Aldehydes and Ketones
By James Ashenhurst
The Simple Two-Step Pattern For Seven Key Reactions of Aldehydes and Ketones
Last updated: March 26th, 2019
“There are just so many reactions! I can’t remember all the mechanisms!!” – distressed organic chemistry student
Yes, yes there are a lot of reactions, particularly in second semester organic chemistry. But there is good news on this front: there is a tremendous amount of repetition in these reactions.
The Simple Two-Step Pattern For Seven Key Reactions Of Aldehydes And Ketones
For instance, what if I told you that there was a simple, two-step pattern behind seven different reactions that each work for aldehydes and ketones? By learning this key pattern, you’d therefore know the mechanism for 7*2 = 14 different reactions.
That would be useful, right?
The two steps are the following:
- Addition of a nucleophile to an aldehyde or ketone
- Protonation of the negatively charged oxygen with acid (often called “acidic workup”)
Here’s the general case for the reaction. I’ve drawn an aldehyde here, but everything I will say here also applies to ketones.
What bonds form, and what bonds break?
Hopefully you can see that a C–O (π) bond is being broken, a C–Nu bond is being formed, and an O–H bond is formed also.
Any mechanism we draw has to account for these bond-forming and bond-breaking events.
- Step 1 is addition of a nucleophile to the electrophilic carbonyl carbon. This forms C–Nu and breaks C–O (π), resulting in a negatively charged oxygen.
- Step 2 is addition of an acid (“protonation”), which results in formation of the O–H bond. This is generally done after the reaction with the nucleophile is complete – otherwise the acid would destroy the nucleophile, sometimes in violent fashion (e.g. LiAlH4 is not something you’d want to bring in close proximity to acid).
The General Mechanism
Here’s the general mechanism:
That’s it for the general example. Now let’s get to specifics.
This two-step pattern is behind the following seven reactions:
Again, although aldehydes are pictured here, the reaction applies equally well to ketones. So this represents fourteen reactions that proceed through this two step mechanism.
These types of mechanistic patterns are a little bit like Hollywood movies: there’s only so many different kinds of plot elements, and they repeat. If you’re familiar with the Hero’s Journey, you’ll recognize a lot of similarities between Star Wars: A New Hope and Happy Gilmore, even though the latter film is ostensibly about a hockey goon turned professional golfer. Likewise, the number of discrete mechanistic steps you will learn in organic chemistry could be counted on your fingers and toes.
Hope you find this useful.
Appendix: The Seven Reactions In Detail
Wait. You want specifics? Like, each reaction written out individually, with a general example, a specific example, and then a mechanism?
That sounds like overkill. But this is MOC. Overkill is what we do here.
Here’s each of those seven reactions treated individually.
1. The Grignard Reaction
The Grignard reaction is the addition of an organomagnesium compound to a carbonyl species. Recall that carbon is significantly more electronegative (2.5) than magnesium, so the partial negative charge is on carbon. In this example I used R-MgBr, although other halides (Cl, I) also work. Also, in the acid workup step I showed the spectator anion for H3O+ which is generally not necessary, but I like to balance the charges so you can see all the byproducts.
For our purposes, essentially the same as the Grignard reaction for aldehydes and ketones.
3. Reduction of Aldehydes and Ketones with Sodium Borohydride
In the borohydride anion (BH4–) it’s important to remember that hydrogen has a higher electronegativity (2.2) than boron (2.0). This means that although boron has the negative “formal” charge, the partial charges are on hydrogen. Hence, it’s the hydrogen that acts as a nucleophile [technically, “hydride” (H–) ].
The mechanistic pattern is the same – addition to carbonyl carbon, followed by protonation of oxygen.
In practice, reduction with NaBH4 is often run at low temperature with methanol as a solvent, with the subsequent workup step being addition of a mild acid such as NH4Cl to ensure full protonation of the alkoxide.
4. Reduction of Aldehydes and Ketones With LiAlH4
Everything I said above with respect to NaBH4 applies to LiAlH4 which is also a source of nucleophilic hydride. On paper, NaBH4 and LiAlH4 are equally effective in performing the reduction of an aldehyde or ketone to an alcohol. In practice, LiAlH4 is a much stronger reductant that will also reduce esters and carboxylic acids to alcohols. NaBH4 will not. Using LiAlH4 to reduce an aldehyde or ketone is like using a sledgehammer to kill a fly.
5. Addition of cyanide ion to aldehydes and ketones
Addition of cyanide ion (CN –) to aldehydes and ketones will result in a cyanohydrin. On paper, this also follows the two-step sequence of addition-protonation, although in practice the reaction can be run in the presence of a proton source such as H2O; unlike Grignards and some hydrides, cyanide ion is only weakly basic and will not be irreversibly destroyed by protonation. [In practice, however, care must be taken not to lower the pH too much; that may result in the formation of deadly HCN gas. ]
A related process, the Strecker synthesis of amino acids, begins with the addition of cyanide ion to an imine.
6. Addition of hydroxide ion to aldehydes to form hydrates (“geminal diols”)
Hydroxide ion will add to aldehydes or ketones to form hydrates, the mechanism of which also follows the two-step pattern. In practice, this doesn’t involve a separate workup step; hydroxide ion would be administered with at least some water as a co-solvent.
One thing to know about hydrates, however; they aren’t easily isolated, except for cases where the carbonyl is adjacent to an electron withdrawing group, such as in the case of chloral hydrate (a solid)
7. Addition of alkoxides to aldehydes and ketones to form hemiacetals
Last example. Addition of alkoxides to aldehydes and ketones will result in the formation of a hemiacetal.
There’s probably nobody reading by this point, but I would just remind those who are left of the tremendous importance of breaking down reactions into key steps of bonds formed/bonds broken and paying attention to how they build up into patterns. They save you a lot of work!