Amines and Nitrogen Compounds Questions and Answers

Answer: A

The major product is (A): the CH₂NH₂ group is acetylated, giving CH₂NHCOCH₃, with acetic acid as by-product.

Acetic anhydride acylates the most nucleophilic nitrogen. The benzylic amine nitrogen has a free lone pair and is strongly nucleophilic. The amide nitrogen of CONH₂ is not: its lone pair is delocalised into the C=O group, so it does not react under these mild conditions. One equivalent of acetic anhydride therefore gives the N-acetylated amine and CH₃COOH.

(B) acetylates the wrong nitrogen. (C) acetylates both and also gives water, which is not a by-product of acylation by an anhydride. (D) again acetylates the amide nitrogen while leaving the more basic amine as a salt.

Answer: C

The products are (C): M is N-methyl-2-pyridone and N is 2-methoxypyridine.

2-Hydroxypyridine exists mainly as its tautomer 2-pyridone, and its anion is ambident, able to react at nitrogen or oxygen. With CH₃I and NaOH, the anion is alkylated at the softer, more nucleophilic nitrogen, giving the N-methyl pyridone M.

Diazomethane methylates acidic O-H groups: it takes the proton from the OH form to give a methyldiazonium ion, which the resulting oxygen anion attacks directly, so the product N is the O-methyl ether, 2-methoxypyridine.

(B) swaps the two products. (A) and (D) put a methyl group on a ring carbon, which neither reagent does.

Answer: D

The products are (D): W is pyridine N-oxide and X is benzaldehyde together with pyridine.

Peracetic acid oxidises the pyridine nitrogen to give pyridine N-oxide, W. The N-oxide oxygen is nucleophilic and is benzylated by benzyl bromide, giving an N-benzyloxypyridinium salt. Sodium hydroxide then removes a benzylic proton; the electrons move to form a C=O bond while the weak N-O bond breaks, releasing pyridine and leaving benzaldehyde. The N-oxide has acted as an oxidant, converting the benzyl halide to the aldehyde.

A peracid does not hydroxylate the ring, so W is not 4-hydroxypyridine, which rules out (A) and (B). (C) would need benzylation at a ring carbon, but the benzyl group goes onto oxygen, and the base then fragments that intermediate.

Answer: A

The products are (A): Y is vinyl isocyanate and Z is benzyl N-vinylcarbamate.

Sodium azide replaces the chloride of acryloyl chloride to give acryloyl azide. On heating, the acyl azide undergoes the Curtius rearrangement: nitrogen gas is lost and the vinyl group migrates from carbon to nitrogen, giving vinyl isocyanate, CH₂=CH-N=C=O (Y). Benzyl alcohol then adds across the C=N bond of the isocyanate to give the carbamate CH₂=CH-NH-C(O)-OCH₂Ph (Z).

(B) assumes azide adds to the C=C bond (conjugate addition) instead of replacing the chloride, and ignores the heating step. (C) is a 1,3-dipolar cycloaddition product, a triazoline. (D) keeps the acyl azide after heating, which is the condition that makes it rearrange, and then has benzyl alcohol add to the alkene instead of reacting with the isocyanate.

Answer: A

The starting materials are (A): tryptamine, benzaldehyde, acetic acid and cyclohexyl isocyanide.

A four-component reaction in methanol at room temperature giving a bis-amide is the Ugi reaction: an amine and an aldehyde form an imine, the isocyanide carbon adds to the protonated imine, the carboxylate traps the resulting nitrilium ion, and a Mumm rearrangement moves the acyl group onto nitrogen.

Mapping P back onto these parts: the nitrogen carrying the indolylethyl group and the acetyl group comes from tryptamine (amine) and acetic acid (acid); the CH(Ph) carbon comes from benzaldehyde; and the C(O)NH-cyclohexyl amide comes from cyclohexyl isocyanide.

The options with cyclohexanecarbonitrile fail because a nitrile (C≡N bonded through carbon) cannot play the isocyanide's role. The options using indole-3-acetaldehyde and benzylamine would put a benzyl group on the amide nitrogen and the indolylmethyl group on the α-carbon, a different product.

