Post on 17-Dec-2015
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Alkyl Halides
Boiling Points
The size of –Br and –CH3 about the same but bromo compounds boil higher due to greater polarizability; more dispersion forces.
Reaction of elemental halogen and alkanes yielding haloalkanes
H3C – H + X – X H3C – X + H - X
Reaction Characteristics
•Requires heat or light to initiate.
•Fluorine is explosive. Reactivity decreases fluorine to chlorine to bromine to iodine which does not react.
Bonds being broken. Bonds being made.
H3C - H 439 kJ
Cl - Cl 247 H3C - Cl 351 H – Cl 431
Br - Br 192 H3C - Br 301 H - Br 368
Bromination: H = 439 +192 – 301 – 368 = -38 kJ less exothermic
Bond Dissociation Energies. Energy Required to Break Bond.
Chlorination: H = 439 + 247 -351 – 431 = -96 kJ exothermic
Regioselectivity
Starting Point in Analysis: Random Substitution. Assume that all hydrogens are equally likely to be replaced by X. There are 8 H in the molecule. Equally likely to be replaced if random….
H3C
H2C
CH3
X2
X
X+
Random substitution
2/8 = 25% 6/8 = 75%
X
X
Regioselectivity
H3C
H2C
CH3
X2
X
X+
Secondary Primary
Random substitution 2/8 = 25% 6/8 = 75%
X = Cl, experimental 57% 43%
X = Br, experimental 92% 8%
X
X
Now include experimental results for X = Cl, Br.
•Replacement of secondary H is favored over primary H. Generally order of reactivity is tertiary > secondary > primary > methyl.
•Bromination displays greater selectivity than does chlorination.
Reactivities of Hydrogens
• For chlorination
tertiary:secondary:primary = 5:4:1• For bromination
tertiary:secondary:primary = 1600:80:1
Quantitative Analysis: Deviations from random replacement quantified by assigning a Reactivity to each kind of Hydrogen (primary, secondary, or primary).
Substitution at a carbon proportional to (# hydrogens at C) x (Reactivity of the H’s)
Predict product mixtureFor chlorination of 2,2,4,4-tetramethylpentane predict the product mixture.
H3C CH2
CH3
H3C
H3C
CH3
CH3H3C C
H2
CH3
H3C
H3C
CH2Cl
CH3
Cl2
UV light
+H3C CH CH3
H3C
H3C
CH3
CH3
Cl
18 x 1 = 18 2 x 4 = 8
Number of hydrogens leading to this product
Reactivity of those hydrogens
Expected ratio 9 : 4
Expected fraction 9/13 4/13
The life of the chain depends on the ongoing presence of the highly reactive Cl atoms and alkyl radicals. Eliminating these species ends chains.
4. 2 Cl. Cl – Cl
5. 2 R. R – R
6. R. + Cl. R - Cl
Heat
or light
Chain Reaction Mechanism
1. Cl – Cl 2 Cl.
2. R – H + Cl. R. + H - Cl
3. R. + Cl – Cl R – Cl + Cl.
Repeat 2, 3, 2, 3,….
Cha
in s
teps
.
Chlorine atom. Highly reactive, only seven electrons
in valence shell
Weak Cl-Cl bond may be broken by heat or light.
Hydrogen to be abstracted. Trade bonds: R-H for H-Cl
Alkyl radical, only seven electrons around the C,
highly reactive alkyl radical.
Trade bonds: weak Cl-Cl for a
stronger C-Cl
Regenerates the Cl atom used in step 2
Ter
min
atio
n st
eps.
Initi
atio
n
Energetics of the Chain Steps
Chlorination of ethane
Step 2, abstraction of the H, controls the regioselectivity of the reaction. Isothermic or slightly exothermic for Cl; endothermic for Br.
Step 3, attachment of the halogen, controls the stereochemistry, which side the halogen attaches. Exothermic
Step 2 and Bond Dissociation Energies, breaking bonds…
More highly substituted radicals are easier to make.
This gives rise to regioselectivity = non-random replacement.
Now bromination.
Compare chlorination and bromination of 2-methyl propane. Bromination is more Regioselective. Examine Step 2 only for
regioselectivity.
BDE, reflecting different radical stabilities
Slightly exothermic
Endothermic
H-Cl is a more stable bond than H-Br.
Step 2 is exothermic for chlorination. Endothermic for Br
First chlorination. Two kinds of H.
Step 2 Transition State Energetics, Cl vs Br
chlorination
Exothermic, tertiary radical more stable
Early transition states, little difference in energies of
activation, rates of abstraction and regioselectivity
bromination
Endothermic, but tertiary radical still more
stable by same amount.
Late transition states, larger difference in
energies of activation, rates of abstraction and
regioselectivity.
R…H……….Cl
R……….H..Br
Halogenation of 2-methylpropane yields two differerent radicals, primary and tertiary.
Now Step 3: Stereochemistry
HStep 2X
Alkyl radical, sp2 hybridization, one electron in the p orbital.
X
XX
Step 3
X
X X
Mirror objects. If a chiral carbon has been produced we get both configurations.
Step 3 has these characteristics
•Determines stereochemistry
•Is exothermic
•Is fast, not rate determining
CH3
H CH3
CH3
H H
C6H14
Cl2
C6H13Cl
Simple Example: monochlorination of 2-methylbutane
Observations:
Optically inactive molecule (can show reflection plane) and products will be optically inactive.
First, look carefully at molecule
Next, organize approach, label the carbons.
aa’
b
c
From a and a’.
From b
d
H
H3C CH3
CH3
H H
CH2Cl
H CH3
CH3
H H
CH3
H CH2Cl
CH3
H H
CH3
Cl CH3
CH3
H H
From cCH3
H CH3
CH3
Cl H
CH3
H CH3
CH3
H Cl
From dCH3
H CH3
CH2Cl
H H
Four optically inactive fractions if distilled.
