Amines - Organic Compounds Containing Nitrogen · Study notes

CHEClassification & Preparation

Classification, Nomenclature & Structure

Amines are classified $1^\circ$ , $2^\circ$ , $3^\circ$ by how many H atoms of ammonia are replaced by carbon groups (count substituents on N, not on the carbon as in alcohols). ISC also expects correct IUPAC names (alkanamine / N-substituted) and the pyramidal, $sp^3$ geometry of the nitrogen with its compressed bond angle.

The three classes (substituents on N)

R\text{-}NH_2\ (1^\circ) \qquad R_2NH\ (2^\circ) \qquad R_3N\ (3^\circ)

one, two, or three of ammonia's H atoms replaced by carbon groups

Quaternary ammonium salt

R_4N^+X^-

all four N bonds to carbon; N carries a positive charge

Classification counts groups on N: $1^\circ$ has one ( $R\text{-}NH_2$ ), $2^\circ$ has two ( $R_2NH$ ), $3^\circ$ has three ( $R_3N$ ); e.g. $C_6H_5NHCH_3$ is $2^\circ$ .
IUPAC naming: amines are alkanamines, e.g. $C_2H_5NH_2$ is ethanamine, $(CH_3)_3N$ is N,N-dimethylmethanamine, $C_6H_5NH_2$ is aniline (benzenamine).
Unsymmetrical $2^\circ$ / $3^\circ$ amines name the largest chain as parent; other groups are N-substituents, e.g. $C_6H_5NHCH_3$ is N-methylaniline.
The locant 'N' (capital, italic) shows a substituent is on nitrogen rather than on the carbon chain, e.g. N,N-dimethyl- means two methyls on N.
Nitrogen in amines is $sp^3$ hybridised with a pyramidal shape: three bond pairs plus one lone pair on N.
Bond angle ( $\approx 107^\circ$ in $NH_3$ , slightly larger in trialkylamines) is less than the tetrahedral $109.5^\circ$ because the lone pair repels the bond pairs more strongly, compressing the H-N-H / C-N-C angle.
Counting the lone pair, N is surrounded by four electron domains, so the electron-pair geometry is tetrahedral but the molecular shape (atoms only) is pyramidal.
Aniline's N is flatter (closer to planar) than methylamine's because the lone pair is partly delocalised into the ring, favouring more $sp^2$ -like character and wider angles at N.
A chiral $3^\circ$ amine with three different groups cannot be resolved because rapid pyramidal (umbrella) inversion at N interconverts the two forms.
Quaternary ammonium salts ( $R_4N^+X^-$ ) have all four N valencies to carbon and bear a positive charge; they are ionic and water-soluble.
Trap: amine class depends on N substitution, whereas alcohol/alkyl-halide class depends on the carbon bearing the functional group.

Where the marks go

Classifying like alcohols — by the carbon attached — instead of by the number of carbon groups on nitrogen (e.g. $(CH_3)_3C\text{-}NH_2$ is a $1^\circ$ amine even though the carbon is $3^\circ$ ).
Calling $C_6H_5NHCH_3$ a $1^\circ$ amine; with two carbon groups on N it is $2^\circ$ (N-methylaniline).
Saying the amine bond angle is $109.5^\circ$ — lone-pair/bond-pair repulsion compresses it to about $107^\circ$ .
Describing the molecular shape of N as tetrahedral; the shape (atoms only) is pyramidal, with tetrahedral electron-pair geometry.
Forgetting the 'N-' locant in names of N-substituted amines, or writing aniline as 'phenylamine only' without recognising N-substituents in derivatives.

How the board asks it

Structure / namingalkanamine and N-substituent naming rules
Give the IUPAC name of (i) $(CH_3)_3N$ , (ii) $C_6H_5NHCH_3$ and (iii) $CH_3CH_2CH(NH_2)CH_3$ .
Identify / classifycounting carbon groups on nitrogen
Classify each of the following as a $1^\circ$ , $2^\circ$ or $3^\circ$ amine: $(CH_3)_3C\text{-}NH_2$ , $C_6H_5NHCH_3$ and $(C_2H_5)_2NH$ .
Structure / namingpyramidal $sp^3$ nitrogen geometry
Draw the structure of trimethylamine and state the hybridisation, shape and approximate $C\text{-}N\text{-}C$ bond angle at the nitrogen atom.
Distinguishamine class set by substitution on N, not on carbon
Explain, with reasons, why $(CH_3)_3C\text{-}NH_2$ is classified as a primary amine although $(CH_3)_3C\text{-}OH$ is a tertiary alcohol.
Give reasonsumbrella (pyramidal) inversion at nitrogen
Account for the fact that a tertiary amine bearing three different groups on nitrogen cannot be resolved into optical isomers, whereas the analogous carbon compound can.

Practise this topic

NextPreparation (Gabriel, Hofmann, Reduction)

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.