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Stereochemistry SN1 reaction. Chiral, optically active.

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Presentation on theme: "Stereochemistry SN1 reaction. Chiral, optically active."— Presentation transcript:

1 Stereochemistry SN1 reaction. Chiral, optically active.
Planar, chirality lost, achiral, optically inactive. Chiral carbons generated, racemic mixture, optically inactive.

2 Stereochemistry, SN1 at a chiral center.
racemization Frequent complication: the Leaving Group will tend to block approach of the nucleophile leading to more inversion than retention for the SN1

3 Stereochemistry SN2, Inversion at a Chiral Center
Inversion, frequently (but not always) the R,S designator changes Examples Here is the inversion motion!

4 Another Example The chiral center will undergo inversion.
The non-reacting chiral C will not change. How to understand the configurations: simply replace the Br with the OCH3 (retention). Now swap any two substituents (here done with H and OCH3) on the reacting carbon to get the other configuration (inversion). Done.

5 Stereochemistry, SN2 Two things happening here:
1)Substitution of iodide, 127I, with labeled iodide, 131I. 2) Change in stereochemistry Substitution Recall iodide a good nucleophile, acetone an aprotic solvent resulting in highly reactive iodide ion. SN2 Stereochemistry: Inversion Inverted configuration 5

6 Comparison of SN1 and SN2 mechanisms. Substitution vs
Comparison of SN1 and SN2 mechanisms. Substitution vs. Loss of Optical Activity Stereochemistry: RI represents the R configuration of the alkyl iodide; RI represents the S configuration. Substitution: I is the normal 127I isotope; I is the tagged 131I iodine isotope. If racemization: “SN1” I - RI RI RI RI RI RI RI RI RI RI Only 20% reacts 20% substituted, 20% racemized, 20 % of optical purity lost (80% optically pure). Rate of Loss of optical activity = Rate of substitution. 100% optically pure If inversion: SN2 I - RI RI RI RI RI RI RI RI RI RI Only 20% reacts 20% substituted, 40% racemized, 40% optical purity lost (60% optically pure). Rate of loss of optical activity = 2 x Rate of substitution. 6

7 Effect of Structure of the Haloalkane on Rates
Recall Stability of resulting carbocation, hyperconjugation SN1 Ease of ionization CH3X CH3CH2X (CH3)2CHX (CH3)3CX Methyl primary secondary tertiary Rate of SN1 Reactions 7

8 Now for SN2 SN2 Steric Hinderance, difficulty of approach for nucleophile CH3X CH3CH2X (CH3)2CHX (CH3)3CX Methyl primary secondary tertiary Rate of Reactions Summary: Methyl, primary use SN2 mechanism due to steric ease. Tertiary uses SN1 mechanism due to stability of carbocations Secondary utilizes SN1 and/or SN2 – depending on solvent and nucleophile. 8

9 Recall: Resonance Stabilization of Carbocations
Allylic and benzylic carbocations are stabilized by resonance. SN2 Both SN1 9

10 Leaving Group Recall that the leaving group becomes more negative.
Generally, the best leaving groups are groups that can stabilize that negative charge: weak bases; conjugate bases of strong acids. Base Strength Example: 10

11 Nuc - + R-X  [Nuc----R----X] -  Nuc-R + X -
Solvents Polar solvents stabilize ions, better stabilization if the charge is compact. Polar Protic solvents stabilize both anions (nucleophiles) and cations (carbocations). Accelerate SN1 reactions where charge is generated in the Rate Determining Step. R-X  [R X -]  R X - Polar aprotic solvents usually stabilize cations more effectively than anions (nucleophiles). Anions (nucleophiles) are left highly reactive. Accelerates SN2 reactions where an anion (nucleophile) is a reactant. Nuc R-X  [Nuc----R----X] -  Nuc-R + X - Stabilized. Stabilized. Not Stabilized. Note that it is the energy of the transition state relative to the reactant which affects the rate of the forward reaction (but not the equilibrium). 11

12 Rearrangements for SN1: 1,2 Shift
Recall carbocations can rearrange (1,2 shift) to yield a more stable carbocation. Occurs in SN1 – but not SN2 – reactions. Initial Ionization in protic solvent. 1,2 shift converting 2o carbocation to 3o benzylic Nucleophile attacks. Deprotonate to to yield ether Next elimination… 12

13 Return to b elimination: competes with nucleophilic substitution.
SN1 and/orSN2 The competition: Zaitsev Rule, prefer to form the more substituted alkene (more stable). 13

14 Mechanistic Possibilities to eliminate the H+ and X-
Possible Sequences for bond making/breaking… Regard the alkyl halide as an acid. First remove H+ producing a carbanion , then in a second step remove X- producing the alkene. or First remove X- producing a carbocation, then in a second step remove H+ yielding the alkene. E1 Remove H and X in one step to yield the alkene. E2 14

15 There are two idealized mechanisms for -elimination reactions
E1 mechanism: at one extreme, breaking of the R-Lv bond to give a carbocation is complete before reaction with base to break the C-H bond only R-Lv is involved in the rate-determining step (as in SN1) E2 mechanism: at the other extreme, breaking of the R-Lv and C-H bonds is concerted (same time) both R-Lv and base are involved in the rate-determining step (as in SN2) 15

16 E1 Mechanism ionization of C-Br gives a carbocation intermediate
proton loss from the carbocation intermediate to a base (for example, the solvent) gives the alkene 16

