Ochem : 1 vs. 100

1. Ochem : 1 vs. 100 By: Joseph Kim, Ryan King, Steven Ma, Vishnavi Reddy, Diane Shin

2. What type of substitution reaction will a 1° alkyl halide undergo? • SN1 • E1 • SN2 • E2

3. What type of substitution reaction will a 1° alkyl halide undergo? • SN1 • E1 • SN2 • E2 WHY: it can’t be SN1 because a 1° carbocation is very unstable and doesn’t form

4. 2.

5. 2. SN2 Hydroxide is a strong nucleophile.

6. All of the following are associated with an SN2 reaction EXCEPT: • Nucleophile concentration affecting the reaction rate • Inversion of Stereochemistry • Rearrangement of alkyl groups • Backside Attack

7. All of the following are associated with an SN2 reaction EXCEPT: • Nucleophile concentration affecting the reaction rate • Inversion of Stereochemistry • Rearrangement of alkyl groups • Backside Attack There will be no rearrangement of alkyl groups in the SN2 reaction because there is NO Carbocation being formed. One can only have rearrangement with carbocation formation.

8. 4.

9. 4. Halides attached directly to phenyl rings will not undergo SN1 or SN2. Thus, no reaction will occur.

10. What type of product(s) is/are formed with an SN1 reaction? Why? • Optically pure • Racemic mixture • A mixture of diasteriomers and enantiomers • Conformational Isomers

11. What type of product(s) is/are formed with an SN1 reaction? Why? • Optically pure • Racemic mixture • A mixture of diasteriomers and enantiomers • Conformational Isomers Due to the carbocation intermediate that forms in an SN1 reaction, a racemic mixture forms since the nuc could attack from the top or bottom. In an SN2 reaction, the product is optically pure and 100% inversion of configuration is seen because of backside attack

12. What type of solvent is required for an SN1 reaction? Why? • Polar protic • Nonpolar • Polar aprotic • Polar

13. What type of solvent is required for an SN1 reaction? Why? • Polar protic • Nonpolar • Polar aprotic • Polar Polar protic solvent because it helps stabilize the carbocation intermediate

14. 7.

15. 7. NaSH is a good nucleophile, weak base. Also, a primary alkyl halide is present. Thus SN2 occurs causing an inversion of configuration

16. Which of the following conditions would SN1 produce the major product? • (CH3)3C-Br / H20 / Δ • (CH3)2CHCH2-Cl / CH3OH / @ low temp • CH3CH2CHClCH3/CH3OH / @ low temp • CH3CH2-Br / H2O / Δ

17. Which of the following conditions would SN1 produce the major product? • (CH3)3C-Br / H20 / Δ • (CH3)2CHCH2-Cl / CH3OH / @ low temp • CH3CH2CHClCH3/CH3OH / @ low temp • CH3CH2-Br / H2O / Δ Both compounds “b” and “d” are primary alkyl halides and will not be able to undergo SN1 even with heat. Although compound “a” can undergo SN1, heat favors elimination. Therefore, the answer is “c”: a secondary alkyl halide that is reacted at a low temperature.

18. 9.

19. 9. Weak nucleophile, solvent is polar protic and tertiary carbon. Sn1

20. Out of the following compounds, which will undergo SN2 most readily • (CH3)2CHCH2CH2CH2-Br • (CH3)3C-I • (CH3)2CHCH2CH2CH2-I • (CH3)3CCH2CH2-Cl

21. Out of the following compounds, which will undergo SN2 most readily • (CH3)2CHCH2CH2CH2-Br • (CH3)3C-I • (CH3)2CHCH2CH2CH2-I • (CH3)3CCH2CH2-Cl The alkyl halide that is less sterically hindered and has the best leaving group will undergo SN2 most readily. Both compounds “a” and “c” are the least sterically hindered; however, compound “c” has the better leaving group. Therefore, “c” is the correct answer.

22. Which of the reactions has a transition state with no intermediate formed? • CH3CH2O- + CH3CH2-Br  CH3CH2CH2OCH2OCH3 + Br- • CH3CH2OH + (CH3)3C-Br CH3CH2OC(CH3)3 + HBr • (CH3)3CSH + (CH3)2CHOH2+(CH3)3CS+HCH(CH3)2 + H2O • NH3+ (CH3CH2C)3CBr (CH3CH2)3NH3+ + Br –

23. Which of the reactions has a transition state with no intermediate formed? • CH3CH2O- + CH3CH2-Br  CH3CH2CH2OCH2OCH3 + Br- • CH3CH2OH + (CH3)3C-Br CH3CH2OC(CH3)3 + HBr • (CH3)3CSH + (CH3)2CHOH2+(CH3)3CS+HCH(CH3)2 + H2O • NH3+ (CH3CH2C)3CBr (CH3CH2)3NH3+ + Br – Because this is the only reaction with a 1° carbocation, along with a strong nucleophile, it will be the one to undergo SN2. By undergoing SN2, there will be no intermediate, as it occurs in ONE STEP.

24. 12.

25. 12. This reaction undergoes SN2 because cyanide is a strong nucleophile and the solvent is polar aprotic.

26. 13.

27. 13. 1° alkyl halide, polar aprotic solvent, strong nuc SN2

28. Which of the following reactions would occur the fastest?

29. Which of the following reactions would occur the fastest? D) would occur the fastest because it will undergo an SN2 reaction, as it has a 1° carbocation, along with a strong nucleophile. The other reactions might undergo an SN1/SN2 reaction, however, the 1° carbocation is still the least hindered, and thus quickest to be attacked.

30. Give a stereochemical structure of the product from the reaction between (S)-2-bromo-2-pentanol and NaN3 in DMSO and heat.

31. Give a stereochemical structure of the product from the reaction between (S)-2-bromo-2-pentanol and NaN3 in DMSO and heat.

32. 16.

33. 16. Here we are finding the disconnect to get the starting material. The product is made by a Williamson ether synthesis, which is an SN2 reaction. The left side of the ether is primary while the right is tertiary. Thus the disconnect is on the left side of the ether. Since SN2 only inverts the stereocenter directly on the halide, only the chlorine side will invert while the alcohol side retains its configuration.