1. Sample Problem
4. A mixture of 1.6 g of methane and 1.5 g of ethane are chlorinated for a short time. The moles of methyl chloride produced is equal to the number of moles of ethyl chloride. What is the reactivity of the hydrogens in ethane relative to those in methane? Show your work.
Solution:
Recall: The amount of product is proportional to the number of hydrogens that can produce it multiplied by their reactivity.
Number of hydrogens leading to methyl chloride =
1.6g * (1 mol/16 g) * (4 mol H/1 mol methane) = 0.40 mol H
Number of hydrogens leading to ethyl chloride =
1.5 g * (1 mol/30 g) * (6 mol H/ 1 mol ethane) = 0.30 mol H
0.40 mol H * Rmethane = mol H * Rethane
Rethane/Rmethane = 0.4/0.3 = 1.3

2. How do we form the orbitals of the pi system…
First count up how many p orbitals contribute to the pi system. We will get the same number of pi molecular orbitals.
Three overlapping p orbitals. We will get three molecular orbitals. 2

3. Anti-bonding, destabilizing. Higher Energy
If atomic orbitals overlap with each other they are bonding, nonbonding or antibonding
Anti-bonding, destabilizing.
Higher Energy
But now a particular, simple case: distant atomic orbitals, on atoms not directly attached to each other. Their interaction is weak and does not affect the energy of the system. Non bonding
If atoms are directly attached to each other the interactions is strongly bonding or antibonding. Bonding, stabilizing the system. Lower energy. 3

4. Molecular orbitals are combinations of atomic orbitals.
They may be bonding, antibonding or nonbonding molecular orbitals depending on how the atomic orbitals in them interact.
Example: Allylic radical
Two antibonding interactions.
Only one weak, antibonding (non-bonding) interaction.
All bonding interactions. 4

5. Allylic Radical: Molecular Orbital vs Resonance
Molecular Orbital. We have three pi electrons (two in the pi bond and the unpaired electron). Put them into the molecular orbitals.
Note that the odd electron is located on the terminal carbons.
Resonance Result
Again the odd, unpaired electron is only on the terminal carbon atoms. 5

6. But how do we construct the molecular orbitals of the pi system
But how do we construct the molecular orbitals of the pi system? How do we know what the molecular orbitals look like?
Key Ideas:
For our linear pi systems different molecular orbitals are formed by introducing additional antibonding interactions. Lowest energy orbital has no antibonding, next higher has one, etc.
2 antibonding interactions
1 weak antibonding
Interaction, “non-bonding”
Antibonding interactions are symmetrically placed.
0 antibonding interactions
This would be wrong. 6

7. Another example: hexa-1,3,5-triene
Three pi bonds, six pi electrons.
Each atom is sp2 hybridized.
Have to form bonding and antibonding combinations of the atomic orbitals to get the pi molecular orbitals.
Expect six molecular orbitals.
# molecular orbitals = # atomic orbitals
Start with all the orbitals bonding and create additional orbitals. The number of antibonding interactions increases as we generate a new higher energy molecular orbital. 7

8. Nucleophilic Substitution and b-elimination8

9. Substitution Process
Nucleophiles have a pair of electrons which are used to form a bond to the electrophile.
A Leaving Group departs making room for the incoming nucleophile.
Note that the nucleophile converts a lone pair into a bond and becomes more positive by +1
Note that the bond from C to the Leaving Group is collapsed into a lone pair on the Leaving Group which becomes more negative by -1.
Nucleophiles can also frequently function as Lewis bases.
The electrophile can function as Lewis acid. 9

10. b-Elimination
Instead of substitution a base can remove both the leaving group and an adjacent hydrogen creating a pi bond. Recall dehydrohalogenation.
A pi bond is created. 10

11. Competition between Nucleophilic Substitution and b-elimination.
First the nucleophilic substitution. The ethoxide attacks the carbon bearing the bromine.
Note the change in charges on the nucleophile and the Leaving group 11

12. Now the b-elimination.
Now the b elimination. The ethoxide (base) attacks the hydrogen on a carbon adjacent to the carbon bearing the Br (b).
Since we are using Br as the leaving group this could also be called a dehydrohalogenation. 12

