Organometallics on Ketones - Online Tutor, Practice Problems & Exam Prep

Organometallic compounds, such as Grignard reagents (R-MgBr) and organolithiums (R-Li), are powerful nucleophiles that undergo nucleophilic addition to carbonyl groups in ketones and aldehydes. The R^- from these organometallics attacks the electrophilic carbon, forming a tetrahedral intermediate. This intermediate typically undergoes protonation to yield an alcohol product with a new alkyl group attached. Understanding this mechanism is crucial for mastering reactions involving carbonyl compounds and organometallics in organic synthesis.

Nucleophilic Addition on Ketones and Aldehydes

Now that we understand nucleophilic addition to carbonyls, I want to focus on one of the most important nucleophiles that can react in nucleophilic addition: organometallics. I know organometallics can be intimidating. They look strange with a lot of atoms, including metals. However, these are some of the easiest compounds to work with in terms of pushing arrows because organometallics are really strong nucleophiles. We will do the same things with them as we would normally do with anything that has a negative charge. Organometallics participate in many different reactions. If you want to thoroughly review organometallics, you should go back to that section of your text or Clutch videos. I have plenty of videos on them. But what I'm going to focus on right now is just a refresher of what they do to ketones and aldehydes.

Remember that we can use the letters R and M to represent any organometallic because there's always going to be some kind of R component, some kind of alkyl group and some kind of metal. The two most common in this section are Grignards and organolithiums. A Grignard could be represented as R-Mg-Br. I'll write it down. Then there's organolithium, very straightforward. These molecules are extremely similar in how they react because they both have an ionic bond with an extremely strong dipole toward the R. In your periodic table, the further you go towards fluorine, the more electronegative elements become. Well, carbon is actually pretty close to fluorine, whereas these group 1 and 2 metals are some of the least electronegative atoms possible. This leads to a very ionic bond towards the carbon. So ionic, in fact, that we can write these things ionized. We can write it as R^- Mg⁺Br. The same thing with organolithium, we write it as R^- Li⁺. As such, these are just both a source of R^-.

When that R^- sees the carbonyl, we are going to carry out nucleophilic addition. We will attack the partial positive carbon, move the electrons up, and what we will get is a tetrahedral intermediate, O^- R₁ R₂, with my new R group added. This tetrahedral intermediate would normally stay there but is usually followed by a protonation step. I don't know exactly what the protonation step will be but I'm going to put some kind of protonation would occur and you would get your substituted alcohol, O H R₁ R₂ and the R that we added. Just a reminder: you should have worked with organometallics before in this course. This is what you need to know for this section on how to add an organometallic to a ketone or an aldehyde. You will always get an alcohol product and a new R group attached to that tetrahedral carbon in the middle.

That being said, why don't you guys take a look at the molecule below, try to do the whole reaction, draw the mechanism, and then I'll show you the answer. Go for it.

Organometallic on Aldehyde

For this reaction, the first thing to do is draw your charges. You have to remember that your C₆H₅ is going to be negatively charged and your MgBr is going to be positively charged. I know this looks strange but in actuality, what we're really dealing with is almost like having a benzene nucleophile. We pretty much have a negative charge just on that benzene ring, and then MgBr is hanging out like a cation, just like a spectator ion. We're going to get our nucleophilic addition. We're going to attack and what we're going to get is a tetrahedral intermediate. That tetrahedral intermediate will have an O negative. It will have my new benzene ring attached. The fourth group is that there was a hydrogen there. It used to be a hydrogen. It's still there. Nothing changed. Usually, guys, if you're stopping at the tetrahedral intermediate, let's say that the second reagent wasn't there, then many times the way that you might see it in the text or your professor might write it is that the MgBr will actually attach to the O because one is negative and one is positive. Your TI, tetrahedral intermediate, a lot of times will just be written like that. But we do have a protonating agent in this case. It's NH₄Cl. This is a very common protonating agent for Grignards and for organometallics. At first glance, it actually looks really weird because you got NH₄Cl. It almost looks like nitrogen has 5 bonds. But that would break nitrogen's octet. It can only have 4 bonds. How are 5 things attached to nitrogen? It's not really like that. What it is is that you've got your nitrogen with your 4 H's. That nitrogen is going to have a formal charge because it's unhappy with the number of valence electrons it has. It wants 5 and it has 4. Then the Cl negative will be hanging out associated with the NH₄ positive but it's not actually going to be attached by a bond. This protonating agent is going to get rid of the negative charge in the O and give us our eventual alcohol. Our alcohol looks like this with our benzene ring, and you've got your NH₃. It doesn't really matter. These other byproducts are not really of our concern, but your MgBr really is just hanging out with a positive charge. You could attach it to the Cl if you wanted. The Cl has a negative, so that would be fine too. Awesome guys. That was it for that mechanism. Let's move on to the next topic.

