# Regioselectivity in radical chlorination of a carboxylic acid

From an online textbook, the first step in the preparation of an amino acid:

Can we be sure that the chlorine will connect exactly at that spot, in the alpha-position? Won't we get an acyl chloride instead? Or a molecule where Cl is connected in the beta-position, or further down the line?

• By what mechanism would you expect to get the acid chloride? The $\sigma _{C-H}$ bond $\alpha$ to the carbonyl is going to be the weakest bond in terms of homolytic cleavage. – SendersReagent Apr 20 '16 at 17:52
• @SendersReagent could you please tell why that sigma bond will be the weakest one, generating a free H atom. – CowperKettle Apr 20 '16 at 18:06
• Sure, I'll put it in an answer since it will go into more detail than I feel is appropriate for a comment. – SendersReagent Apr 20 '16 at 18:10
• This turned out to be a pretty nice thread. Three answers approaching the problem in a different way each and equally useful. Favoriting. – M.A.R. Apr 20 '16 at 18:47
• @PhMgBr Agreed, and it's a perfect example of "Theory can only get you so far, what actually happens in the lab is most important." – SendersReagent Apr 20 '16 at 21:38

I am going to say: no, this does not work in practice. In fact it does not even work in theory.

In simple theory you would say that that the α C-H bond is the weakest because the resulting radical is stabilised by conjugation into the C=O. But under radical conditions, we don't observe α-chlorination in the lab. Is there a better way to explain this?

The problem is twofold. Firstly, the chlorine radical, $\ce{Cl.}$, is much less stable and therefore extremely unselective in radical chlorinations. In physical organic chemistry, we quote something called the Hammond postulate, which says that in this case, the transition state for hydrogen abstraction resembles the starting material more than the product. Therefore, the stability of the product does not significantly affect the activation energy, and the rate, of the hydrogen abstraction.

On top of that, it is what we would consider an "electrophilic radical"; it has a very low-energy SOMO, and therefore has a tendency to abstract a more electron-rich hydrogen atom. Essentially, it means that the radical behaves more like an electrophile than a nucleophile. In the transition state for hydrogen abstraction, the chlorine wants to have a significant build-up of negative charge on itself (remember, the product of hydrogen abstraction is going to be $\ce{HCl}$). It therefore goes for the hydrogen that is most willing to give it this negative charge, and that is the most electron-rich hydrogen atom.

## An example

Consider the radical chlorination of propionic acid, i.e. $\ce{R} = \ce{CH3}$ in your diagram above.

The α-hydrogen, compared to the β-hydrogen, is comparatively electron-poor because of the electron withdrawal by the $\ce{-COOH}$ group. So, it turns out that the rate of abstraction of the α-hydrogen is much slower: $k_\alpha/k_\beta = 0.03$ (Moody & Whitham, Reactive Intermediates, p 12). Here's an article which discusses the selectivity observed in the chlorination of propionic acid under different conditions: Tetrahedron, 1970, 26, 5929.

As written in Carey & Sundberg 5th ed., Part A, p 1022:

Radical chlorination shows a substantial polar effect. Positions substituted by EWG [electron-withdrawing groups] are relatively unreactive toward chlorination, even though the substituents are capable of stabilizing the radical intermediate. [...] Because the chlorine atom is highly reactive, the reaction is expected to have a very early TS [transition state] and the electrostatic effect predominates over the stabilizing effect of the substituent on the intermediate. The electrostatic effect is the dominant factor in the kinetic selectivity of the reaction and the relative stability of the radical intermediate has relatively little influence.

As an example, they use the chlorination of butyronitrile, which is the same as propionic acid except that the EWG is now $\ce{-CN}$ instead of $\ce{-COOH}$:

As @Loong mentions in the comments, in the specific case of propionic acid, there is a further problem: there are three β-hydrogens for it to abstract, and only two α-hydrogens. Therefore, even in the absence of everything else I have described, we would expect there to be 1.5 times as much chlorination at the β-position. This certainly does not help at all with the selectivity! The importance of statistical factors is most clearly seen in the radical chlorination of simple alkanes, where all the hydrogens are pretty much similar. I shan't go into detail here, but you can look at this answer (by Loong again), and there are examples on Clayden 2nd ed., pp 987-8.

