Question

Why is CH₃–NH₂ a stronger base than (CH₃)₃N in aqueous solution?

Answer 1

Step 1: What controls basic strength in water.
An amine is basic because nitrogen can donate its lone pair and accept a proton. In water two effects act together: the electron-pushing (+I) effect of alkyl groups, and how well water can stabilise the protonated cation by hydrogen bonding (solvation).

Step 2: Compare the inductive effect.
\((CH_3)_3N\) has three methyl groups pushing electrons onto nitrogen, while \(CH_3NH_2\) has only one. By this effect alone the tertiary amine should look more basic.

Step 3: Now look at solvation.
After accepting a proton, \(CH_3NH_2\) becomes \(CH_3NH_3^+\), which has three N-H bonds that water can hydrogen bond to. \((CH_3)_3N\) becomes \((CH_3)_3NH^+\), which has only one N-H bond. So the cation from \(CH_3NH_2\) is much better hydrated and stabilised.

Step 4: Steric effect too.
The three bulky methyl groups in \((CH_3)_3N\) crowd the nitrogen, making it harder for a proton and for water molecules to approach.

Step 5: Combine the effects.
In water, the better solvation (and less steric hindrance) of \(CH_3NH_3^+\) outweighs the larger inductive push in \((CH_3)_3N\). So \(CH_3NH_2\) is the stronger base in aqueous solution.

Answer: In water \(CH_3NH_2\) is more basic than \((CH_3)_3N\) because its protonated cation \(CH_3NH_3^+\) has more N-H bonds and is far better stabilised by hydrogen bonding (solvation), and because the bulky methyls in \((CH_3)_3N\) hinder both protonation and solvation.

Answer 2

Step 1: Identify the first reaction.
\(CH_3CONH_2\) (acetamide) treated with NaOBr (that is \(Br_2 + NaOH\)) undergoes the Hofmann bromamide degradation. In this, the amide loses its carbonyl carbon and gives a primary amine with one carbon fewer.

Step 2: Work out A.
Acetamide has two carbons \((CH_3CONH_2)\). Losing the carbonyl carbon leaves a one-carbon amine, which is methylamine. So \[ A = CH_3NH_2\ (\text{methylamine}). \]

Step 3: Identify the second reaction.
\(C_6H_5COCl\) (benzoyl chloride) reacts with a base. With an amine (or ammonia) present as the base, this is benzoylation, forming an amide. The general case (Schotten-Baumann type) gives a benzamide.

Step 4: Work out B.
Benzoyl chloride reacting with ammonia/amine base gives benzamide. So \[ B = C_6H_5CONH_2\ (\text{benzamide}). \]

Step 5: State both products.
A is methylamine and B is benzamide.

Answer: \(A = CH_3NH_2\) (methylamine), formed by Hofmann bromamide degradation of acetamide. \(B = C_6H_5CONH_2\) (benzamide), formed by reaction of benzoyl chloride with a base (benzoylation).

Answer 3

Step 1: Note the formula.
Compound X is \(C_3H_9N\). This formula fits several amines: n-propylamine, isopropylamine (both primary), N-methylethylamine (secondary) and trimethylamine (tertiary).

Step 2: Recall the Hinsberg test.
Hinsberg reagent is benzenesulphonyl chloride \((C_6H_5SO_2Cl)\). It reacts differently with the three classes of amine, which lets us tell them apart.

Step 3: Behaviour of each class.
A primary amine gives a sulphonamide that has an acidic N-H, so it dissolves in alkali (KOH). A secondary amine gives a sulphonamide with no N-H, so it stays insoluble in alkali. A tertiary amine does not react at all.

Step 4: Use the clue.
X gives a product that is insoluble in alkali. That matches a secondary amine, because its sulphonamide has no acidic N-H to dissolve.

Step 5: Pick the secondary amine.
The only secondary amine with formula \(C_3H_9N\) is N-methylethylamine, \(CH_3-NH-C_2H_5\).

Answer: X is N-methylethylamine, \(CH_3NHC_2H_5\) (a secondary amine), because its Hinsberg product is a sulphonamide with no acidic N-H and is therefore insoluble in alkali.

Answer 4

Step 1: Plan the conversion.
We need to change \(-NH_2\) of aniline into \(-C\equiv N\) of benzonitrile. This is done in two steps: first make a diazonium salt, then replace the diazonium group with a cyanide group.

Step 2: Make the diazonium salt.
Treat aniline with nitrous acid (\(NaNO_2 + HCl\)) at a low temperature of 273 to 278 K (0 to 5 degree C). This gives benzenediazonium chloride. \[ C_6H_5NH_2 \xrightarrow{NaNO_2/HCl,\ 273-278\,K} C_6H_5N_2^+Cl^- \]

Step 3: Why diazotisation works.
The diazonium group \((-N_2^+)\) is an excellent leaving group, so many nucleophiles can replace it easily.

Step 4: Sandmeyer reaction.
Treat the diazonium salt with cuprous cyanide \((CuCN/KCN)\). The \(-N_2^+\) group is replaced by the cyano \((-CN)\) group, releasing nitrogen gas. \[ C_6H_5N_2^+Cl^- \xrightarrow{CuCN/KCN} C_6H_5CN + N_2 \]

Step 5: Final product.
The product is benzonitrile, \(C_6H_5CN\).

Answer: Aniline is first diazotised with \(NaNO_2/HCl\) at 273 to 278 K to give benzenediazonium chloride, which is then treated with CuCN/KCN (Sandmeyer reaction) to give benzonitrile \(C_6H_5CN\).

Answer 5

Step 1: Why aniline is hard to nitrate directly.
The \(-NH_2\) group is a very strong activating, electron-releasing group. It makes the benzene ring so reactive that nitration is hard to control and would give a mixture with a lot of unwanted product.

Step 2: The oxidation problem.
Nitration uses strong nitric acid, which is also a strong oxidising agent. Since aniline is easily oxidised, direct nitration can destroy part of it by oxidation, lowering the yield.

Step 3: Protect the amino group.
To solve both problems, aniline is first acetylated by treating it with acetic anhydride. This changes \(-NH_2\) into \(-NHCOCH_3\) (acetanilide).

Step 4: Effect of acetylation.
The acetamido group \((-NHCOCH_3)\) is much less strongly activating than \(-NH_2\), because the nitrogen lone pair is partly tied up with the carbonyl. So the ring is calmed down and is no longer easily oxidised, allowing controlled nitration mainly at the para position.

Step 5: Remove the protection.
After nitration, the acetyl group is removed by hydrolysis to get back the free \(-NH_2\) group, giving p-nitroaniline cleanly.

Answer: The \(-NH_2\) group is acetylated first because it converts the strongly activating, easily oxidised \(-NH_2\) into the milder \(-NHCOCH_3\) group. This controls the high reactivity, prevents oxidation by nitric acid, and allows clean nitration; the amino group is restored afterwards by hydrolysis.