Download Handout 3

PHCM 331 – Organic and Medicinal/Pharmaceutical Chemistry I
Handout # 3
Winter 2015 / 2016
The objectives of the 3rd handout are to know about:
• Functional groups: Definition, Most common functional groups neutral, acidic, and
basic functional groups, pKa values
• Some Typical Reactions of Hydrocarbons
• Radical Substitution Reactions, course, stability, and reactivity of radicals
• Some free radical reaction mechanisms, Halogenation of alkanes, NBS alllylic substitution,
• Free radical addition to alkenes (HBr / Peroxide addition), Radical polymerization of
alkenes
• Addition Reactions of Alkenes; Hydrohalogenation, Halogenation, Halohydrin
Formation , Alcohols from, Syn addition of hydrogen
• Addition to alkynes
• Addition to dienes
•
Relative stabilities of conjugated and isolated dienes.
1
1.5 Functional groups
1.5.1 Definition
It is the part of a molecule where most of its chemical
reactions occur. Alkanes pKa 50 do not have a functional
group. It is the part that effectively determines the
compound’s chemical and most of its physical properties.
1.5.2 Most common functional groups; examples
H
Alkene
R
C C
H
Alcohol
Amino acid
pKa 25
R
CH3OH
Ether
Alkyne
pKa 40
pKa 16
R O R
pKa alpha COOH 2.4
2
H
O
R C O
R C O R
Aldehyde
Carboxylic ester
R
C O
R
Ketone
O
R C OH
Carboxylic acid
R NH2
Amine
O
R C NH2
Amide
3
3
Average values of pKa
Alkanes
51
Amines
33-35
Esters
23-25
Ketones and
aldehydes
18-20
Alcohols
15-19
b
-Ketoesters
11
-Diketones
10
b
Carboxylic acids
3-5
4
1.6
Some Typical Reactions of Hydrocarbons
1.6.1 Substitution radical Reactions
In a substitution reaction one atom is replaced by another
Br
Introduction
Br
Addition reaction
room
+ Br2 temperature
H
H
HH H H
HH
H
H
H
H
HH
HH H H
HH
H
H
HH
1, 2-Dibromocyclohexane
no reaction
+ Br2
however
+ Br2
h or 
i.e. energy
H
H
H H H Br
HH
H substitution
H
reaction
HH
Bromocyclohexane
5
Mechanism: How a given reaction takes place?
In a chemical reaction some bonds are broken and others are formed via
movement of electrons.
In explaining a mechanism, movement of a pair of electrons is
represented as follows:
Movement of one electron as follows:
RH
an alkane
Cl2
h
RCl product
R
Intermediate
(free radical)
1.6.2 course, stability, and reactivity of radicals
stability of intermediate
Why?
o
3
>
o
o
2
>
1
> CH3
stability increases
energy increases
The energy needed to break C  H bond depends on the type of C atom i.e. tert(3), sec-(2), primary-(1) or methyl.
6
The higher the amount of energy needed to break C  H bond, the less stability of
the intermediate (free radical) hence, the greater the tendency of the intermediate
to recombine with hydrogen i.e.
CH3
H
CH3 + H
Highly unstable
More than 100 k cal/mol needed to break the C-H bond
H
H
H
H C
C
C
H
H
H
CH3
H
98 k cal/mol
CH3
C
H
CH3
Bond dissociation
energy = 91 k cal/mol
only
77
7
1.6.3 Halogenation of Alkanes
CH4 + Cl2
Mechanism:
Cl
Cl
Cl
hor
heat 
Cl
C
CH3Cl + HCl
light
(h
Chain Initiation step
Cl + Cl
H
H
Cl + H
(monosubstitution only)
H
H
(all H are equivalent)
C
H + HCl
H
methyl free radical
(unstable, one ē short)
H
H C Cl
H
Cl + CH4
H
Cl
H
C
chain
propagation
steps
Cl + Cl
H
CH3 + HCl
etc.....
