Difficulty: Intermediate | Prerequisites: Organic Chemistry I (nucleophiles, electrophiles, SN1/SN2 mechanisms, leaving groups)
In most of Organic Chemistry I, carbon plays the electrophile: it is the atom that accepts electrons. Metals flip that role. When a metal bonds to carbon, it pushes electron density onto the carbon, turning it into a nucleophile, an electron-pair donor that can attack electrophilic centres. This is the foundation of organometallic chemistry and it opens up entire families of carbon-carbon bond-forming reactions (Grignard, Gilman, acetylide alkylations) that are not possible with carbon acting as an electrophile alone.
If you are comfortable with nucleophilic substitution (SN1/SN2), leaving groups and the idea of partial charges (delta-plus, delta-minus), you have what you need. If those feel shaky, revisit them first.
Metals bonded to carbon make that carbon nucleophilic, which is the opposite of its usual electrophilic behaviour. Grignard reagents (RMgBr) use this to attack carbonyls and epoxides, building new C-C bonds. Acetylide ions (formed with NaNH2) do the same via SN2 on alkyl halides. Knowing which reagent works with which substrate is the core exam skill here.
Organometallic compound
A molecule containing a carbon-metal bond (e.g. C-Mg, C-Li, C-Cu). The metal's electropositive nature pushes electron density onto carbon, making it nucleophilic.
In simple terms: the metal is so willing to give up electrons that the carbon it is bonded to becomes electron-rich.
Nucleophile
An electron-pair donor that attacks an electron-poor (electrophilic) centre. In organometallic reactions, the carbon bonded to the metal acts as the nucleophile.
Think of it as the species that "brings" electrons to a reaction.
Electrophile
An electron-pair acceptor, typically a carbon bearing a partial positive charge (e.g. a carbonyl carbon, or a carbon bonded to a halogen leaving group).
In simple terms: the atom that wants electrons.
Grignard reagent (RMgBr / RMgX)
An organometallic reagent formed by inserting magnesium into a C-halogen bond. The resulting R-MgBr species has a nucleophilic carbon that attacks carbonyls and epoxides. Reactions are run in THF and finished with an acid quench (H3O+).
Think of it as a carbon with a negative charge, courtesy of the magnesium.
Acetylide ion (RC triple-bond C minus)
The conjugate base of a terminal alkyne, formed by deprotonating the sp-hybridised C-H with a strong base such as NaNH2. The resulting carbanion is a powerful nucleophile for SN2 alkylation.
In simple terms: you strip the proton off the end of a triple bond, and the electrons left behind make that carbon a nucleophile.
THF (tetrahydrofuran)
A cyclic ether solvent used for Grignard and organolithium reactions. It coordinates to the metal without reacting with the organometallic reagent.
Hydronium quench (H3O+)
The acid workup step that protonates the alkoxide intermediate after a Grignard addition, converting it to an alcohol product.
Metals are cations: they form positively charged ions and readily give up electrons. When a metal bonds to carbon, the bonding electrons sit closer to the carbon (the more electronegative partner), giving it a partial or full negative charge.
In an organolithium (R-Li), the C-Li bond is so polar it is essentially ionic. The carbon carries a formal negative charge.
In a Grignard reagent (R-MgBr), the C-Mg bond is also highly polar. The carbon is the nucleophilic end.
Compare with NaCl: the same logic applies. In organometallics, we are doing with carbon what nature does with chlorine in table salt, except now the carbon is the anion.
The key mental shift: carbon has always been the electrophile in earlier coursework (think of carbonyl carbons, alkyl halides). Organometallics reverse that. Carbon bonded to a metal donates electrons and moves towards the electrophile.
This is a two-step sequence that extends a carbon chain through a triple bond.
Step 1: Form the acetylide ion
Start with a terminal alkyne (H-C≡C-R).
Treat with NaNH2 (sodium amide). The NH2 minus strips the terminal proton (the sp C-H), leaving the electrons behind on carbon.
Product: the acetylide ion (Na+ plus RC≡C:⁻), which is a strong nucleophile, along with ammonia (NH3) as a byproduct.
Step 2: SN2 alkylation
The acetylide ion attacks an alkyl halide (e.g. CH3-Cl) via an SN2 mechanism.
Chlorine is the leaving group. The more electronegative Cl pulls electron density towards itself (delta-plus on carbon), making that carbon electrophilic.
The acetylide's lone pair attacks that carbon, displaces the halide, and forms a new C-C bond.
Product: an internal alkyne (RC≡C-CH3) plus Cl⁻.
Worked example from the lecture
Starting material: CH3-C≡C-H
Treat with NaNH2. Intermediate: CH3-C≡C:⁻ Na⁺ plus NH3.