Answer: B

The major product is (B), 3-benzoylisoquinoline-1,4-diol.

Step 1 is a Gabriel alkylation: potassium phthalimide displaces bromide from phenacyl bromide to give N-phenacylphthalimide. Step 2 is the Gabriel-Colman rearrangement. Methoxide opens one imide carbonyl, and the enolate of the now acidic CH₂ (flanked by nitrogen and the benzoyl group) attacks the remaining ester carbonyl, closing a six-membered ring. The result is the isoquinoline skeleton with OH groups at C1 and C4 (the enol and lactam tautomers drawn as diols) and the benzoyl group at C3.

(A) is a quinoline, with the nitrogen attached to the benzene ring, which this ring expansion cannot give since nitrogen stays between the two former carbonyl carbons. (C) and (D) are indole and oxindole products with an N-phenacyl group, which would need loss of a carbon from the phthalimide.

Answer: C

The major product is (C), the tetralin nitrile with the gem-dimethyl group at the far ring-fusion side.

The oxime carbon is bonded to a quaternary CMe₂ carbon. In strong acid the protonated oxime does not undergo a normal Beckmann rearrangement; instead it undergoes a Beckmann fragmentation: the C-C bond to the CMe₂ carbon breaks, because that releases a stable tertiary carbocation, and the oxime carbon becomes an aryl nitrile (Ar-C≡N).

The cation sits at the end of a three-carbon chain on the ring carbon next to the new CN group. It is captured intramolecularly by the benzene ring (Friedel-Crafts alkylation) at the free ortho position on the far side of that chain, closing a six-membered ring. The product is a 1,1-dimethyltetralin with the CN on the aromatic carbon beside the CH₂-bearing ring fusion, as drawn in (C).

(D) places the CMe₂ next to the CN, which would need the cation to attack the carbon already carrying the nitrile. (A) and (B) are the lactams of an ordinary Beckmann rearrangement (alkyl or aryl migration), which the fragmentation to a tertiary cation outcompetes.

Answer: C

The order of basicity is II \(\gt\) I \(\gt\) III.

II is 2-imidazoline, a cyclic amidine. Protonating its C=N nitrogen gives a cation whose positive charge is shared equally by both nitrogens through resonance, so it is a strong base, with a conjugate acid pK_a around 11, far above the other two.

I is imidazole. Its pyridine-like nitrogen is protonated to give the symmetric imidazolium ion, which is also stabilised by resonance between the two nitrogens, so imidazole (conjugate acid pK_a about 7) is a stronger base than pyridine.

III is pyridine. Its lone pair sits in an sp² orbital and its conjugate acid gains no such resonance sharing, so it is the weakest base of the three (pK_a about 5.2). Orders that put pyridine above imidazole or the imidazoline anywhere but first are wrong.

Answer: B

The order of acidity is II \(\gt\) III \(\gt\) I.

These are conjugate acids, so the weaker the parent base, the stronger the acid. Pyrrolidine (I) is a saturated amine with a localised sp³ lone pair, a strong base, so its cation is the weakest acid. In III the nitrogen sits next to a C=C bond, so its lone pair is partly delocalised (an enamine) and the base is weaker. In pyrrole (II) the lone pair is part of the aromatic sextet; protonating nitrogen destroys aromaticity, so pyrrole is an extremely weak base and II is the strongest acid.

Any order placing I first reverses the effect of delocalising the lone pair, and III \(\gt\) II overlooks the loss of aromaticity in II.

Answer: B

The answer is (B): Y is the N,N-bis(2-ethoxycarbonylethyl)aniline, Z is 1-phenylpiperidin-4-one.

With excess ethyl acrylate, the aniline nitrogen adds twice in a Michael reaction, giving the tertiary amine with two CH₂CH₂CO₂Et arms (Y). Sodium ethoxide then drives a Dieckmann condensation: the carbon next to one ester attacks the other ester, closing a six-membered ring and forming a β-keto ester. Hydrolysis with NaOH, acidification and heating then loses CO₂ (decarboxylation) and leaves the 4-piperidone.

(A) and (C) show only a single addition, which excess acrylate rules out. (D) has the right Y but closes a five-membered ring, which does not form here.