Question 1

What is the product formed when toluene (C6H5CH3) is treated with chromyl chloride (CrO2Cl2)?

Accepted Answer

benzaldehyde. Toluene with chromyl chloride (Etard reaction) oxidises the methyl group only as far as the aldehyde, giving benzaldehyde, not phenylacetic acid.

Question 2

o-Xylene is first treated with HNO3 to give compound X, which on reaction with phenol in the presence of H2SO4 forms product Y. What is Y?

Accepted Answer

Phenolphthalein. Oxidation/nitration handling of o-xylene gives phthalic acid (X), which as its anhydride condenses with two molecules of phenol under H2SO4 to form phenolphthalein (Y).

Question 3

What products are formed when a benzene ring containing a formyl group (–CHO) and a chloro substituent is treated with 50% KOH?

Accepted Answer

A benzene ring containing –CH2OH and –Cl substituents, along with a benzene ring containing –COO− and –Cl substituents. An aromatic aldehyde with no alpha-hydrogen undergoes the Cannizzaro reaction with concentrated (50%) KOH: one molecule is reduced to the alcohol (-CH2OH) and one is oxidised to the carboxylate (-COO-). The ring-bound chlorine is not displaced, so the products are a ring bearing -CH2OH and -Cl and a ring bearing -COO- and -Cl.

Question 4

Which of the following compounds contains a carboxyl group (–COOH)?

Accepted Answer

Aspirin. Aspirin (acetylsalicylic acid) retains a free carboxyl group -COOH on the ring. Picric acid (trinitrophenol) and ascorbic acid (an enediol lactone) have no -COOH, and barbituric acid has only amide/keto groups.

Question 5

Which of the following compounds is converted to the corresponding hydrocarbon on reduction with zinc and hydrochloric acid?

Accepted Answer

Butan-2-one. Zinc and HCl (Clemmensen-type) reduces a carbonyl group to a methylene, giving the hydrocarbon. Of the choices only butan-2-one (a ketone) is reduced this way (to butane); acids, esters and amides are not converted to hydrocarbons under these conditions.

Question 6

Why do carboxylic acids not undergo the typical addition–elimination reactions shown by aldehydes and ketones?

Accepted Answer

Because the carboxylate ion is stabilised by resonance. In carboxylic acids the carbonyl carbon is deactivated toward nucleophilic addition because the -COOH group is resonance stabilised (the lone pair on the -OH oxygen delocalises onto C=O), reducing the electrophilicity of the carbonyl carbon. Hence they resist the addition reactions typical of aldehydes/ketones.

Question 7

Which reagent can convert phenylmethyl ketone directly into ethylbenzene in a single step?

Accepted Answer

Zn-Hg/HCl. Acetophenone (PhCOCH3) is converted to ethylbenzene (PhCH2CH3) in one step by Clemmensen reduction (Zn-Hg/HCl), which reduces the carbonyl directly to a methylene. LiAlH4/NaBH4 stop at the alcohol.

Question 8

In the following sequence of reactions, acetic acid is converted into product C:

CH3COOH + PCl5 → A
A —(C6H6 / anhyd. AlCl3)→ B
B —(C2H5MgBr / dry ether)→ C

What is the structure of product C?

Accepted Answer

CH3–C(OH)(C2H5)C6H5. The correct option is right because the sequence of reactions indicates that acetic acid is first converted to an acyl chloride, which then undergoes Friedel-Crafts alkylation with benzene to form a product that contains both a benzene ring and an ethyl group attached to a central carbon, resulting in the structure CH3–C(OH)(C2H5)C6H5.

Question 9

Which reagent can convert C6H5CH=CHCHO into C6H5CH=CHCH2OH without affecting the carbon–carbon double bond?

Accepted Answer

NaBH4. NaBH4 is a selective reducing agent that can reduce aldehydes to alcohols without affecting carbon–carbon double bonds, making it suitable for converting the given compound into the desired alcohol.

Question 10

When acetaldehyde undergoes condensation with semicarbazide, the product formed is semicarbazone. Which of the following represents its structure?

Accepted Answer

CH3CH=N–NHCONH2. The correct option, CH3CH=N–NHCONH2, accurately represents the structure of semicarbazone formed from the condensation of acetaldehyde and semicarbazide, featuring the characteristic semicarbazone functional group with the appropriate bonding and functional groups.

Question 11

Which reagent can be used to convert benzoic acid into ethyl benzoate?