Add CH3CH2MgBr (here acting as the electrophilic partner in a separate step). Product: CH3-C≡C-CH2CH3.
The key point: the acetylide is the nucleophile, the alkyl halide (or similar electrophile) is the substrate. SN2 rules apply, so primary substrates work best. Secondary and tertiary substrates will give elimination instead.
A Grignard reagent is a powerful nucleophile and a strong base. When it encounters an acid (H3O+, water, alcohols, or any protic source), it is simply protonated. No useful C-C bond forms.
The Grignard (e.g. CH3MgBr) acts as the nucleophile: the carbon carries the electrons.
An acid is always an electrophile in this context: it wants electrons.
The oxygen in H3O+ is very electronegative ("greedy" for electrons) and pulls electron density towards itself, leaving the protons delta-plus.
The nucleophilic carbon grabs a proton from the acid.
Product: an alkane (H3C-H) plus MgBr+ and water.
This is why Grignard reactions must be run under strictly anhydrous conditions (no water, no protic solvents). Even trace moisture will destroy the reagent by protonating it before it can do anything useful. THF is the solvent of choice because it is aprotic.
Epoxides are strained three-membered rings containing an oxygen. That ring strain makes them unusually reactive towards nucleophiles.
Mechanism
The Grignard's nucleophilic carbon attacks the less substituted (less sterically hindered) carbon of the epoxide. This is the regioselectivity rule: attack the side with less stuff on it.
The C-O bond breaks, relieving ring strain. Oxygen picks up the MgBr as an alkoxide intermediate.
An acid quench (H3O+) protonates the alkoxide to give the final alcohol product.
Example from the lecture
CH3CH-CH2 (epoxide) plus CH3MgBr in THF, followed by H3O+ quench, gives CH3CH(OH)CH2CH3 plus water.
The intermediate is an alkoxide: the oxygen still carries a negative charge (bonded to MgBr) until the acid workup.
Another worked example
An organolithium (R-Li) works the same way with epoxides. Li and MgBr both make the bonded carbon nucleophilic. The nucleophilic carbon attacks the less hindered side of the epoxide, then H3O+ quench delivers the alcohol.
Product: a longer-chain alcohol with an OH group two carbons away from where the new C-C bond formed.
This is the most important Grignard reaction for exams. The nucleophilic carbon of the Grignard attacks the electrophilic carbonyl carbon (C=O), forming a new C-C bond.
Mechanism
The carbonyl carbon is electrophilic because oxygen pulls electron density away from it (delta-plus on C, delta-minus on O).
The Grignard's nucleophilic carbon attacks the carbonyl carbon. The pi bond breaks, and oxygen takes on a negative charge (alkoxide intermediate).
Acid quench (H3O+) protonates the alkoxide to give an alcohol.
What type of alcohol do you get?
Aldehyde (RCHO) plus RMgBr gives a secondary alcohol.
Ketone (RCOR') plus RMgBr gives a tertiary alcohol.
Formaldehyde (HCHO) plus RMgBr gives a primary alcohol.
Retrosynthetic analysis: choosing the right Grignard and carbonyl
This is a common exam question. You are given a target alcohol and asked to identify which Grignard reagent and which carbonyl would produce it.
The approach: look at the carbon bearing the OH in the product. The bonds to that carbon came from two sources, the Grignard and the carbonyl. You need to figure out which bond was already there (from the carbonyl) and which was formed (from the Grignard).
Worked example from the lecture
Target: CH3CH2-CH(OH)-CH3
This is a secondary alcohol. The carbon bearing OH has two alkyl groups attached: an ethyl (CH3CH2) and a methyl (CH3). One of those came from the Grignard, the other from an aldehyde.
Option A: CH3CH2-CHO (propanal) plus CH3MgBr. The Grignard adds CH3 to the aldehyde carbon. This gives the correct product. Correct.
Option B: CH3CH2-CO-CH3 (a ketone) plus CH3MgBr. This would add a third alkyl group, giving a tertiary alcohol. Wrong product.
Option C: CH3CH2CH2-CHO (butanal) plus CH3MgBr. The carbon count is wrong: this gives a five-carbon chain. Wrong product.
The lesson: work backwards from the product. Identify the OH-bearing carbon, disconnect one of the groups attached to it (that is the Grignard piece), and what remains is the carbonyl partner.
