Stereochemistry, S N 1 at a chiral center. Frequent complication: the Leaving Group will tend to block approach of the nucleophile leading to more inversion than retention for the S N 1 racemization
Stereochemistry S N 2, Inversion at a Chiral Center Inversion, frequently (but not always) the R,S designator changes Examples Here is the inversion motion!
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 OCH 3 (retention). Now swap any two substituents (here done with H and OCH 3 ) on the reacting carbon to get the other configuration (inversion). Done.
Stereochemistry, S N 2 Substitution Recall iodide a good nucleophile, acetone an aprotic solvent resulting in highly reactive iodide ion. Inverted configuration S N 2 Stereochemistry: Inversion Two things happening here: 1)Substitution of iodide, 127 I, with labeled iodide, 131 I. 2) Change in stereochemistry
Comparison of S N 1 and S N 2 mechanisms. Substitution vs. Loss of Optical Activity RI RI RI RI RI Stereochemistry: RI represents the R configuration of the alkyl iodide; RI represents the S configuration. Substitution: I is the normal 127 I isotope; I is the tagged 131 I iodine isotope. I - If racemization: “S N 1” 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. RI RI RI RI RI I - If inversion: S N 2 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. 100% optically pure
Effect of Structure of the Haloalkane on Rates CH 3 X CH 3 CH 2 X (CH 3 ) 2 CHX (CH 3 ) 3 CX Methyl primary secondary tertiary SN1SN1 Recall Stability of resulting carbocation, hyperconjugation Ease of ionization Rate of S N 1 Reactions
Now for S N 2 CH 3 X CH 3 CH 2 X (CH 3 ) 2 CHX (CH 3 ) 3 CX Methyl primary secondary tertiary SN2SN2 Steric Hinderance, difficulty of approach for nucleophile Rate of Reactions Summary: Methyl, primary use S N 2 mechanism due to steric ease. Tertiary uses S N 1 mechanism due to stability of carbocations Secondary utilizes S N 1 and/or S N 2 – depending on solvent and nucleophile.
Recall: Resonance Stabilization of Carbocations Allylic and benzylic carbocations are stabilized by resonance. SN2SN2SN1SN1Both
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:
Solvents Polar solvents stabilize ions, better stabilization if the charge is compact. Polar Protic solvents stabilize both anions (nucleophiles) and cations (carbocations). Accelerate S N 1 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 S N 2 reactions where an anion (nucleophile) is a reactant. Nuc - + R-X [Nuc----R----X] - Nuc-R + X - 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). Not Stabilized.
Rearrangements for S N 1: 1,2 Shift Recall carbocations can rearrange (1,2 shift) to yield a more stable carbocation. Occurs in S N 1 – but not S N 2 – reactions. Initial Ionization in protic solvent. 1,2 shift converting 2 o carbocation to 3 o benzylic Nucleophile attacks. Deprotonate to to yield ether Next elimination…
Return to elimination: competes with nucleophilic substitution. Zaitsev Rule, prefer to form the more substituted alkene (more stable). The competition: S N 1 and/or S N 2
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 or Remove H and X in one step to yield the alkene. E2
There are two idealized mechanisms for - elimination reactions E1 mechanism: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 S N 1) E2 mechanism: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 S N 2)
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
Rate Determining Step; formation of the carbocation. Energy Profile for E1 mechanism, carbocations. Alkyl Halide (E1) Alkene + HX Alkyl Halide (Addition) Alkene + HX Reaction can occur in either direction…..
E2 Mechanism breaking of the R-Lv and C-H bonds is concerted Needs Strong Base
Kinetics of E1 and E2 E1 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
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, (CH 3 ) 3 CO -. 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 S N 2: In competition of S N 2 vs E2: steric bulk in either the alkyl halide or the base/nucleophile prevents the S N 2 reaction and favors the E2.
E2 is most favorable (lowest activation energy) when H and Lv are anti and coplanar Stereochemistry of E2 A B D E E DA B
Examples of E2 Stereochemistry cis Major product. Zaitsev product trans Only product Anti-Zaitsev Explain both regioselectivity and relative rates of reaction. But Faster reaction Slower reaction
In order for the H and the Cl to be anti, both must be in axial positions First the cis isomer. Reactive Conformation; H and Cl are anti to each other Iso-propyl groups is in more stable equatorial position. Dominant conformation is reactive conformation. 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.
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 Now the trans Unreactive conformation Reactive but only with the H on C 6 Most of the compound exists in the unreactive conformation. Slow reaction. Anti Zaitsev
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… H & Br not anti yet! Now anti and we can see where the pi bond will be.
Alternative Approach: CAR The H and Br will be leaving: just indicate by disks. Meso or Racemic?? Anti Geometry CRCR A Relationship works in both directions. Should get cis isomer. Note: As we have said before it may take some work to characterize a compound as “racemic” or “meso”. This may be recognized as one of the enantiomers of the racemic mixture. A C R
ionization 1 o, 2 o, 3 o polar solvents, weak nucleophiles, weak bases 1 o strong, bulky bases 2 o strong bases 3 o strong bases 1 o good nucleophiles, aprotic solvents 2 o good nucleophiles but also poor bases, aprotic solvents 3 0 SN2SN2 SN1SN1 E2 E1 Rearrange ? 1 o 2 o heat, more hindered 3 o heat, more hindered 1 o 2 o lower hinderance, better nucleophile than base 3 o lower hinderance, better nucleophile than base
Recall Halohydrins and Epoxides Creation of Nucleophile Internal S N 2 reaction with inversion Creation of good leaving group. Attack by poor nucleophile
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 Bis(2-chloroethyl)sulfide (a sulfur mustard gas) Bis(2-chloroethyl)methylamine (a nitrogen mustard gas)
–the reason is neighboring group participation by the adjacent heteroatom –proton transfer to “solvent” completes the reaction Good nucleophile.
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. 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 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!! From an old quiz
We have to put the molecule in the correct conformation. = Reactive conformation reached by 180 rotation around C-C bond And then the ring is opened by attack of water S and Cl are eclipsed, not anti. 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.
Two modes of attack by water. And… Again note the ring structure is achiral and that we must, of course, produce optically inactive product. Enantiomers, racemic mixture