The correct match of natural product in Column II with class in Column I is :Column IColumn II(a) Terpene(i) Strychnine(b) Alkaloid(ii) Cortisone(c) Steroids(iii) Chitin(d) Polysaccharide(iv) Camphor

The correct match is (a-iv) (b-i) (c-ii) (d-iii).Camphor is a bicyclic monoterpene ketone (terpene, a-iv).Strychnine is an indole alkaloid (b-i).Cortisone is a corticosteroid hormone with the steroid nucleus (c-ii).Chitin is a polysaccharide of N-acetylglucosamine (d-iii).Option B calls strychnine a terpene, option C calls chitin a steroid, and option D calls chitin a terpene; each misclassifies at least two compounds.

The correct match of the natural product in column II with class in column I is :Column IColumn II(a) Vitamin(i) Glycogen(b) Terpene(ii) Nicotine(c) Polysaccharide(iii) Ascorbic acid(d) Alkaloid(iv) Carvone

The correct match is (a-iii) (b-iv) (c-i) (d-ii).(a) Vitamin - (iii) ascorbic acid, vitamin C.(b) Terpene - (iv) carvone, a monoterpene ketone (C10) of spearmint and caraway oils.(c) Polysaccharide - (i) glycogen, the branched glucose polymer that stores energy in animals.(d) Alkaloid - (ii) nicotine, the nitrogenous base of tobacco.The first option makes carvone a vitamin and ascorbic acid a terpene. The second calls glycogen a vitamin and ascorbic acid a polysaccharide. The third swaps the terpene and the alkaloid, calling nicotine a terpene and carvone an alkaloid, although carvone contains no nitrogen.

Structure and Bonding in Organic Compounds Questions and Answers

Q: The correct order of acidity of the following molecules is :

The order is (IV) \(\lt\) (III) \(\lt\) (II) \(\lt\) (I): toluene \(\lt\) 4-nitrophenol \(\lt\) 4-methylbenzoic acid \(\lt\) benzoic acid.Toluene's methyl C-H is barely acidic (pKa about 41). 4-Nitrophenol is a phenol made more acidic by the para nitro group, which delocalises the phenoxide charge (pKa about 7.2), but it is still weaker than a carboxylic acid. Benzoic acid has pKa 4.20; a para methyl group donates electrons, destabilising the carboxylate slightly, so 4-methylbenzoic acid is a little weaker (pKa 4.37).The other orders put the phenol above the carboxylic acids, or rank toluene above the phenol, which the pKa values contradict.

Q: Which of the following will form the most stable complex with TiCl4 ?

The most stable complex is formed by (D), the amide N,N-diethylpropanamide.TiCl4 is a hard Lewis acid that binds the carbonyl oxygen. The more electron-rich that oxygen, the stronger the complex. In an amide, the nitrogen lone pair is delocalised into the carbonyl (N-C=O ↔ N+=C-O-), placing substantial negative charge on oxygen and making amides the most Lewis-basic carbonyl compounds.Ketones and aldehydes lack such a strong donor, and in an ester the alkoxy oxygen donates less effectively than nitrogen. The order of basicity is amide \(\gt\) ketone \(\gt\) aldehyde, ester, so the amide forms the most stable TiCl4 adduct.

Q: Which one among the following arrows is the correct representation of resonance ?

Resonance is shown by the double-headed arrow (B), ↔.Resonance structures are not different molecules in equilibrium; they are alternative drawings of one molecule whose true structure is a hybrid of them. The single straight arrow with a head at each end is reserved for linking such contributing structures.The other arrows mean something else. (A) is a pair of curved arrows of the kind used to show electrons moving in a mechanism. (C), the pair of half-headed arrows, denotes a reversible chemical equilibrium between distinct species. (D), the hollow arrow, is the retrosynthetic arrow, read as 'can be made from'.

Q: Which of the structures given below do NOT follow isoprene rule ?

The structure that does not follow the isoprene rule is D.The isoprene rule says a terpenoid carbon skeleton is built from 2-methylbutane (C5) units joined head to tail. Written from the carboxyl carbon, D is HOOC-C(CH3)=CH-CH2-CH2-CH=C(CH3)2: the first methyl sits on C-2, directly beside the COOH. In a head-to-tail chain such as geranic acid the methyl branches fall on C-3 and C-7, so D has its first branch one carbon too close to the carboxyl end and the units do not link head to tail.A is menthol and B is camphor, both monoterpenes. C is a chain with methyl branches at the spacing the rule requires (as in farnesoic acid), so A, B and C all follow the rule.