Reaction | Reagents | Conditions | Product |
|---|---|---|---|
Acetylide alkylation | 1. NaNH2, 2. R-X (1° alkyl halide) | - | Internal alkyne |
Grignard + acid | RMgBr + H3O+ | THF | Alkane (protonation only, no C-C bond) |
Grignard + epoxide | RMgBr + epoxide, then H3O+ | THF | Alcohol (chain extended by 2+ carbons) |
Grignard + aldehyde | RMgBr + RCHO, then H3O+ | THF | Secondary alcohol |
Grignard + ketone | RMgBr + RCOR', then H3O+ | THF | Tertiary alcohol |
Grignard + formaldehyde | RMgBr + HCHO, then H3O+ | THF | Primary alcohol |
Grignard reactions are one of the most widely used tools in pharmaceutical synthesis. Building complex drug molecules requires forming C-C bonds in a controlled way, and Grignard additions to carbonyls are a workhorse method for doing exactly that. Acetylide alkylations are used to install alkyne groups that can later be reduced to alkenes or alkanes, giving chemists precise control over the carbon skeleton of a molecule.
Students often think the Grignard reagent is the electrophile. It is not. The carbon bonded to Mg is the nucleophile. The carbonyl (or epoxide, or acid) is the electrophile.
Students forget that Grignard reagents are destroyed by water, alcohols, and any protic source. If the question includes an -OH or -NH group on the substrate, the Grignard will deprotonate it instead of performing a useful addition. Always check for acidic protons.
Students mix up the acid quench step with the reaction itself. The H3O+ comes after the Grignard has already attacked the carbonyl. It is a separate workup step, not part of the C-C bond-forming mechanism.
Students assume any base will form an acetylide. You need a base stronger than the terminal alkyne's pKa (~25). NaNH2 works. NaOH does not.
Retrosynthetic Grignard problems (given a target alcohol, identify the carbonyl and the Grignard) are a near-certainty on exams. Practise these until they feel mechanical.
Know the difference in product when a Grignard reacts with an aldehyde (secondary alcohol) versus a ketone (tertiary alcohol) versus formaldehyde (primary alcohol).
Acetylide alkylation questions test whether you remember that SN2 rules apply: the alkyl halide must be primary (or methyl). Secondary and tertiary substrates give elimination.
Expect a question that tests whether a Grignard will react or give "no reaction" based on the substrate. If there is no carbonyl or epoxide, the Grignard has nothing useful to attack.
True or false: A Grignard reagent can react with water without being destroyed. (False. Water protonates and destroys the Grignard.)
Fill in the blank: The acetylide ion is formed by treating a terminal alkyne with ____. (NaNH2)
True or false: A Grignard plus a ketone gives a secondary alcohol. (False. It gives a tertiary alcohol.)
Fill in the blank: Grignard reactions are run in ____ solvent and quenched with ____. (THF; H3O+)
True or false: The Grignard's carbon attacks the more substituted side of an epoxide. (False. It attacks the less substituted side.)
Q: What is the product when CH3MgBr reacts with CH3CHO in THF, followed by H3O+ quench?
A: CH3CH(OH)CH3, a secondary alcohol. The Grignard's methyl group attacks the aldehyde carbon, then the acid quench protonates the alkoxide.
Q: You want to synthesise 1-pentyne (CH3CH2CH2-C≡CH extended by one carbon to give an internal alkyne). What two reagents do you need?
A: First, treat 1-pentyne with NaNH2 to form the acetylide ion. Second, add a primary alkyl halide (e.g. CH3Br) for SN2 alkylation. Product: CH3CH2CH2-C≡C-CH3.
Q: A Grignard reagent (CH3CH2MgBr) is added to a substrate that contains both a ketone and an -OH group. What happens?
A: The Grignard is protonated by the -OH group (it acts as an acid). No useful C-C bond formation occurs. The Grignard is destroyed before it can attack the carbonyl.
Q: Identify the Grignard and carbonyl needed to make CH3CH2CH(OH)CH2CH3.
A: The OH-bearing carbon has an H, an ethyl group, and a propyl group attached. Disconnect one alkyl group as the Grignard: CH3CH2MgBr (Grignard) plus CH3CH2CHO (propanal) gives the target secondary alcohol after H3O+ quench. Alternatively, CH3CH2CH2MgBr plus CH3CHO works too.
Q: Why does a Grignard attack the less substituted carbon of an epoxide?
A: Steric reasons. The less substituted carbon has fewer groups blocking the nucleophile's approach, making it the more accessible site for backside attack.
This material connects directly to nucleophilic substitution (SN2) from Organic Chemistry I: the acetylide alkylation is an SN2 reaction, so all the rules about primary substrates, backside attack and inversion of stereochemistry apply. Grignard additions to carbonyls connect to the carbonyl chemistry unit that follows, where you will see other nucleophiles attacking C=O bonds (e.g. hydride reductions with NaBH4 and LiAlH4). The retrosynthetic thinking practised here (working backwards from the product) is the same skill you will use throughout the rest of the course.
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