Question 1

Lactic acid loses its optical activity when the hydroxyl group is replaced by hydrogen because

Accepted Answer

the chiral center present in the molecule is eliminated. Lactic acid CH3-CHOH-COOH becomes CH3-CH2-COOH when OH is replaced by H, eliminating the asymmetric (chiral) carbon - hence loss of optical activity. The cause is removal of the chiral centre, not a mere change of configuration.

Question 2

Which is the most suitable technique for separating a mixture of sugars?

Accepted Answer

Chromatography. A mixture of sugars (non-volatile, similar properties) is best separated by chromatography. Sugars do not sublime, so the stored 'sublimation' is wrong.

Question 3

Which statement correctly compares the staggered and eclipsed conformations of ethane?

Accepted Answer

The staggered form of ethane is more stable than the eclipsed form because the staggered form has no torsional strain.. The staggered conformation of ethane is more stable than the eclipsed one because it is free of torsional (eclipsing) strain. The stored statement (eclipsed more stable) is backwards.

Question 4

Which statement correctly describes a common feature shared by optical and geometrical isomerism?

Accepted Answer

Both of these types are classified under stereoisomerism.. Optical and geometrical isomerism are both subclasses of stereoisomerism - that is their shared feature. One does not require the presence of the other, so the stored statement is wrong.

Question 5

Which substituent lowers the reactivity of a benzene ring toward electrophilic substitution, yet still mainly directs a new group to the ortho and para positions?

Accepted Answer

–Cl. -Cl deactivates the ring (electron-withdrawing -I) yet directs o,p (lone-pair resonance donation). -NO2 deactivates but is meta-directing, so the stored answer is wrong.

Question 6

Among the following substances, which has the greatest affinity for a proton?

Accepted Answer

NH3. Ammonia (NH3) has a lone pair of electrons on the nitrogen atom, which allows it to effectively donate this pair to bond with a proton, making it a stronger base compared to the other options.

Question 7

When chlorobenzene (C6H5Cl) undergoes sulphonation, which product is obtained?

Accepted Answer

A mixture of o- and p-Chlorobenzenesulphonic acids is produced. In chlorobenzene the -Cl group is an ortho/para director, so electrophilic sulphonation gives a mixture of o- and p-chlorobenzenesulphonic acids (para predominating).

Question 8

Which of the following compounds can also exist as a meso form?

Accepted Answer

2,3-Dichlorobutane. 2,3-Dichlorobutane (CH3-CHCl-CHCl-CH3) has two identical stereocentres, so it has a meso form with an internal mirror plane. 2,3-Dichloropentane has two different stereocentres (the two halves are not equivalent), so it cannot be meso.

Question 9

When silver benzoate is treated with bromine, the product obtained is

Accepted Answer

C6H5Br. Silver benzoate with bromine undergoes the Hunsdiecker reaction: C6H5COOAg + Br2 -> C6H5Br + CO2 + AgBr. The product is bromobenzene, C6H5Br.

Question 10

Epichlorohydrin corresponds to which of the following compounds?

Accepted Answer

3-Chloro-1,2-epoxypropane. Epichlorohydrin is 3-chloro-1,2-epoxypropane (chloromethyloxirane), ClCH2-CH(-O-)CH2 with an epoxide ring. It is not a simple chloropropane or chloropropanol.

Question 11

Zerevitinov’s method for estimating active hydrogen in an organic compound relies on its reaction with which reagent?

Accepted Answer

CH3MgI. Zerevitinov’s method utilizes CH3MgI, a Grignard reagent, to react with active hydrogen in organic compounds, allowing for the quantification of hydrogen atoms that can participate in such reactions.

Question 12

Osmium tetroxide is commonly employed as a reagent for

Accepted Answer

converting olefins into cis-diols. Osmium tetroxide adds across a C=C double bond via a cyclic osmate ester, delivering both -OH groups to the same face. Hydrolysis gives a cis (syn) 1,2-diol from the olefin.

