Here’s the summary for today’s post:
So far in this series we’ve introduced organometallic compounds and said that their carbons tend to be nucleophilic. We’ve learned how to make them from alkyl, alkenyl or aryl halides (along with some ways not to make them!) and saw that they are very strong bases.
So what now? Well, we want to know what we can actually DO with these things, right?
So let’s get started!
Reminder: Grignard Reagents Are Nucleophiles
I know we’ve said this many times in this series already, but it bears repeating: Grignard reagents are nucleophiles, meaning that they (specifically the carbon bound to Mg) will readily donate a pair of electrons to an appropriate electrophile.
We’ve already seen them react with weak acids, where the electrophile is electrophilic hydrogen (H+ ).
More interesting and useful for us is a variety of carbon based electrophiles, because we can thus combine Grignard reagents with those species to form new carbon-carbon bonds. And since carbon-carbon bonds constitute the “backbone” of molecules in organic chemistry, it turns out that this class of reactions is very useful. As a matter of fact, it won its discoverer, Victor Grignard, the Nobel Prize for Chemistry back in 1912.
For our purposes, the key carbon-based electrophiles that Grignard reagents react with are epoxides, aldehydes, ketones, and esters. Let’s go through them in turn.
1. Addition of Grignard Reagents To Epoxides
Epoxides (“oxiranes” if you are an IUPAC stickler) are 3-membered cyclic ethers which possess considerable ring strain. As we’ve seen, this ring strain makes them somewhat “spring loaded” toward attack by nucleophiles, which will result in formation of a new bond to carbon and opening of the ring.
Negatively charged nucleophiles (such as Grignards) tend to react with epoxides in a manner similar to the SN2 reaction: attack occurs at the least substituted carbon of the epoxide. Here’s an example:
Note the bonds that formed and broke here: we formed a new C-C bond (between carbons A and B), and broke a C-O bond (between carbon A and the oxygen). This resulted in a negatively charged oxygen (alkoxide): to produce final alcohol product, we typically quench the reaction with a source of acid, forming O–H.
Here’s how the reaction works. The hard thing is to recognize that the nucleophile is the pair of electrons in the C-Mg bond: remember from previous posts that carbon is strongly δ- (nucleophilic) because of its greater electronegativity as compared to magnesium. It might be helpful to imagine the Grignard reagent below as CH3CH2– . Other than that the reaction is fairly straightforward if you’ve seen an SN2 reaction before: we simultaneously form C-C and break C-O.
Note that this reaction also forms an “alkoxide”. In order to obtain our neutral alcohol product at the end, we must perform second step: a “workup” (“quench”) with a source of acid. This is written a variety of ways – H+, H3O+, H2O, or just “acid workup”. This step occurs after our key Grignard reaction, for what should be obvious reasons – Grignard reagents are destroyed by acid.
Another thing to keep in mind is stereochemistry of the epoxide.Consistent with an SN2 reaction, if the reaction occurs at a secondary carbon, we will observe inversion of configuration:
2. Reaction of Grignards With Aldehydes and Ketones
A second class of important electrophiles that react with Grignards (and arguably THE most important class of electrophiles) is aldehydes and ketones. If you haven’t covered the reactions of these functional groups yet, a short summary would be this: the carbonyl carbon is an electrophile, and when nucleophiles react at this carbon, it’s accompanied by cleavage of the C-O pi bond (π bond). More on that mechanism (“addition”) in a second.
Here are some examples of reactions of Grignards with aldehydes and ketones. Note that in each case we are forming a new bond between the carbonyl carbon (labelled A) and the carbon bound to magnesium (labelled B), and we are breaking the C-O pi bond in the process.
So how does this reaction work?
Let’s get familiar with a VERY important mechanism called “addition” (sometimes called, “1,2-addition”). This is by far the most important reaction of the carbonyl group, and if you give yourself a dollar for every time you will see variations of it in Org 2, you can buy yourself a vacation to Hawaii by the end of the semester. Not really, but almost.
Note that this reaction also forms an “alkoxide”. In order to obtain our neutral alcohol product at the end, we must perform a “workup” (“quench”) with a source of acid, forming O-H.
