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Chapter 4: Alcohols and Alkyl Halides



Summary | Alcohols | Alky halides | Nucleophilic Substitution Reactions | Radical Substitution Reactions | Preparations of Alkyl Halides | Self Assessment | Quiz |


Radical Substitution

Chapter 4: Alcohols and Alkyl Halides

Radical Substitution

Radical substitution of alkanes using X2
Summary
Radicals
A radical is a species that contains unpaired electrons.
Typically formed by a homolytic bond cleavage as represented by the fishhook curved arrows:
fishhook arrows showing homolytic cleavage of Br2
Stability:
The general stability order of simple alkyl radicals is: (most stable) 3o > 2o > 1o > methyl (least stable)

radical stability order

This is because alkyl groups are weakly electron donating due to hyperconjugation and inductive effects.
Note that this is the same order as for carbocations. Resonance effects can further stabilize radicals when present.

Structure:
 

Alkyl radicals are sp2 hybridized, planar systems at the radical C center. 
The p-orbital that is not utilized in the hybrids contains the single electron. 
radical geometry

Reactivity:
As they have an incomplete octet, radicals are excellent electrophiles and react readily with nucleophiles.
Alternatively, loss of H. can generate a p bond

Rearrangements:
Unlike carbocations, radicals do not tend to undergo rearrangements

Reactions involving radicals: Radical halogention of alkanes

Radical Substitution Mechanism

Unlike the large majority of reactions that you will see in your organic chemistry course, radical mechanism require that fishhook curly arrows that represent the motion of a single electron are used. These can be a little more confusing and more difficult to master. We suggest you get to grips with normal curved arrows first.

However, the mechanism for the bromination of methane is shown below, but the mechanism for chlorination or higher alkanes in the same. Note that it contains three distinct type of steps, depending on the net change in the number of radicals that are present.
 

RADICAL CHAIN MECHANISM
FOR REACTION OF METHANE WITH Br2
Step 1 (Initiation)
Heat or uv light cause the weak halogen bond to undergo homolytic cleavage to generate two bromine radicals and starting the chain process.
Step 2 (Propagation)
(a) A bromine radical abstracts a hydrogen to form HBr and a methyl radical, then
(b) The methyl radical abstracts a bromine atom from another molecule of Br2 to form the methyl bromide product and another bromine radical,  which can then itself undergo reaction 2(a) creating a cycle that can repeat.
Step 3 (Termination)
Various reactions between the possible pairs of radicals allow for the formation of ethane, Br2 or the product, methyl bromide. These reactions remove radicals and do not perpetuate the cycle.

 

Selectivity

There are two components to understanding the selectivity of radical halogenations of alkanes:

R-H
The strength of the R-H varies slightly depending on whether the H is 1o, 2o or 3o. The following table shows the bond dissociation energy, that is the energy required to break the bond in a homolytic fashion, generating R and H.
 
Type
R-H
kJ/mol kcal/mol Note how the bonds get weaker as we move down the table, so the Ralso get easier to form, with 3o  being the easiest.
CH3-H
435
104
1o
CH3CH2-H
410
98
2o
(CH3)2CH-H
397
95
3o
(CH3)3C-H
380
91

Halogen radical, X.


The selectivity of the radical reactions can be predicted mathematically based on a combination of an experimentally determined reactivity factor, Ri, and a statistical factor, nHi. In order to use the equation shown below we need to look at our original alkane and look at each H in turn to see what product it would give if it were to be susbtituted. This is an exercise in recognizing different types of hydrogen, something that will be important later.
 

equation for calculating % yield from radical halogeantions
%Pi = % yield of product "i"
nHi = number of H of type "i"
Ri = reactivity factor for type "i"
Si = sum for all types 

Reactivity factors, Ri
Br
Cl
1o
1
1
2o
82
3.9
3o
1640
5.2

What do the reactivity factors indicate ? Well as an example of the conclusions we could make:

Lets work an example, say propane, CH3CH2CH3
How many different monochlorides can be produced by radical chlorination ?
1-chloropropane and 2 -chloropropane, so we have two types of H, the 6 x 1o in -CH3 and the 2 x 2o in-CH2- .
Use the diagrams below to highlight this if you are unsure.
1o CH
2o CH
Reset
propane
1-chloropropane
2-chloropropane

Now for the calculations, so plugging the values into the equations we get (the reactivity factors Ri are in the table above):

% 1-chloropropane = 100 x (6 x 1) / (6 x 1 + 2 x 3.9) = 100 x 6 / 13.8 = 43.5 %  (experimental = 44 %)

% 2-chloropropane = 100 x (2 x 3.9) / (6 x 1 + 2 x 3.9) = 100 x 7.8 / 13.8 = 56.5 % (experimental = 56 %)

What about bromination of propane ?

Most of the process is the same, all we have to do is change the reactivty factors.

% 1-bromopropane = 100 x (6 x 1) / (6 x 1 + 2 x 82) = 100 x 6 / 170 = 3.5 %  (experimental = 4 %)

% 2-bromoopropane = 100 x (2 x 82) / (6 x 1 + 2 x 82) = 100 x 164 / 170 = 96.5 % (experimental = 96 %)

Note that the results match well with experimental values and that they illustrate the high regioselectivity of the bromination reaction for the 2o radical, whereas in the chlorination the number of 1o H dictates the regioslectivity.

There are other examples in the Self Assessment.

Radical Halogenation of Alkanes

Radical substitution of alkanes using X2
Reaction type: Radical Substitution

Summary:

RADICAL CHAIN MECHANISM
FOR REACTION OF METHANE WITH Br2
Step 1 (Initiation)
Heat or uv light cause the weak halogen bond to undergo homolytic cleavage to generate two bromine radicals and starting the chain process.
Step 2 (Propagation)
(a) A bromine radical abstracts a hydrogen to form HBr and a methyl radical, then 
(b) The methyl radical abstracts a bromine atom from another molecule of Br2 to form the methyl bromide product and another bromine radical,  which can then itself undergo reaction 2(a) creating a cycle that can repeat.
Step 3 (Termination)
Various reactions between the possible pairs of radicals allow for the formation of ethane, Br2 or the product, methyl bromide. These reactions remove radicals and do not perpetuate the cycle.

More highly brominated by-products are possible if methyl bromide reacts with a bromine radical in the same fashion as methane does. Can you draw the cycle that leads to the formation of dibromomethane ?

 

 



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