Difficulty: Intermediate to Advanced | Prerequisites: Parts 1 and 2 of these notes, SN1/SN2/E2 mechanisms, carbonyl electrophilicity
Organometallic reagents (Grignard and alkyl lithium) are the single most important tool for building carbon-carbon bonds in this course. This section introduces how to make them, why they behave the way they do, and how they react with carbonyls to produce alcohols. The chapter closes with retrosynthesis, a strategic way of thinking backwards from a target molecule to plan a multi-step synthesis. If you master this material, you have the foundation for every synthesis problem from here through the end of the course.
Grignard reagents (RMgX) and alkyl lithium reagents (RLi) contain a carbon with strong nucleophilic and basic character. They react with carbonyls to form new C-C bonds, producing alcohols after workup. Because they are also strong bases, they react with protic sources (water, alcohols, amines) before they can do anything else, so solvent choice is critical. Retrosynthesis is the process of working backwards from the target molecule to identify the key bond-forming steps and the starting materials needed.
Grignard reagent (RMgBr or RMgCl)
An organometallic compound made by inserting magnesium metal (Mg⁰) into a carbon-halogen bond, in ether solvent (Et2O or THF). The carbon bonded to magnesium carries a partial negative charge (anionic carbon) and is both a strong nucleophile and a strong base.
Alkyl lithium reagent (RLi)
An organometallic made by lithium-halogen exchange: an alkyl halide reacts with lithium metal (Li⁰) in hexanes. Like Grignard reagents, the carbon is effectively anionic. In simple terms, it is the lithium version of a Grignard, made by a different method but with similar reactivity.
Anionic carbon
The carbon bonded to the metal in an organometallic reagent. It bears significant negative charge because of the large electronegativity difference between carbon and the metal (Mg or Li). This is what makes it nucleophilic and basic.
Retrosynthesis
The process of working backwards from a target molecule to identify simpler precursors and the reactions needed to assemble them. Uses a special open arrow (⟹) pointing from product to starting materials.
Retrosynthetic disconnection
The mental act of breaking a bond in the target molecule to identify the nucleophile and electrophile that would form it in the forward direction.
Grignard reagents:
Made by metal insertion: R-X + Mg⁰ → RMgX (in Et2O or THF)
Works with sp3 and sp2 C-X bonds
Commonly uses Cl, Br, or I as the halide
The bond between C and Mg is ionic in character but is often drawn as covalent for convenience
Alkyl lithium reagents:
Made by lithium-halogen exchange: R-X + Li⁰ → RLi (in hexanes)
Also works with sp3 and sp2 C-X bonds
Same ionic/covalent drawing convention as Grignard reagents
Synthesis of organometallics is faster than an acid-base (ABC) reaction. This means the metal insertion or exchange happens before the organometallic has a chance to react with anything else in the flask, provided the solvent is appropriate.
Strong bases (this property takes priority; ABC happens first)
The anionic carbon will deprotonate any sufficiently acidic proton: water, alcohols, terminal alkynes, amines, thiols, carboxylic acids.
This is why protic solvents destroy Grignard and alkyl lithium reagents.
Strong nucleophiles
The anionic carbon attacks electrophilic centres, especially carbonyls.
Compatible with sp3 and sp2 C-X bonds
Halogens are good leaving groups in the metal insertion step.
Must be used in ether solvents (THF, Et2O)
The ether oxygen coordinates to the metal cation via dipole interactions, stabilising and solubilising the organometallic.
Ether solvents also avoid ABC reactions (no acidic protons).
Reaction 1: Acid-base chemistry (ABC)
ABC is fast and favourable, so it happens first whenever a proton source is present.
Grignard + alcohol → hydrocarbon + alkoxide (the Grignard is destroyed)
RLi + ketone with an acidic alpha-H → deprotonation, not addition (if the proton is more accessible)
PhMgBr + a compound with both a carbonyl and an N-H or O-H → reacts with the acidic proton first
This means: you must protect any acidic protons before attempting a Grignard addition, or you must account for the stoichiometry.
Reaction 2: Bimolecular elimination (E2)
Organometallics are strong bases, so with secondary or tertiary alkyl halides, E2 elimination competes with or dominates over substitution.
Example: RLi + a 2° alkyl chloride in toluene → alkene (E2 product), not substitution.
Reaction 3: Addition to carbonyls (the key reaction)
The anionic carbon of the organometallic attacks the electrophilic carbonyl carbon.
