Free Radical Chemistry and the Preparation of Alkyl Halides (Ch. 10, Part 1)
Introduction to Free Radical reactions (section 5.3)
Homolytic bond cleavage, each atom gets one of the bonding electrons
Both are left with an odd electron:
A:B
A. + B.
= “free radicals”
Since free radicals are reactive, further reactions will occur (chain reaction)
Types of steps in the mechanism of a radical substitution reaction:
Initiation:
Formation of the initial radical, usually by application of energy to
a compound with a weak or unstable σ bond:
hν or ∆
2 Cl
Examples: Cl2
Propagation:
Ex:
Cl .
.
Ex:
Cl .
2 . OH
HO – OH
Step produces a new radical plus a substition product
Occurs when a radical collides with a stable molecule
Many different propagation steps possible in each mechanism,
so different products may form during these steps
+
CH3 +
Termination:
hν or ∆
.
H – CH3
HCl +
.
Cl – Cl
CH3Cl
+
CH3
Cl .
2 radicals combine to form a stable bond
+
.
CH3
CH3Cl
Radical halogenation is a common reaction used to make alkyl halides
“Monochlorination” = replacement of a single H in an alkane with Cl
Problem: What are the monochlorination products of 2-methylpentane?
Radical halogenation of alkanes: a closer look at product isomer distribution
Cl2
Simple case:
CH4
CH3Cl
After initiation step, Cl . radical collides with a methane molecule and breaks bond
Cl . +
.
CH3 +
H - CH3
Cl - Cl
HCl +
CH3Cl
.
CH3 (alkyl radical)
+
Cl .
With larger alkanes, there are several positions where loss of H . could occur
Products of free-radical halogenations therefore can be mixtures of isomers
Key consideration: Are all positions equally likely to react to form radicals?
Example: 3-methylpentane
Factors to consider:
• Number of hydrogens at each position
• Relative stabilities of each type of radical
Reactivity factors:
The relative likelihood of radical formation at a given
position in the molecule (1o, 2o or 3o)
The relative stabilities of alkyl radicals have the same trend as carbocations:
.
.
.
R – CR – R >
R – CH – R > R – CH2 >
· CH3
3o
2o
1o
methyl
So at room temperature, the relative reactivities of C – H toward Cl .
depend on structure, and are given below:
3o (5.0)
>
2o (3.5)
>
1o (1.0)
To determine the relative amounts of each isomer product that forms:
Relative amount 1o = (number of 1o H) x (reactivity factor for 1o)
Relative amount 2o = (number of 2o H) x (reactivity factor for 2o)
Relative amount 3o = (number of 3o H) x (reactivity factor for 3o)
Divide each relative amount by the sum of all isomers (x 100%) to obtain the
product distribution in the form of a percentage.
For 3-methylpentane, possible products:
Calculation of percent distribution for each one:
Food for thought: Why is radical chlorination an effect way to prepare
chlorocyclopentane?
Bromination vs. Chlorination
Reactivity factors for Br . (at 125oC):
1600 :
82
:
1
The selectivity of Br . is much higher than Cl . but the reactivity is less
Because the abstraction of H by Br . is endothermic, the transition states
resemble the radical. So radical stability has a larger effect on reactivity.
What's the percent distribution of products of bromination of 3-methylpentane?
2. Radical halogenation of alkenes as a route to alkyl halides (RX)
A. Use of peroxide for primary alkyl halides (note: reaction is not in McMurry)
Bromine free radicals are generated by HBr in the presence of H2O2
These will react with alkenes to give non-Markovnikov products:
Ex:
CH3 – CH = CH2
HBr, peroxide
CH3 – CH2– CH2Br
Mechanism: Free radical reacts with π-bonding e- of alkene, forming alkyl
radical:
light
Initiation:
H–O–O–H
2 H – O·
Propagation:
·Br
Q.
A.
H – O· + H – Br
+
CH2 = CH – CH3
·Br + H2O
.
Br – CH2 – CH – CH3
Why doesn't · CH2 - CHBr - CH3 form?
Consider relative stability of 1o, 2o, 3o radicals
The more stable alkyl radical forms, then reacts with HBr to produce RX:
.
Br – CH2 – CH – CH3 +
H – Br
Br – CH2 – CH2 – CH3 + · Br
General considerations:
• Addition occurs in a "non-Markovnikov" fashion: the electrophile ends up
on the more substituted carbon.
• Reaction is useful for preparing primary alkyl bromides
Some conclusions about alkyl halide preparation:
 Any of the C – H bonds in an alkane can be replaced with C - X
 To prepare 3o alkyl halides from alkanes, bromination is most effective;
far more 3o product, less 1o and 2o are formed
 To prepare 1o alkyl halides use an alkene rather than an alkane, add
peroxide with HBr
B. Allylic Radical Bromination: Use of NBS
When the radical is on an allylic C, adjacent to a C=C, resonance stabilizes it:
.
.
H2C = CH – CH – R
H2C – CH = CH – R
Allylic radicals form easily; bond energy of allylic C – H < alkyl C – H
N-bromosuccinimide (NBS) can be used to substitute Br into allylic position
O
H
H
H
H
Br
N Br
Br 2
NBS
O
Resonance means that more than one product is possible as electrons rearrange.
