2-Butanamine on reaction with excess CH3I and Ag2O, H2O followed by heating at 150°C, gives the following major product :

The major product is 1-butene.This is the Hofmann elimination. Excess CH3I methylates the amine exhaustively to the quaternary salt CH3CH2CH(CH3)N(CH3)3+ I−; moist Ag2O exchanges iodide for hydroxide; and on heating, hydroxide removes a β-hydrogen in an E2 reaction with trimethylamine as the leaving group.The bulky, positively charged -N(CH3)3+ group makes the base take the most accessible, most acidic β-hydrogen, which is on the terminal CH3, so the less substituted alkene (Hofmann product) dominates. 2-Butene is the more substituted Zaitsev product that a halide elimination would favour. Butane would need a reduction, and 2-butanol a substitution, neither of which occurs.

Amines and Nitrogen Compounds Questions and Answers

Q: Which of the following is not a natural amino acid ?

The one that is not a natural (protein) amino acid is (B), piperidine-2-carboxylic acid (pipecolic acid).The twenty amino acids coded by DNA and used to build proteins include proline (A), the five-membered cyclic amino acid, tryptophan (C), with its indole side chain, and tyrosine (D), with its p-hydroxyphenyl side chain.(B) is the six-membered ring homologue of proline. It does occur in living things, as a product of lysine breakdown and in some plants, but it is not one of the protein amino acids, which is the sense of 'natural amino acid' the paper uses. Its ring is one carbon larger than proline's, which is the detail to look for.

Q: Predict the major product of the following reaction :

The product is (A), the 1,4-disubstituted 1,2,3-triazole with the ester at C4.Step 1 is an SN2 displacement: azide replaces the bromide, giving the uracil-CH2-N3 azide. Step 2 is the copper-catalysed azide-alkyne cycloaddition (CuAAC, 'click' reaction). CuSO4 with sodium ascorbate generates Cu(I), which forms a copper acetylide from methyl propiolate and directs the cycloaddition so that only the 1,4-regioisomer forms.(B), the 1,5-isomer, is what ruthenium catalysts give, and a mixture of 1,4 and 1,5 is what the uncatalysed thermal reaction gives. (C) is a non-aromatic dihydrotriazole, and (D) wrongly introduces a C=C at the methylene linker; the triazole product is aromatic and the CH2 link is untouched.

Q: The major product of the following reaction is :

The product is (B), methylenecyclopentane.H2O2 oxidises the tertiary amine to its N-oxide, and heating brings about the Cope elimination: a concerted, syn elimination through a five-membered cyclic transition state in which the N-oxide oxygen removes a β-hydrogen.The β-hydrogens available are on the methyl group and on the two ring CH2 groups. Like the Hofmann elimination, the Cope elimination favours the less substituted alkene, here formed by removing a methyl hydrogen, giving the exocyclic methylene. (A), the more substituted endocyclic alkene, would be the Zaitsev product, which this elimination does not favour. (C) and (D) have the double bond away from the carbon that carried nitrogen, which an elimination cannot produce.

Q: Rank the following molecules in decreasing order of basicity :

The order is I \(\gt\) II \(\gt\) III \(\gt\) IV: piperidine \(\gt\) thiomorpholine \(\gt\) morpholine \(\gt\) N-methylmorpholine.Measured pKa values of the conjugate acids are about 11.1 for piperidine, 9.0 for thiomorpholine, 8.4 for morpholine and 7.4 for N-methylmorpholine.Piperidine (I) has nothing withdrawing electrons from nitrogen. In II and III a ring heteroatom two carbons away pulls electron density from N by induction; oxygen is more electronegative than sulfur, so morpholine (III) is weaker than thiomorpholine (II). N-methylmorpholine (IV) is a tertiary amine; in water, its protonated form is less well solvated by hydrogen bonding than that of the secondary amine, so it is the weakest base here.The orders that put III or I last reverse these effects.

Q: Which of the following compounds is an alkaloid ?

The alkaloid is morphine.Alkaloids are naturally occurring, mostly plant-derived compounds containing basic nitrogen, usually in a heterocyclic ring, and they typically have strong physiological effects. Morphine, from the opium poppy, has a tertiary amine nitrogen in its pentacyclic skeleton and is the classic example.The other three contain no nitrogen. Tocopherol (vitamin E) is a chromanol with a long isoprenoid side chain, and carvone (in spearmint and caraway oils) and camphor are monoterpene ketones.