Accepted Answer

Dry hydrogen chloride in ethanol. Benzoic acid is converted to ethyl benzoate by Fischer esterification: heating with ethanol in the presence of an acid catalyst such as dry HCl gas (or conc. H2SO4). Ethanol alone or ethyl chloride will not esterify it efficiently.

Question 12

When a mixture of sodium benzoate is heated with soda-lime, the product formed is

Accepted Answer

benzene. Heating sodium benzoate with soda lime (NaOH/CaO) removes CO2 (decarboxylation), giving benzene.

Question 13

Among the following carbonyl compounds, which one undergoes nucleophilic addition most readily?

Accepted Answer

CH3—C(=O)—H. Nucleophilic addition is fastest for the least hindered, least stabilised carbonyl. Acetaldehyde (CH3CHO) has only one +I alkyl group and no conjugation, so it reacts fastest; acetone has two alkyl groups, and the benzaldehyde/acetophenone carbonyls are deactivated by resonance with the ring.

Question 14

The tetrahedral intermediate shown below collapses to form:
CH3—CH2—C(OH)(OCH3)—Cl

Accepted Answer

methyl propanoate and HCl. The tetrahedral intermediate undergoes a rearrangement where the hydroxyl group is converted into a methoxy group, leading to the formation of methyl propanoate, while releasing HCl as a byproduct.

Question 15

Which of the following compounds does not give either a semicarbazone or an oxime?

Accepted Answer

CH3CONHCH3. The compound CH3CONHCH3, which is an amide, does not react to form a semicarbazone or an oxime because these reactions typically occur with carbonyl compounds (aldehydes and ketones), while amides do not have the necessary reactive carbonyl structure to participate in these reactions.

Question 16

Which of the following is unable to reduce Fehling's solution?

Accepted Answer

Ethanoic acid. Ethanoic acid, being a carboxylic acid, does not have the ability to act as a reducing agent to reduce Fehling's solution, unlike the other options which contain aldehyde functional groups capable of reducing it.

Question 17

The transformation of acetaldehyde to ethyl acetate in the presence of aluminum ethoxide is known as

Accepted Answer

Tischenko reaction. The Tischenko reaction involves the conversion of acetaldehyde to ethyl acetate through the action of an aluminum alkoxide catalyst, which facilitates the esterification process by promoting the coupling of two aldehyde molecules.

Question 18

Nitrobenzene is obtained from benzene by treating it with a mixed acid of concentrated HNO3 and concentrated H2SO4. In this nitration mixture, nitric acid functions as a/an:

Accepted Answer

base. In the nitrating mixture, conc. H2SO4 is the stronger acid and protonates HNO3; HNO3 thus acts as a base, accepting a proton before losing water to give the nitronium ion NO2+.

Question 19

Which of the following nitrogen-containing compounds gives a primary amine (R–NH2) on treatment with bromine and concentrated KOH, i.e., under Hofmann rearrangement conditions?

Accepted Answer

RCONH2. The Hofmann bromamide degradation requires a primary (unsubstituted) amide: RCONH2 + Br2 + 4KOH -> RNH2 + 2KBr + K2CO3 + 2H2O, giving an amine with one fewer carbon. N-substituted amides and the others do not undergo it.

Question 20

Which of the following compounds yields propylamine on hydrolysis?

Accepted Answer

CH3–C(=O)–NHCH2CH2CH3. CH3-CO-NH-CH2CH2CH3 (N-propylacetamide) on hydrolysis gives acetic acid plus propylamine (CH3CH2CH2NH2). Nitrile hydrolysis would give an acid, not an amine, so it is the amide that yields propylamine.

Question 21

Which compound is formed when o-xylene is oxidized with HNO3, and then on reaction with phenol in the presence of H2SO4 gives phenolphthalein?

Accepted Answer

Phthalic anhydride. Phthalic anhydride is produced when o-xylene undergoes oxidation with HNO3, and this compound can subsequently react with phenol in the presence of H2SO4 to form phenolphthalein, making it the correct answer.

Question 22

For the substitution reaction R–C(=O)–X + Nu− → R–C(=O)–Nu + X−, the rate is highest when the leaving group X is:

Accepted Answer

–Cl. Reactivity of acyl derivatives toward nucleophilic acyl substitution follows leaving-group ability: acid chloride > anhydride > ester > amide. -Cl is the best leaving group, so X = -Cl gives the highest rate.

Question 23

In the reaction sequence shown below:
Toluene KMnO4 A SOCl2 B H2/Pd, BaSO4 C
what is the final compound C?

