The correct match of natural products in Column-I with class in Column-II is :Column-IColumn-II(P) Cholesterol(i) Polysaccharide(Q) Cellulose(ii) Steroid(R) Quinine(iii) Vitamin(iv) Alkaloid

The correct match is (P)-(ii), (Q)-(i), (R)-(iv).Cholesterol is a steroid, built on the four-ring cyclopentanoperhydrophenanthrene nucleus.Cellulose is a polysaccharide, a linear polymer of β-1,4-linked D-glucose units.Quinine, from cinchona bark, is an alkaloid, a nitrogenous base with a quinoline and a quinuclidine ring.Options A and D call cholesterol a vitamin or a polysaccharide, and option B calls cellulose an alkaloid and quinine a polysaccharide, so only C matches all three.

Structure and Bonding in Organic Compounds Questions and Answers

Q: An organic compound is heated with Na2O2 then boiled with HNO3. The solution is then treated with ammonium molybdate. The yellow precipitate obtained is due to the presence of the element

The yellow precipitate shows phosphorus.Heating the compound with sodium peroxide oxidises any phosphorus to phosphate. After boiling with nitric acid, ammonium molybdate gives a canary-yellow precipitate of ammonium phosphomolybdate, (NH4)3PO4·12MoO3, which is the test for phosphorus.Nitrogen and sulphur are detected through the sodium fusion (Lassaigne's) extract, nitrogen as Prussian blue and sulphur with sodium nitroprusside or lead acetate. Carbon is not tested this way, and molybdenum comes from the reagent itself.

Q: When 0.25 g of an organic compound is completely oxidised with excess oxygen, 0.22 g of CO2 is produced. What is the percentage of carbon in the compound?

The compound contains 24 % carbon.All the carbon ends up in CO2, and 44 g of CO2 contains 12 g of carbon.\[ \begin{aligned} \text{mass of C} &= \frac{12}{44} \times 0.22 = 0.06 \text{ g} \\ \%\ \text{C} &= \frac{0.06}{0.25} \times 100 = 24 \end{aligned} \]48 % is twice the true value; the key step is that only 12/44 of the mass of CO2 is carbon.

Q: In Carious method, 0.40 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is (Atomic mass of Ag = 108 g mol-1 and Br = 80 g mol-1).

The percentage of bromine is 20 %.AgBr has molar mass 108 + 80 = 188 g mol-1, so a fraction \(\frac{80}{188}\) of it is bromine:\[ \begin{aligned} \text{mass of Br} &= \frac{80}{188} \times 0.188 = 0.08\ \text{g} \\ \%\ \text{Br} &= \frac{0.08}{0.40} \times 100 = 20\ \% \end{aligned} \]The numbers are chosen so that 0.188 g of AgBr is exactly 0.001 mol. 24 % and 25 % come from using the mass of AgBr or of silver in the wrong place.

Q: Which of the following method/s is/are used in the estimation of nitrogen in organic compounds?(i) Carius method(ii) Dumas method(iii) Silver salt method(iv) Kjeldhal method

Nitrogen is estimated by the Dumas and Kjeldahl methods, (ii) and (iv).In the Dumas method the compound is heated with copper(II) oxide, and the nitrogen gas set free is collected and measured. In the Kjeldahl method the nitrogen is converted to ammonium sulphate, the ammonia released is absorbed in a known amount of acid, and the acid left over is titrated.The Carius method estimates halogens, sulphur and phosphorus, and the silver salt method is used to find the molar mass of an organic acid. So every option that includes (i) or (iii) is wrong.

Q: The technique used to purify liquids having very high boiling points and decompose at or below their boiling point is

The technique is distillation under reduced pressure.A liquid boils when its vapour pressure equals the pressure above it. Lowering the pressure lowers the boiling point, so a liquid that would decompose at its normal boiling point can be distilled at a lower temperature without decomposing. Glycerol is purified this way, and the method is used to concentrate sugar-cane juice.Ordinary and fractional distillation work at atmospheric pressure, at which the compound would decompose. Steam distillation suits steam-volatile liquids that are immiscible with water, and differential extraction separates a compound from a solution using a second solvent.

Q: When sodium fusion extract is treated with sodium nitropruside, the appearance of violet colour is due to the formation of

The violet colour is due to [Fe(CN)5NOS]4-.In Lassaigne's test for sulphur, sulphide ions in the sodium fusion extract react with sodium nitroprusside:S2- + [Fe(CN)5NO]2- → [Fe(CN)5NOS]4-The 2- charge of nitroprusside plus the 2- of sulphide gives 4- for the violet complex. [Fe(CN)5NO]2- is the nitroprusside ion itself, before reaction, and the other charges do not balance.

