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Organic Chemistry 202-SNU-RE • CEGEP, Montreal

Revision sheet: amines (202-SNU-RE)

Amines are the only family in the course where the class is counted on the heteroatom and not on the carbon that bears it. This difference, harmless in appearance, decides everything that follows, from the Hinsberg test to the formation of imines.

This sheet leaves the course in your notes and gathers what decides an exam: the availability of the lone pair, the carbon count in each synthesis, and the reagents that identify a class in a single manipulation.

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The thread of the chapter

The basicity of an amine is not read off the molecule but off the AVAILABILITY of its lone pair: conjugated with a ring or a carbonyl, it no longer picks up a proton.

This chapter is part of 202-SNU-RE, Organic Chemistry

The essentials

Classes are counted on nitrogen

  • • Primary RNH2RNH_{2}, secondary R2NHR_{2}NH, tertiary R3NR_{3}N, and QUATERNARY ammonium salt R4N+R_{4}N^{+}.
  • • Count the carbons carried by NITROGEN, unlike alcohols, where you look at the carbon bearing the −OH-OH.
  • • Nitrogen is sp3sp^{3} and pyramidal, its lone pair occupying the fourth vertex.
  • • It INVERTS like an umbrella, with a barrier near 2525 kJ/mol, which rules out isolating the two forms of an amine whose nitrogen is the only stereogenic center.
Nprimary1 carbonNsecondary2 carbonsNtertiary3 carbonscount the carbons carried by NITROGEN
One, two or three bonds leaving nitrogen: this count alone gives the class, without looking at the neighbouring carbons.

A tertiary alcohol carries three carbons on ITS CARBON; a tertiary amine carries three on ITS NITROGEN. The same word, two different counts, and it is the most frequent mistake of the chapter.

Basicity and its values

  • • pKapK_{a} of the conjugate acids: ammonium ion 9.259.25; methylammonium 10.6410.64; dimethylammonium 10.7310.73; trimethylammonium 9.809.80.
  • • Aromatic cases: anilinium 4.634.63; pyridinium 5.235.23; para-nitroanilinium 1.001.00; protonated amide about −0.5-0.5.
  • • The LARGER the pKapK_{a} of the conjugate acid, the more basic the amine.
  • • In the gas phase the order follows the number of alkyl groups; in solution it partly reverses, the solvation of the cation counting as much as the donor effect.
036912pKaanilinium 4.63pyridinium 5.23ammonium 9.25methylammonium 10.64increasingly basic amine
On the pKapK_{a} scale, anilinium and pyridinium sit four to six units below the aliphatic ammonium ions: their amines are much less basic.

Trimethylammonium is more acidic than dimethylammonium (pKapK_{a} of 9.809.80 against 10.7310.73), so trimethylamine is less basic than dimethylamine: three alkyl groups hinder the solvation of the cation formed.

Why aniline is a weak base

  • • Its lone pair is CONJUGATED with the aromatic ring: it takes part in the π\pi cloud and is no longer available.
  • • Mobilizing it to pick up a proton costs the loss of this resonance, hence a gap of about six pKapK_{a} units.
  • • Same logic for the acidity of phenol, and for the non-basicity of the amide.
  • • An electron-withdrawing group on the ring makes the effect worse still: 1.001.00 for para-nitroanilinium.
NHHaniline:conjugated lone pair,weak baseNHHH⁺methylamine:available lone pair,strong base
The lone pair of aniline is committed to the ring and is no longer available; that of an aliphatic amine stays on nitrogen and picks up a proton.

Three families, a single line of reasoning: as soon as a lone pair is delocalized, it stops being basic or nucleophilic. This is the sentence that answers half the questions of the chapter.

