Difficulty: Intermediate | Prerequisites: Functional groups, nucleophilic addition, acid-base chemistry
Amines are nitrogen-containing organic compounds that show up everywhere in biology and pharmaceutical chemistry. This material sits near the end of Organic Chemistry I and draws on carbonyl chemistry, nucleophilic substitution, and acid-base concepts you covered earlier in the course. You need to be comfortable with arrow-pushing mechanisms and carbocation/carbanion intermediates before diving in. If you missed the carbonyl addition unit, revisit that first.
Amines react as nucleophiles and bases. The key reactions to predict products for include reductive amination (making C-N bonds from carbonyls), catalytic reduction of nitro groups, exhaustive methylation followed by Hofmann elimination (breaking rings or making alkenes), protonation to form ammonium salts, and reduction of nitriles with LiAlH4 to give primary amines.
Reductive amination
The formation of a C-N bond by condensing an amine with a carbonyl compound (aldehyde or ketone) to form an imine or iminium ion, followed by reduction (typically with NaBH3CN). Think of it as: gluing a nitrogen onto a carbonyl carbon, then locking the bond by adding hydrogen.
Imine (Schiff base)
A compound with a C=N double bond, formed when a primary amine reacts with an aldehyde or ketone and loses water. In simple terms, this is the intermediate that forms before the reducing agent does its job in reductive amination.
NaBH3CN (sodium cyanoborohydride)
A mild, selective reducing agent that reduces iminium ions but leaves unreacted aldehydes and ketones alone. Think of it as: a picky version of NaBH4 that only reduces after the nitrogen has already attacked.
Exhaustive methylation (Hofmann elimination sequence)
A three-step process: (1) treat an amine with excess CH3I to form a quaternary ammonium salt, (2) convert the iodide to hydroxide with Ag2O/H2O, (3) heat to induce E2 elimination. In simple terms, you overload the nitrogen with methyl groups until it becomes a leaving group, then kick it out to form an alkene.
Quaternary ammonium salt
A nitrogen bearing four organic groups and a positive charge, with a counterion. Think of it as: nitrogen has used all four of its bonds on carbon groups, so it is permanently charged and can no longer act as a nucleophile or base.
Gabriel synthesis
A method for making primary amines using phthalimide as a protected nitrogen nucleophile. The nitrogen in phthalimide is deprotonated by base, alkylated with an alkyl halide, then freed by hydrolysis. Think of it as: a way to guarantee you get a primary amine without over-alkylation.
Catalytic hydrogenation of nitro groups
Reduction of -NO2 to -NH2 using H2 gas over a palladium on carbon (Pd/C) catalyst. In simple terms, you swap out the oxygens on nitrogen for hydrogens.
LiAlH4 (lithium aluminium hydride, LAH)
A strong reducing agent that converts nitriles (C≡N) to primary amines (CH2NH2). Think of it as: the heavy-duty reducer that forces two equivalents of H across the triple bond to nitrogen.
Primary amine + aldehyde + NaBH3CN/MeOH gives a secondary amine
Example: PhNH2 + PhCHO with NaBH3CN in MeOH produces PhCH2NHPh (N-benzylaniline)
The carbonyl oxygen leaves as water; the reducing agent delivers H to the C=N intermediate
Solvent is typically methanol (MeOH) or ethanol
ArNO2 + H2/Pd/C gives ArNH2 (a primary aromatic amine)
Example: PhNO2 reduced to PhNH2 (aniline)
This is a clean, high-yielding route to aromatic amines
The Pd/C catalyst surface adsorbs H2 and delivers it to the nitro group
Step 1: Amine + excess CH3I produces a quaternary ammonium iodide salt
Step 2: Ag2O/H2O converts the iodide counterion to hydroxide
Step 3: Heating causes E2 elimination, with hydroxide as the base
For cyclic amines, this process opens the ring:
A cyclic secondary amine (e.g. pyrrolidine derivative) undergoes ring-opening elimination to give an aminoalkene plus trimethylamine, AgI, and water as byproducts
The least substituted alkene (Hofmann product) is typically favoured because the bulky NMe3 leaving group steers the base toward the less hindered hydrogen
Primary amine + excess CH3I (without the Ag2O/heat steps) simply gives the quaternary ammonium iodide salt
Example: RNH2 + xs CH3I produces RN(CH3)3+ I-
Primary amine + HBr gives an ammonium bromide salt
Example: RNH2 + HBr produces RNH3+ Br-
This is straightforward acid-base chemistry: the lone pair on nitrogen picks up the proton
R-C≡N + (1) LiAlH4, (2) H2O gives RCH2NH2 (a primary amine)
Two equivalents of hydride are delivered across the triple bond
The aqueous workup (H2O) frees the amine from the aluminium complex
This extends the carbon chain by one carbon compared to the starting material if you count the nitrile carbon
Students often assume NaBH3CN and NaBH4 are interchangeable. They are not. NaBH4 would reduce the aldehyde directly before the imine has a chance to form. NaBH3CN is selective for the iminium ion.