This forms a new C-C bond (the single most important outcome of this reaction).
The product (after aqueous acid workup) is an alcohol.
Addition is possible because the carbonyl carbon is electrophilic (resonance gives it partial positive charge).
Addition products by carbonyl type:
Formaldehyde (H2CO) + RMgBr → primary alcohol (one new C-C bond, then -CH2OH)
Aldehyde (RCHO) + R'MgBr → secondary alcohol
Ketone (R2CO) + R'MgBr → tertiary alcohol
Ester: addition happens, but is covered later in the course
Workup step: After the addition, the alkoxide intermediate is quenched with H2O/HCl (or H2O/H2SO4) to give the free alcohol. This is the same principle as LiAlH4 workup.
Significance of Grignard additions: This is C-C bond formation. Before this chapter, you could change functional groups, but you could not easily build new carbon-carbon bonds. Grignard (and RLi) additions to carbonyls are the primary tool for constructing carbon skeletons.
Retrosynthesis is thinking backwards from the product to figure out how to make it. The retrosynthetic arrow (⟹) points from the target to simpler precursors.
Step 1: Identify the hardest bond to form in the target molecule. This is usually a C-C bond, and often there is only one reaction you know that forms it. That bond is your key step.
All reactions happen around functional groups. The functional group is the reactive point of the molecule.
Step 2: Identify the nucleophile and electrophile needed to form that bond.
Consider selectivity. If both SN1 and SN2 are needed at different points, think about which must come first to avoid problems (unstable carbocations, bulky nucleophiles causing E2).
Step 3: Figure out how to synthesise the nucleophile and electrophile from available starting materials.
Step 4: Repeat until you reach the given starting materials.
Multiple answers are possible. Different disconnections lead to different synthetic routes. Compare them by considering reactivity and selectivity to decide which is best. Exams award points for the best synthesis.
Target: A tertiary ether (Me-C(Me2)-O-Me) from isobutane, methane, and water.
Step 1: The hardest bond is the C-O bond. Two options for forming it:
Option 1: Alcohol nucleophile (tert-butanol → tert-butoxide) + MeBr electrophile → SN2. Problem: the tert-butoxide is bulky and would cause E2 instead of SN2 at a primary leaving group.
Option 2: MeOH nucleophile + tert-butyl bromide electrophile → SN1. This works because the tertiary substrate forms a stable carbocation, and MeOH is a weak nucleophile suitable for SN1.
Step 3 (best option, Option 2):
Isobutane → radical bromination (Br2, hv, CBr4) → tert-butyl bromide (electrophile)
Methane → radical bromination → CH3Br → NaOH/H2O → MeOH (nucleophile)
MeOH + tert-butyl bromide → SN1 → target ether
Always consider whether your nucleophile and electrophile are compatible with the mechanism you need (SN1 vs. SN2 vs. E2).
Remember that converting weak nucleophiles to strong ones (or vice versa) through ABC is a single extra step. Deprotonate an alcohol with NaH to get a strong alkoxide nucleophile for SN2.
No more than three steps per reaction arrow on exams. Draw out intermediates.
Use your reaction notebook to keep track of all known transformations.
Grignard reactions are used extensively in the pharmaceutical and fine chemicals industries to build complex carbon frameworks. The total synthesis of many drug molecules, including steroids and terpenes, relies on strategic C-C bond formation through organometallic additions to carbonyls. Retrosynthetic analysis, formalised by E.J. Corey (Nobel Prize, 1990), is the standard planning tool used by synthetic chemists worldwide.
Students often forget that Grignard reagents react with any acidic proton before they react with a carbonyl. If a substrate has both an -OH and a C=O, the Grignard will deprotonate the -OH first, consuming one equivalent without forming the desired C-C bond.
A common error is using a protic solvent (MeOH, H2O) with a Grignard or RLi reagent. This destroys the organometallic immediately.
Students sometimes attempt Grignard addition to a substrate that also has a secondary or tertiary alkyl halide, expecting substitution. The organometallic is more likely to cause E2 elimination.
In retrosynthesis, students often pick the first disconnection that comes to mind and do not evaluate alternatives. The exam rewards the best synthesis, so considering multiple retrosynthetic options and comparing them is essential.
⚠️ Grignard addition to carbonyls is the primary way to form C-C bonds in this course. Know the product for each carbonyl type (formaldehyde → 1° alcohol, aldehyde → 2° alcohol, ketone → 3° alcohol).