8
Cl2
Cl + Cl
[Cl
H3C + CH3
CH3CH3
CH3 + Cl
CH3Cl
Some examples
1) CH3CH3 + Cl2
2)
CH3CH2CH3

Cl]
[H3C CH3]
Chain
Termination
Steps
CH3CH2Cl
Cl2/h
CH3CHCH3 + CH3CH2CH2Cl
Cl
Major product
Minor product
The relative yield of products is dictated by the relative intermediates stabilities.
in case 2 CH3CHCH3 is more stable than CH3CH2oCH2
CH3
(2o)
(1 )
CH3
3)CH3 CH CH3 + Br2 h
CH3
CH3
C
Br
CH3
CH3
CH3
CHCH2Br Minor product
CH3
in case 3 CH3 C CH3 is more stable than CH3CHCH2
(3o)
(1o)
9
1.6.4 Addition Reactions
General Equation
C
C
A
A
C
B
B
product
reagent
An Alkene
C
The addition to alkenes is electrophilic.
Mechanism.
A
B
A+ + Belectrophile nucleophile
A
C
C
A+
C
C
carbocation (intermediate)
A
A
B-
C
C
C
C
B saturated compound
10
i.e. Addition to a double bond proceeds via two steps:
1An electrophile starts the reaction to produce an
intermediate.
2- A nucleophile reacts with the intermediate to yield a neural
molecule (product).
1.6.4.1 Hydrohalogenation of Alkenes
X
RCH
CHR\ + HX
an alkene
e.g.
CH2
RCH2CHR\
Hydrogen halide
CH2 + HBr
CH3CH2Br
11
However
CH3CH CH2 + HBr
CH2CHCH3 + CH3CH2CH2Br
Br
Mechanism: (Electrophilic addition; H+ then Br-)
CH3CH CH2 + H+
CH3CHCH3 or CH3CH2
CH2
trace if any
major
more stable
less stable
(20) intermediate (10) intermediate
Br
Br
CH3CHCH3
Br
CH3CH2CH2
CH3CHCH3 Major product
CH3CH2CH2Br
Minor product
12
Problem:
What is the major product when HCl reacts with
3-Methyl-2-Pentene?
CH3
CH3
CH3CH2CCH2CH3
CH3CH CCH2CH3 + HCl
Cl
1.6.4.2 Halogenation of Alkenes (Br2 and Cl2)
F2 is too reactive)
I2 is too slow
General Equation
RCH
CHR\ + Cl2
an alkene
CCl4
(CCl4 is a solvent)
(i.e does not react)
Cl
RCH
CHR\
Cl
1,2- Dichloroalkane
13
This is an electrophilic anti addition reaction. Why? Bridge!
Mechanism: One chlorine (or Bromine) atom acts as the
electrophile.
Br
C
C
Br
C
Br
C
Br
an alkene
Br
- stable intermediate since the
charge is scattered over more than one atom C C
Br
carbocation
Br
Br
C
C
Br
C
Br
C
Br
C
C
Bridge
14
example
Br2
H
organic solvent
(absence of H2O)
Br
cyclopentene
Br
not
H
H
H
Br
Br
trans isomer cis isomer
1.6.4.3 Halohydrin Formation (Addition of X and OH)
C
Br
C
an alkene
Br2/H2O
C
C
i.e. anti addition
OH
Mechanism is similar to 1.7.2.2 however, water molecule is the
nucleophile rather than bromide ion
15
Br
C
Br
C
Br
- Br -
C
C
H
Br
O
C
H
C
C
OH
Br
Br2/CCl4
Br
Br2/H2O
CH3CHCHCH3
CH3CHCHCH3
OH
1.6.4.4 Alcohols from Alkenes
Types of Alcohols
CH3CH2OH
H
exists in
higher concentration
than Br -
problems
CH3CH CHCH3
-H +
O
H
CH3CH CHCH3
C
Br
Br
(Hydration i.e. addition of H2O)
(1) alcohol [Ethyl alcohol; Ethanol]
CH3CHCH3
OH
CH3
CH3
C
OH
CH3
(2) alcohol [Isopropyl alcohol; Isopropanol]
(3) alcohol [tert-Butyl alcohol; tert-Butanol]
16
Markovnikov’s Rule: on addition to double bonds, H is
attached to the carbon with the most hydrogens.