Accepted Answer

C6H5CHO. The reaction sequence involves the oxidation of toluene to form benzaldehyde (C6H5CHO) using KMnO4, followed by conversion of the resulting carboxylic acid to an acyl chloride with SOCl2, and finally, the reduction step does not affect the aldehyde, confirming that C is benzaldehyde.

Question 24

In Cannizzaro reaction given below
2PhCHO —OH−→ PhCH2OH + PhCO2−
the slowest step is:

Accepted Answer

the transfer of hydride to the carbonyl group. The transfer of hydride to the carbonyl group is the slowest step because it involves the formation of a new bond and is typically the rate-determining step in reactions involving nucleophilic addition to carbonyl compounds.

Question 25

An organic compound A upon reacting with NH3 gives B. On heating B gives C. C in presence of KOH reacts with Br2 to give CH3CH2NH2. A is:

Accepted Answer

CH3CH2COOH. The correct option is CH3CH2COOH because it is a carboxylic acid that can react with ammonia to form an amide, which upon heating can lead to an amine. The final product, CH3CH2NH2, indicates that the original compound must have a two-carbon chain to yield the correct amine after the reactions.

Question 26

In the sequence of reactions,
CH3COOH —LiAlH4→ A
A —PCl5→ B
B —alcoholic KOH→ C,
what is the compound C?

Accepted Answer

Ethene. The sequence starts with acetic acid (CH3COOH) being reduced by LiAlH4 to form ethanol (A). When ethanol reacts with PCl5, it forms ethyl chloride (B), which upon treatment with alcoholic KOH undergoes elimination to produce ethene (C).

Question 27

A sodium salt of an organic acid, denoted as X, gives brisk effervescence on treatment with concentrated H2SO4. X also forms a white precipitate with acidified aqueous CaCl2, and this precipitate decolourises acidic KMnO4 solution. Identify X.

Accepted Answer

Na2C2O4. X = Na2C2O4 (sodium oxalate). With conc. H2SO4 it gives brisk effervescence (CO + CO2), with acidified CaCl2 it forms insoluble white CaC2O4, and oxalate is a reducing agent that decolourises acidic KMnO4. Sodium formate fails because calcium formate is water-soluble (no white precipitate).

Question 28

What is the principal product formed in the following sequence?
R–C≡N
(i) AlH(i-Bu)2
(ii) H2O

Accepted Answer

RCHO. DIBAL-H (AlH(i-Bu)2) delivers one hydride to the nitrile to give an imine-aluminium intermediate; aqueous work-up hydrolyses it to the aldehyde. Hence R-C#N -> R-CHO.

Question 29

What is the principal product formed in the following reaction?

CH3CH = CHCO2CH3
LiAlH4 →

Accepted Answer

CH3CH = CHCH2OH. LiAlH4 reduces the -CO2CH3 ester group to a primary alcohol but does not reduce the carbon-carbon double bond, giving CH3CH=CHCH2OH.

Question 30

An organic compound A upon reacting with NH3 gives B. On heating, B gives C. C in presence of KOH reacts with Br2 to give CH3CH2CH2NH2. A is -

Accepted Answer

CH3CH2CH2COOH. Propylamine CH3CH2CH2NH2 (3 carbons) comes from the Hofmann degradation of a 4-carbon amide. So C = CH3CH2CH2CONH2, B is the same amide, and A = CH3CH2CH2COOH (butanoic acid), which forms the amide with NH3.

Question 31

Compound (A), C8H9Br, gives a white precipitate when warmed with alcoholic AgNO3. Oxidation of (A) gives an acid (B), C8H6O4. (B) easily forms anhydride on heating. Identify the compound (A).

Accepted Answer

o-bromobenzyl bromide with a methyl group on the ring. The correct option is o-bromobenzyl bromide with a methyl group on the ring because the presence of the ortho-bromine and the methyl group allows for the formation of a carboxylic acid upon oxidation, which can subsequently form an anhydride, consistent with the properties described.

Question 32

In the reaction CH3COOH —LiAlH4→ A —PCl5→ B —Alc. KOH→ C the product C is -

Accepted Answer

Ethylene. The reaction sequence involves the reduction of acetic acid (CH3COOH) to ethanol, which is then dehydrated by PCl5 to form an alkene. The presence of alcoholic KOH facilitates the elimination of water, resulting in the formation of ethylene.

