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 = 0.30 mol H * Rethane Rethane/Rmethane = 0.4/0.3 = 1.3

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

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

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

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

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

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

Nucleophilic Substitution and b-elimination 8

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

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

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

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

Summary. 13

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

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

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

Energy Profile, SN2 17

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

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

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

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

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

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

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

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

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

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

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