Provide the major product for the following reaction. 

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Provide the major product for the following reaction dealing with the Grignard reagent. 

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 Provide the major product for the following reaction dealing with the Grignard reagent.

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Provide the major product for the following reaction dealing with the Grignard reagent.

Hey everyone, let's take a look at the following practice question. Here it states, Provide the major product for the following reaction dealing with the Grignard reagent. If we take a look here, we have our cyclohexane and on it we have a bromine and a hydroxyl group. Now, remember that we can't properly create a Grignard reagent if there's an acidic group present. These acidic groups can be in the form of OH, NH, SH, or a triple bonded carbon with a hydrogen. These groups are too acidic for a Grignard reagent to be created. What would occur is that they would be deprotonated if they were present. What we have to do here in this example is we have to protect this OH group. Protecting a hydroxyl group before making a Grignard reagent is a common occurrence in organic chemistry. So, in order to do this, we use TSCL over our triethylamine. Now, here TSCL looks like this; it is three methyl groups connected to a silicon, connected to a chlorine. This is a trimethyl silane and this repr

Here’s what students ask on this topic:

What are Grignard reagents and how do they react with ketones?

Grignard reagents are organometallic compounds with the general formula R-MgBr, where R is an alkyl or aryl group. They are strong nucleophiles due to the highly polarized carbon-magnesium bond. When Grignard reagents react with ketones, the R- group attacks the electrophilic carbonyl carbon, forming a tetrahedral alkoxide intermediate. This intermediate is then protonated, typically by water or an acid, to yield a tertiary alcohol. The overall reaction can be summarized as:

$R_{2}C\left(O\right)+R_1\mathrm{MgBr}\to R_{2}C\left(\mathrm{OH}\right)R_1$

How do organolithium compounds react with aldehydes?

Organolithium compounds, represented as R-Li, are strong nucleophiles due to the highly polarized carbon-lithium bond. When they react with aldehydes, the R- group attacks the electrophilic carbonyl carbon, forming a tetrahedral alkoxide intermediate. This intermediate is then protonated, typically by water or an acid, to yield a secondary alcohol. The overall reaction can be summarized as:

$R\mathrm{CHO}+R\mathrm{Li}\to R\mathrm{CH}\left(\mathrm{OH}\right)R$

What is the mechanism of nucleophilic addition of organometallics to carbonyl compounds?

The mechanism of nucleophilic addition of organometallics to carbonyl compounds involves several steps:

1. The nucleophilic R- group from the organometallic reagent attacks the electrophilic carbonyl carbon, forming a tetrahedral alkoxide intermediate.
2. The electrons from the carbon-oxygen double bond are pushed onto the oxygen, creating an alkoxide ion.
3. The alkoxide intermediate is then protonated, typically by water or an acid, to yield an alcohol product.

The overall reaction can be summarized as:

$R2C\left(O\right)+RM\to R2C\left(\mathrm{OH}\right)R$

What are the common types of organometallic reagents used in reactions with ketones and aldehydes?

The two most common types of organometallic reagents used in reactions with ketones and aldehydes are Grignard reagents and organolithium compounds. Grignard reagents have the general formula R-MgBr, where R is an alkyl or aryl group. Organolithium compounds are represented as R-Li. Both types of reagents are strong nucleophiles due to the highly polarized carbon-metal bonds, making them effective in nucleophilic addition reactions with carbonyl compounds to form alcohols.

Why are organometallic compounds considered strong nucleophiles?

Organometallic compounds are considered strong nucleophiles because of the highly polarized carbon-metal bonds. In these compounds, the carbon atom is bonded to a metal such as magnesium or lithium, which are much less electronegative than carbon. This creates a significant dipole with a partial negative charge on the carbon, making it highly nucleophilic. This strong nucleophilicity allows organometallic compounds to effectively attack electrophilic carbonyl carbons in ketones and aldehydes, facilitating nucleophilic addition reactions.