⚠️ The ABC-first rule is heavily tested. Expect a substrate with both an acidic proton and a carbonyl, and be asked what happens.
⚠️ Retrosynthesis problems appear on every exam from this chapter onward. Practise working backwards from products.
⚠️ Solvent choice matters. Grignard/RLi in ether solvents only. NaBH4 in protic. LiAlH4 in aprotic. Getting the solvent wrong loses full marks.
⚠️ Synthesis problems often stipulate "from alkanes with 5 carbons or fewer" or similar constraints. Read the instructions carefully.
True or false: A Grignard reagent can be prepared in methanol.
Fill in the blank: The reaction of a Grignard reagent with formaldehyde, followed by workup, gives a _________ alcohol.
True or false: Organometallic reagents react with acidic protons before they attack carbonyls.
Fill in the blank: In retrosynthesis, the first bond to identify is the _________ bond to form.
True or false: Alkyl lithium reagents are made by metal insertion of lithium into a C-X bond.
Answers: 1. False (MeOH is protic and would destroy the Grignard). 2. Primary. 3. True. 4. Hardest. 5. False (they are made by lithium-halogen exchange; Grignard reagents use metal insertion).
Q: What product forms when phenylmagnesium bromide (PhMgBr) reacts with acetone (CH3COCH3) in THF, followed by H2O/HCl workup?
A: A tertiary alcohol: 2-phenyl-2-propanol. The phenyl group from the Grignard adds to the carbonyl carbon of the ketone, forming a new C-C bond. Workup protonates the alkoxide.
Q: A substrate has both a free -OH group and a ketone carbonyl. You treat it with one equivalent of PhMgBr. What happens?
A: The Grignard deprotonates the -OH first (ABC reaction), producing an alkoxide and benzene (PhH). The carbonyl is untouched because the Grignard was consumed by the acidic proton. You would need two equivalents: one to deprotonate, one to add.
Q: Outline a retrosynthesis for cyclopentyl-butyl ether starting from cyclopentane and 1-bromobutane.
A: Target: cyclopentyl-O-butyl. Disconnection at the C-O bond gives cyclopentanol (or cyclopentoxide, the nucleophile) and 1-bromobutane (the electrophile, primary, good for SN2). Cyclopentanol comes from cyclopentane via radical bromination then SN2 with NaOH/H2O. Convert cyclopentanol to cyclopentoxide with NaH. Forward: cyclopentane → (Br2, hv, CBr4) → bromocyclopentane → (NaOH, H2O) → cyclopentanol → (NaH, THF) → cyclopentoxide → (1-bromobutane) → cyclopentyl-butyl ether.
Q: Why must Grignard reagents be prepared and used in ether solvents?
A: Two reasons. First, ether solvents have no acidic protons, so they do not destroy the organometallic through ABC. Second, the ether oxygen coordinates to the Mg cation through dipole interactions, stabilising and solubilising the Grignard reagent in solution.
Q: In the retrosynthesis strategy, why is it important to consider multiple disconnections rather than just using the first one that comes to mind?
A: Different disconnections lead to different nucleophile/electrophile pairs, which require different mechanisms (SN1, SN2, Grignard addition). Some combinations are better than others in terms of selectivity and feasibility. For example, one disconnection might require an SN2 with a bulky nucleophile (which would fail), while another uses SN1 with a stable carbocation (which works). Exams award marks for the best synthesis.
Grignard additions to carbonyls reappear with esters, acid chlorides, and epoxides in later chapters, each giving different product types.
Retrosynthesis becomes the standard planning method for every synthesis problem in the rest of the course and in CHEM 2520.
The ABC-first behaviour of organometallics connects to protecting group strategy introduced in later chapters.
Grignard reagent, RMgBr, RMgCl, organometallic, alkyl lithium, RLi, anionic carbon, metal insertion, lithium-halogen exchange, C-C bond formation, nucleophilic addition, carbonyl addition, formaldehyde, aldehyde, ketone, primary alcohol, secondary alcohol, tertiary alcohol, acid-base chemistry, E2 elimination, ether solvent, THF, diethyl ether, retrosynthesis, retrosynthetic analysis, disconnection, forward synthesis, target molecule, starting materials, selectivity, CHEM 2510, Chapter 8, organic chemistry, Baldwin, Ohio State