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Organometallic — controlled formylation

Bouveault aldehyde synthesis

How the Bouveault aldehyde synthesis works, in 3 steps. Each step shows the structures and the curved arrows that move the electrons.

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In short. A 3-step mechanism. Typical reagents: n-BuLi or R–MgX (1 equiv.), DMF; then H₃O⁺. Conditions: THF or Et₂O, –78 to 0 °C; aqueous acid workup. Bonds formed: C–C; aldehyde C=O (at workup); bonds broken: C=O π; C–N (at workup). Stereochemistry: None (aldehyde product). Rate and selectivity: Addition fast at low temperature; intermediate stable until workup.

At a glance

Reagentsn-BuLi or R–MgX (1 equiv.), DMF; then H₃O⁺
Solvent / conditionsTHF or Et₂O, –78 to 0 °C; aqueous acid workup
Bonds formedC–C; aldehyde C=O (at workup)
Bonds brokenC=O π; C–N (at workup)
StereochemistryNone (aldehyde product)
Rate / selectivityAddition fast at low temperature; intermediate stable until workup

Step 1: Organolithium adds to DMF

+LiONOLiN1 · ORGANOLITHIUM ADDS TO DMFn-butyllithiumNUCLEOPHILEDMFELECTROPHILEtetrahedral alkoxideINTERMEDIATE

An organolithium or Grignard reagent attacks DMF's carbonyl carbon — the same nucleophilic acyl substitution mechanism as any amide addition — forming a tetrahedral alkoxide intermediate.

Step 2: Li⁺ chelated; intermediate does not collapse

chelationOLiNOLiN2 · LI⁺ CHELATED; INTERMEDIATE DOES NOT COLLAPSEtetrahedral intermediateINTERMEDIATELi chelateINTERMEDIATELi⁺ held by O⁻ and NMe₂ keeps the ring intact, so no second R adds.

Just as in Weinreb amide addition, this tetrahedral intermediate is stabilized rather than collapsing immediately, here because the developing alkoxide and the adjacent dimethylamino group can chelate the metal cation, holding the intermediate together until aqueous workup.

Step 3: Aqueous workup releases the aldehyde

H₃O⁺ workup+OHNHONH3 · AQUEOUS WORKUP RELEASES THE ALDEHYDEprotonated hemiaminalINTERMEDIATEpentanalC₅H₁₀OPRODUCTdimethylamineC₂H₇NBYPRODUCTThe O lone pair re-forms C=O as NMe₂ leaves.

Hydrolysis on workup collapses the stabilized intermediate, expelling dimethylamine and revealing the aldehyde — an older, simpler formylation strategy than the Weinreb approach, giving direct access to an aldehyde from any organometallic nucleophile without over-addition to the alcohol.

Abbreviations and terms

R
R group. Any carbon substituent (alkyl, aryl…) whose identity does not matter for the mechanism.
THF
Tetrahydrofuran. Cyclic ether solvent; its oxygen lone pairs solvate Li⁺/Mg²⁺ so organometallics stay dissolved and reactive.
Et2O
Diethyl ether. Low-boiling ether solvent, classic for Grignard reagents.
DMF
N,N-Dimethylformamide. Polar aprotic solvent – dissolves ionic nucleophiles but does not H-bond to them, speeding SN2 and SNAr.
n-BuLi
n-Butyllithium. Very strong base/nucleophile (pKa of butane ≈ 50); deprotonates alkynes and does lithium–halogen exchange.

How it compares with related reactions

Bouveault aldehyde synthesisGrignard addition to acetaldehydeGrignard addition to an ester (2 equivalents → tertiary alcohol)
Reagentsn-BuLi or R–MgX (1 equiv.), DMF; then H₃O⁺CH₃MgBr in Et₂O or THF; then H₃O⁺Ester (methyl benzoate), 2+ equiv. CH₃MgBr, then H₃O⁺ workup
ConditionsTHF or Et₂O, –78 to 0 °C; aqueous acid workupDry ether, 0 °C → 25 °C; strictly anhydrousDry Et₂O or THF, 0 °C → 25 °C; then aqueous NH₄Cl or dilute acid
StereochemistryNone (aldehyde product)Planar C=O: both faces attacked (racemic if new stereocentre)Product has two identical groups: no stereocentre at carbinol C
Rate / selectivityAddition fast at low temperature; intermediate stable until workupFast with aldehydes; RMgX is also a strong base — incompatible with acidic HKetone intermediate is more reactive than the ester, so no stop at the ketone

Test yourself

Try to answer each question before opening it.

Which reagents and conditions are typical for the Bouveault aldehyde synthesis?

n-BuLi or R–MgX (1 equiv.), DMF; then H₃O⁺. Conditions: THF or Et₂O, –78 to 0 °C; aqueous acid workup.

Which bonds are formed and which are broken?

Formed: C–C; aldehyde C=O (at workup). Broken: C=O π; C–N (at workup).

What is the stereochemical outcome?

None (aldehyde product)

What controls the rate or the selectivity?

Addition fast at low temperature; intermediate stable until workup

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