Q: A prodrug is :

A prodrug is an inactive drug that is transformed in the body to an active metabolite.A prodrug is administered in an inactive or weakly active form and converted by metabolism, usually by enzymes such as esterases or by oxidation in the liver, into the active drug. The approach improves absorption, stability, taste or targeting. Examples are enalapril, hydrolysed to enalaprilat, and levodopa, decarboxylated to dopamine in the brain.A 'prototype' (printed 'protype') is the first or model member of a drug class, and the oldest member is not necessarily that. A drug stored in tissues and released slowly describes a depot or tissue reservoir effect, not a prodrug.

Q: The correct order of the acidity between pairs of compounds given below is :

The order is P > Q and S > R.All four are active methylene compounds whose CH2 proton is acidic because the anion is shared over two carbonyl groups. Approximate pKa values in water: pentane-2,4-dione (P) 9; dimethyl malonate (Q) 13; dimedone (R) 5.2; Meldrum's acid (S) 4.8.P is more acidic than Q because ester carbonyls are less electron-withdrawing than ketone carbonyls: the alkoxy oxygen donates into the C=O and stabilises the anion less. S is more acidic than R, an anomaly explained by the rigid ring of Meldrum's acid, which holds the ester groups in a conformation that strongly stabilises the enolate. The options with Q > P or R > S reverse one of these.

21.

The correct order of acidity of the following molecules is :

Four benzene derivatives labelled I to IV: benzoic acid (I), 4-methylbenzoic acid (II), 4-nitrophenol (III) and toluene (IV)

A.(IV) \(\lt\) (III) \(\lt\) (II) \(\lt\) (I)
B.(III) \(\lt\) (IV) \(\lt\) (II) \(\lt\) (I)
C.(IV) \(\lt\) (II) \(\lt\) (I) \(\lt\) (III)
D.(IV) \(\lt\) (I) \(\lt\) (III) \(\lt\) (II)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The order is (IV) \(\lt\) (III) \(\lt\) (II) \(\lt\) (I): toluene \(\lt\) 4-nitrophenol \(\lt\) 4-methylbenzoic acid \(\lt\) benzoic acid.

Toluene's methyl C-H is barely acidic (pK_a about 41). 4-Nitrophenol is a phenol made more acidic by the para nitro group, which delocalises the phenoxide charge (pK_a about 7.2), but it is still weaker than a carboxylic acid. Benzoic acid has pK_a 4.20; a para methyl group donates electrons, destabilising the carboxylate slightly, so 4-methylbenzoic acid is a little weaker (pK_a 4.37).

The other orders put the phenol above the carboxylic acids, or rank toluene above the phenol, which the pK_a values contradict.

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22.

Which of the following will form the most stable complex with TiCl₄ ?

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The most stable complex is formed by (D), the amide N,N-diethylpropanamide.

TiCl₄ is a hard Lewis acid that binds the carbonyl oxygen. The more electron-rich that oxygen, the stronger the complex. In an amide, the nitrogen lone pair is delocalised into the carbonyl (N-C=O ↔ N⁺=C-O^-), placing substantial negative charge on oxygen and making amides the most Lewis-basic carbonyl compounds.

Ketones and aldehydes lack such a strong donor, and in an ester the alkoxy oxygen donates less effectively than nitrogen. The order of basicity is amide \(\gt\) ketone \(\gt\) aldehyde, ester, so the amide forms the most stable TiCl₄ adduct.

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23.

The correct match of natural product in Column II with class in Column I is :

Column I	Column II
(a) Terpene	(i) Strychnine
(b) Alkaloid	(ii) Cortisone
(c) Steroids	(iii) Chitin
(d) Polysaccharide	(iv) Camphor

A.(a-iv) (b-i) (c-ii) (d-iii)
B.(a-i) (b-ii) (c-iv) (d-iii)
C.(a-iv) (b-i) (c-iii) (d-ii)
D.(a-iii) (b-iv) (c-ii) (d-i)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The correct match is (a-iv) (b-i) (c-ii) (d-iii).

Camphor is a bicyclic monoterpene ketone (terpene, a-iv).
Strychnine is an indole alkaloid (b-i).
Cortisone is a corticosteroid hormone with the steroid nucleus (c-ii).
Chitin is a polysaccharide of N-acetylglucosamine (d-iii).

Option B calls strychnine a terpene, option C calls chitin a steroid, and option D calls chitin a terpene; each misclassifies at least two compounds.