Question 13

Which alkyl halide is required to prepare 5-methyl-2-hexanone from acetoacetic ester?

Accepted Answer

(CH3)2CHCH2Br. Acetoacetic ester after alkylation, hydrolysis and decarboxylation gives a methyl ketone CH3-CO-CH2-R. For 5-methyl-2-hexanone, CH3-CO-CH2-CH2-CH(CH3)-CH3, the fragment R is -CH2-CH(CH3)2 (isobutyl), so the required alkyl halide is isobutyl bromide, (CH3)2CHCH2Br.

Question 14

When two moles of ethyl acetate undergo self-condensation in the presence of sodium ethoxide, which product is formed?

Accepted Answer

Acetoacetic ester. Two molecules of ethyl acetate undergo Claisen self-condensation with NaOEt to give ethyl 3-oxobutanoate (ethyl acetoacetate), i.e. acetoacetic ester.

Question 15

Which compound is obtained when malonic ester is heated with urea?

Accepted Answer

Barbituric acid. Diethyl malonate condenses with urea (base catalysed) with loss of two ethanol molecules to form the cyclic ureide barbituric acid (malonylurea). This is the classic synthesis of barbiturates.

Question 16

Which starting compound gives the precursor used in the nitration that produces the powerful explosive RDX?

Accepted Answer

urotropine. RDX (cyclonite) is prepared by nitration of hexamethylenetetramine (urotropine), the condensation product of formaldehyde and ammonia. So urotropine is the precursor.

Question 17

What kind of isomerism is shown by C6H5C≡N and C6H5N≡C?

Accepted Answer

Functional isomerism. C6H5-C#N is benzonitrile (a nitrile) and C6H5-N#C is phenyl isocyanide; same molecular formula but different functional groups, so they are functional isomers.

Question 18

Match each compound in List I with the reaction it characteristically shows in List II. Choose the correct matching.

List I (Compounds)
(1) CH3CH2CH2NH2
(2) CH3C≡CH
(3) CH3CH2COOCH3
(4) CH3CH(OH)CH3

List II (Reactions)
(i) undergoes alkaline hydrolysis
(ii) with alcoholic KOH and CHCl3 g

Accepted Answer

(ii) (iii) (i) (iv). (1) Propylamine gives the carbylamine (foul odour) reaction with CHCl3/alc. KOH -> (ii); (2) propyne, a terminal alkyne, gives a white precipitate with ammoniacal AgNO3 -> (iii); (3) the ester undergoes alkaline hydrolysis -> (i); (4) isopropanol, a secondary alcohol, gives turbidity with Lucas reagent after ~5 min -> (iv). Order: (ii)(iii)(i)(iv).

Question 19

A nitroalkane is treated with HONO and forms a product that is insoluble in alkali but becomes blue when alkali is added. Which nitroalkane is it?

Accepted Answer

(CH3)2CHNO2. With HNO2, a secondary nitroalkane forms a pseudonitrole, R2C(NO)(NO2), which is blue and insoluble in alkali. Primary nitroalkanes give nitrolic acids (red in alkali) and tertiary give no reaction; hence the compound is (CH3)2CHNO2.

Question 20

When ethyl isocyanide is hydrolysed in an acidic medium, which products are formed?

Accepted Answer

ethylamine salt and methanoic acid. Acidic hydrolysis of an isocyanide (R-NC) gives a primary amine and formic acid. Ethyl isocyanide therefore yields ethylamine (as its salt) and methanoic (formic) acid.

Question 21

In aqueous solution, identify the nature of each compound in List I and match it with List II.

List I
I. Acetamide
II. Benzonitrile
III. Triethylamine
IV. Phenol

List II
A. Acidic
B. Basic
C. Neutral

Accepted Answer

I–C; II–C; III–B; IV–A. Acetamide and benzonitrile are essentially neutral in water, triethylamine is basic, and phenol is weakly acidic. Thus I-C, II-C, III-B, IV-A.