Prepare, identify, eliminate

  • • PREPARATIONS: direct alkylation of NH3NH_{3}, to be avoided because of polyalkylation; GABRIEL through phthalimide, clean; reduction of a nitrile, which GAINS a carbon; reduction of an aromatic nitro compound; reductive amination; Hofmann REARRANGEMENT of an amide, which LOSES a carbon.
  • • HINSBERG TEST: primary gives a sulfonamide with an acidic N−HN-H, SOLUBLE in sodium hydroxide; secondary gives a sulfonamide with no N−HN-H, INSOLUBLE; tertiary does NOT react.
  • • NITROUS ACID in the cold: aliphatic primary gives unusable dinitrogen; secondary gives a nitrosamine; AROMATIC primary gives a diazonium salt, stable between 00 and 55 degrees Celsius.
  • • HOFMANN ELIMINATION: exhaustive methylation, exchange to the hydroxide, then heating. The LEAST substituted alkene is obtained.

Two routes change the number of carbons, and the question picks them precisely for that: the nitrile adds one, the Hofmann rearrangement removes one.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Counting the class of an amine as for an alcohol

1.5 points, and the whole Hinsberg test is misread

What not to write

“The carbon bearing the nitrogen is secondary, so it is a secondary amine.”

What to write

“The carbons carried by NITROGEN are counted: here there is only one, so it is a PRIMARY amine.”

NH₂OHpropan-2-amineN bonded to 1 C: PRIMARYpropan-2-olC of OH bonded to 2 C: SECONDARY
Same skeleton: the nitrogen of propan-2-amine carries only ONE carbon, so it is primary; the carbon of the −OH-OH carries two, so the alcohol is secondary.

Why: The class describes the environment of the functional heteroatom. For an alcohol, oxygen carries only one carbon in every case, which forces you to look at the carbon; for nitrogen, you can count directly.

2. Trying to isolate the two enantiomers of a chiral amine

2 points on the stereochemistry question

What not to write

“Nitrogen carries three different groups and a lone pair, so the amine is chiral and can be resolved.”

What to write

“Nitrogen INVERTS millions of times per second: the two forms interconvert and cannot be separated.”

NEtMePrNEtMePr25 kJ/moltwo mirror images, constantly interconverting
The lone pair flips from one side to the other like an inside-out umbrella: the two forms, mirror images of each other, interconvert constantly and cannot be separated.

Why: The inversion barrier is of the order of 2525 kJ/mol, easily crossed at room temperature. Only a quaternary ammonium, which has no lone pair left, can be resolved.

3. Treating aniline as an ordinary amine

2 points, and a poorly designed acid-base extraction

What not to write

“Aniline is a primary amine, so its conjugate acid has a pKapK_{a} close to 1010.”

What to write

“Anilinium has a pKapK_{a} of 4.634.63: aniline is about a million times less basic than an aliphatic amine.”

Why: The lone pair is committed to the aromatic ring. This is also what allows an aniline to be separated from an aliphatic amine by a wash at controlled pHpH.

4. Expecting the nitrogen of an amide to be basic

2 points, and an acid-base reaction that was made up

What not to write

“The amide carries a nitrogen, so it is basic like an amine.”

What to write

“Its lone pair is delocalized toward the carbonyl: the protonated amide has a pKapK_{a} close to −0.5-0.5, so it is not basic at all.”

CCH₃ONHHamide: the N lone pair moves toward C=O,it is no longer basic
In the amide, the nitrogen lone pair is pulled toward the carbonyl (red arrows): it is no longer available to pick up a proton.

Why: This is the same mechanism as for aniline, only stronger: the carbonyl is a much hungrier acceptor than the aromatic ring.

5. Preparing a primary amine by direct alkylation

2 points, and a derisory yield in practice

What not to write

“Heat ammonia with the bromoalkane and the primary amine is obtained.”

What to write

“The amine formed is more nucleophilic than ammonia: it reacts in turn, and a mixture results, all the way to the quaternary ammonium.”

NH₃RBrRNH₂RBrR₂NHRBrR₃NRBrR₄N⁺each amine formed is more nucleophilic than the previous onea mixture of all four is obtained
Each amine formed attacks the bromoalkane in turn: the cascade runs down to the quaternary ammonium, as a mixture.