Q: Complete the following reaction by suggesting the appropriate substrates (I & II) :

The dipeptide Boc-Ala-Val-OMe is built from Boc-L-alanine (I) and L-valine methyl ester (II), option A.To join alanine to valine with DCC/HOBt, the amino group of alanine and the carboxyl group of valine must be blocked: alanine is protected as N-Boc, leaving its COOH free to be activated, and valine as its methyl ester, leaving its NH2 free to attack. The starting amino acids are drawn as the L forms (substituent hashed on a lower carbon, wedged on an upper carbon), and the product keeps the same drawing, so I and II must keep it too. In A both stereocentres match the product and the starting materials.In B and D the alanine carbon is drawn with the methyl hashed on an upper carbon, which is the mirror image (D-alanine). C carries the residues the wrong way round, Boc-valine and alanine methyl ester, which would give Boc-Val-Ala-OMe.

Q: The major product of the following reaction is :

The keyed product is the open-chain amino ketone CH3COCH2CH(NH2)CH3, option D.The weakest bond of an isoxazole is the N-O bond, and H2 over Pd/C cleaves it. Opening the ring gives a β-amino enone (an enaminone), and with the hydrogenation continuing the C=C and C=N are reduced as well, so the carbonyl and an amine on a saturated chain remain. That is D, 4-aminopentan-2-one. The options A, B and C all keep the N-O bond, which a Pd/H2 reduction does not leave intact: B is the unchanged starting material, and A and C are ring-saturated isoxazolines that would themselves be opened by the same conditions.The enaminone intermediate is not offered, so D is the best of the choices; whether the reduction stops at the enaminone depends on conditions, and the key reads it as going all the way.

51.

Which of the following is not a natural amino acid ?

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The one that is not a natural (protein) amino acid is (B), piperidine-2-carboxylic acid (pipecolic acid).

The twenty amino acids coded by DNA and used to build proteins include proline (A), the five-membered cyclic amino acid, tryptophan (C), with its indole side chain, and tyrosine (D), with its p-hydroxyphenyl side chain.

(B) is the six-membered ring homologue of proline. It does occur in living things, as a product of lysine breakdown and in some plants, but it is not one of the protein amino acids, which is the sense of 'natural amino acid' the paper uses. Its ring is one carbon larger than proline's, which is the detail to look for.

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52.

Predict the major product of the following reaction :

6-(Bromomethyl)-5-methyluracil treated with 1. NaN3, DMF, rt; 2. methyl propiolate with CuSO4.5H2O (2 mol%), sodium ascorbate (5 mol%), tBuOH/H2O

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The product is (A), the 1,4-disubstituted 1,2,3-triazole with the ester at C4.

Step 1 is an S_N2 displacement: azide replaces the bromide, giving the uracil-CH₂-N₃ azide. Step 2 is the copper-catalysed azide-alkyne cycloaddition (CuAAC, 'click' reaction). CuSO₄ with sodium ascorbate generates Cu(I), which forms a copper acetylide from methyl propiolate and directs the cycloaddition so that only the 1,4-regioisomer forms.

(B), the 1,5-isomer, is what ruthenium catalysts give, and a mixture of 1,4 and 1,5 is what the uncatalysed thermal reaction gives. (C) is a non-aromatic dihydrotriazole, and (D) wrongly introduces a C=C at the methylene linker; the triazole product is aromatic and the CH₂ link is untouched.

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53.

The major product of the following reaction is :

1-Methylcyclopentyl dimethylamine treated with H2O2, then heated

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The product is (B), methylenecyclopentane.

H₂O₂ oxidises the tertiary amine to its N-oxide, and heating brings about the Cope elimination: a concerted, syn elimination through a five-membered cyclic transition state in which the N-oxide oxygen removes a β-hydrogen.

The β-hydrogens available are on the methyl group and on the two ring CH₂ groups. Like the Hofmann elimination, the Cope elimination favours the less substituted alkene, here formed by removing a methyl hydrogen, giving the exocyclic methylene. (A), the more substituted endocyclic alkene, would be the Zaitsev product, which this elimination does not favour. (C) and (D) have the double bond away from the carbon that carried nitrogen, which an elimination cannot produce.

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54.

Rank the following molecules in decreasing order of basicity :

Four six-membered amines: I piperidine; II thiomorpholine; III morpholine; IV N-methylmorpholine

A.\(\mathrm{I} \gt \mathrm{II} \gt \mathrm{III} \gt \mathrm{IV}\)
B.\(\mathrm{III} \gt \mathrm{IV} \gt \mathrm{II} \gt \mathrm{I}\)
C.\(\mathrm{I} \gt \mathrm{III} \gt \mathrm{II} \gt \mathrm{IV}\)
D.\(\mathrm{III} \gt \mathrm{II} \gt \mathrm{IV} \gt \mathrm{I}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The order is I \(\gt\) II \(\gt\) III \(\gt\) IV: piperidine \(\gt\) thiomorpholine \(\gt\) morpholine \(\gt\) N-methylmorpholine.

