Download + ∂ - CHEM171 – Lecture Series Seven : 2012/05

CHAPTER 7: REACTION MECHANISMS
Reaction mechanisms involve the movement of electrons
1-electron
2-electrons
BOND BREAKING AND BOND MAKING
HOMOLYTIC FISSION: formation of free radicals
X - Y → X• + Y•
If we split the carbon-carbon bond in ethane homolytically, then
two methyl radicals will result.
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H3C
CH 3 → 2 CH3•
C2H6 → 2 CH3•
HETEROLYTIC FISSION: two electrons in the bond go to one of
the atoms.
X
Y
→ X+ + Y -
Have a cation and an anion forming.
If a carbon-carbon bond breaks in this fashion then we have a
carbocation and a carbanion resulting
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BOND FORMATION:
Cl -
+CH
CH 3Cl
3
The chloride ion is termed a nucleophile and the carbocation is
an electrophile.
Reactive centre – encourages either nucleophilic or electrophilic
attack.
Functional groups will give rise to reactive centres within
molecules. For example, if we have a ketone containing the
functional group:
∂O
C∂
+
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Nu
C
Nu-
O
C
O
C
O-
Nu-
This is called a concerted reaction mechanism where bonds are
forming and breaking at the same time
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CH 3
CARBOCATIONS
H3C
C
CH 3
Cl
CH 3
H3C
C
+ Cl-
CH 3
The chloride ion “leaves” the molecule and is therefore simply
called a leaving group.
Some carbocations are more stable than others. This comes from
the so-called inductive effect.
C
CH 3
The inductive effect is the shifting of electron density
within a σ-bond towards a positive centre in a
molecule. If a substituent group is electron-donating it
has a positive inductive effect and if it is electron
withdrawing it is said to have a negative inductive
effect.
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FREE RADICAL SUBSTITUTION OF ALKANES
Step1
Cl2 + hν  2 Cl•
Initiation: Generation of free radicals:
H
H
Cl
H
C
H
Cl
H
+
C
H
H
H
Step 2
Propagation
Cl
Cl
CH3
Cl
+ HCl
CH 3
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Step 3
Termination
For a free radical substitution to stop, free radicals must react
with each other.
Cl
Cl
Cl
Cl
Cl
CH 3
Cl
CH 3
CH 3
CH 3
C2 H6
Consider the free radical substitution of larger alkanes:
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Cl
H
CH 2CH2 CH2CH3
CH 2CH2 CH2CH3
+ HCl
CH 3
Cl
H
CHCH2CH2CH3
CH 3CHCH2CH2 CH3 + HCl
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ELECTROPHILIC ADDITION REACTIONS OF ALKENES
C
C
+
XY
X
Y
C
C
Addition reactions of alkenes proceed via a carbocation
intermediate.
H
C
H
H
H
C
H
H
Br
H
C
C
H
H
H3C
H + Br-
CH 2Br
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H3 C
H
C
H
C
H
H
H
H3C
Br
H
C
C
H
H
H + Br-
H
C
C
H
H
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CH 3 + Br-
H3 C
CH3
C
H3 C
C
H
H
Br
C
H
H
H3 C
C
H
H
H3 C
C
H
Br-
H
H
C
H
Br
C
H
Br-
*
C
H
H
H
CH3
CH3
C
H
+ Br-
H
H3 C
CH3
H
C
H3 C
CH3
C
CH3
*C
H
Br
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OZONOLYSIS
R
R
C
C
R
H3 C
H3 C
O3
R
*
C
x
C
H
CH3
R
R
C
O
+
O
C
R
O3
R
R
C*
O +
O
R
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x
C
R
R
CH3
H
C
O
O
C
CH2CH 3
H3 C
O3
H
CH3
C
H3 C
H
or
C
CH2CH 3
CH2CH 3
C
H3 C
C
CH3
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ORGANIC REDOX CHEMISTRY
The oxidation states in between +4 and –4 will be
encountered for carbon compounds which contain both
oxygen and hydrogen.