H 3C
CH2
NBS
H 3C
CH2
Br
+
CH2
H 3C
Br
Similarly, NBS can be used to brominate a benzylic carbon (adjacent to benzene):
Predict the products of the reaction of each with NBS:
(c)
Polar substitution reactions using alkyl halides (RX) - Ch. 10, Part 2
Remember that C – X bonds are polar: key to polar reactions of alkyl halides
Transforming alcohols to alkyl halides by polar substitution (10.6)
Alkyl halides and alcohols have in common a polar bond between C and functional group:
R – CH2 – Br
R – CH2 – Cl
R – CH2 – OH
The OH group of alcohols is a poor leaving group compared to Br, Cl, or I so it is harder to
replace - we'll see in Ch. 11 that halides are easy to replace
Methods for preparation of 3o alkyl halides
- Radical substitution at the 3o C of an alkane
- Reaction of 3o alcohols with HCl, HBr or HI
Polar substitution by SN1 mechanism (Ch. 11)
will form 3o alkyl halides (as shown at right)
This is unlikely with 1o or 2o ROH due to
unstable intermediate
CH 3
OH
HBr
ether
Preparation of 1o and 2o alkyl halides: SN2 substitution of alcohols:
•
•
1o & 2o alcohols require a bit more “encouragement” to get rid of the OH group
Halogenating reagents help “activate” the removal of the OH group
• These reagents reduce the risk of undesired elimination reactions.
• Reactions proceed with inversion
CH 3
Br
Some advantages to preparing alkyl halides from alcohols
• The alcohols are often inexpensive and readily available
• "Targeted" placement of the halide group where the -OH group was
• These reactions usually produce only the desired major product, not
mixtures like you would get by radical halogenation
•
•
•
Use of Organometallic Reagents in Coupling Reactions
Concept: Metals can “activate” carbon towards SN2 substitution
 Alkyl halides have a partial positive charge on the C
 Reaction of alkyl halides with certain metals results in “insertion” of metal
 In the new carbon-metal bond, the carbon bears a partial negative charge
Grignard reagent: R3C - Mg - Br
Gilman reagent: R2 CuLi
 The carbon becomes nucleophilic
 An earlier example: Nucleophilic C is generated when acetylene is
deprotonated to make acetylide ions (R-C=C:-)
 Like acetylides, organometallics react with R-X to form new C-C bond
Grignard Reagents (10.7)
•
•
•
Used to form new C – C bonds
Used in reactions with carbonyl compounds (more on this in CHM252)
Used in substitution reactions with alcohols to make ethers
Magnesium inserts itself into the C – X bond of 1o, 2o or 3o alkyl halides:
CH3 – Br
+
Mg
CH3 – Mg – Br
Dry ether
bromomethane
•
Methylmagnesium bromide
(a Grignard reagent)
Grignard reagents react readily with any electrophile which makes them useful,
but unwanted side reactions can occur with any source of H+
Since they behave like bases, they react readily with H2O to replace X with H:
CH3CH2CH2-Mg-Br
•
H2O, H+
CH3CH2CH3
+ MgBrOH
They also react with R-X in coupling reactions:
CH3MgBr +
CH3CH2CH2Br
CH3CH2CH2CH3
Organolithium/copper reagents: Gilman reagents (10.8)
Lithium is small and strongly electropositive. It can replace a halogen atom, turning the
carbon atom directly bonded to it strongly basic
2 Li
CH3CH2CH2CH2-Br
CH3CH2CH2CH2–Li + LiBr
Pentane
•
•
•
The resulting organolithium reagents are very strong nucleophiles
They can be used in many of the same reactions as Grignard reagents
RLi reagents are used in the preparation of the Gilman reagent, a milder nucleophile
Gilman reagents:
Further reaction of RLi with copper produces R2CuLi
ether
2 CH3CH2Li + CuI
(CH3CH2)2CuLi
+
LiI
A Gilman reagent
Use of a Gilman reagent in a coupling reaction with an alkyl halide:
Alkyl bromides, iodides or chlorides can react with Gilman or Grignard reagents
in carbon-skeleton building substitution reactions (SN2 mechanism, Ch. 11)
Alkyl halides and organic oxidation and reduction (10.9):
Formation of Grignard or Gilman reagents can be considered an organic
reduction since the electron density on the carbon atom increases.
Reductions increase the e- density on C by forming C – H, C – M (metal)
bonds or by breaking C – O, C – N, or C – X bonds
Oxidations decrease e- density on C (formation of C-O, C-N or C-X bonds)
so reactions that form alkyl halides (R-H to R-X) are oxidations
Choice of halogens in making & using alkyl halides in substitution rxns
• Our focus has been primarily on alkyl bromides, chlorides & iodides
because they contain a good leaving group and are readily available
commercially
• Alkyl fluorides are not readily prepared by the same methods; for example,
fluorine radicals are extremely reactive & unpredictable so radical
fluorination is impractical
• Alkyl fluorides also are less useful as starting reagents; F is a poor leaving
group and thus not easily substituted out
Fill in the missing products:
Identify the missing reagents