3. Reaction of Grignard Reagents With Esters
Esters are close relatives of aldehydes and ketones: they consist of a carbonyl group directly attached to an OR group. As you might expect, they react with Grignards in a similar fashion to aldehydes and ketones: with formation of a new C-C bond and breakage of a C-O (pi bond).
However, there’s a twist with the reaction of esters that isn’t present with aldehydes and ketones. Look carefully: what’s different?
Note that in both cases we added two equivalents of our Grignard reagent to the ester, forming a tertiary alcohol.
Wait a minute – how did this happen?!
This reaction incorporates the second most important mechanism of carbonyls (next to “addition”), namely, “elimination“. In fact “elimination” is the exact reverse of “addition” ! Let’s walk through it. There are 4 steps.
- In the first step, the Grignard performs an addition reaction on the ester, forming C-C and breaking C-O (pi), giving us an intermediate with a negatively charged oxygen. We’ve seen this type of reaction before in the addition of Grignards to aldehydes and ketones.
- Now comes the new step: elimination (sometimes, “1,2 elimination”). This intermediate has a reasonably good leaving group (OCH2CH3 in the case below). What happens next is reformation of the C-O pi bond with expulsion of the leaving group (CH3CH2O– in the case below). In other words, we form C–O π and break a C–O single bond. The new product is a ketone.
[Why might “elimination” happen for esters but not for aldehydes or ketones? Because we are expelling only a moderately strong base (RO–) as compared with the extremely strong bases H– or R– (which would be the leaving groups were this to happen with aldehydes or ketones)]. You can think of “basicity” as a proxy for “stability”. Alkoxides, which are expelled in this elimination reaction, are reasonably stable by themselves. Hydrides and most carbanions are more than 1025 more unstable. Plus, the elimination reaction begins with an alkoxide intermediate. We are essentially just swapping one alkoxide for another. [Note 1] ]
- But wait! There’s more! After Step 2, we have a new ketone. As we’ve seen before, Grignards will react quickly with ketones in yet another addition reaction [Step 3]. Here, as in Step 1, we form C–C and break C–O (pi). The result is a tertiary alkoxide (the conjugate base of a tertiary alcohol).
[Wait, you might ask. If we just use one equivalent of Grignard reagent, is it possible to get the reaction to stop at the ketone stage? The short answer is “no”. [See note 3 for the long answer]]
- Finally, protonation of this tertiary alkoxide yields the tertiary alcohol (Step 4).
Here’s the graphical walkthrough:
That does it for the key reactions of Grignard reagents you’ll see in most Org 1 and Org 2 courses.
In the next post we’ll talk about yet another way to screw up formation of Grignard reagents, and it involves the reactions in this post.
Next Post: Protecting Groups In Grignard Reactions
Note 1: Although alkoxides (RO–, the conjugate base of alcohols, pKa 16-18) are not on anyone’s list of Great Leaving Groups, they are some 25 orders of magnitude better leaving groups than hydrides (H–, the conjugate base of hydrogen, pKa 40) and more than 30 orders of magnitude better than alkyl groups (R- , the conjugate base of alkanes, pKa 50). Thus, when the alkoxide intermediate is formed in Step 1, there is not any deep energetic penalty for the C-O pi bond to reform and for RO- to be expelled: after all, we are simply replacing a strong base (the O- ) with one of comparable basicity.
[Note 2] Why are ketones more reactive towards Grignard reagents than esters? This requires understanding the phenomenon of pi donation. The lone pair on oxygen donates electron density into the carbonyl carbon. This is worthy of a separate post, but here’s the bottom line:
[Note 3] Alas, no. Using 1 equivalent of Grignard will result in 0.5 equivalents of a tertiary alcohol and 0.5 equivalents of the starting ester. The reason why is that Step 2 [elimination] is quite fast! Once elimination occurs, we will have ketone in the presence of an ester. For interesting reasons [see Note 2] ketones are more reactive than esters toward Grignard reagents, which means they will be consumed more quickly.