CO2 : C in the +4 oxidation state
CH4 : C in the -4 oxidation state
Addition of hydrogen results in reduction
Addition of oxygen results in oxidation
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INTER-CONVERSION OF ALCOHOLS/ ALDEHYDES/ KETONES/
CARBOXYLIC ACIDS
OXIDATION
1o alcohol oxidized to an aldehyde
H
R
C
OH
CrO3
H
H+
C
O
C
O
R
H
2o alcohol oxidized to a ketone
H
R
C
R'
OH
CrO3
R
H+
R'
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If we treat a 1o alcohol with excess CrO3 in acid, a carboxylic
acid is formed.
H
R
C
O
OH
CrO3/H+
R
excess
C
OH
H
An aldehyde or ketone is reduced to the respective alcohols
using a mild reducing agent, sodium borohydride, NaBH4.
Carboxylic acids can be reduced to an aldehyde using NaBH4
or directly to the alcohol using a stronger reducing agent,
lithium aluminium hydride, LiAlH4
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ADDITION REACTIONS OF ALKENES
Hydration
H3 C
CH3
C
C
H3 C
H2O/H+
H3C
H
OH
H
C
C
H
CH 3 CH 3
Hydrogenation
H
H
C
H
C
H
H2
catalyst
H
H
H
C
C
H
H
H
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Bromination
H
H
C
H
C
H
Br2
H
Br
Br
C
C
H
H
H
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ELIMINATION REACTIONS OF ALKYL HALIDES
H
H
Br
C
C
H
CH 3
CH 3
H
-HBr
CH3
C
C
H
CH3
Base- catalysed reaction
H
H
C
C
H
CH 3
OH3C
C
Br
CH3
H
CH3
C
H
+ Br- + (CH3 )3COH
C
CH3
CH 3
CH 3
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NUCLEOPHILIC SUBSTITUTION REACTIONS OF ALKYL HALIDES
Nu-
+
C
X
C
Nu
+
X-
SN1 Reactions
SN1 reactions proceed via a carbocation intermediate.
R
R''
C
R'''
R
X
R''
C
+
X-
R'''
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R
R''
C
R
+
Nu-
R''
C
R'''
R'''
CH 3
H3C
C
CH 3
Cl
H3C
CH 3
C
CH 3
+
OH-
H3C
CH 3
C
C3 H7
C
CH 3
CH 3
C2 H5
Nu
CH 3
Br
C2 H5
C
+ BrC3H 7
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OH + Cl-
OH
C 2H5
CH 3
C2 H5
C
CH3
C3 H7
C
OHC3H 7
OH
H3C
C
C2H 5
C3 H7
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SN2 Reactions
H
H
H
C
Br
H
C
H
Br- +
H3C
H
H
-
CH 3
C
I
H
+ Br -
Br
C
H
I
H
CH 3
Br
C
H
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+ IH
ACIDS AND BASES
Carboxylic Acids
RCOOH + H2O ⇌ RCOO- + H3O+
O-
O
R
C
R
O-
C
O
O
R
C
O
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Alcohols
R – O – H + base  R – O- + base – H+
O
H
O-
+ NH2-
NH 3 +
H
R
O
H+
H
R
O
H
oxonium ion
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CONDENSATION REACTION
CH3CH2COOH + CH3CH2OH  CH3CH2COOCH2CH3 + H2O
TERMINAL ALKYNES
R – C ≡ C – H, weakly acidic and will react with a strong base
such as sodium amide:
R
C
C
NH2-
H
R
C
C
+ NH 3
R
R
C
C
C
H
X
R
C
C
CH 2R + X -
H
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O
R
C
C
O
C
O
H+
R
C
C
C
OH
REAGENTS AND REACTION CONDITIONS
Most organic reactions need to have the following specified.
•
•
•
•
solvent
temperature
catalyst
reagents to be used
R
C
C
R' + H2
?
