Amines

The classified amines are:

- Primary amine: one alkyl/aryl group attached to nitrogen, e.g. **$\mathrm{RNH_2}$
- Secondary amine: two alkyl/aryl groups attached to nitrogen, e.g. $\mathrm{R_2NH}$ or $\mathrm{RNH R'}$
- Tertiary amine: three alkyl/aryl groups attached to nitrogen, e.g. $\mathrm{R_3N}$**

So the given amines are classified accordingly as primary, secondary, or tertiary based on how many hydrogens of ammonia are replaced.

The molecular formula **$\mathrm{C_4H_{11}N}$ has the following isomeric amines:

1. $\mathrm{CH_3CH_2CH_2CH_2NH_2}$
2. $\mathrm{CH_3CH_2CH(NH_2)CH_3}$
3. $\mathrm{(CH_3)_2CHCH_2NH_2}$
4. $\mathrm{CH_3CH_2NHCH_2CH_3}$
5. $\mathrm{CH_3NHCH_2CH_2CH_3}$
6. $\mathrm{CH_3NHCH(CH_3)_2}$
7. $\mathrm{(CH_3)_3CNH_2}$
8. $\mathrm{(CH_3)_2NCH_2CH_3}$**

These include primary, secondary and tertiary amines.

The IUPAC names of the isomers of **$\mathrm{C_4H_{11}N}$ are:

- butan-1-amine
- butan-2-amine
- 2-methylpropan-1-amine
- 2-methylpropan-2-amine
- N-ethylethanamine
- N-methylpropan-1-amine
- N-methylpropan-2-amine
- N,N-dimethylethanamine**

The different isomeric amines of **$\mathrm{C_4H_{11}N}$ show:

- Chain isomerism: due to different carbon skeletons, e.g. butan-1-amine and 2-methylpropan-1-amine.
- Position isomerism: due to different position of the amino group, e.g. butan-1-amine and butan-2-amine.
- Metamerism: in secondary and tertiary amines because alkyl groups on either side of nitrogen may differ.

So the main types are chain isomerism and position isomerism; metamerism** is also present in some pairs.

The conversions are:

### (i) Benzene into aniline
1. First nitrate benzene to nitrobenzene:

$$\mathrm{C_6H_6 \xrightarrow{conc.\ HNO_3/H_2SO_4} C_6H_5NO_2}$$

2. Reduce nitrobenzene to aniline:

$$\mathrm{C_6H_5NO_2 \xrightarrow{Sn/HCl\ or\ Fe/HCl} C_6H_5NH_2}$$

### (ii) Benzene into N,N-dimethylaniline
1. Convert benzene to aniline as above.
2. Methylate aniline with excess methyl iodide:

$$\mathrm{C_6H_5NH_2 + 2CH_3I \xrightarrow{Na_2CO_3} C_6H_5N(CH_3)_2}$$

So the required product is N,N-dimethylaniline.

For increasing basic strength, aromatic amines are weakest, then benzylamine, then tertiary aliphatic amine, then primary and secondary methyl amines. Thus the order is:

$$\mathrm{C_6H_5NH_2 < C_6H_5CH_2NH_2 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH}$$

This matches the chapter’s answer for the corresponding intext question.

The acid-base reactions are:

### (i)
$$\mathrm{CH_3CH_2CH_2NH_2 + HCl \rightarrow CH_3CH_2CH_2NH_3^+Cl^-}$$
Product: propylammonium chloride or propan-1-aminium chloride.

### (ii)
$$\mathrm{(C_2H_5)_3N + HCl \rightarrow (C_2H_5)_3NH^+Cl^-}$$
Product: triethylammonium chloride.

Amines act as bases and form ammonium salts with acids.

Aniline on exhaustive methylation with excess methyl iodide in the presence of sodium carbonate forms the quaternary ammonium salt, benzyltrimethylammonium iodide type only if the amine were benzylamine; for aniline, the final alkylation product is N,N,N-trimethylanilinium iodide.

The sequence is:

$$\mathrm{C_6H_5NH_2 + CH_3I \rightarrow C_6H_5NHCH_3}$$
$$\mathrm{C_6H_5NHCH_3 + CH_3I \rightarrow C_6H_5N(CH_3)_2}$$
$$\mathrm{C_6H_5N(CH_3)_2 + CH_3I \rightarrow [C_6H_5N(CH_3)_3]^+I^-}$$

With sodium carbonate solution, the free base is regenerated/maintained during methylation. The final product is the quaternary ammonium salt [N,N,N-trimethylanilinium iodide].