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24.

Which one among the following arrows is the correct representation of resonance ?

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

Resonance is shown by the double-headed arrow (B), ↔.

Resonance structures are not different molecules in equilibrium; they are alternative drawings of one molecule whose true structure is a hybrid of them. The single straight arrow with a head at each end is reserved for linking such contributing structures.

The other arrows mean something else. (A) is a pair of curved arrows of the kind used to show electrons moving in a mechanism. (C), the pair of half-headed arrows, denotes a reversible chemical equilibrium between distinct species. (D), the hollow arrow, is the retrosynthetic arrow, read as 'can be made from'.

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25.

The correct match of the natural product in column II with class in column I is :

Column I	Column II
(a) Vitamin	(i) Glycogen
(b) Terpene	(ii) Nicotine
(c) Polysaccharide	(iii) Ascorbic acid
(d) Alkaloid	(iv) Carvone

A.(a-iv) (b-iii) (c-i) (d-ii)
B.(a-i) (b-iv) (c-iii) (d-ii)
C.(a-iii) (b-ii) (c-i) (d-iv)
D.(a-iii) (b-iv) (c-i) (d-ii)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The correct match is (a-iii) (b-iv) (c-i) (d-ii).

(a) Vitamin - (iii) ascorbic acid, vitamin C.
(b) Terpene - (iv) carvone, a monoterpene ketone (C₁₀) of spearmint and caraway oils.
(c) Polysaccharide - (i) glycogen, the branched glucose polymer that stores energy in animals.
(d) Alkaloid - (ii) nicotine, the nitrogenous base of tobacco.

The first option makes carvone a vitamin and ascorbic acid a terpene. The second calls glycogen a vitamin and ascorbic acid a polysaccharide. The third swaps the terpene and the alkaloid, calling nicotine a terpene and carvone an alkaloid, although carvone contains no nitrogen.

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26.

Which of the structures given below do NOT follow isoprene rule ?

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The structure that does not follow the isoprene rule is D.

The isoprene rule says a terpenoid carbon skeleton is built from 2-methylbutane (C₅) units joined head to tail. Written from the carboxyl carbon, D is HOOC-C(CH₃)=CH-CH₂-CH₂-CH=C(CH₃)₂: the first methyl sits on C-2, directly beside the COOH. In a head-to-tail chain such as geranic acid the methyl branches fall on C-3 and C-7, so D has its first branch one carbon too close to the carboxyl end and the units do not link head to tail.

A is menthol and B is camphor, both monoterpenes. C is a chain with methyl branches at the spacing the rule requires (as in farnesoic acid), so A, B and C all follow the rule.

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27.

A prodrug is :

A.A protype member of a class of drugs
B.The oldest member of a class of drugs
C.An inactive drug that is transformed in the body to an active metabolite
D.A drug that is stored in the body tissues and is gradually released in the circulation

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

A prodrug is an inactive drug that is transformed in the body to an active metabolite.

A prodrug is administered in an inactive or weakly active form and converted by metabolism, usually by enzymes such as esterases or by oxidation in the liver, into the active drug. The approach improves absorption, stability, taste or targeting. Examples are enalapril, hydrolysed to enalaprilat, and levodopa, decarboxylated to dopamine in the brain.

A 'prototype' (printed 'protype') is the first or model member of a drug class, and the oldest member is not necessarily that. A drug stored in tissues and released slowly describes a depot or tissue reservoir effect, not a prodrug.

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28.

The correct order of the acidity between pairs of compounds given below is :

Four structures: P is pentane-2,4-dione, Q is dimethyl malonate, R is 5,5-dimethylcyclohexane-1,3-dione and S is 2,2-dimethyl-1,3-dioxane-4,6-dione

A.Q > P and R > S
B.P > Q and R > S
C.P > Q and S > R
D.Q > P and S > R

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The order is P > Q and S > R.

All four are active methylene compounds whose CH₂ proton is acidic because the anion is shared over two carbonyl groups. Approximate pK_a values in water: pentane-2,4-dione (P) 9; dimethyl malonate (Q) 13; dimedone (R) 5.2; Meldrum's acid (S) 4.8.

P is more acidic than Q because ester carbonyls are less electron-withdrawing than ketone carbonyls: the alkoxy oxygen donates into the C=O and stabilises the anion less. S is more acidic than R, an anomaly explained by the rigid ring of Meldrum's acid, which holds the ester groups in a conformation that strongly stabilises the enolate. The options with Q > P or R > S reverse one of these.

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