Measured pK_a values of the conjugate acids are about 11.1 for piperidine, 9.0 for thiomorpholine, 8.4 for morpholine and 7.4 for N-methylmorpholine.

Piperidine (I) has nothing withdrawing electrons from nitrogen. In II and III a ring heteroatom two carbons away pulls electron density from N by induction; oxygen is more electronegative than sulfur, so morpholine (III) is weaker than thiomorpholine (II). N-methylmorpholine (IV) is a tertiary amine; in water, its protonated form is less well solvated by hydrogen bonding than that of the secondary amine, so it is the weakest base here.

The orders that put III or I last reverse these effects.

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55.

2-Butanamine on reaction with excess CH₃I and Ag₂O, H₂O followed by heating at 150°C, gives the following major product :

A.1-butene
B.2-butene
C.butane
D.2-butanol

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The major product is 1-butene.

This is the Hofmann elimination. Excess CH₃I methylates the amine exhaustively to the quaternary salt CH₃CH₂CH(CH₃)N(CH₃)₃⁺ I⁻; moist Ag₂O exchanges iodide for hydroxide; and on heating, hydroxide removes a β-hydrogen in an E2 reaction with trimethylamine as the leaving group.

The bulky, positively charged -N(CH₃)₃⁺ group makes the base take the most accessible, most acidic β-hydrogen, which is on the terminal CH₃, so the less substituted alkene (Hofmann product) dominates. 2-Butene is the more substituted Zaitsev product that a halide elimination would favour. Butane would need a reduction, and 2-butanol a substitution, neither of which occurs.

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56.

Which of the following compounds is an alkaloid ?

A.Morphine
B.Tocopherol
C.Carvone
D.Camphor

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The alkaloid is morphine.

Alkaloids are naturally occurring, mostly plant-derived compounds containing basic nitrogen, usually in a heterocyclic ring, and they typically have strong physiological effects. Morphine, from the opium poppy, has a tertiary amine nitrogen in its pentacyclic skeleton and is the classic example.

The other three contain no nitrogen. Tocopherol (vitamin E) is a chromanol with a long isoprenoid side chain, and carvone (in spearmint and caraway oils) and camphor are monoterpene ketones.

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57.

Complete the following reaction by suggesting the appropriate substrates (I & II) :

L-alanine on protection gives I and II, which couple with DCC and HOBt to give Boc-Ala-Val-OMe; L-valine is drawn below the arrow

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The dipeptide Boc-Ala-Val-OMe is built from Boc-L-alanine (I) and L-valine methyl ester (II), option A.

To join alanine to valine with DCC/HOBt, the amino group of alanine and the carboxyl group of valine must be blocked: alanine is protected as N-Boc, leaving its COOH free to be activated, and valine as its methyl ester, leaving its NH₂ free to attack. The starting amino acids are drawn as the L forms (substituent hashed on a lower carbon, wedged on an upper carbon), and the product keeps the same drawing, so I and II must keep it too. In A both stereocentres match the product and the starting materials.

In B and D the alanine carbon is drawn with the methyl hashed on an upper carbon, which is the mirror image (D-alanine). C carries the residues the wrong way round, Boc-valine and alanine methyl ester, which would give Boc-Val-Ala-OMe.

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58.

The major product of the following reaction is :

3,5-Dimethylisoxazole treated with H2 over Pd/C

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The keyed product is the open-chain amino ketone CH₃COCH₂CH(NH₂)CH₃, option D.

The weakest bond of an isoxazole is the N-O bond, and H₂ over Pd/C cleaves it. Opening the ring gives a β-amino enone (an enaminone), and with the hydrogenation continuing the C=C and C=N are reduced as well, so the carbonyl and an amine on a saturated chain remain. That is D, 4-aminopentan-2-one. The options A, B and C all keep the N-O bond, which a Pd/H₂ reduction does not leave intact: B is the unchanged starting material, and A and C are ring-saturated isoxazolines that would themselves be opened by the same conditions.

The enaminone intermediate is not offered, so D is the best of the choices; whether the reduction stops at the enaminone depends on conditions, and the key reads it as going all the way.

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