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Amine synthesis

Gabriel synthesis of a primary amine

How the Gabriel synthesis of a primary amine 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: Potassium phthalimide; R–X; then N₂H₄. Conditions: DMF, 60–90 °C; then EtOH reflux. Bonds formed: C–N σ bond; primary amine R–NH₂; bonds broken: C–X; two C(O)–N bonds. Stereochemistry: Inversion at a chiral primary/secondary carbon (SN2). Rate and selectivity: SN2: methyl > 1° ≫ 2°; no 3° or aryl halides.

At a glance

ReagentsPotassium phthalimide; R–X; then N₂H₄
Solvent / conditionsDMF, 60–90 °C; then EtOH reflux
Bonds formedC–N σ bond; primary amine R–NH₂
Bonds brokenC–X; two C(O)–N bonds
StereochemistryInversion at a chiral primary/secondary carbon (SN2)
Rate / selectivitySN2: methyl > 1° ≫ 2°; no 3° or aryl halides

Step 1: Phthalimide anion attacks R–Br (SN2)

+ONOBr1 · PHTHALIMIDE ANION ATTACKS R–BR (SN2)phthalimide anionC₈H₄NO₂⁻NUCLEOPHILEbromoethaneC₂H₅BrELECTROPHILESN2The imide N–H (pKa ≈ 8.3) is removed by KOH or K₂CO₃ beforehand.

Potassium phthalimide's nitrogen — freed of its acidic proton by base — acts as a nucleophile in a standard SN2, displacing halide from an alkyl halide.

Step 2: N-alkylphthalimide: no over-alkylation

+ONOBr2 · N-ALKYLPHTHALIMIDE: NO OVER-ALKYLATIONN-ethylphthalimideC₁₀H₉NO₂INTERMEDIATEbromideBr⁻LEAVING GROUPN lone pair delocalisedover two C=OUnlike amine + RX, the product cannot alkylate again.

This gives an N-alkylphthalimide. Because the nitrogen is now doubly flanked by carbonyls, it's not nucleophilic or basic enough to over-alkylate — unlike a direct amine alkylation, which is hard to stop at one substitution.

Step 3: Hydrazinolysis frees the amine

+EtOH, Δ+ONONH₂NH₂NH₂ONHNHO3 · HYDRAZINOLYSIS FREES THE AMINEN-ethylphthalimideC₁₀H₉NO₂SUBSTRATEhydrazineH₄N₂NUCLEOPHILEethylamineC₂H₇NPRODUCTphthalhydrazideC₈H₆N₂O₂BYPRODUCTattackA second N–C=O attack by the other NH₂ closes the stable six-membered cyclic hydrazide.

Hydrazine attacks both phthalimide carbonyls in sequence, releasing the alkyl group as a clean, unpolluted primary amine plus a cyclic hydrazide byproduct.

Abbreviations and terms

R
R group. Any carbon substituent (alkyl, aryl…) whose identity does not matter for the mechanism.
X
Halogen / leaving group. Cl, Br or I (sometimes any leaving group).
DMF
N,N-Dimethylformamide. Polar aprotic solvent – dissolves ionic nucleophiles but does not H-bond to them, speeding SN2 and SNAr.
EtOH
Ethanol. Polar protic solvent.
K2CO3
Potassium carbonate. Mild inorganic base.
KOH
Potassium hydroxide. Strong aqueous/alcoholic base.
H2
Hydrogen gas. Reductant for catalytic hydrogenation.
SN2
Bimolecular nucleophilic substitution. One concerted step with backside attack and inversion; rate = k[RX][Nu].
Δ
Heat. Reaction is run at elevated temperature.
pKa
Acid-dissociation constant (–log). Lower pKa = stronger acid; used to judge which proton a base removes.

How it compares with related reactions

Gabriel synthesis of a primary amineDelépine reactionAza-Wittig reaction
ReagentsPotassium phthalimide; R–X; then N₂H₄Hexamethylenetetramine (HMTA), then HCl/EtOH (or H₃O⁺)R–N=PPh₃ (from R–N₃ + PPh₃, Staudinger), aldehyde or ketone
ConditionsDMF, 60–90 °C; then EtOH refluxCHCl₃ or EtOH, reflux; then ethanolic HClTHF or toluene, 25–110 °C, anhydrous
StereochemistryInversion at a chiral primary/secondary carbon (SN2)Inversion at a stereogenic primary/benzylic carbon (SN2)Imine E/Z set at ring collapse; no stereocentre needed
Rate / selectivitySN2: methyl > 1° ≫ 2°; no 3° or aryl halidesSN2 step rate-limiting; works for benzylic, allylic, α-halo carbonylsCycloreversion is fast; driven by strong P=O bond (≈ 130 kcal/mol)

Test yourself

Try to answer each question before opening it.

Which reagents and conditions are typical for the Gabriel synthesis of a primary amine?

Potassium phthalimide; R–X; then N₂H₄. Conditions: DMF, 60–90 °C; then EtOH reflux.

Which bonds are formed and which are broken?

Formed: C–N σ bond; primary amine R–NH₂. Broken: C–X; two C(O)–N bonds.

What is the stereochemical outcome?

Inversion at a chiral primary/secondary carbon (SN2)

What controls the rate or the selectivity?

SN2: methyl > 1° ≫ 2°; no 3° or aryl halides

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