17 Rate Determining Step; formation of the carbocation.
Energy Profile for E1 mechanism, carbocations. Rate Determining Step; formation of the carbocation. Reaction can occur in either direction….. Alkyl Halide  (E1) Alkene + HX Alkyl Halide  (Addition) Alkene + HX 17

18 E2 Mechanism breaking of the R-Lv and C-H bonds is concerted
Needs Strong Base 18

19 E2 19

20 Kinetics of E1 and E2 E1 mechanism E2 mechanism
reaction occurs in two steps the rate-determining step is carbocation formation involving only RLv the reaction is 1st order in RLv and zero order in base E2 mechanism reaction occurs in one step involving both RLv and the base. reaction is 2nd order; first order in RLv and 1st order in base 20

21 Regioselectivity of E1/E2
E1: major product is the more stable alkene (more substituted, more resonance) E2: with strong base, the major product is the more stable (more substituted, more resonance) alkene Special notes about sterically hindered bases such as tert-butoxide, (CH3)3CO -. E2 – anti-Zaitsev: with a strong, sterically hindered base the major product is often the less stable (less substituted) alkene. Reason: hydrogens on less substituted carbons are more accessible. Also E2 vs SN2: In competition of SN2 vs E2: steric bulk in either the alkyl halide or the base/nucleophile prevents the SN2 reaction and favors the E2. 21

22 Stereochemistry of E2 E2 is most favorable (lowest activation energy) when H and Lv are anti and coplanar D E A D E B A B 22

23 Examples of E2 Stereochemistry
Explain both regioselectivity and relative rates of reaction. Faster reaction Major product. Zaitsev product cis But Slower reaction Only product Anti-Zaitsev trans 23

24 Reactive Conformation; H and Cl are anti to each other
Principles to be used in analysis Stereochemical requirement: anti conformation for departing groups. This means that both must be axial. Dominant conformation: ring flipping between two chair conformations, dominant conformation will be with iso propyl equatorial. First the cis isomer. Reactive Conformation; H and Cl are anti to each other In order for the H and the Cl to be anti, both must be in axial positions Iso-propyl groups is in more stable equatorial position. Dominant conformation is reactive conformation. 24

25 Now the trans In the more stable chair of the trans isomer, there is no H anti and coplanar with Lv, but there is one in the less stable chair Reactive but only with the H on C 6 Unreactive conformation Most of the compound exists in the unreactive conformation. Slow reaction. Anti Zaitsev 25

26 Example, Predict Product
Problem!: Fischer projection diagram represents an eclipsed structure. Task: convert to a staggered structure wherein H and Br are anti and predict product. We will convert to a Newman and see what we get… Now anti and we can see where the pi bond will be. H & Br not anti yet! 26

27 Alternative Approach: CAR
Anti Geometry A A The H and Br will be leaving: just indicate by disks. CR C < -- > R Note: As we have said before it may take some work to characterize a compound as “racemic” or “meso”. Relationship works in both directions. Should get cis isomer. Meso or Racemic?? This may be recognized as one of the enantiomers of the racemic mixture. 27

28 E1 or E2 (Carbocation) 28

29 1o good nucleophiles, aprotic solvents
2o good nucleophiles but also poor bases, aprotic solvents 30 1o 2o lower hinderance, better nucleophile than base 3o lower hinderance, better nucleophile than base SN2 SN1 ionization Rearrange ? 1o, 2o, 3o polar solvents, weak nucleophiles, weak bases 1o strong, bulky bases 2o strong bases 3o strong bases E1 E2 1o 2o heat, more hindered 3o heat, more hindered 29

30 Recall Halohydrins and Epoxides
Internal SN2 reaction with inversion Creation of Nucleophile Creation of good leaving group. Attack by poor nucleophile 30

31 Neighboring Group Effect
Mustard gases contain either S-C-C-X or N-C-C-X what is unusual about the mustard gases is that they undergo hydrolysis rapidly in water, a very poor nucleophile C l S N Bis(2-chloroethyl)sulfide (a sulfur mustard gas) Bis(2-chloroethyl)methylamine (a nitrogen mustard gas) 31

32 proton transfer to “solvent” completes the reaction
the reason is neighboring group participation by the adjacent heteroatom proton transfer to “solvent” completes the reaction Good nucleophile. 32

33 5. Provide a clear, unambiguous mechanism to explain
From an old quiz 5. Provide a clear, unambiguous mechanism to explain the following stereochemical results. Complete structures of intermediates, if any, should be shown. Use curved arrow notation consistently. Here is the crux of the matter: how can the non-reacting carbon change its configuration??? Further it does not always change but only if configuration of the reacting carbon changes!! We got a mixture of enantiomers, a racemic mixture. Something strange is happening!! Expect sulfur to attack the C-Cl, displacing the Cl and forming a three membered ring. Like this… But we have to be careful with stereochemistry 33

34 We have to put the molecule in the correct conformation.
S and Cl are eclipsed, not anti. = Reactive conformation reached by 180 rotation around C-C bond And then the ring is opened by attack of water But let’s pause for a moment. Our reactant was optically active with two chiral carbons. Recall the problem: If reaction occurs only at the C bearing the Cl the other should remain chiral! Hmmmm? But now notice that the intermediate sulfonium ion is achiral. It has a mirror plane of symmetry. Only optically inactive products will result. 34

35 Two modes of attack by water.
Enantiomers, racemic mixture And… Again note the ring structure is achiral and that we must, of course, produce optically inactive product. 35

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