Why: Each alkylation makes nitrogen richer, hence more reactive. The Gabriel synthesis exists precisely to block nitrogen after the first alkylation.

6. Forgetting the carbon gained by the nitrile route

2 points, and the product named is not the one asked for

What not to write

“1-Bromopropane gives, via the nitrile, propylamine.”

What to write

“The cyanide ion BRINGS a carbon: butylamine is obtained, with four carbons.”

Br123CN⁻N1234LiAlH₄NH₂12341-bromopropane3 Cbutanenitrile4 Cbutan-1-amine4 C
The carbon of the cyanide enters the chain: three carbons at the start, four in the nitrile and in the amine.

Why: The carbon of the nitrile group becomes a carbon of the chain after reduction. Counting the carbons before and after is the quickest check of any amine synthesis.

7. Forgetting the carbon lost in the Hofmann rearrangement

2 points, the mirror image of the previous pitfall

What not to write

“The four-carbon amide gives the four-carbon amine.”

What to write

“The Hofmann rearrangement LOSES the carbonyl carbon: the amine obtained has three.”

ONH₂1234Br₂NaOHNH₂123+CO₃²⁻butanamide: 4 Cpropan-1-amine: 3 Cthe carbonyl C
The Hofmann rearrangement drives out the carbonyl carbon as carbonate: the four-carbon amide gives the three-carbon amine.

Why: The alkyl group migrates from the carbonyl carbon to nitrogen, and the original carbon leaves as carbonate in basic medium. The skeleton shortens by one unit.

8. Applying Zaitsev to the Hofmann elimination

2 points, and it is precisely the question that tells the two rules apart

What not to write

“The major alkene is the more substituted one, as usual.”

What to write

“The Hofmann elimination gives the LEAST substituted alkene: the very bulky leaving group forces removal of the most accessible proton.”

N+heatingammonium, HO⁻but-1-ene: 95%but-2-ene: 5%
With the bulky ammonium, the proton removed is the one on the terminal methyl: the LEAST substituted alkene wins, the opposite of Zaitsev.

Why: The quaternary ammonium is an enormous leaving group. The base therefore reaches a terminal hydrogen more easily than an internal one, which reverses the usual rule.

9. Believing that every amine gives a diazonium salt

1.5 points on the recognition question

What not to write

“Nitrous acid turns any primary amine into a usable diazonium salt.”

What to write

“Only AROMATIC primary amines give a diazonium that is stable between 00 and 55 degrees; the aliphatic one decomposes at once, releasing dinitrogen.”

NN+stable between 0 and 5 °CR−N₂⁺R⁺ + N₂ (immediate evolution)aromatic:aliphatic:
The aromatic diazonium keeps in the cold; the aliphatic one loses its dinitrogen at once and leaves only a carbocation.

Why: The aryldiazonium cation is stabilized by the aromatic ring, the alkyldiazonium is not and immediately loses its dinitrogen. This difference is what makes the reaction useful in aromatic synthesis.

Which method to choose

Recognizing the class of an amine in the laboratory

Apply the Hinsberg test, then read the solubility.

primaryprecipitate,dissolved by NaOHsecondaryprecipitate,stays in NaOHtertiaryno reactiontube 1: benzenesulfonyl chloride; tube 2: then NaOH
Three outcomes of the Hinsberg test: the precipitate dissolves in sodium hydroxide (primary), it stays (secondary), or nothing forms (tertiary).

  • If a precipitate forms and dissolves in sodium hydroxide → PRIMARY amine: the sulfonamide keeps an acidic N−HN-H

    Example: propylamine

  • If a precipitate forms and stays insoluble in sodium hydroxide → SECONDARY amine: the sulfonamide has no N−HN-H left

    Example: diethylamine

  • If no reaction → TERTIARY amine: nitrogen has no hydrogen left to give up

    Example: triethylamine

  • If nitrous acid in the cold gives off plenty of gas → ALIPHATIC primary amine

    Example: dinitrogen is released immediately

  • If nitrous acid gives a solution that is stable in the cold → AROMATIC primary amine, and you are holding a diazonium salt

    Example: usable for an aromatic synthesis

Hinsberg answers the class, nitrous acid answers whether the amine is aliphatic or aromatic. The two tests complement each other and do not answer the same question.

