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S N 1 and S N 2 Reactions SN1SN1 SN2SN2 Rate =k[RX] =k[RX][Nuc: - ] Carbocation intermediate? Y N Stereochemistry mix Inversion of configuration Rearrangement.

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Presentation on theme: "S N 1 and S N 2 Reactions SN1SN1 SN2SN2 Rate =k[RX] =k[RX][Nuc: - ] Carbocation intermediate? Y N Stereochemistry mix Inversion of configuration Rearrangement."— Presentation transcript:

1 S N 1 and S N 2 Reactions SN1SN1 SN2SN2 Rate =k[RX] =k[RX][Nuc: - ] Carbocation intermediate? Y N Stereochemistry mix Inversion of configuration Rearrangement ~H, ~ CH 3 possible No rearrangements

2 S N 1 and S N 2 Reactions SN1SN1SN2SN2 Substrate3°>2°>”1°”CH 3 X>1°>2° NucleophileUnimportant, but usually weak Strong and unhindered Leaving groupExcellentBetter than nucleophile SolventPolar and ionizing Polar aprotic

3 S N 1 and S N 2 Reactions  Do not occur with vinyl halides or aryl halides. Consider the carbocation formed for S N 1. Consider the backside attack for S N 2.

4 Elimination Reactions  The substitution reaction mechanisms you have learned are just part of the picture.  In the S N 1 and S N 2 reactions, the species that acts as a nucleophile may also act as a base and abstract a proton. This causes the elimination of HX and the formation of an alkene.  An elimination reaction can occur along with the substitution reaction.

5 Elimination B: - is a species acting as a base. Since HX is lost, this particular reaction is called a dehydrohalogenation.

6 E1 Reactions  E1 = elimination, unimolecular  Rate = k[substrate] (a first order process)  The rate-limiting step is the formation of the carbocation, the same as for S N 1 (that’s why E1 competes).

7 E1 Reaction Mechanism  Step 1: The substrate forms a carbocation intermediate (rate-limiting step).  Step 2: Methanol acts as a base and removes H + (fast step).

8 E1 Reaction Profile rate = k[(CH 3 ) 3 Br] k = Ae -E A (step 1)/RT

9 E1 Occurs with and Competes with S N 1  When bromocyclohexane is heated with methanol, two products are possible.  Can you draw the mechanism that leads to each product?

10 E1 Reactions  E1 reactions are exothermic.  E1 reactions occur in at least two steps and compete with S N 1 reactions. The first step is slow. It is the formation of the carbocation intermediate. The second step is fast. It is the abstraction of H + from the carbocation by the base.

11 Factors Affecting E1 Reactions  Structure of the substrate Can a stable carbocation be formed?  Strength of the base  Nature of the leaving group  The solvent in which the reaction is run. Must be able to stabilize the carbocation and the LG (which is usually an ion).

12 Factors Affecting E1 Reactions - Structure of the Substrate  The most important factor influencing the rate of E1 reactions is the stability of the carbocation formed.  Stability of carbocation: 3° > 2° >1°  Relative rates for E1: 3°>2°≈1°(resonance)

13 Factors Affecting E1 Reactions - Strength of the Base  The rate is not much affected by the strength of the base. Weak bases will work.

14 Factors Affecting E1 Reactions - the Leaving Group  The LG should be good.

15 Factors Affecting E1 Reactions - Solvent Effects  The solvent must be capable of dissolving both the carbocation and the leaving group.  E1 reactions require highly polar solvents that strongly solvate ions.  Typical solvents: water, an alcohol, acetone (to help the alkyl halide to dissolve), or a mix.

16 Rearrangements in E1 Reactions  The carbocation can undergo a structural rearrangement to produce a more stable species. hydride shift (~H) methyl shift (~CH 3 )  If ionization would lead to a 1° carbocation, look for a rearrangement to occur.  Ionization rates can be increased by the addition of reagents such as AgNO 3 (how?); however, Ag is not cheap.

17 A Hydride Shift Can Occur in Either S N 1 or E1 Reactions ~H CH 3 OH -H +  What would the E1 products be? SN1SN1 -Br -

18 A Methyl Shift Can Occur in Either S N 1 or E1 Reactions ~CH 3 CH 3 OH -H +  What would the E1 product be? SN1SN1

19 Zaitsev’s Rule  When two or more elimination products are possible, the product with the more substituted double bond will predominate.

20 Zaitsev’s Rule  When two or more elimination products are possible, the product with the more substituted double bond will predominate. Alkyl groups are electron-donating and contribute electron density to the π bond. Bulky groups can be more widely separated.

21 E1 Reactions - Summary  The structure of the carbocation is the most important factor: Relative rates for E1: 3°>2°.  The base is typically weak or moderate in strength.  The LG should be good.  The solvent should be polar and protic to stabilize the carbocation and LG.  Products can exhibit rearrangements and will follow Zaitsev’s Rule.


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