Transcript Document

What is the major product of the following reaction?
Predict the major product of the following reaction:
2-methylbutane + Br2/light energy ?
A) 1-bromo-2-methylbutane
B) 2-bromo-2-methylbutane
C) 2-bromo-3-methylbutane
D) 1-bromo-3-methylbutane
CH3
H3C
Br
H2
C CH3
Which of the following alkanes will give more than one
monochlorination product upon treatment with chlorine and light?
A) 2,2-dimethylpropane
B) cyclopropane
C) ethane
D) 2,3-dimethylbutane
CH3CH3
H3C
CH3CH3
Cl2
CH3
hv
H3C
CH3
+
Cl
CH3CH3
H3C
CH2Cl
Which of the following is the major product of the chlorination of
methane if a large excess of methane is used?
A) CH3Cl
B) CH2Cl2
C) CH3CH2Cl
D) CCl4
Consider the reaction of 2-methylpropane with a halogen. With
which halogen will the product be almost exclusively 2-halo-2methylpropane?
A) F2
B) Cl2
C) Br2
D) I2
CH3
H3C
CH3
Br2
hv
CH3
H3C
CH3
Br
Cl2
hv
CH3
+
CH3 ClH
2C
H3C
Cl
CH3
CH3
The alllyl carbon is the carbon adjacent to a double bond.
The allylic radical is more stable than a 3 radical.
•Because allylic C—H bonds are weaker than other sp3 hybridized
C—H bonds, the allylic carbon can be selectively halogenated
using NBS in the presence of light or peroxides.
•NBS contains a weak N—Br bond that is homolytically cleaved
with light to generate a bromine radical, initiating an allylic
halogenation reaction.
•Propagation then consists of the usual two steps of radical
halogenation.
•NBS also generates a low concentration of Br2 needed
in the second chain propagation step (Step [3] of the
mechanism).
•The HBr formed in Step [2] reacts with NBS to form Br2,
which is then used for halogenation in Step [3] of the
mechanism.
Thus, an alkene with allylic C—H bonds undergoes two
different reactions depending on the reaction conditions.
How is the vicinal dibromide formed?
•Halogens add to  bonds because halogens are polarizable.
•The electron rich double bond induces a dipole in an approaching
halogen molecule, making one halogen atom electron deficient and
the other electron rich (X+—X–).
•The electrophilic halogen atom is then attracted to the nucleophilic
double bond, making addition possible.
Why does a low concentration of Br2 (from NBS) favor allylic
substitution (over ionic addition to form the dibromide)?
•The key to getting substitution is to have a low concentration of
bromine (Br2).
•The Br2 produced from NBS is present in very low
concentrations.
•A low concentration of Br2 would first react with the double bond
to form a low concentration of the bridged bromonium ion.
•The bridged bromonium ion must then react with more bromine
(in the form of Br¯) in a second step to form the dibromide.
•If concentrations of both intermediates—the bromonium ion and
Br¯ are low (as is the case here), the overall rate of addition is
very slow, and the products of the very fast and facile radical
chain reaction predominate.
Predict the products.
Br
NBS
hv
H2 C
C
H
CH3
Br2
H2C
H
C
CH3
Br Br
hv
NBS
H2C
C
H
CH2Br
Halogenation at an allylic carbon often results in a mixture of
products. Consider the following example:
•A mixture results because the reaction proceeds by way of a
resonance stabilized radical.
Predict the products.
H3C
C
H
C
H
CH3
NBS
BrH2C
hv
C
H
C
H
H2C
C
H
CH3 +
H
C
CH3
Br
Br
H2C
H2 C
C
H2 C
CH3
CH3
NBS
hv
CH CH3
H2C
C
+
H2C
CH3
BrH2C
HC
C
H2C
CH3
CH3
NBS
CH2
CH2 +
hv
Br
CH2Br
Radical Additions to Double Bonds
•HBr adds to alkenes to form alkyl bromides in the presence of
heat, light, or peroxides.
•The regioselectivity of the addition to unsymmetrical alkenes is
different from that in addition of HBr in the absence of heat,
light or peroxides.
•The addition of HBr to alkenes in the presence of heat, light or
peroxides proceeds via a radical mechanism.
Hydrohalogenation—Electrophilic Addition of HX
•The mechanism of electrophilic addition consists of two successive
Lewis acid-base reactions. In step 1, the alkene is the Lewis base
that donates an electron pair to H—Br, the Lewis acid, while in step
2, Br¯ is the Lewis base that donates an electron pair to the
carbocation, the Lewis acid.
•With an unsymmetrical alkene, HX can add to the double
bond to give two constitutional isomers, but only one is
actually formed:
•This is a specific example of a general trend called
Markovnikov’s rule.
•Markovnikov’s rule states that in the addition of HX to an
unsymmetrical alkene, the H atom adds to the less
substituted carbon atom—that is, the carbon that has the
greater number of H atoms to begin with.
•Note that in the first propagation step, the addition of Br•
to the double bond, there are two possible paths:
1.Path [A] forms the less stable 1° radical.
2.Path [B] forms the more stable 2° radical.
•The more stable 2° radical forms faster, so Path [B] is
preferred.
•The radical mechanism illustrates why the regioselectivity of HBr addition is different depending on the
reaction conditions.
•HBr adds to alkenes under radical conditions, but HCl and HI do
not. This can be explained by considering the energetics of the
reactions using bond dissociation energies.
•Both propagation steps for HBr addition are exothermic, so
propagation is exothermic (energetically favorable) overall.
•For addition of HCl or HI, one of the chain propagating steps is
quite endothermic, and thus too difficult to be part of a repeating
chain mechanism.
H3C
C
H
HBr
CH2
H3C
H
C
CH3
Br
H3C
C
H
C
CH3
HBr
ROOR
CH3
HCl
hv
H3 C
CH3
H
C CH
CH3
Br
Cl