Q: The correct order of acidity for the following compounds is :

The order of acidity is III > II > I > IV.III, 4-cyanobenzoic acid, is a carboxylic acid (pKa about 3.5); the electron-withdrawing CN makes it stronger than benzoic acid. II, 4-nitrophenol, is a phenol whose anion is stabilised by resonance onto the para nitro group, pKa about 7.2. I, 4-methoxyphenol, is a phenol weakened by the electron-donating methoxy group, pKa about 10.2. IV, 4-nitrobenzyl alcohol, is an alcohol: its OH is on a CH2, so the anion cannot delocalise into the ring and its pKa is about 15.Any order that puts the phenol II above the carboxylic acid III, or the alcohol IV above a phenol, contradicts these values, which rules out the other three options.

Q: Which one of the following has the highest melting point ?

The highest melting point belongs to p-chlorophenol.Approximate melting points: p-chlorophenol about 43 °C, m-chlorophenol and m-bromophenol about 33 °C, and o-bromophenol about 5 °C, a liquid at room temperature.Two effects explain the pattern. Para isomers are symmetrical and pack more efficiently in the crystal, so they melt higher than meta isomers. Ortho halophenols form an intramolecular hydrogen bond between the OH and the neighbouring halogen, which reduces hydrogen bonding between molecules, so they melt lowest.

11.

An organic compound is heated with Na₂O₂ then boiled with HNO₃. The solution is then treated with ammonium molybdate. The yellow precipitate obtained is due to the presence of the element

A.nitrogen
B.sulphur
C.phosphorus
D.carbon
E.molybdenum

Exam Year(s)

KEAM - 2026,

Answer: C

The yellow precipitate shows phosphorus.

Heating the compound with sodium peroxide oxidises any phosphorus to phosphate. After boiling with nitric acid, ammonium molybdate gives a canary-yellow precipitate of ammonium phosphomolybdate, (NH₄)₃PO₄·12MoO₃, which is the test for phosphorus.

Nitrogen and sulphur are detected through the sodium fusion (Lassaigne's) extract, nitrogen as Prussian blue and sulphur with sodium nitroprusside or lead acetate. Carbon is not tested this way, and molybdenum comes from the reagent itself.

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12.

When 0.25 g of an organic compound is completely oxidised with excess oxygen, 0.22 g of CO₂ is produced. What is the percentage of carbon in the compound?

A.40 %
B.24 %
C.48 %
D.30 %
E.18 %

Exam Year(s)

KEAM - 2026,

Answer: B

The compound contains 24 % carbon.

All the carbon ends up in CO₂, and 44 g of CO₂ contains 12 g of carbon.

\[ \begin{aligned} \text{mass of C} &= \frac{12}{44} \times 0.22 = 0.06 \text{ g} \\ \%\ \text{C} &= \frac{0.06}{0.25} \times 100 = 24 \end{aligned} \]

48 % is twice the true value; the key step is that only 12/44 of the mass of CO₂ is carbon.

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13.

In Carious method, 0.40 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is (Atomic mass of Ag = 108 g mol^-1 and Br = 80 g mol^-1).

A.30 %
B.25 %
C.35 %
D.24 %
E.20 %

Exam Year(s)

KEAM - 2026,

Answer: E

The percentage of bromine is 20 %.

AgBr has molar mass 108 + 80 = 188 g mol^-1, so a fraction \(\frac{80}{188}\) of it is bromine:

\[ \begin{aligned} \text{mass of Br} &= \frac{80}{188} \times 0.188 = 0.08\ \text{g} \\ \%\ \text{Br} &= \frac{0.08}{0.40} \times 100 = 20\ \% \end{aligned} \]

The numbers are chosen so that 0.188 g of AgBr is exactly 0.001 mol. 24 % and 25 % come from using the mass of AgBr or of silver in the wrong place.

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14.

Which of the following method/s is/are used in the estimation of nitrogen in organic compounds?

(i) Carius method
(ii) Dumas method
(iii) Silver salt method
(iv) Kjeldhal method

A.(i) only
B.(ii) and (iii)
C.(iii) and (iv)
D.(ii) and (iv)
E.(i) and (iii)

Exam Year(s)

KEAM - 2026,

Answer: D

Nitrogen is estimated by the Dumas and Kjeldahl methods, (ii) and (iv).

In the Dumas method the compound is heated with copper(II) oxide, and the nitrogen gas set free is collected and measured. In the Kjeldahl method the nitrogen is converted to ammonium sulphate, the ammonia released is absorbed in a known amount of acid, and the acid left over is titrated.

The Carius method estimates halogens, sulphur and phosphorus, and the silver salt method is used to find the molar mass of an organic acid. So every option that includes (i) or (iii) is wrong.

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15.

The technique used to purify liquids having very high boiling points and decompose at or below their boiling point is

A.steam distillation
B.distillation
C.differential extraction
D.distillation under reduced pressure
E.fractional distillation

Exam Year(s)

KEAM - 2026,

Answer: D

The technique is distillation under reduced pressure.

A liquid boils when its vapour pressure equals the pressure above it. Lowering the pressure lowers the boiling point, so a liquid that would decompose at its normal boiling point can be distilled at a lower temperature without decomposing. Glycerol is purified this way, and the method is used to concentrate sugar-cane juice.

