Free Radical Reactions
Selectivity In Free Radical Reactions
Last updated: December 7th, 2022 |
Free Radical Chlorination: Selectivity
This post is all about the selectivity of free-radical halogenation: what does “selectivity” mean, anyway? And how do we calculate it? It’s often said that chlorination is less “selective” than bromination. What does that mean? (More on that topic in the next post).
Table of Contents
- Chlorination Of Propane Does Not Lead To A Statistical Mixture Of Products. Why Not?
- Secondary C–H Bonds Are Weaker Than Primary C–H Bonds, And Result In More Stable Free Radicals
- How To Quantify “Selectivity” In Free-Radical Chlorination
- A Puzzle: What Is The Selectivity For 2-Bromopropane In This Reaction?
- (Advanced) References and Further Reading
1. Chlorination Of Propane Does Not Lead To A “Statistical” Mixture Of 1-Chloropropane And 2-Chloropropane. Why Not?
In the last post we showed some examples of how different isomers might be formed in free-radical halogenation of alkanes. And I left off with a question.
If we performed a free-radical chlorination on propane with 1 equiv Cl2 under normal conditions (25°C, initiated with light), what be the expected ratio of 1-chloropropane and 2-chloropropane?
Notice that there are six primary (“methyl”) hydrogens (removal of any of which would give rise to 1-chloropropane) and two secondary (“methylene”) hydrogens. That’s a 3:1 ratio. IF halogenation was statistically random, we would therefore expect a 75:25 ratio of 1-chloropropane to 2-chloropropane.
That’s not what happens!
Instead, experiment tells us that we obtain a 55:45 ratio of 2-chloropropane to 1-chloropropane. So clearly there is more to this product distribution than pure randomness.
Why might this be?
Recall our earlier discussion of free radical stabilities. We observed that free radicals increase in stability as the number of carbon substituents increases, from methyl (least stable) < primary < secondary < tertiary (most stable).
As we’d also mentioned earlier, this is also the ratio of C–H bond strengths: CH4 > R–CH3 > R–CH2–R > R3C–H . The more stable the free radical that is left behind, the weaker will be the C–H bond strength.
2. Secondary C–H Bonds Are Weaker Than Primary C–H Bonds. Breaking A Secondary C–H Bond Results In A More Stable Free Radical
Let’s turn to propane.
In the free radical chlorination reaction, chlorine radical may abstract a hydrogen from either from one of the methyl groups of propane, or from the methylene.
All else being equal, which of these should be the easiest process?
Removing the secondary hydrogen, giving us a secondary free radical!
It’s for this reason that C–H bonds are not broken randomly (and hence we don’t get a 75:25 ratio). The stability of the free radical we create is a very important factor.
3. How To Quantify “Selectivity” In Free Radical Chlorination
This deviation from a statistical mixture is referred to as selectivity. We say here that chlorination is more selective for the secondary carbon than for the primary carbon.
So how do we quantify selectivity?
If there were equal numbers of methylene (CH2) hydrogens and methyl (CH3) hydrogens, it would be simple: 55:45 in this case, or 1.22 favoring methylene.
However, it’s not quite that simple. We have to adjust for the statistical bias in favor of the methyl groups.
Recall that there are three times as many methyl hydrogens as methylene hydrogens. In order to make this a true ratio, we need to divide the number we obtained for methyl by 3. This gives us a ratio of 55:15 , or 3.66.
So for this reaction, the free radical chlorination of propane at 25°C, chlorine is 3.66 times more selective for secondary hydrogens than for primary hydrogens.
If we were examining a reaction where three different types of hydrogens were present (or even more) we would likewise adjust each yield by a factor equal to the number of hydrogens of each type, and then compare directly.
4. A Puzzle: What Is The Selectivity For 2-Bromopropane In This Reaction?
Let’s leave off with yet another puzzle. Look at the exact same reaction, except using Br2 instead of Cl2 . Here, we observe a ratio of 97% 2-bromopropane to 1-bromopropane.
What’s the selectivity for 2-bromopropane in this reaction?
And why might the selectivity for bromine be much higher than that for chlorine?
Answer next time!
Next Post: Selectivity In Free Radical Reactions – Bromine vs. Chlorine
(Advanced) References and Further Reading
- The unusual and the unexpected in an old reaction. The photochlorination of alkanes with molecular chlorine in solution
K. U. Ingold, J. Lusztyk, and K. D. Raner
Accounts of Chemical Research 1990, 23 (7), 219-225
This account has a lot of useful information on the selectivity of the reaction of chlorine radicals with organic substrates. K. U. Ingold was the son of eminent chemist Prof. C. K. Ingold.