Question 33

Sodium salt of an organic acid 'X' produces effervescence with conc. H2SO4. 'X' reacts with the acidified aqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of KMnO4. 'X' is

Accepted Answer

Na2C2O4. The correct option, Na2C2O4, is a sodium salt of oxalic acid, which reacts with concentrated H2SO4 to produce carbon dioxide, causing effervescence. Additionally, it forms a white precipitate with CaCl2 due to the formation of calcium oxalate, and it can reduce KMnO4, indicating its ability to act as a reducing agent.

Question 34

A compound 'X' on treatment with Br2/NaOH, provided C3H9N, which gives positive carbylamine test. Compound 'X' is:
(1) CH3CH2CH2CONH2
(2) CH3CON(CH3)2
(3) CH3CH2COCH2NH2
(4) CH3COCH2NHCH3

Accepted Answer

CH3CH2CH2CONH2. Br2/NaOH (Hofmann bromamide degradation) converts a primary amide RCONH2 to RNH2 with one fewer carbon. To obtain C3H9N that gives a positive carbylamine test (a primary amine, propan-1-amine), X must be butanamide CH3CH2CH2CONH2. The other choices are not primary amides and cannot undergo this reaction to give a primary amine.

Question 35

Which of the following compounds will show the maximum enol content?

Accepted Answer

CH3COCH2COCH3. Pentane-2,4-dione (CH3COCH2COCH3) is a symmetric 1,3-diketone whose enol is stabilized by intramolecular H-bonding and conjugation, giving the maximum enol content among these compounds.

Question 36

In the following reaction carbonyl compound + MeOH ⇌ acetal. Rate of the reaction is the highest for:

Accepted Answer

Propanal as substrate and methanol in excess. The reaction rate is highest when propanal is used as the substrate with methanol in excess because the presence of excess methanol shifts the equilibrium towards the formation of the acetal, thus enhancing the reaction rate.

Question 37

The increasing order of the following compounds towards HCN addition is:
(i) o-methoxybenzaldehyde
(ii) o-nitrobenzaldehyde
(iii) m-methoxybenzaldehyde
(iv) m-nitrobenzaldehyde

Accepted Answer

(i) < (iii) < (iv) < (ii). The correct order reflects the influence of substituents on the reactivity of the carbonyl group towards HCN addition. Electron-donating groups like methoxy increase nucleophilicity, while electron-withdrawing groups like nitro decrease it, leading to the observed increasing order of reactivity.

Question 38

An organic compound [A], molecular formula C10H20O2 was hydrolyzed with dilute sulphuric acid to give a carboxylic acid [B] and an alcohol [C]. Oxidation of [C] with CrO3 – H2SO4 produced [B]. Which of the following structure are not possible for [A] ?

Accepted Answer

CH3CH2CH2COOCH2CH2CH3. The structure CH3CH2CH2COOCH2CH2CH3 cannot produce a carboxylic acid and an alcohol upon hydrolysis because it lacks a suitable functional group that would yield a corresponding alcohol upon oxidation, making it incompatible with the requirements of the reaction.

Question 39

The compound A in the following reactions is:
A
(i) CH3MgBr/H2O
(ii) Conc. H2SO4/Δ
→
B
(i) O2
(ii) Zn/H2O
→ C + D
C
(i) Conc. KOH
(ii) Δ
→ C6H5COO−K+ + C6H5CH2OH
D
Ba(OH)2
Δ
→ CH3–C(CH3)=CH–CO–CH3

Accepted Answer

(2) C6H5–CH2–CO–CH3. The correct option is C6H5–CH2–CO–CH3 because the reactions indicate that compound A undergoes a Grignard reaction followed by acid hydrolysis, leading to the formation of a benzyl alcohol and a ketone, which aligns with the structure of option (2).

Question 40

The increasing order of the reactivity of the following compounds in nucleophilic addition reaction is: Propanal, Benzaldehyde, Propanone, Butanone

Accepted Answer

Benzaldehyde < Butanone < Propanone < Propanal. Reactivity to nucleophilic addition: aldehydes > ketones, and more/larger alkyl groups (steric + +I) reduce it; benzaldehyde is least reactive due to ring conjugation. Increasing order: Benzaldehyde < Butanone < Propanone < Propanal.

Question 41

Given below are two statements:
Statement I: The nucleophilic addition of sodium hydrogen sulphite to an aldehyde or a ketone involves proton transfer to form a stable ion.
Statement II: The nucleophilic addition of hydrogen cyanide to an aldehyde or a ketone yields amine as final product.

