Transcript Alkanes - City University of New York

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.
If atomic orbitals overlap with each other they are bonding, nonbonding or antibonding
sigma type anti-bonding
pi type anti-bond
Anti-bonding, destabilizing.
Higher Energy
non-bonded
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.
sigma type bonding
pi type bond
or
or
or
or
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 (nonbonding) interaction.
All bonding interactions.
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.
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.
Antibonding interactions are
symmetrically placed.
2 antibonding interactions
1 weak antibonding
Interaction, “non-bonding”
0 antibonding interactions
This would be
wrong.
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.
Nucleophilic Substitution and
b-elimination
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.
b-Elimination
Instead of substitution a base can remove both the leaving group and an adjacent
hydrogen creating a pi bond. Recall dehydrohalogenation.
H
base
+ H-Lv
Lv
A pi bond is created.
Competition between Nucleophilic
Substitution and b-elimination.
OEt
Br
nucleophilic
substitution
+ Br-
+ Na+ C2H5OH
First the nucleophilic substitution.
The ethoxide attacks the carbon bearing
the bromine.
Note the change in charges on the
nucleophile and the Leaving group
Now the b-elimination.
Br
+ Na+ C2H5OH
elimination
+ C2H5OH + Br-
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.
Summary.
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.
Negative Nucleophile
H
H
H
H
H
C
N
H
H
Neutral Nucleophile
H2N
Br
C
H
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
H
H
H
H
H
H
Br
C
N
H3N
Br
C
H
H
H
Br
N: from 0 to +1; Br: from 0 to -1
Negative Nucleophile, positive leaving group
H
H
H
H
C
Br
OH2
H
Br: from -1 to 0; O: from +1 to 0
Br
OH2
C
H
Two Nucleophilic Substitution Mechanisms: SN1 & SN2
SN2 mechanism: substitution, nucleophilic, 2nd order
Backside
attack
Examine important points….
Hydrogens flip to the
other side. Inversion
of configutation
Look at energy profile next…
Energy Profile, SN2
Now the alternative mechanism: SN1
SN1 reaction: substitution, nucleophilic, first order.
CH3OH +
(CH3)3C-Br
CH3OC(CH3)3
+
H + + Br-
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.
Step 3, lesser importance, deprotonation of the ether.
Next, energy profile….
Slow step to form
carbocation. Rate
determining.
Energy Profile of SN1, two steps.
Examine important points…..
Fast step to form product.
Carbocation, sp2
Kinetics: SN1 vs. SN2
SN1, two steps.
SN2, one step.
Effect of Nucleophile on Rate:
Structure of Nucleophile
SN1: Rate Determining Step
does not involve nucleophile.
Choice of Nucleophile: No Effect
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.
SN2: Rate Determining Step
involves nucleophile. Choice of
nucleophile affects rate.
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
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
CH3CN
acetone
Increasing polarity
toluene
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
Methanol, protic solvent,
stabilizing the fluoride ion,
reducing its nucleophilicity.
Small, compact anions (like fluoride
ion) are especially well stabilized and
have reduced nucleophilicity. Iodide ion
is large diffuse charge and less
stabilization occurs.
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.
Summary for Halide Ions
basicity
Protic
solvents.
Protic solvent solvation
Nucleophilicity in protic solvents
Iodide ion
But in
aprotic
solvents.
Bromide ion
Chloride ion Fluoride ion
basicity
Protic solvent solvation
Nucleophilicity in aprotic solvents