Negatively charged ion is considered a nucleophile. However, neutral
H2O molecule (or ROH) may react as a nucleophile also using lone
pair’s electrons of oxygen atom.
Addition of X2 or X & OH: anti-addition (trans product).
Acidic water addition of H, OH follows Mark.Rule (with possible
rearrangement ).(hydride or methyl shift)
Oxymercuration reaction is anti addition of H, OH that follows
Mark.Rule (However, no rearrangement occurs).
Hydroboration Method
For the preparation of all types of alcohols 1°, 2°, 3°
-Syn addition i.e. both H and OH are added to the same face of double bond.
- Anti-Markovnikov’s addition
-NO possible rearrangement.
17
17
A)
The acidic water method
H2O
C
C
H /H2O
No reaction because there
is no electrophile.
C C
[H+  trace of inorganic acid e.g. HCl]
H OH
Mechanism:
H
CH2 CH2 H
+
(nucleophile)
O
(electrophile)
CH3CH2
H
ps. the electrophile may add to any of the two C atoms as both are equivalent
CH3CH2
O
H
H
-H+
(i.e. H+ is a
catalyst)
CH3CH2OH
(1) alcohol since OH is attached to a (1)
carbon atom.
18
However
CH3CH CH2
H+/H2O
OH
CH3CHCH3
(Mark. Rule)
(2°) alcohol
also
CH3
CH3C
CHCH2CH3
H+/H2O
CH3
CH3CCH2CH2CH3
OH
(Mark. Rule)
(3°) alcohol
19
Rearrangement Problem:
In some cases, addition of water to alkenes is not a simple
process.
e.g.
CH3
CH3
CH3
H / H2O
CH3CCH CH2
CH3C CHCH3 minor
CH3C CHCH3
major
CH3
CH3 OH
OH CH3
This is due to 1,2-Methyl shift that is similar to the hydride shift.
In 1,2-Methyl shift a methyl group could move to an adjacent
carbon atom to increase the stability of intermediate.
CH3
CH3
H
CH3C CHCH3 carbocation intermediate
CH3CCH CH2
o
+ ve charge on (2 ) carbon atom
CH3
CH3
To increase the stability of the intermediate, methyl group shifts as follows:
CH3
1,2-methyl
CH3C
CHCH3
CH3
shift
CH3
CH3C CHCH3
CH3
o
more stable intermediate since +ve charge is located on (3 ) C atom
This intermediate leads to the formation of the major product
20
B) Oxymercuration Method
To avoid rearrangement problem, the following reagents are used instead of acidic water.
Reagents
1) Mercuric acetate in water
O
O
CH3C O
Hg
O CCH3 (abbreviated) AcO
O
Hg
OAc Ac
CH3C
2) Sodium Borohydride: NaBH4 (strong reducing reagent)
21
The reaction is anti addition of H, OH that follows Mark.Rule (no
rearrangements).
OH drives from water and H, (produced from NaBH4) replaces the
electrophile HgOAc after it adds.
Mechanism:
C
C
AcO
HgOAc

HgOAc
C
C
-OAc
Stable intermediate, Hg helps
stablizing ve charge
The actual picture of the intermediate:
as a result water
Hg OAc
C
C
attacks the opposite side i.e. (Anti) addition
22
HgOAc
H
O
HgOAc
C
C
H
C
H
C
HgOAc
-H +
C
O
H
H OH
C
H+/H2O
(rearrangement)
CH3
CH3C
CH
CH3
CH2
NaBH4
C
OH
Conclusion:
C
CH3
CH3 C CH CH3
OH CH3
1) Hg(OAc)2/H2O
2) NaBH4
(no rearrangement)
CH3
CH3C
CHCH3
OH
CH3
23
C)
Hydroboration Method (Nobel Prize 1979)
For the preparation of all types of alcohols 1°, 2°, 3°by
choosing the right alkene
-Syn addition i.e. both H and OH are added to the same
face of the double bond.
