Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 9: Elimination Reactions of Alkyl Halides: Competition between Substitutions and Eliminations.

Similar presentations


Presentation on theme: "Chapter 9: Elimination Reactions of Alkyl Halides: Competition between Substitutions and Eliminations."— Presentation transcript:

1 Chapter 9: Elimination Reactions of Alkyl Halides: Competition between Substitutions and Eliminations

2 Goals After this chapter, you should be able to: Predict products of E2 and E1 reactions Determine stereochemistry of E2/E1 Products Determine whether S N 2, S N 1, E1 or E2 will occur

3 What is an S N 2 Reaction? S N 2 mechanism; S for substitution, N for nucleophilic and 2 because two molecules collide at the critical point in the reaction.

4 Review: An S N 2 Reaction

5 Stereochemistry of Inversion If the nucleophile and the leaving group are both high in the R/S priority order, this means that an R alkyl halide gives an S product, and vice-versa

6 Energy of Inversion

7

8 With S N 2, Size of Substituent Groups Matters Relative Reactivity Toward S N 2 tertiary < secondary < primary < methyl

9 Kinetics of Nucleophilic Substitution Rate = k[RBr][Nu - ] Second order kinetics

10 Effect of Bond Strength of the Leaving Group on S N 2 Reactivity Since the carbon-halogen bond strength increases up the periodic table the relative S N 2 reactivity of the alkyl halide is: RF < RCl < RBr < RI TosO - is a better leaving group than I - OH -, NH 2 -, and RO - are worse than F -

11 Nucleophilicity: CH 3 CO 2 (-) < Cl (-) < Br (-) < N 3 (-) < CH 3 O (-) < CN (-) < I (-) < SCN (-) < CH 3 S (-)

12 Nucleophilicity Parallels basicity H 2 O < C 2 H 3 O 2 - < OH - Increases down the periodic table I - < Cl - < F - Anions are more nucleophilic than neutral compounds The solvent matters!

13 Solvent Effects Consider KBr as a nucleophile source Protic solvents with –OH, -NH slow S N 2 rxn These solvents cluster around the nucleophile lowering the effective nucleophilicity Polar aprotic solvents speed S N 2 These solvents cluster around the metal ion of the salt freeing the nucleophile to be nucleophilic.

14 Characteristics of S N 2 Reactions Single Step Mechanism Inversion of configuration S N 2 reactions are generally reliable only when the alkyl halide is primary Halogen is generally Cl or Br since C-F bond is too strong C-I bond is weak and compounds are unstable

15 An S N 2 Reaction

16 S N 1 Reactions S N 1 reactions proceed by a two step mechanism First: Leaving group leaves giving a carbocation Second: Nucleophile attacks carbocation

17 Review: An S N 1 Reaction

18 S N 1 Reactions

19 Leaving Groups OH - < NH 2 - <RO - F - < Cl - < Br - < I < TosO - Susceptibility to leaving

20 Evidence for S N 1 Kinetics The reaction rate is only dependent upon the concentration of the substance with the leaving group R-X  R + + X - is a slow = rate determining Racemic mixtures are usual Carbocation formation Rate = k[R-X] where X is leaving group

21 S N 1 Reaction Rates Depend on stability of the carbocation More stable carbocation=faster reaction - CH 3 < 1° <  2° < 3° Relative Stability of Carbocation

22 The Nucleophile and S N 1 NO EFFECT!

23 Energy for S N 1

24 Solvent Effects on S N 1 Polar solvents stabilize the intermediate carbocation.

25 Summary S N 1 Fastest with Compounds that form stable carbocation Good leaving group Nucleophiles that are not basic to prevent competing elimination reactions Polar solvents

26 An S N 1 Reaction

27 Elimination Reactions Zaitsev’s Rule: Base induced elimination reactions generally give the more highly substituted double bond alkene product

28 An E2 Reactions

29 E2 Reactions Single step attack of nucleophile on hydrogen on carbon adjacent to the carbon containing the leaving group.

30 E2 Kinetics The rate of the reaction is dependent upon the concentration of the compound containing the leaving group and the nucleophile base. Rate = k[RX][Base]

31 Geometry of E2 All atoms involved are in same plane The hydrogen and leaving group are anti

32 Cycloalkane E2: What do you expect?

33 E2 Reaction

34 An E2 Reactions

35 Zaitsev’s Rule Limitations Don’t use for conjugated double bonds. You can trick the reaction into favoring the least substituted alkene by using a Bulky base.

36 Zaitsev’s Rule Limitations Don’t use for conjugated double bonds. You can trick the reaction into favoring the least substituted alkene by using a Bulky base.

37 E1 Reactions First step is identical to S N 1 – Elimination of the leaving group giving a carbocation First step is slow and rate determining Second step is the attack of a hydrogen on a carbon adjacent to the carbocation Racemic mixtures are usual

38 The E1 Reaction

39 E1 Kinetics Rate = k[R-X] E

40 NucleophileAnionic Nucleophiles (Weak Bases) (RS -, SCN -, I -, Br -, N 3 -, CN - etc.) Anionic Nucleophiles (Strong Bases: HO -, RO - ) Neutral Nucleophiles (H 2 O, ROH, RSH, R 3 N) Alkyl Group Primary RCH 2 - Rapid S N 2 substitution. The rate may be reduced by substitution of  -carbons, as in the case of neopentyl. Rapid S N 2 substitution. E2 elimination may also occur. e.g. ClCH 2 CH 2 Cl + KOH ___ > CH 2 =CHCl S N 2 substitution. (N  S >>O) Secondary R 2 CH- S N 2 substitution plus E2 elimination (depends on basicity of the nucleophile). The rate of substitution may be reduced by branching at the  -carbons, and this will increase elimination. E2 elimination will dominate. S N 2 substitution. (N  S >>O) In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, S N 1 and E1 products may be formed slowly. Tertiary R 3 C- E2 elimination will dominate with most nucleophiles (even if they are weak bases). No S N 2 substitution due to steric hindrance. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, S N 1 and E1 products may be expected. E2 elimination will dominate. No S N 2 substitution will occur. In high dielectric ionizing solvents S N 1 and E1 products may be formed. E2 elimination with nitrogen nucleophiles (they are bases). No S N 2 substitution. In high dielectric ionizing solvents S N 1 and E1 products may be formed. Allyl H 2 C=CHCH 2 - Rapid S N 2 substitution for 1° and 2°-halides. For 3°-halides a very slow S N 2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, S N 1 and E1 products may be observed. Rapid S N 2 substitution for 1° halides (note there are no  hydrogens. E2 elimination will compete with substitution in 2°-halides, and dominate in the case of 3°-halides. In high dielectric ionizing solvents S N 1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give S N 2 substitution in the case of1° and 2°-halides. 3°- halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents S N 1 and E1 products may be formed. Water hydrolysis will be favorable for 2° & 3°-halides. Benzyl C 6 H 5 CH 2 - Rapid S N 2 substitution for 1° and 2°-halides. For 3°-halides a very slow S N 2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, S N 1 and E1 products may be observed. Rapid S N 2 substitution for 1° halides (note there are no  hydrogens. E2 elimination will compete with substitution in 2°-halides, and dominate in the case of 3°-halides. In high dielectric ionizing solvents S N 1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give S N 2 substitution in the case of1° and 2°-halides. 3°- halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents S N 1 and E1 products may be formed. Water hydrolysis will be favorable for 2° & 3°-halides.


Download ppt "Chapter 9: Elimination Reactions of Alkyl Halides: Competition between Substitutions and Eliminations."

Similar presentations


Ads by Google