8. Analogy / Scenario / Action 1 5 3 2 4 Turned inside out due to wind

9. 1 5 3 2 4 Diagram for reference Hydrogen atom Carbon atom Leaving group Nucleophile

10. Want to know more… (Further Reading) ‏ Definitions Graphs/Diagram (for reference) ‏ Animation Area Test your understanding (questionnaire) ‏ Lets Learn! Concepts Lets Sum up (summary) ‏ Instructions/ Working area Choose the Nucleophile/ substrate Interactivity options Sliders(IO1) ‏ / Input Boxes(IO2) ‏ /Drop down(IO3) ‏ (if any) ‏ Play/pauseRestart Output result of interactivity (if any) ‏ What will you learn Credits Ethyl Chloride Methyl Chloride N- Butyl Chloride Isopropyl Chloride

11. Master layout or diagram Make a schematic diagram of the concept Explain to the animator about the beginning and ending of the process. Draw image big enough for explaining. In above image, identify and label different components of the process/phenomenon. (These are like characters in a film) ‏ Illustrate the basic flow of action by using arrows. Use BOLD lines in the diagram, (minimum 2pts.) ‏ In the slide after that, provide the definitions of ALL the labels used in the diagram 5 3 2 4 1 INSTRUCTIONS SLIDE

12. Master layout or diagram You may have multiple master layouts.  In this case, number the master layout. ( e.g. Master layout 1) ‏  Each Master layout should be followed by the stepwise description of the animation stages related to it. 5 3 2 4 1 INSTRUCTIONS SLIDE

13. In this IDD, all atoms will be represented by spheres of different colours for e.g. Carbon - black, Hydrogen - white, Nucleophile – green leaving group - dark pink as shown in the figure The bonds between 2 atoms (i.e. line joining them) will be represented by cylinders with atoms at each end. Carbon atom Hydrogen atoms Leaving group For the animator

14. Video link for animation of Methyl chloride (for animator) ‏ http://www.bluffton.edu/~bergerd/classes/cem221/sn- e/SN2_content.html

15. 1 5 3 2 4 Nucleophile approaches Methyl Chloride (CH 3 Cl) ‏ Step 1 for Methyl Chloride DescriptionAudio Narration Show the nucleophile (shown above in green) approaching the molecule The nucleophile approaches in a direction opposite to the leaving group bond. Direction of motion of Nucleophile CH 3 Cl

16. 1 5 3 2 4 Direction of motion of Nucleophile Step 2 for Methyl Chloride Rotation of Hydrogen atoms and bonds DescriptionAudio Narration As the nucleophile goes close to the molecule, the hydrogen atoms (shown in white) and bonds (shown in black) flip forward as shown by arrows (refer to video link) The molecule re-orients itself to the trigonal bipyrimidal transition state CH 3 Cl

17. 1 5 3 2 4 Nucleophile forming the bond Leaving group’s bond is weakening Step 3 for Methyl Chloride DescriptionAudio Narration Bond 1 appears (fade-in) in a green colourThe bond between the leaving group and the substrate weakens. Bond 2 becomes red (fade-out)‏

18. 1 5 3 2 4 1 5 3 2 4 DescriptionAudio Narration Bond 1 appears (fade-in) in a green colour (same size as Bond 1) The leaving group leaves along the direction of it’s bond. The molecular re-orients itself from the trigonal bipyrimidal transition state to the final tetrahedral shape Bond 2 becomes red (fade-out)‏ Step 4 for Methyl Chloride Retreat of the Leaving group

19. 1 5 3 2 4 1 5 3 2 4 Step 5 for Methyl Chloride DescriptionAudio Narration The bond (red colour in previous slide) between leaving group and molecule breaks totally. The bond between the substrate and leaving group breaks. Nucleophile forms bond with Methyl Chloride

