Week 4. Even harder stuff!

Transcription

1 Week 4 Even harder stuff!

2 Focus: SN1 and SN2 Two organic reactions Learn about two basic pathways for how these reactions happen Focus on stereochemistry

3 Focus: SN1 and SN2 You need a couple of things for a reaction Nucleophile: atom or molecule that is electron rich, looks for a position charge. It s a Lewis Base: uses its e- to form a bond Electrophile: atom or molecule that is electron poor, looking for e- charge, Lewis Acid: accepts e- to form a bond. Leaving Group: atom that leaves a molecule, taking a pair of electrons with it R-L Nuc R-Nuc L R = any type of molecule

4 Example H Nucleophile H C Cl + I product H Electrophile Carbon is e-poor because of the polarization of the Cabon-Chloride bond (chlorine is more electronegative, so it s greedy and taking a lot of the electron density, leaving carbon more positive than normal) In order for the nucleophile to come in and form a bond with the central carbon, Chlorine needs to leave! (remember the octet rule, and carbon forms only 4 full bonds) Leaving groups are recognizable as they are usually halide groups! Notice the iodine? Yes, it is a halide, but it s a good nucleophile and will stick to the carbon!

5 Unimolecular nucleophilic substitution reactions (SN1) acetate ion tert-butylchloride H 3 C C Cl Nucleophile O O C O H 3 C C OC Cl Electrophile Leaving Group For this reaction, you have a Leaving Group, nucleophile and electrophile! Reaction depends only on the concentration of t-butyl chloride, it s independent of the acetate ion. First order rate law: rate = k [ t-bucl ] k=constant, [ ] = concentration of molecule The first step to this reaction involved t-bucl this is called the slow step of a reaction or the rate-determining step or RDS As was determined by the 1 st order rate law only t-bucl is involved in the first step HENCE why this is called a UNIMOLECULAR reaction

6 Step 1: Bromine Leaves Step 2: Nucleophile comes in to attack the positive C Here the extra proton is just coming off to form an OH group If you were curious, it s called a hydrolysis reaction because water comes in as a nucleophile!

7 Step 1 Focus Step 1: Leaving Group comes off It comes off primarily because it s more stable in solution than it is stuck to a molecule As it leaves, it takes a pair of electrons By taking a pair of electrons it leaves a positively charged carbon called a carbocation Carbo = carbon Cation = positively charged ion

8 Step 1 Focus Transition State * H 3 C C Cl As Chlorine leaves, it takes the electrons in that bond, so it becomes more negatively charged, while carbon becomes positively charged

9 SN1 mechanism H 3 C C Cl slow H 3 C C Cl Step 1 Arrows show movement of e- This is the carbocation It s a reactive intermediate But it s stabilized by the electron donating groups, the methyls H 3 C C O O C fast O H 3 C C OC Step 2

10 Energy Diagram

11 If the Leaving Group is At a chiral center? Then, we care about the stereochemistry of the SN1 reaction! When the carbocation is formed, the Nuc can attack from the front or back! So you can get both types of enantiomers in an SN1 rxn.

12 Here s another way to see this Racemic means both (S) and (R) enantiomers

13 SN1 Summary First step is the rate determining step Leaving Group comes off the electrophile Forms a carbocation You can usually tell because it is stabilized by electron donating groups like methyls! Then, the nucleophile attacks It can attack frontside and backside!

14 Bimolecular nucleophilic substitution reactions (SN2) H 2 C Cl OH H 2 C OH Cl For this reaction, you have a Leaving Group, nucleophile and electrophile! Reaction depends on the concentrations of both starting materials! Second order rate law: rate = k [ EtCl ] [ OH ] Both chloroethane and Hydroxide must be present in the first/slowest step of this reaction HENCE why it s called Bimolecular!!

15 SN2 Transition State This carbon in the transition state does have 5 bonds but it s ok, because they are just partial bonds between OH and Br, NOT full bonds. All at the same, partial bond is forming the C-OH bond while partial bond is breaking between C-Cl

16 Energy Diagram

17 SN2 stereochemistry IF Leaving Group is at a chiral center Nucleophile can only do a backside attack because L.G. hasn t left yet. No frontside attack! Forms only one enantiomer!