Students sometimes draw the product of amine + HBr as a covalent N-Br bond. The product is an ionic ammonium salt (NH3+ Br-), not a bromo-amine.
Students forget that exhaustive methylation requires excess CH3I. If you use just one equivalent, you get a higher-substituted amine, not the quaternary salt.
Hofmann elimination favours the less substituted alkene, not the more substituted one (which would be the Zaitsev product). The bulky trimethylammonium leaving group is the reason.
⚠️ Predicting the correct product of reductive amination (secondary amine, not alcohol or unreduced imine) is a near-certain exam question.
⚠️ Know the difference between what happens when you treat an amine with excess CH3I alone (quaternary salt) versus the full three-step Hofmann elimination sequence (alkene + NMe3).
⚠️ Be able to identify when Hofmann (less substituted) versus Zaitsev (more substituted) products form. The Hofmann elimination always gives the Hofmann product.
⚠️ Amine salt formation (protonation by HBr, HCl) is tested as a quick product-prediction question. Do not overthink it.
True or false: NaBH3CN can reduce a standalone aldehyde under reductive amination conditions. (False, it is selective for the iminium ion.)
Fill in the blank: PhNO2 + H2/Pd/C produces ____. (PhNH2, aniline.)
True or false: Treating a primary amine with HBr forms a covalent N-Br bond. (False, it forms an ammonium bromide salt.)
Fill in the blank: The Hofmann elimination favours the ____ substituted alkene. (Less substituted.)
True or false: LiAlH4 followed by H2O can convert a nitrile to a primary amine. (True.)
Q: What is the major product when benzylamine (PhCH2NH2) reacts with acetone and NaBH3CN in MeOH?
A: N-isopropylbenzylamine. The amine condenses with acetone to form an imine, which is reduced to give a secondary amine.
Q: An unknown amine is treated with excess CH3I, then Ag2O/H2O, then heated. The products include an alkene, trimethylamine, water, and AgI. What type of amine was the starting material?
A: A secondary amine (or a cyclic secondary amine if ring-opening occurred). The three-step Hofmann elimination sequence requires at least a secondary amine to reach a quaternary salt that can eliminate.
Q: Why does reductive amination use NaBH3CN rather than NaBH4?
A: NaBH3CN is a milder reducing agent that selectively reduces the iminium ion intermediate without reducing the starting aldehyde or ketone. NaBH4 would reduce the carbonyl before the amine could attack it.
Q: What is the product when a nitrile (RCN) is treated with LiAlH4 followed by aqueous workup?
A: A primary amine, RCH2NH2. Two equivalents of hydride reduce the C≡N triple bond.
Q: Write the product of cyclohexylamine + HCl.
A: Cyclohexylammonium chloride (C6H11NH3+ Cl-). The nitrogen lone pair is protonated by the acid.
Reductive amination connects directly to carbonyl chemistry (nucleophilic addition to aldehydes and ketones) and to imine formation, which you covered earlier. The Hofmann elimination ties back to E2 elimination and regioselectivity (Hofmann vs Zaitsev). Amine salt formation is acid-base chemistry at its simplest.
Reductive amination is one of the most common reactions in pharmaceutical synthesis. Many drugs contain amine groups formed this way. Catalytic hydrogenation of nitro groups is used industrially to produce aniline, a precursor to dyes and polyurethanes.
Reductive amination, NaBH3CN, sodium cyanoborohydride, imine, Schiff base, iminium ion, Hofmann elimination, exhaustive methylation, quaternary ammonium salt, Gabriel synthesis, phthalimide, nitro reduction, Pd/C, catalytic hydrogenation, LiAlH4, LAH, nitrile reduction, amine salt, ammonium salt, protonation of amines, organic chemistry I, amine reactions, amine product prediction