Question 22

When aniline (C6H5NH2) is treated with carbon disulphide in the presence of HgCl2 and heat, which product is formed?

Accepted Answer

phenyl isothiocyanate. Heating aniline with carbon disulphide gives a dithiocarbamic acid intermediate, which is desulphurised by HgCl2 to form phenyl isothiocyanate (C6H5-N=C=S) - the Hofmann mustard-oil reaction.

Question 23

Which solvent is most suitable for taking out a butter stain from fabric?

Accepted Answer

C2H5OC2H5. Butter is a fat/grease, which dissolves best in a non-polar organic solvent. Diethyl ether (C2H5OC2H5) is the most suitable choice; water and ethanol are too polar to remove the grease.

Question 24

When the shade of a dye becomes deeper, the absorption band moves toward the red end of the spectrum. What is this shift called?

Accepted Answer

bathochromic shift. A bathochromic (red) shift moves the absorption band to longer wavelengths, deepening the colour. A hypsochromic shift is the opposite (blue, shorter wavelength); hyperchromic/hypochromic refer to intensity, not position.

Question 25

Which one of the following constitutes a group of the isoelectronic species?

Accepted Answer

NO⁺, C2^−, CN^−, N2. Isoelectronic species have equal electron counts. NO+ , C2^2-, CN- and N2 each have 14 electrons, so they form the isoelectronic group (option as written shows a typo C2^- for C2^2-). The stored set contains O2 (16 e), so it cannot be isoelectronic with CO/N2.

Question 26

Which amongst the following is the strongest acid?

Accepted Answer

CH(CN)3. In CH(CN)3 the three electron-withdrawing -CN groups delocalize/stabilize the negative charge of the conjugate carbanion most effectively (pKa ~ -5), making it by far the strongest acid among the four.

Question 27

Identify the compound that does not show chirality.

Accepted Answer

1-chloropentane. 1-chloropentane (ClCH2-CH2-CH2-CH2-CH3) has no stereocentre, so it is achiral. The other three each contain a carbon attached to four different groups and are therefore chiral.

Question 28

What kind of isomerism is exhibited by 2,3-dichlorobutane?

Accepted Answer

Optical isomerism. 2,3-dichlorobutane has two stereocentres bearing identical groups, giving a pair of enantiomers plus a meso form, so it shows optical isomerism (option index 2).

Question 29

Arrange the following free radicals in order of increasing stability:

Accepted Answer

(CH3)2CH < (CH3)3C < (C6H5)2CH < (C6H5)3C. Radical stability increases with hyperconjugation/resonance: (CH3)2CH (2-degree alkyl) < (CH3)3C (3-degree alkyl) < (C6H5)2CH (benzylic, 2 rings) < (C6H5)3C (triphenylmethyl, 3 rings, most resonance-stabilized). Correct order: (CH3)2CH < (CH3)3C < (C6H5)2CH < (C6H5)3C.

Question 30

For 2-fluoroethanol, which sequence correctly represents the increasing order of stability of its three principal conformations—eclipsed, anti, and gauche?

Accepted Answer

Eclipsed, anti, gauche. In 2-fluoroethanol the gauche conformer is stabilized by an intramolecular O-H...F hydrogen bond (and the gauche effect), making it the most stable; the eclipsed form is the least stable. Increasing stability: eclipsed < anti < gauche.

Question 31

For the conjugate bases shown below, with R = CH3, which sequence gives the order of increasing basic strength?

Accepted Answer

RCOO− < HC≡C− < NH2 < R. Conjugate-acid acidity: RCOOH (pKa ~5) > HC≡CH (~25) > NH3 (~38) > CH4 (~50). Since stronger acid gives weaker base, increasing base strength is RCOO- < HC≡C- < NH2- < R- (CH3-).

Question 32

A solution of (−)-1-chloro-1-phenylethane in toluene undergoes slow racemization when a small quantity of SbCl5 is added, because it forms a:

Accepted Answer

carbocation. The addition of SbCl5 facilitates the formation of a carbocation by stabilizing the positive charge, allowing for the racemization process to occur as the molecule can interconvert between enantiomers.