20. Animation for Ethyl chloride

21. 1 5 3 2 4 Step 1 for Ethyl Chloride Nucleophile approaches Ethyl Chloride DescriptionAudio Narration Show the nucleophile (shown above in green) approaching the molecule The nucleophile approaches in a direction opposite to the leaving group bond Direction of motion of Nucleophile C 2 H 5 Cl

22. 1 5 3 2 4 Step 2 for Ethyl Chloride Repulsion by hydrogen atoms DescriptionAudio Narration The red line shown above slowly appears (fade-in) as the nucleophile moves close to the hydrogen atom. Due to the extra methyl group, steric repulsion takes place when the nucleophile approaches. The nucleophile moves slowly towards the hydrogen atom C 2 H 5 Cl

23. 1 5 3 2 4 Step 2 for Ethyl Chloride (contd..)‏ Repulsion by hydrogen atoms DescriptionAudio Narration The nucleophile moves back and the red line disappears. Due to the extra methyl group, steric repulsion takes place when the nucleophile approaches. C 2 H 5 Cl

24. 1 5 3 2 4 1 5 3 2 4 FOR THE ANIMATOR: This hydrogen atom points to the left Now the same hydrogen atom points to the right A B Rotate by 180 0 along the axis as mentioned

25. 1 5 3 2 4 Step 3 for Ethyl Chloride Re-alignment of methyl group DescriptionAudio Narration Show the rotation from position “A” to position “B” as shown in slide 24. The extra methyl group at the carbon atom, thus re-aligns itself by rotating about the free sigma bond. The red line shown in slide 22 disappears (fade-out) as the molecule re-aligns itself in this step Due to this, there is decreased hindrance to the nucleophile’s approach Refer to slide 23 - During the rotation, show the entire molecule of ethyl chloride

26. 1 5 3 2 4 1 5 3 2 4 Nucelophile approaches the carbon atom Step 4 for Ethyl Chloride DescriptionAudio Narration The nucleophile moves in the direction of the carbon (center) and forms the green bond (fade-in) and red color bond becomes thinner (fade-out) ‏ Thus the trigonal bipyramidal intermediate is formed

27. 1 5 3 2 4 1 5 3 2 4 This group of atoms has already rotated in the previous slide Rotation of hydrogen atoms and methyl group Step 5 for Ethyl Chloride DescriptionAudio Narration The hydrogen atoms and methyl group rotates as shown above (refer to video link) ‏ The molecular re-orients itself from the trigonal bipyrimidal transition state to the final tetrahedral shape

28. 1 5 3 2 4 1 5 3 2 4 Leaving group moves away. Step 6 for Ethyl Chloride DescriptionAudio Narration Green color bond becomes thicker (fade-in) and shorter The bond between the substrate and leaving group breaks, forming the product. Red color bond becomes longer and Disappears (fade-out) ‏

29. Animation for Isopropyl Chloride

30. 1 5 3 2 4 Step 1 for Isopropyl Chloride Nucleophile approaches Isopropyl Chloride DescriptionAudio Narration The nucleophile is moving in the direction as shown. The nucleophile approaches in a direction opposite to the leaving group bond Direction of motion of Nucleophile C 3 H 7 Cl

31. 1 5 3 2 4 Step 2 for Isopropyl Chloride Repulsion by hydrogen atoms DescriptionAudio Narration The nucleophile comes close to hydrogen atoms and red bond appears (fade-in) ‏ Due to the extra methyl groups, a lot of steric repulsion takes place when the nucleophile approaches. In most cases, the reaction does not proceed beyond this. But in special cases, when the nucleophile is very strong and the solvent, temperature and other factors are favourable, the reaction proceeds. The nucleophile (green colour) is pushed behind The red coour bonds fade out C 3 H 7 Cl

32. 1 5 3 2 4 1 5 3 2 4 C1 C2 C3 FOR THE ANIMATOR: Rotation by 180 0

33. 1 5 3 2 4 Step 3 for Isopropyl Chloride Re-alignment of methyl groups DescriptionAudio Narration Show the rotation as explained in slide 32 The methyl groups re-orient as much as they can to a conformation which allows the nucleophile to move The red line shown in slide 31 disappears (fade-out) as the molecule re-aligns itself in this step Refer to slide 32 - During the rotation, show the entire molecule of isopropyl chloride