Which synthesis route, according to the carbon count

Compare the starting skeleton with that of the desired product.

  • If the product has the SAME number of carbons as the start → Gabriel synthesis, or reductive amination

    Example: phthalimide blocks nitrogen after a single alkylation

  • If the product has ONE carbon MORE → substitution by cyanide, then reduction of the nitrile

    Example: three carbons give a four-carbon amine

  • If the product has ONE carbon FEWER → Hofmann rearrangement of the corresponding amide

    Example: four carbons give a three-carbon amine

  • If you start from an aromatic nitro compound → reduction, which gives the corresponding aniline directly

    Example: nitrobenzene gives aniline

  • If you want a specific secondary or tertiary amine → reductive amination, controlling the stoichiometry

    Example: direct alkylation would give a mixture

The carbon count is the first sort: on its own it eliminates two thirds of the possible routes, before even looking at the conditions.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Justifying a difference in basicity

When to use it: The question asks you to compare two amines, or to explain why aniline is weak.

  1. 1 Write the two conjugate acids and give their pKapK_{a}.
  2. 2 Examine the AVAILABILITY of the lone pair in each amine: free, conjugated with a ring, or conjugated with a carbonyl.
  3. 3 Add the effect of substituents, donating or withdrawing.
  4. 4 Conclude by giving the gap in pKapK_{a} units, hence in orders of magnitude.

Concluding sentence

“The lone pair of aniline is conjugated with the ring: anilinium has a pKapK_{a} of 4.634.63 against 10.6410.64 for methylammonium, a gap of six orders of magnitude.”

The trap: Comparing the amines with each other instead of their conjugate acids: it is the pKapK_{a} of the ammonium ion that is tabulated.

Marking: 1 mark for the pKapK_{a} values cited, 1 mark for the availability of the lone pair, 1 mark for the numerical conclusion.

Choosing and writing up an amine synthesis

When to use it: The question gives a substrate and asks for a specific amine.

  1. 1 Count the carbons of the substrate and of the product, and write down the difference.
  2. 2 Choose the route that matches this difference, and rule out the others in one sentence.
  3. 3 Write the steps with their reagents, specifying the sensitive conditions.
  4. 4 Check the class obtained: primary, secondary or tertiary.

Concluding sentence

“The product has as many carbons as the substrate and must be a clean primary amine: the Gabriel synthesis is used, not direct alkylation of ammonia.”

The trap: Proposing direct alkylation: it gives a mixture of all the classes, and the marker expects this reservation to be stated.

Marking: 1 mark for the carbon count, 1 mark for the route chosen, 1 mark for the steps, 0.5 mark for the class checked.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The typical problem, taken apart

Three routes to an amine, only one correct carbon count

You have 1-bromopropane, with three carbons. You want to prepare propylamine, also with three carbons.

Examine the three classic routes and say which one is suitable.

Br123which route??NH₂1231-bromopropane: 3 Cpropan-1-amine: 3 C
The start and the target, carbons numbered: it is by comparing these two counts that the route is chosen.

Step 1

Cyanide route: 1-bromopropane plus CN−CN^{-} gives butanenitrile, with FOUR carbons.

Why

The carbon of the nitrile group is added to the chain. After reduction, butylamine is obtained: this is not the product asked for.

Step 2

Hofmann rearrangement: you would have to start from an amide with FOUR carbons to obtain an amine with three.

Why

This route removes the carbonyl carbon. It is excellent, but it requires a different starting material from the one available.

Step 3

Direct alkylation of ammonia: it would give a mixture of propylamine, dipropylamine, tripropylamine and quaternary salt.