- Syntheses from Natural Gas Hydrocarbons Identity of Monochlorides from Chlorination of Simpler Paraffins
H. B. Hass, E. T. McBee, and Paul Weber
Industrial & Engineering Chemistry 1935, 27 (10), 1190-1195
One of the oldest papers on free-radical chlorination of hydrocarbons. Interestingly, the authors state, “Carbon skeleton rearrangements do not occur during either photochemical or thermal chlorinations if pyrolysis temperatures are avoided; every possible monochloride derivable without such rearrangement is always formed”.
- The Photochlorination of 2-Methylpropane-2-d and α-d1-Toluene; the Question of Free Radical Rearrangement or Exchange in Substitution Reactions
Herbert C. Brown and Glen A. Russell
Journal of the American Chemical Society 1952, 74 (16), 3995-3998
This paper by Nobel Laureate H. C. Brown quantifies the reactivity of a H vs. a D atom in the same position in free-radical chlorination. As he says, “By comparison of the relative reactivity of deuterium and hydrogen atoms of 2-methylpropane-2-d (1.00:2.5) and of the tertiary and primary hydrogen atoms of isobutane (1.0:4.5), a tertiary deuterium and tertiary hydrogen atom of isobutane have a relative reactivity of 1.0: 1.8 measured in the photochlorination reaction at – 15°”.
- The Liquid Phase Photochlorination and Sulfuryl Chloride Chlorination of Branched-chain Hydrocarbons; the Effect of Structure on the Relative Reactivities of Tertiary Hydrogen in Free Radical Chlorinations
Glen A. Russell and Herbert C. Brown
Journal of the American Chemical Society 1955, 77 (15), 4031-4035
Sulfuryl chloride (SO2Cl2) can also be used as a chlorinating agent and reacts with alkyl radicals in a chain reaction. The SO2Cl radical is less reactive than the Cl radical and is therefore more selective.
- The Competitive Halogenation of Cyclohexane and Aralkyl Hydrocarbons; Evidence as to the Nature of the Transition States in Halogenation Reactions
Glen A. Russell and Herbert C. Brown
Journal of the American Chemical Society 1955, 77 (17), 4578-4582
The introduction states “toluene is 60 times as reactive as cyclohexane toward bromine atoms, whereas toward chlorine atoms cyclohexane is 11 times as reactive as toluene”. Prof. Brown explains this using Hammond’s postulate – bromination of toluene involves a late transition state (closer to the benzyl radical), and chlorination of toluene an early transition state, where the C-H bond is still mostly intact.
- Electron transference in the attack of atoms or radicals upon carbon-hydrogen bonds
G. A. Russell
Tetrahedron 1959, 5 (1), 101-104
This paper makes a similar argument to ref. #5. More reactive radicals (i.e. those with higher electron affinities) will abstract the most electron-rich hydrogen atom in the molecule, and the transition state will be stabilized by ‘electron transference’. Less reactive radicals are more selective, and the transition state will resemble the radical intermediate more, so stability of the resulting radical dictates reaction path.
- Highly selective radicals. Chlorovinylation of hexane
Frederick F. Rust and Charles S. Bell
Journal of the American Chemical Society 1970, 92 (18), 5530-5531
n-hexane reacts with trichloroethylene in a radical chain reaction with high selectivity – over 80% of the reaction is at the 2-position.
- Chlorination of 2,3-dimethylbutane: A quantitative organic chemistry experiment
Journal of Chemical Education 1969, 46 (9), 610
This paper describes an experiment suitable for undergraduate organic chemistry laboratory sections. Chlorine radicals can be generated from SO2Cl2 or NCS (N-chlorosuccinimide) and analysis of the product distribution from reaction with 2,3-dimethylbutane, which has multiple types of hydrogen atoms, demonstrates the selectivity for tertiary over primary sites.
8 thoughts on “Selectivity In Free Radical Reactions”
Which secondary hydrogen is replaced by chlorine in case of pentane?
You would get mixtures of replacement at C2, C3, and C4. There are twice as many hydrogens that would generate 2-chloropentane than would generate 3-chloropentane, so your dominant product would be 2-chloropentane, but you’d get a mixture of 1-chlropentane, 2-chloropentane, and 3-chloropentane.
why free radical reactions are non-selective?
Chlorination is relatively non-selective, but bromination is quite selective. See this post. https://www.masterorganicchemistry.com/2013/10/31/selectivity-in-free-radical-reactions-bromine-vs-chlorine/
For the chlorination of propane in the presence of u.v light, what is the name of the longest possible
non-chlorinated alkane product that can form (as a side product).
Think about possible termination steps.
Is it propane?