## The solution

If you wanted to selectively chlorinate propionic acid at the α-position, you would use a polar mechanism that goes via the enol. The paper above has some examples of conditions that work. All of them involve suppressing the radical reaction by adding radical scavengers.

• The paper you cited uses $\ce{Cl2}$ under dark conditions. I'll look at it further, but can it be applied directly if the OP is referencing photolytic conditions? – SendersReagent Apr 20 '16 at 18:51
• @SendersReagent I don't feel like there would be a difference whether the $\ce{Cl.}$ is generated photolytically or via radical initiators. Once generated there is no reason to think that hv would make it prefer the alpha position. – orthocresol Apr 20 '16 at 18:53
• This paper might contain the information, but I can't read French: onlinelibrary.wiley.com/doi/10.1002/bscb.19520610705/abstract Reaxys says "Irradiation.im UV-Licht". I doubt the result is any different. – orthocresol Apr 20 '16 at 18:57
• @SendersReagent Well, I'm not 100% sure on whether the $\sigma(\ce{C-H})$ is affected. Although I personally don't see a reason for it to be, I don't profess to know everything so if there is I'd be happy to hear it :) – orthocresol Apr 20 '16 at 19:07
• Note that the reaction of the chlorine radical is also affected by a statistical factor since there are one acid-H, two α-H, and three β-H atoms in propionic acid. – Loong Apr 20 '16 at 19:25

Yes, we can :-D

Note the $h\nu$ in the reaction. It is a radical reaction. Irradiation will cleave the $\ce{Cl-Cl}$ bond and yield chlorine radicals ($\ce{Cl*})$. These will abstract hydrogen atoms (not protons) from the carboxylic acid to form a carbon centered radical (and $\ce{HCl}$). The favoured and position for this radical is the $\alpha$ carbon, where the radical is stabilized by interaction with the $\ce{-C=O}$ bond.

Update

Chances are even better in the presence of reagents that would facilitate the formation of an acid halide, as mentioned in the question. Catalytic amounts of phosphorus might be sufficient. This type of transformation, usually performed with bromine, is known as the Hell-Vollhardt-Zelinsky reaction.

• Debatable. Alkylradicals are electron-deficient, and carboxygroup is in general electron-withdrawing. I would ask a reference for this claim. – permeakra Apr 20 '16 at 20:43
• @permeakra In this case, the carbonyl wouldn't be electron withdrawing (as far as resonance is concerned). You would have $\pi _{C-O} \ce{->} \text{SOMO}_C$, I think? – SendersReagent Apr 20 '16 at 21:51

The $\sigma _{C-H}$ bond $\alpha$ to the carbonyl is going to be the weakest bond in terms of homolytic cleavage.

The $\ce{C-H}$ bond is weakened due to hyperconjugation. Hyperconjugation goes beyond the "intro organic" explanation of "adjacent $\ce{C-C}$ bonds stabilizing carbocations." Essentially, electron-rich $\sigma$ bonds, or bonds where the electrons are fairly central within the bond (in this case, the $\ce{C-H}$ bond) will donate electron density into an adjacent empty or partially-empty $\pi$ or $\pi ^\star$ orbital. This will strengthen the $\ce{\sigma _{C-C}}$ bond of the two adjacent carbons, but will weaken the $\ce{\sigma _{C-H}}$ by reducing electron density in its $\sigma$ bond and will also slightly lower the bond order of the carbonyl due to donation into $\pi ^\star _{C-O}$.

In the image below, you can see the molecular orbital diagram for hyperconjugation between a $\ce{C-H}$ bond and an adjacent alkene.

When looking at a carbonyl, this picture is essentially the same, except the orbital on the right would be ${\pi ^\star _{C-O}}$. Essentially, in the $\ce{H-C-C=O}$ system, the $\ce{H-C}$ and $\pi ^\star _{C-O}$ bonds are weakened/lengthened and the $\ce{C-C}$ bond is shortened and strengthened.

Because some electron density is lost from $\ce{\sigma _{C-H}}$, it becomes weaker and is, therefore, the easiest bond to break by the resultant chlorine radical.