Why

Each alkylation makes nitrogen more nucleophilic, hence more reactive than the starting ammonia. The yield of primary amine is low and the separation is painful.

Step 4

Gabriel synthesis: potassium phthalimide plus 1-bromopropane, then hydrolysis or hydrazinolysis.

Why

The nitrogen of phthalimide carries only one acidic hydrogen and is blocked after a single alkylation. Polyalkylation becomes impossible, and the amine released is clean.

Step 5

Carbon check: 33 at the start, 33 at the end.

Why

Phthalimide contributes no carbon to the final chain, since it is entirely removed at the hydrolysis. It is this count that validates the choice.

Step 6

Class check: nitrogen carries only one carbon, so it is indeed a PRIMARY amine.

Why

The class is read off the nitrogen of the product as drawn, never off the bearing carbon. Two independent checks, and the answer is locked.

The conclusion, written out

“Only the Gabriel synthesis is suitable: it keeps the three carbons and gives a pure primary amine, where cyanide would add one and direct alkylation would give a mixture.”

The classic mistake on this problem: Choosing the cyanide route because it seems simpler: it gives butylamine, with four carbons.

Learn by heart

  • • The class of an amine is counted on NITROGEN; that of an alcohol on the bearing CARBON.
  • • Nitrogen inverts, barrier near 2525 kJ/mol: an amine that is chiral only through its nitrogen cannot be resolved.
  • • pKapK_{a} of the conjugate acids: ammonium 9.259.25, methylammonium 10.6410.64, anilinium 4.634.63, protonated amide −0.5-0.5.
  • • A CONJUGATED lone pair is no longer basic: aniline, amide, same reasoning.
  • • The nitrile ADDS a carbon, the Hofmann rearrangement REMOVES one, Gabriel changes none.
  • • Hinsberg: primary soluble in sodium hydroxide, secondary insoluble, tertiary no reaction.
  • • Hofmann elimination: the LEAST substituted alkene, unlike Zaitsev.

Frequently asked questions

How do you determine the class of an amine?

By counting the carbons carried by the nitrogen itself. One gives a primary amine, two a secondary, three a tertiary, and four a quaternary ammonium ion. This is the opposite of the rule for alcohols, where you count the carbons neighbouring the carbon that bears the hydroxyl group.

Why is aniline much less basic than an ordinary amine?

Because the lone pair on its nitrogen delocalizes into the aromatic ring and is no longer really available. Picking up a proton would mean giving up this stabilization. The conjugate acid of aniline thus has a pKa of about 4.6 against about 10.6 for methylammonium, a gap of six orders of magnitude.

Why avoid direct alkylation of ammonia?

Because the primary amine formed is more nucleophilic than the starting ammonia: it reacts in turn with the halide, and so does the next product, all the way to the quaternary ammonium ion. The result is a mixture that is hard to separate. The Gabriel synthesis blocks nitrogen after a single alkylation and avoids this problem.

Which routes change the number of carbons?

Two. Substitution by the cyanide ion followed by a reduction adds one carbon, that of the nitrile group, which becomes a carbon of the chain. The Hofmann rearrangement of an amide removes one, the carbonyl carbon, which leaves as carbonate in basic medium. All the other routes keep the skeleton.

What does the Hofmann elimination give?

The least substituted alkene, contrary to Zaitsev's rule. The leaving group is a quaternary ammonium ion, which is very bulky, so the base reaches a terminal hydrogen more easily than one inside the chain. It is one of the rare cases where steric hindrance decides the regioselectivity.

Practise it

Corrected exercises: Amines, 202-SNU-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

  • 15 corrected exercises
  • 150 points
  • 225 minutes
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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-snu-amines. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

See also

All revision sheets All exercise sets Corrected exercises: carboxylic acids and their derivativesCorrected exercises: aldehydes and ketonesCorrected exercises: aromatic compoundsCorrected exercises: alkyl halidesCorrected exercises: alcohols, phenols and ethersCEGEP tutoringLes amines, fiche de révision (in French)

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