34. 1 5 3 2 4 1 5 3 2 4 Nucelophile approaches the carbon atom Step 4 for Isopropyl Chloride Description (fig.1 to fig.2) ‏ Audio Narration The nucleophile moves in the direction of the carbon (fig.1.) ‏ Thus the trigonal bipyramidal intermediate is formed forms the green bond (fade-in) and red color bond becomes thinner (fade- out) fig.2. Show a little rotation as shown above (refer to video link) ‏ fig.1. fig.2.

35. 1 5 3 2 4 1 5 3 2 4 Rotation of methyl groups Step 5 for Isopropyl Chloride DescriptionAudio Narration The hydrogen atoms and methyl group rotates as shown above (refer to video link) ‏ The hindrance is reduced when the methyl groups re-align to the trigonal bipyramidal intermediate

36. 1 5 3 2 4 1 5 3 2 4 Leaving group moves away. Step 6 for Isopropyl Chloride DescriptionAudio Narration Green color bond becomes thicker (fade-in) and shorter The leaving group leaves along the direction of it’s bond. The molecule re-orients itself from the trigonal bipyrimidal transition state to the final tetrahedral shape. The bond between the substrate and leaving group breaks, forming the product. Red color bond becomes longer and Disappears (fade-out) ‏

37. Animation for n- butyl chloride

38. 1 5 3 2 4 Step 1 for n- butyl chloride DescriptionAudio Narration The nucleophile is moving in the direction as shown. (do not show arrows) ‏ The nucleophile approaches in a direction anti-parallel to the leaving group bond Direction of motion of Nucleophile Nucleophile approaches n- butyl chloride C 4 H 9 Cl

39. 1 5 3 2 4 Step 2 for n- butyl chloride DescriptionAudio Narration The three red lines appear (fade in), lines become thicker as the nucleophile comes closer. It finally comes to a stop (do not show the thin red lines seen on the atoms) Nucleophile tries to form a bond C 4 H 9 Cl

40. 1 5 3 2 4 Step 3 for n- butyl chloride DescriptionAudio Narration When the nucleophile moves away from the substrate, the lines fade-out slowly and they become thinner as well. Go to step 2 and repeat step 3 The reaction is not possible in this case, the nucleophile moves towards the substrate but it cannot react with it. As the steric hindrance increases the reaction begins to proceed by SN 1 reaction mechanism. Repulsion by hydrogen atoms C 4 H 9 Cl

41. Summary The SN2 reaction is a single-step displacement of a leaving group by a nucleophile. During the transition state, the bond to the nucleophile forms at the same time that the bond to the leaving group breaks. Therefore the nucleophile is required to approach from the back, and configuration at carbon is inverted.

42. 1) For the following molecules, the order of reactivity molecules towards SN2 reaction is a) CH 3 Cl, b) (CH 3 ) 3 Cl, c) C 3 H 7 Cl, d)C 2 H 5 Cl i) b>c>d>a ii) a<b<c<d iii) d>>c=b>a iv) b<c<d<a 2) Which is likely to undergo SN2 reaction more easily? a) CH 3 Br b) CH 3 F i) a ii) b iii) both equally react iv) not enough data to predict 3) Why the configuration of molecule gets inverted during an SN2 reaction? i) No change in configuration occurs. ii) It is always not necessary that configuration is inverted. iii) nucleophile attacks from the back side of leaving group. iv) Nucleophile is not able to provide energy to flip the substrate. 4) Aromatic compounds usually do not undergo SN2 mechanism because i) Aromatic groups are bulky ii) ground state is more stable. iii) transition state can not be formed iv) none of the above. Questionnaire:

43. Links for further reading Books: Advanced Organic Chemistry, Morrison and Boyd, Sixth edition, Prentice Hall of India, New Delhi