Question 33

Arrange the following carbocations in decreasing order of stability:
I. CH2=CH–CH2+
II. CH3–CH2–CH2+
III. C6H5–CH2+

Accepted Answer

III > I > II. C6H5-CH2+ (benzyl) is most stabilised by ring resonance, CH2=CH-CH2+ (allyl) is resonance-stabilised, and CH3-CH2-CH2+ is an unstabilised primary cation. Order: III > I > II (index 3).

Question 34

Which of the following alkenes can show geometrical isomerism?

Accepted Answer

1-Phenyl-2-butene. Geometrical isomerism needs each double-bond carbon to carry two different groups. 1-Phenyl-2-butene (Ph-CH2-CH=CH-CH3) has H plus a different group on each sp2 carbon -> shows cis/trans. The others have a terminal =CH2 or two identical groups (=CPh2), so no geometrical isomerism.

Question 35

What is the absolute configuration of the compound represented below?

CO2H
|
H — C — OH
|
H — C — Cl
|
CH3

Accepted Answer

(2S, 3R). At C2 priorities OH>C3>CO2H>H; with H on a horizontal bond the apparent rotation is clockwise, inverted gives S. At C3 priorities Cl>C2>CH3>H; apparent counter-clockwise, inverted gives R. Configuration is (2S, 3R).

Question 36

When a single molecule of HBr reacts with a single molecule of 1,3-butadiene at 40°C, the major product formed is

Accepted Answer

1-bromo-2-butene formed under thermodynamic control. The correct option is right because at higher temperatures, the reaction favors the formation of the more stable product, which in this case is 1-bromo-2-butene. This product is formed under thermodynamic control, indicating that the reaction has reached equilibrium and the more stable product predominates.

Question 37

Among the five structural isomers of hexane, which one produces only two distinct monochloro derivatives on chlorination?

Accepted Answer

2,3-dimethylbutane. 2,3-dimethylbutane has a symmetrical structure that allows for only two unique positions for chlorine substitution during chlorination, resulting in just two distinct monochloro derivatives.

Question 38

Which gas escaped from the storage tank at the Union Carbide plant during the Bhopal gas disaster?

Accepted Answer

Methyl isocyanate. Methyl isocyanate was the toxic gas that leaked from the storage tank at the Union Carbide plant, leading to the catastrophic Bhopal disaster, as it is a highly hazardous chemical used in pesticide production.

Question 39

Which one of the following substances has the highest proton affinity ?

Accepted Answer

NH3. Ammonia (NH3) has the highest proton affinity among the given options due to its ability to stabilize the positive charge on the resulting ammonium ion (NH4+) through effective resonance and inductive effects, making it more favorable for protonation compared to the other substances.

Question 40

For alkyl halides containing a halogen atom X, which sequence represents the decreasing order of reactivity toward an S_N2 reaction?

Accepted Answer

RCH2X > R2CHX > R3CX. SN2 proceeds by backside attack, so steric crowding slows it down. Reactivity therefore decreases from primary to tertiary: RCH2X > R2CHX > R3CX.

Question 41

On monochlorination of 2-methylbutane, how many chiral compounds can be formed?

Accepted Answer

4. Monochlorination of 2-methylbutane gives 1-chloro-2-methylbutane and 2-chloro-3-methylbutane as the only products with a stereocentre. Each exists as a pair of enantiomers, so the number of chiral compounds formed is 2 x 2 = 4.

Question 42

Arrange the following compounds in order of decreasing acidity:
(I) p-chlorophenol
(II) p-cresol
(III) p-nitrophenol
(IV) p-methoxyphenol

Accepted Answer

III > I > II > IV. The correct order of acidity is determined by the electron-withdrawing and electron-donating effects of the substituents on the phenol ring. p-Nitrophenol (III) is the most acidic due to the strong electron-withdrawing nitro group, while p-methoxyphenol (IV) is the least acidic because the methoxy group donates electrons, reducing acidity.