Chiral carbon
Chiral carbon
Example
H(S) (R)
H3C CH3 Cl2
H H
Ha a’
b b’
c
H(S) (R)
ClH2C CH3
H H
H
a
H(S) (R)
H3C CH2Cl
H H
H
a’
H(R) (R)
H3C CH3
Cl H
H
H(S) (R)
Cl CH3
H3C H
Hb
Diastereomers both sides used.
H(S) (S)
H3C CH3
H Cl
H b’
H(S) (R)
H3C Cl
H CH3
H
Diastereomers both sides used.
(s)
H(R) (S)
H3C CH3
H H
Cl
(r)
Cl(R) (S)
H3C CH3
H H
H
meso meso
b b’enantiomers
enantiomers
First get stereochemical relationships between carbons. enantiomers
Can get relative amounts made of each using reactivities of 1:4:5
3 x 1 3 x 1
1 x 5 / 2 1 x 5 / 2 1 x 5 / 2 1 x 5 / 2
2 x 4 / 2 2 x 4 / 2
Distillation would yield 5 optically inactive fractions.
c c
Allylic Systems
Allylic C – H bonds, weak easily broken.
Removal of H produces the allylic radical.
Vinyl C – H bonds, difficult to
break.
CH3CH2 – H: 411 kJ/mol (101 kCal/mol)
Now the allylic radical…
Resonance in Allylic RadicalResonance provides the stabilization.
The pi system is delocalized.
Odd electron located on alternate carbons, C1 and C3, not C2.
Allylic Substitution
H2C
H
HH
H2C CH2 +
Br
H BrBr Br
heat
Allylic C-H,
372 kJBr-Br
192 kJ Allylic C-Br
247 kJ
H-Br 368 kJ
H = -247 - 386 – (-372 – 192) = -51kJ
Mechanism
initiation
Br2 2Brheat
Chain steps
Weakest C-H bond selected, highest
reactivity
372 kJ368 kJ
H2C CH2
H2C CH2
Br2
H2C CH2 + H2C CH2
BrBr
+ Br
H2C
H
HH Br H2C CH2
+ H Br
H2C CH2
Termination: usual combining of radicals
We have a Problem: seem to have two possible reactions for an alkene with Br2.
1. Addition to the double bond yielding a dibromide.
Br2
Br
Br
2. Substitution at allylic position.
Br2
Br
+ HBr
And/or
Competing Reactions: Addition vs Substitution
Alkene reacts preferentially with Br atoms if present.
Favored by high Br atom concentrations.
High temperature favors Br2 2Br and thus substitution.
Alkene reacts directly with Br2
Happens at low Br atom concentrations.
Low temperature keeps Br concentration low and thus favors addition.
BrBr
Br Br2 Br., not Br2
Addition Substitution
Produced from Br2 at
high temperature
Convenient allylic bromination
For allylic substitution to occur we need both bromine atoms and Br2
Br: R-H + Br. R. + H-Br
Br2: R. + Br2 R-Br + Br.
Both Br and Br2 can be supplied from Br2 at high temperature or from NBS (N-bromosuccinimide).
NBS
Allylic Rearrangement
NBS
Br
Expect bromination of but-2-ene to yield 1-bromo but-2-ene by replacing allylic hydrogen.
But get rearranged product as well….
+
Br
Major product, more stable with subsituted double bond.
Mechanism of Rearrangement
H
HH Br
CH2
+ H Br
CH2
Two different sites of reactivity
CH2
CH2
Br2
CH2 + CH2
BrBr
+ Br
More highly substituted alkene (more stable, recall hydrogenation data) is
the major product
H
HH Br
CH2
+ H Br
CH2
The actual radical is a blending of these two structures.
Secondary radicals are more stable than primary. This predicts most of the radical character at the secondary carbon, favoring this structure. But…
But more highly substituted alkenes are more stable. This predicts most radical character at primary carbon favoring this structure. This appears to be the dominant factor leading to dominant product. But also….
An interesting competition is occurring. Consider the allylic radical…
Results of Calculation of Spin Densities in radical formed from 1-butene
Blue is unpaired electron density. More at primary than secondary.
Anti Markovnikov addition of HBr
Only with HBr, not HCl, HI
HBr
ionic condition, polar solvents
H
Br
Markovnikov
Br
H
anti-Markovnikovradical conditions, peroxides, light,heat
HBr
Mechanism of anti-Markovnikov Radical Addition of HBr
.BrBr
RO-OR 2RO .
peroxide
RO . + H-Br ROH + Br .
Initiation via peroxide to generate Br .
Br
H
Br H
Br
.Br
Chain Steps
Contrast radical and ionic addition of HBr
Radical Ionic
HH
BrBr
Br
Br
HBr
Not a Chain Process
Common Concept: More stable intermediate formed, secondary radical or secondary carbocation
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.
Sample Problem
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
Anti-bonding, destabilizing.Higher Energy
pi type anti-bond sigma type anti-bonding
If atoms are directly attached to each other the interactions is strongly bonding or antibonding. Bonding, stabilizing the system. Lower 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
non-bonded
pi type bond sigma type bonding
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.
All bonding interactions.
Only one weak, antibonding (non-bonding) interaction.
Two antibonding interactions.
Example: Allylic radical
Allylic Radical: Molecular Orbital vs Resonance
Note that the odd electron is located
on the terminal carbons.
Molecular Orbital. We have three pi electrons (two in the pi bond and the unpaired electron). Put them into the molecular orbitals.
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.
0 antibonding interactions
1 weak antibonding Interaction, “non-bonding”
2 antibonding interactions
Antibonding interactions are symmetrically placed.
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.