Class 12 Chemistry Chapter 9: Amines worksheet with answer key

Answer: (a) 2-phenylethanamine

Gabriel synthesis gives only primary amines, from alkyl halides: (2-bromoethyl)benzene, C₆H₅CH₂CH₂Br, gives C₆H₅CH₂CH₂NH₂. 4-Methylaniline would need an aryl halide, which does not substitute with the phthalimide anion; (b) and (d) are secondary and tertiary.

Answer: (c) pentan-1-amine

Primary amines have two N–H hydrogens for hydrogen bonding and associate most; secondary amines have one; tertiary amines, (a) and (d), have none. So the boiling point order of isomers is 1° > 2° > 3°.

Answer: (a) butan-1-amine

In Hoffmann bromamide degradation the alkyl group migrates from the carbonyl carbon to nitrogen and the carbonyl carbon leaves as Na₂CO₃: the amine, CH₃(CH₂)₃NH₂, has one carbon less than the amide.

Answer: (d) 4-methylaniline

Both groups are on the ring, so the parent is aniline, with –CH₃ at C-4. 2-Methylaniline is the ortho isomer; N-methylaniline has CH₃ on nitrogen; phenylmethanamine, C₆H₅CH₂NH₂, has –NH₂ off the ring.

Answer: (d) N-ethylpropan-1-amine

The longer chain on nitrogen, propyl, gives the parent propan-1-amine; the ethyl group on nitrogen takes the locant N. Ethylpropylamine is the common name, not the IUPAC name; (a) takes the shorter chain as parent; (b) puts the nitrogen on C-2.

(a) CH₃CH₂CH₂–N(CH₃)–CH₂CH₃: propyl is the longest chain, and the ethyl and methyl groups carry the locant N. All three hydrogens of NH₃ are replaced, so it is a tertiary amine (C₆H₁₅N). (b) CH₃CH₂CH(NH₂)CH₂CH₂CH₃, also C₆H₁₅N: the 'e' of hexane is replaced by 'amine' with the locant of the carbon bearing –NH₂, hexan-3-amine; one hydrogen of NH₃ replaced, so primary.

• (a) correct structure, tertiary (1)
• (b) CH₃CH₂CH(NH₂)CH₂CH₂CH₃, hexan-3-amine (1)

In acetanilide the lone pair on nitrogen is in resonance with the carbonyl oxygen: >N–C(=O)–CH₃ ↔ >N⁺=C(–O⁻)–CH₃. Part of the lone pair is thus drawn towards oxygen and is less available for donation to the ring, so –NHCOCH₃ is a milder activator than –NH₂.

• resonance >N–C(=O)CH₃ ↔ >N⁺=C(O⁻)CH₃ shown (1)
• lone pair less available to the ring, so less activation (1)

C₂H₅NH₂ + H⁺ ⇌ C₂H₅NH₃⁺ and NH₃ + H⁺ ⇌ NH₄⁺. The ethyl group pushes electrons towards nitrogen (+I effect), so its unshared pair is more available to a proton; and the same +I effect disperses the positive charge of C₂H₅NH₃⁺, making it more stable relative to the amine than NH₄⁺ is relative to NH₃. The more stable the cation, the more basic the amine.

• +I effect makes the lone pair more available (1)
• the ethylammonium ion is stabilised by charge dispersal, so the amine is more basic (1)

(a) Gabriel synthesis: phthalimide + ethanolic KOH → potassium phthalimide; with CH₃CH₂CH₂Br on heating → N-propylphthalimide; hydrolysis with NaOH(aq) → CH₃CH₂CH₂NH₂ + the disodium salt of phthalic acid. Only a primary amine can form. (b) A mixture of propan-1-amine, N-propylpropan-1-amine, tripropylamine and the quaternary salt (CH₃CH₂CH₂)₄N⁺Br⁻, because each amine formed is itself a nucleophile and reacts further with the halide. (c) 1-Iodopropane: the reactivity of halides with amines is RI > RBr > RCl, so the iodide reacts faster and the chloride slower.

• (a) Gabriel route with the three stages (1)
• (b) mixture of 1°, 2°, 3° amines and quaternary salt, with reason (1)
• (c) 1-iodopropane, RI > RBr > RCl (1)

1. The +I effect of alkyl groups raises the electron density on nitrogen and stabilises the cation; on this alone basicity would be 3° > 2° > 1°, the order seen in the gas phase. 2. In water the cations are also stabilised by hydrogen bonding with water: RNH₃⁺ has three N–H, R₂NH₂⁺ two, R₃NH⁺ one, so on solvation alone the order would be 1° > 2° > 3°. 3. A methyl group causes no steric hindrance to hydrogen bonding, so the poorer solvation of (CH₃)₃NH⁺ puts (CH₃)₃N below CH₃NH₂. The bigger ethyl group does hinder hydrogen bonding, and this changes the balance of the three factors: the observed order in water is (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃.

• inductive effect alone favours 3° > 2° > 1° (1)
• solvation by H-bonding favours 1° > 2° > 3°; CH₃ gives no steric hindrance, so (CH₃)₃NH⁺ least solvated and (CH₃)₃N < CH₃NH₂ (1)
• C₂H₅ hinders H-bonding, changing the balance; order (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃ (1)

First diazotise: Cl–C₆H₄–NH₂ + NaNO₂ + 2HCl → Cl–C₆H₄–N₂⁺Cl⁻ + NaCl + 2H₂O, at 273–278 K. (a) Sandmeyer: with Cu₂Cl₂/HCl → Cl–C₆H₄–Cl + N₂. (b) Let the solution warm to 283 K: the salt is hydrolysed, Cl–C₆H₄–OH + N₂ + HCl. (c) Hypophosphorous acid and water (or ethanol): Cl–C₆H₄–N₂⁺Cl⁻ + H₃PO₂ + H₂O → C₆H₅Cl + N₂ + H₃PO₃ + HCl.

• diazotisation in the cold, then Cu₂Cl₂/HCl for (a) (1)
• (b) warming to 283 K, hydrolysis to the phenol (1)
• (c) H₃PO₂ and water (or ethanol) replaces the diazo group by H (1)