Question 43

Which sequence represents the acids in order of decreasing acidity?

Accepted Answer

NO2CH2COOH > CNCH2COOH > FCH2COOH > ClCH2COOH. Acidity tracks the electron-withdrawing power of the substituent. Approximate pKa values: nitroacetic 1.68, cyanoacetic 2.47, fluoroacetic 2.59, chloroacetic 2.87, giving decreasing acidity NO2CH2COOH > CNCH2COOH > FCH2COOH > ClCH2COOH.

Question 44

A substance having molecular mass 180 undergoes acetylation with CH3COCl, producing a derivative of molecular mass 390. How many amino groups are present in each molecule of the original substance?

Accepted Answer

5. Each acetylation replaces an H with COCH3, a net mass increase of 42. The total increase is 390 - 180 = 210, so 210/42 = 5 groups are acetylated, meaning 5 amino groups.

Question 45

In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is:
(at. mass Ag = 108; Br = 80)

Accepted Answer

24. To find the percentage of bromine in the compound, we first determine the moles of AgBr produced, which is 141 mg divided by the molar mass of AgBr (188 mg/mol). This gives us the moles of bromine, which we then convert to mass and calculate the percentage based on the original mass of the organic compound.

Question 46

A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO3 to give fraction A. The left over organic phase was extracted with dilute NaOH solution to give fraction B. The final organic layer was labell

Accepted Answer

m-chlorobenzoic acid, m-chlorophenol and m-chloroaniline. Fraction A contains m-chlorobenzoic acid, which is acidic and soluble in the NaHCO3 solution. Fraction B contains m-chlorophenol, which is a weak acid and can be deprotonated by dilute NaOH, leaving m-chloroaniline, a neutral compound, in the final organic layer (fraction C). Thus, the correct order of fractions is m-chlorobenzoic acid in A, m-chlorophenol in B, and m-chloroaniline in C.

Question 47

In S_N2 reactions, the correct order of reactivity for the following compounds: CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl is

Accepted Answer

CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl. In S_N2 reactions, the reactivity order is influenced by steric hindrance; primary alkyl halides like CH3Cl react more readily than secondary and tertiary ones due to less steric crowding, making it easier for the nucleophile to attack the carbon atom.

Question 48

In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is:
(at. mass Ag = 108; Br = 80)

Accepted Answer

24. To find the percentage of bromine in the compound, we first determine the moles of AgBr produced, which is 141 mg divided by the molar mass of AgBr (188 mg/mol). This gives us the moles of bromine, which is then converted to mass and expressed as a percentage of the original 250 mg of the organic compound, resulting in 24%.

Question 49

Which of the following compounds will exhibit geometrical isomerism?
(1) 1 - Phenyl - 2 - butene
(2) 3 - Phenyl - 1 - butene
(3) 2 - Phenyl - 1 - butene
(4) 1, 1 - Diphenyl - 1 - propane

Accepted Answer

1 - Phenyl - 2 - butene. 1-Phenyl-2-butene has a double bond between the second and third carbon atoms, allowing for different spatial arrangements of the substituents around the double bond, which is the basis for geometrical isomerism. The other options either lack a double bond or do not have the necessary substituents to create distinct isomers.

Question 50

The absolute configuration of

CO2H
 |
H—C—OH
 |
H—C—Cl
 |
 CH3

is:

Accepted Answer

(2S, 3R). The correct option (2S, 3R) is determined by analyzing the chiral centers at carbon 2 and carbon 3. At carbon 2, the highest priority group is the carboxylic acid (CO2H), which gives it an 'S' configuration, while at carbon 3, the chlorine (Cl) takes priority over the hydroxyl (OH) and methyl (CH3) groups, resulting in an 'R' configuration.

JEE Main Chemistry: Organic Chemistry – Basic Principles and Techniques questions with solutions

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