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H
H
C
H
C
R'
C
R
R'
C
R
R
H
H
H
C
C
H
H
R'
The actual product formed depends upon the experimental
conditions
R
C
R
C
C
R'
C
H
H2
C
Lindlar's catalyst
R'
2 H2
Pd/C
H
C
R
R
R'
H
H
C
C
H
H
R'
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R
CH2
OH
R
CH2
CrO3
H+
OH
R
PCC
CHO
R
CrO3
H+
R
CHO
CH2Cl2
PCC = pyridinium chlorochromate
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COOH
EXAMPLE
A hydrocarbon of unknown structure has the molecular formula
C8H10. On catalytic hydrogenation over the Lindlar catalyst, 1
equivalent of H2 is absorbed. On hydrogenation over a Pd/C
catalyst 3 equivalents of H2 react. Draw a structure for the
hydrocarbon that fits the data.
SOLUTION
R
C
C
R'
C
C
C
H
H
R'
C
R
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H
H
C
H
C
C
CH
C
H
H
C
C
H
C
H
H
H
2-cyclohexen-2-ylethyne
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EXAMPLE
An unknown hydrocarbon, A, has a formula of C6H10. On
catalytic hydrogenation over Pd/C it reacts with only 1
equivalent of H2. A also undergoes reaction with ozone to
give a single product a dialdehyde, B. Suggest a structure
for A and B.
SOLUTION
H
C
H
O
H
H
C
H
C
H
H
C
C
H
C
H
H
H
O
H
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EXAMPLE
How could you prepare 2-methylhexane from an alkyl halide
and an alkyne?
SOLUTION
H
C
C
H
+ NH2-
H
C
C
+
Cl
2H2 Pd/C
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+ NH3
EXAMPLE
An aldehyde A (C5H10O) can be reduced by LiAlH4 to B.
Treatment of B with concentrated H2SO4 gave C. Ozonolysis of
C affords methanal and D (C4H8O). D gave a positive iodoform
test which identifies methylketones. Identify A, B, C and D.
SOLUTION
O
O
+ 3NaOI
C
R
CH3
+ 3NaOH
C
R
CI 3
O
+ NaOH
C
R
CI 3
RCOO-Na+ + CHI3
yellow ppt.
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LiAlH 4
O
O
B
OH
H2SO4
A
O3
methyl ethyl ketone
D
O
methanal
C
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EXAMPLE
A hydrocarbon A adds one equivalent of hydrogen in the
presence of a Pd/C catalyst to form n-hexane. When A is
reacted vigorously with KMnO4 a single carboxylic acid
containing three carbon atoms is isolated. Give the structure and
name of A.
SOLUTION
O
KMnO4
O
OH +
HO
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O
OH
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EXAMPLE
The trans isomer of an unknown alkene U was reacted with ozone
followed by Zn in water. The products of the reaction were V,
C4H8O and W, C3H6O. Only W gave a positive Tollen’s test.
(i) What information is obtained from the Tollen’s test?
(ii) What would you see that will tell you if the test was positive?
(iii) Draw a structure for the trans isomer of U and structures for,
V and W
SOLUTION
(i) Tollen’s test is used to distinguish between an aldehyde and
ketone
(ii) Tollen’s reagent contains the silver ammonia ion, Ag(NH3)2+.
Oxidation of the aldehyde is accompanied by reduction of silver
ion to free silver – in the form of a mirror)
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RCHO
+
Ag(NH3)2+

RCOO-
+
Ag
silver mirror
H
O
O
V
W
U
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EXAMPLE
Compound A, C4H8, reacts with O3 to give methanal and
compound B. Compound B reacts with CrO3 to give compound
C. Compound B also reacts with LiAlH4 to give D. Upon
treatment with 95% H2SO4 at 170oC compound D gives rise to
compound E. E, when treated with HCl gives us the major
product F and also a minor product G. Identify compounds A
to G.
SOLUTION
O3
A
H
HCHO +
B
O
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H
OH
CrO3
B
C O
O
LiAlH 4
H
D
H2SO4
170°C
E
OH
HCl
Cl
F
Cl
major
G
minor
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Markovnikov’s rule: in the addition of an acid to a carboncarbon double bond of an alkene, the hydrogen of the acid
attaches itself to the carbon that already holds the greater
number of hydrogens.
On addition of peroxide- get Anti-Markovnikov.
Cl
HCl
H2O2
Cl
Minor
major
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