Why does basic peroxide react via conjugate addition rather than through direct addition?

The following reaction occurs:

The $\ce{HOO^{-}}$ ion reacts as a nucleophile at carbon number three and consequently the reaction is conjugate addition. However, Why would a peroxide not react through direct addition. I know that if the double bond wasn't there it would attack the carbonyl directly and then a rearrangement would occur (Baeyer-Villiger reaction: http://en.wikipedia.org/wiki/Baeyer%E2%80%93Villiger_oxidation). Why must it react at the alkene?

• I have no way to prove that, but i think there is a chance that the ketone forms a hydrate, deprotonated or not, blocking the most electrophilic platoon. Hence there is only the beta carbon left. Mar 13 '15 at 20:46

The reason that is given by Clayden et al. (chapter 23) is that because the peroxide has two adjecant lone pair on oxygen it exhibits an alpha effect, which is more or less a delocalization interaction between the lone pair orbitals. For reasons not entirely known (the basis of the alpha effect is an open problem in chemistry) this results in an increase in energy of the electrons in the resulting delocalized HOMO, making it more nucleophilic. You can imagine that this delocalized HOMO is also softer than the single localized oxygen lone pair and hence it tends to react at the soft double bond instead of the hard carbonyl site.

Excerpt from Clayden et al.

• Your answer is contradictory. When the peroxide ion is more nucleophilic, why wouldn't it primarily react at the carbon with the highest positive charge? I understand that the alpha effect is not yet elucidated, but how and in what way does Clayden explain the stereoselectivity based on increased nucleophilicity, while it reacts at the very opposite centre? Mar 13 '15 at 20:21
• @Martin Nucleophilicity has nothing to do with it, I just mentioned that because it is at the core of the alpha effect. I attached the excerpt from the book. Also, why would a better nucleophile attack more easily at the carbonyl group? I don't see the relation. Iodine is a good nucleophile, yet rather soft. A bit harsh to immediately downvote without giving me time to comment. I'm just trying to help and answer questions and I think I used an OK source. But ok, it's your vote...
– Jori
Mar 13 '15 at 23:49
• A downvote is just a downvote, it is nothing personal. Also your source is of course ok. The orbital picture, however, I do not agree with. It is a massive stretch to describe any oxygen orbital as sp3, and then there is the problem, that the depicted HOMO has no pi symmetry. I also would not be too sure about that the pi orbital is the HOMO. But it is the weekend now and i don't want to think about chemistry anymore. Mar 14 '15 at 5:59
• @Martin No problem. It was a bit late and I was a little agitated and then seeing those stingy red -2 :) . Efin, there is only so much I can do with an undergraduate text which doesn't cover MO theory thoroughly. This isn't the first time you come and complain about some MO theory effect that I do not understand haha. Could you direct me to some books that cover MO theory more extensively? Anyway, when I have some time next week I'll ask Clayden about it, see what he thinks.
– Jori
Mar 14 '15 at 11:42
• I really think the Clayden is an excellent organic chemistry book, I like it a lot. I know that he is more thorough with MO theory than most other books, and when it comes to the alpha effect, it is absolutely understandable that things get fuzzy. We have a brilliant question on the network reminding us about the very discourse we have with the whole matter: chemistry.stackexchange.com/q/7460/4945 I also removed my vote, the context you posted certainly helped a lot with understanding what is actually meant. (I still don't quite agree, but let's save that for another time.) Mar 16 '15 at 14:41

Why would a peroxide not react through direct addition […]

Nobody says that it doesn't, but apparently that isn't a productive route: there are no products found that would result from such an attack. Consequently, this reaction, which cannot be ruled out, probably is reversible.

I know that if the double bond wasn't there it would attack the carbonyl directly and then a rearrangement would occur […]

True indeed, but the double bond is there.

As a result, attack on the $$\beta$$-carbon of the enone, formation of a hydroperoxy enolate and subsequent formation of the epoxide are seemingly favoured.

• What chemical drawing program do you use? I really like the drawing style.
– Jori
Mar 11 '15 at 23:41
• @Jori I use BKChem under Linux (Ubuntu 14.04), installed from the repositories. In the settings, I enabled automatic cropping of the SVG drawings. For Chemistry.SE, I export to transparent PNG (Cairo engine). Mar 12 '15 at 5:32
• You might want to switch to a non-transparent background, I have experienced that the image hosting platform of SE likes to misinterpret the alpha channel and making the background black instead. Mar 13 '15 at 20:29
• @Martin For small line drawing like the one in my answer, imgur seems to work flawless and does not alter the file. A diff on the hexdump of the original png and the file on imgur shows no difference. It is however true that larger files are compressed and/or converted to jpg. Mar 14 '15 at 6:57

First off, it may be useful to go back and take a look at this earlier question of yours and the answer to it. The concept of hard and soft electrophiles and nucleophiles was discussed. Basically, it was said that

hard nucleophiles are generally small, highly charged atoms. Because of this they tend to be more reactive. They prefer to react with similar (hard) electrophiles. Their reactions are usually kinetically controlled with early transition states and are governed by electrostatic interactions. Soft nucleophiles are generally larger systems with a more diffuse charge. Because of this they tend to be less reactive (more discriminating). They prefer to react with similar (soft) electrophiles

The "take away" is that a hard electrophile prefers to react with a hard nucleophile while a soft electrophile prefers to react with a soft nucleophile.

Let's examine the system at hand, the nucleophile is the hydroperoxide anion ($\ce{HOO^{-}}$). You can draw two resonance structures to describe the anion, delocalizing the negative charge over both oxygens.

$$\ce{H-O-O^{-} <-> H^{+}~ [O-O]^{-2}}$$

Therefore, this anion is a softer nucleophile than $\ce{HO^{-}}$

The carbonyl group is a hard electrophile, while an $\ce{\alpha,\beta}$ unsaturated carbonyl, where the electron density is distributed over 4 atoms is a softer electrophile.

Soft reacts with soft, so would you expect the soft hydroperoxide anion to preferrentially react with a hard (carbonyl; 1,2 addition) or soft ($\ce{\alpha,\beta}$ unsaturated carbonyl; 1,4 addition) electrophile?

• @Jori I've never really understood the basis of the alpha effect, so if you'd like to write up an answer based on it please go ahead.
– ron
Mar 11 '15 at 22:24