- Anti-Markovnikov’s addition due to what is called the steric
effect.
CH3CH2CH CH2 1) BH3/THF
2) H2O2
CH3CH2CH2CH2OH
-NO possible rearrangement.
Reagents
THF
1) B2H6 (Diborane) (tetrahydrofuran)
O
2) H2O2 (Hydrogen peroxide)
H
2BH3
B
OH
H
H
Lewis acid
"the electrophile"
24
Mechanism:
The electrophile attacks the less substituted carbon atom. Why?
The electrophile BH3 coordinates with the solvent THF and the
actual attacking species is of a larger size:
BH3
O
e.g.
O
H
B
H
H
H
H
H
C
C
H
C
H
H
BH3
H
H
H
H B
H
C
H
C
H3C
X
H
B
H
CH3C
C
H
H
H
H
H2O2
HO B
H
H
CH3CH2CH2OH
1
o
alcohol
25
General Examples:
CH3CH2CH
1) BH3/THF
CH2
2) H2O2
CH3
CH3
1) BH3/THF
CHCH3
2) H2O2
CH3C
CH3CH2CH2CH2OH (1°) Alc.
CH3CHCHCH3
(2°) Alc.
OH
CH3
CH3C
C CH3
CH3
1) BH3/THF
2) H2O2
H3C OH
CH3CH C CH3
(3°) Alc.
CH3
26
(Review)
OH
CH3CH CH2
CH3
CH3C
CHCH2CH3
H+/H2O
(Mark. Rule)
CH3CHCH3
(2°) alcohol
CH3
H+/H2O
CH3CCH2CH2CH3
(Mark. Rule)
OH
(3°) alcohol
CH3CH CHCH3
Br
Br2/CCl4
(CCl4 is a solvent)
CH3CHCHCH3
(i.e does not react)
CH3CH CHCH3
Br2/H2O
Br
Br
CH3CHCHCH3
OH
Intermediates may rearrange for more stability by moving a hydride or methyl
group. The move should be 1,2 shift ONLY.
CH3
CH3
1,2-methyl
CH3C
CHCH3
CH3C CHCH3
shift
CH3
CH3
27
1.6.4.5
HBr/Peroxide/ h Free Radical Addition to Alkenes
i.e. Addition of HBr in the presence of an organic peroxide ROOR.
Don’t confuse organic peroxide e.g. CH3OOCH3 with inorganic peroxide
e.g. HOOH (hydrogen peroxide).
Recall:
CH2
CH3CH
CH3CHCH3
HBr
Br
electrophilic addition where H+ adds first followed by the bromide Br(i.e. Markovnikov's Rule)
However in the presence of organic peroxide and light the addition is
free radical .
CH3CH
.
CH2 + HBr
h
RO OR
Why?
.
In this case Br adds first instead of H
e.g.
CH3CH2CH2Br
i.e. Anti Markovnikov's Rule
28
The mechanism
h
RO OR
RO + H
ROH + Br
Br
CH3CH CH2 +Br
free radicals
2RO
CH3CH
(Step 1)
CH2Br + CH2CHCH2
major (more stable)
(Step 2)
Br
minor (less stable)
CH3CHCH2Br + H Br
CH3CH2CH2Br + Br (Step 3)
Steps 2 and 3 are repeated over and over
Conclusion:
Markovnikov's Rule
CH3CHCH3
HBr
Br
CH3CH CH2
HBr/Peroxide
h
CH3CH2CH2Br
Anti Markovnikov's Rule
29
General Information:
H2 gas is used for hydrogenation (reduction) however, under high
pressure and the presence of a catalyst such as Pt, Pd, or Ni.