13. Summary. 13

14. Formal Charges and Nucleophilic Substitution
In the free nucleophile the pair of electrons is a lone pair belongs exclusively to the nucleophile.
In the product, it is a bond and shared.
The result is the nucleophile increases its charge by +1
Conversely the leaving group converts a shared pair of electrons (a bond) into unshared electrons (lone pair). The charge of the leaving group becomes more negative by -1. 14

15. N: from -1 to 0 ; Br: from 0 to -1
Negative Nucleophile
N: from -1 to 0 ; Br: from 0 to -1
Other things being equal, the more basic species will be a better nucleophile. NH2- is a better nucleophile than NH3
Neutral Nucleophile
N: from 0 to +1; Br: from 0 to -1
Negative Nucleophile, positive leaving group
Br: from -1 to 0; O: from +1 to 0 15

16. Two Nucleophilic Substitution Mechanisms: SN1 & SN2
SN2 mechanism: substitution, nucleophilic, 2nd order
Hydrogens flip to the other side. Inversion of configutation
Backside attack
Examine important points….
Look at energy profile next… 16

17. Energy Profile, SN2 17

18. Now the alternative mechanism: SN1
SN1 reaction: substitution, nucleophilic, first order.
Step 1, Ionization,
Rate determining step.
Step 2, Nucleophile reacts with Electrophile.
Note stereochemistry: nucleophile can bond to either side of carbocation. Get both configurations.
Protonated ether. 18

19. Step 3, lesser importance, deprotonation of the ether.
Next, energy profile…. 19

20. Energy Profile of SN1, two steps.
Slow step to form carbocation. Rate determining.
Examine important points…..
Fast step to form product.
Carbocation, sp2 20

21. Kinetics: SN1 vs. SN2
SN1, two steps.
SN2, one step. 21

22. Effect of Nucleophile on Rate: Structure of Nucleophile
SN1: Rate Determining Step does not involve nucleophile. Choice of Nucleophile: No Effect
SN2: Rate Determining Step involves nucleophile. Choice of nucleophile affects rate.
Note the solvent for this comparison: alcohol/water. Talk about it later…
Frequently, better nucleophiles are stronger bases.
Compare
Compare
But compare the halide ions!!
In aq. solution F – more basic than I -. (HI stronger acid.) But iodide is better nucleophile. 22

23. We need to discuss Solvents
Classifications
Polar vs non-polar solvents, quantified by dielectric constant. Polar solvents reduce interaction of positive and negative ions.
Water > EtOH > Acetic acid > hexane 23

24. Solvents. Another Classification
Protic vs aprotic solvents. Protic solvents have a (weakly) acidic hydrogen having a positive charge which stabilize anions.
Alcohols are protic solvents
Aprotic solvents
ROH --- Br HOR
Increasing polarity 24

25. Some solvents can stabilize ions, reducing their reactivity.
Role of Solvents
Some solvents can stabilize ions, reducing their reactivity.
Many nucleophiles are ions, anions.
Protic solvents can stabilize anions. Protic solvents have (weakly) acidic hydrogens bearing a positive charge. Anions may be stabilized
Small, compact anions (like fluoride ion) are especially well stabilized and have reduced nucleophilicity. Iodide ion is large diffuse charge and less stabilization occurs.
Methanol, protic solvent, stabilizing the fluoride ion, reducing its nucleophilicity. 25

26. Halide ion problem
The problem: basicity and nucleophilicity of the halide ions do not parallel each other in protic solvents.
nucleophilicity
basicity
Iodide ion Bromide ion Chloride ion Fluoride ion
Protic solvent solvation
nucleophilicity
The explanation. Fluoride most stabilized in protic solvents reducing its nucleophilicity. 26

27. Polar Aprotic Solvents
Sodium ion stabilized by negative oxygen
Iodide ion, the nucleophile, kept away from the center of positive charge and not stabilized by solvent. 27

28. Summary for Halide Ions
basicity
Protic solvents.
Protic solvent solvation
Nucleophilicity in protic solvents
Iodide ion Bromide ion Chloride ion Fluoride ion
basicity
But in aprotic solvents.
Protic solvent solvation
Nucleophilicity in aprotic solvents 28