Ordinary and fractional distillation work at atmospheric pressure, at which the compound would decompose. Steam distillation suits steam-volatile liquids that are immiscible with water, and differential extraction separates a compound from a solution using a second solvent.

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16.

When sodium fusion extract is treated with sodium nitropruside, the appearance of violet colour is due to the formation of

A.\(\mathrm{[Fe(CN)_5NO]^{2-}}\)
B.\(\mathrm{[Fe(CN)_5NO]^{2+}}\)
C.\(\mathrm{[Fe(CN)_5NOS]^{4-}}\)
D.\(\mathrm{[Fe(CN)_5NOS]^{2-}}\)
E.\(\mathrm{[Fe(CN)_5NOS]^{2+}}\)

Exam Year(s)

KEAM - 2026,

Answer: C

The violet colour is due to [Fe(CN)₅NOS]^4-.

In Lassaigne's test for sulphur, sulphide ions in the sodium fusion extract react with sodium nitroprusside:

S^2- + [Fe(CN)₅NO]^2- → [Fe(CN)₅NOS]^4-

The 2- charge of nitroprusside plus the 2- of sulphide gives 4- for the violet complex. [Fe(CN)₅NO]^2- is the nitroprusside ion itself, before reaction, and the other charges do not balance.

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17.

The correct order of acidity for the following compounds is :

Four para-substituted benzene derivatives labelled I to IV: 4-methoxyphenol (I), 4-nitrophenol (II), 4-cyanobenzoic acid (III) and 4-nitrobenzyl alcohol (IV)

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

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The order of acidity is III > II > I > IV.

III, 4-cyanobenzoic acid, is a carboxylic acid (pK_a about 3.5); the electron-withdrawing CN makes it stronger than benzoic acid. II, 4-nitrophenol, is a phenol whose anion is stabilised by resonance onto the para nitro group, pK_a about 7.2. I, 4-methoxyphenol, is a phenol weakened by the electron-donating methoxy group, pK_a about 10.2. IV, 4-nitrobenzyl alcohol, is an alcohol: its OH is on a CH₂, so the anion cannot delocalise into the ring and its pK_a is about 15.

Any order that puts the phenol II above the carboxylic acid III, or the alcohol IV above a phenol, contradicts these values, which rules out the other three options.

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18.

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

Column-I	Column-II
(P) Cholesterol	(i) Polysaccharide
(Q) Cellulose	(ii) Steroid
(R) Quinine	(iii) Vitamin
	(iv) Alkaloid

A.(P)-(iii), (Q)-(i), (R)-(iv)
B.(P)-(ii), (Q)-(iv), (R)-(i)
C.(P)-(ii), (Q)-(i), (R)-(iv)
D.(P)-(i), (Q)-(iii), (R)-(iv)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The correct match is (P)-(ii), (Q)-(i), (R)-(iv).

Cholesterol is a steroid, built on the four-ring cyclopentanoperhydrophenanthrene nucleus.
Cellulose is a polysaccharide, a linear polymer of β-1,4-linked D-glucose units.
Quinine, from cinchona bark, is an alkaloid, a nitrogenous base with a quinoline and a quinuclidine ring.

Options A and D call cholesterol a vitamin or a polysaccharide, and option B calls cellulose an alkaloid and quinine a polysaccharide, so only C matches all three.

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19.

Amongst the following the rate of cis-trans isomerisation is expected to be highest in :

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The fastest cis-trans isomerisation is expected in B, the cyclopentadienylidene-cyclopropene (calicene) system.

Cis-trans isomerisation needs rotation about the central C=C bond, and the barrier is low when the twisted, dipolar form is stabilised. In calicene, polarising the central bond puts a positive charge on the three-membered ring and a negative charge on the five-membered ring: a cyclopropenium cation (2 π electrons) and a cyclopentadienide anion (6 π electrons), both aromatic. This large contribution of the dipolar form gives the central bond much single-bond character.

In A (two five-membered rings) and C (two three-membered rings) both rings would need the same charge type, so no doubly aromatic dipolar form exists. In D, the cycloheptatrienylidene and cyclopropenylidene rings both prefer to be cations, so again the polarisation is not favourable.

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20.

Which one of the following has the highest melting point ?

A.o-bromophenol
B.p-chlorophenol
C.m-bromophenol
D.m-chlorophenol

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The highest melting point belongs to p-chlorophenol.

Approximate melting points: p-chlorophenol about 43 °C, m-chlorophenol and m-bromophenol about 33 °C, and o-bromophenol about 5 °C, a liquid at room temperature.

Two effects explain the pattern. Para isomers are symmetrical and pack more efficiently in the crystal, so they melt higher than meta isomers. Ortho halophenols form an intramolecular hydrogen bond between the OH and the neighbouring halogen, which reduces hydrogen bonding between molecules, so they melt lowest.

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