Hydrogen-rich molecules are also used as a good source of hydrogen
e.g. Sodium Borohydride: NaBH4 , LiAlH4 (strong reducing reagents)
On the other hand, while O2 is used as oxidizing reagent,
oxygen-rich molecules are also useful for the oxidation of
organic molecules. e.g. O3 , H2O2 (hydrogen peroxide), and
KMnO4 (potassium permanganate)
Lone pair of electrons may be transferred into a single bond
and vice versa i.e. a single bond may be transferred into
lone pair.
30
1.6.4.6 Hydrogenation (reduction) of Alkenes
addition of HH molecule (Syn)
i.e.
C C
H
H
H2 is adsorbed on the
surface of catalyst.
catalyst
RCH
Catalyst
CHR\ + H2 e.g. Pd, Ni or Pt
RCH CHR\
H
H
syn addition
CH3
CH3
CH3
H2/Pd
H
i.e. same side
CH3
H
31
cis-1,2-Dimethylcyclopentane
1.6.5 Addition to Alkynes (H2, X2, HX and H2O)
A)
R
Reduction of Alkynes
H2/Catalyst
C
C
R C C
R
C
C
H
H
cis isomer (less stable)
syn addition
R
Li/NH3
(NH2Li + H)
B)
R
Halogenation
H
R
C
R
C
H
trans isomer (more stable)
anti addition
Br R
R Br
Br2/CCl4 Br
2
R
C C
another equiv.BrC
(one equiv)
R
Br
R
trans-isomer
anti addition
(recall 1.6.4.2)
CBr
Br
32
C) Hydrohalogenation of Alkynes
Br
R C CH
HBr
RC
Br
CH2 HBr R C CH3
D) Addition of water to Alkynes : Tautomerism Br
H O
R C C R
H2SO4/HgSO4
[H+/H2O]
R
C
C R
H
RCH
O
C R\
H
a ketone
an enol
i.e. the product is a mixture of the enol and keto forms.
Tautomer: Isomeric equilibrium of 2 structures that differ in the
arrangement of atoms.
The acidic water method is used in making a ketone from
an alkyne with the same # of C atoms.
33
1.6.6 Addition to Dienes (Electrophilic Addition)
CH2
1
CH
2
CH
3
CH2 HCl
one mole
4
[one equivalent]
Why?
Cl
CH3CHCH
CH2 + CH3 CH CHCH2Cl
1, 2-addition
1, 4-addition
Mechanism (as in 1.7.2.1)
CH2
CH
CH
CH2
H+
CH3
CH CH CH2
CH3 CH
Cl
1, 2-addition
CH CH2 Cl 1, 4-addition
34
1.6.6.1 Relative stabilities of conjugated and isolated dienes.
There are three types of Dienes; conjugated, cumulated, & isolated.
CH2
4
CH CH
3
2
s h orter th a n a s b o n d, h o wev er lo n g er
th a n d. b. Why?
CH2
1, 3-Butadiene (conjugated)
1
The most stable, lowest energy among other dienes.
The p orbital pictures of dienes.
H2C
CH
CH
CH2
H2C CH CH CH2
most of the time
i.e.
sometimes
H2C CH CH CH2
The Resonance Picture
H2C CH CH CH2
CH2
i.e. the 2 central carbon atoms are linked together by a
single bond that has some double bond character.
CH CH CH2
sp3 no p orbitals
CH2
5
CH CH2
4
3
CH CH2 1, 4-Pentadiene (isolated)
2
1
Less stable and of a higher energy than conjugated due to the
absence of resonance.
CH2 C CH2 1, 2-Propadiene (cumulated, allene)
3
2
1
Cumulated diene is the least stable (of highest energy) among
other dienes. Why?
Tutorial (molecular orbital picture)