Accepted Answer

Statement I is true but Statement II is false.. Statement I is correct because the addition of sodium hydrogen sulfite to aldehydes or ketones does involve a proton transfer that leads to the formation of a stable bisulfite adduct. However, Statement II is incorrect because the reaction of hydrogen cyanide with aldehydes or ketones typically forms a cyanohydrin, not an amine.

Question 42

To synthesize 1.0 mole of 2-methylpropan-2-ol from Ethylethanoate ______ equivalents of CH3MgBr reagent will be required. (Integer value)

Accepted Answer

2. CH3COOC2H5 reacts with CH3MgBr: the first equivalent gives a ketone (acetone) and the second adds to it, giving (CH3)3COH = 2-methylpropan-2-ol. Thus 2 equivalents of CH3MgBr are required.

Question 43

Which one of the following reactions will not form acetaldehyde?
(1) CH3CH2OH —Cu, 573 K→
(2) CH3CN —(i) DIBAL-H (ii) H2O→
(3) CH2=CH2 + O2 —Pd(II)/Cu(II)→ H2O
(4) CH3CH2OH —CrO3/H2SO4→

Accepted Answer

(4). Reactions (1) Cu/573K dehydrogenation, (2) CH3CN + DIBAL-H then H2O, and (3) Wacker oxidation of ethene all give acetaldehyde. But (4) ethanol with CrO3/H2SO4 (a strong oxidant) oxidizes all the way to acetic acid, so it does NOT form acetaldehyde.

Question 44

The major product of the following chemical reaction is:
CH3CH2CN
(1) H3O+, Δ
(2) SOCl2
(3) Pd/BaSO4, H2

Accepted Answer

CH3CH2CHO. The reaction conditions indicate that the compound undergoes reduction with palladium on barium sulfate in the presence of hydrogen, which typically leads to the formation of an aldehyde from a nitrile. Therefore, CH3CH2CHO is the expected product as it results from the partial reduction of the nitrile group.

Question 45

The correct sequential addition of reagents in the preparation of 3-nitrobenzoic acid from benzene is:

Accepted Answer

HNO3/H2SO4, Br2/AlBr3, Mg/ether, CO2, H3O+. For 3-nitrobenzoic acid the NO2 and COOH must be meta. Nitrate benzene first (NO2 is a meta-director), then brominate to place Br meta to NO2, then convert C-Br to COOH via Mg/ether, CO2, H3O+. Sequence: HNO3/H2SO4, Br2/AlBr3, Mg/ether, CO2, H3O+. (Brominating first makes Br an o/p-director, giving the para nitro product instead.)

Question 46

47.[4] The following reaction sequence is shown: benzene + HNO3 / H2SO4 → nitrobenzene; nitrobenzene + Br2 / AlBr3 → m-bromonitrobenzene; m-bromonitrobenzene + Mg (Ether) → corresponding Grignard reagent; then CO2 / H+ → 3-nitrobenzoic acid. Which of the following is the correct intermedia

Accepted Answer

Benzene → nitrobenzene → m-bromonitrobenzene → m-nitrophenylmagnesium bromide → 3-nitrobenzoic acid. The correct option accurately follows the sequence of reactions starting from benzene, where nitration produces nitrobenzene, followed by bromination that selectively yields m-bromonitrobenzene due to the meta-directing effect of the nitro group. The subsequent formation of the Grignard reagent and its reaction with carbon dioxide leads to the formation of 3-nitrobenzoic acid, aligning perfectly with the described reaction pathway.

Question 47

A
(C4H8Cl2)
Hydrolysis
373 K
B
(C4H8O)
B reacts with Hydroxyl amine but does not give Tollen's test. Identify A and B

Accepted Answer

2,2-Dichlorobutane and 2-butane-one. B (C4H8O) reacts with hydroxylamine (a carbonyl) but gives no Tollens test, so it is a ketone, namely 2-butanone. The gem-dihalide that hydrolyses to a ketone is 2,2-dichlorobutane. Hence A = 2,2-dichlorobutane and B = 2-butanone.

Question 48

2,4-DNP test can be used to identify:

Accepted Answer

Aldehyde. 2,4-Dinitrophenylhydrazine (Brady's reagent) reacts with the C=O group of aldehydes and ketones to give orange/yellow hydrazone precipitates. It is a test for carbonyl compounds; among the options it identifies the aldehyde. Amines, ethers and halogens give no such reaction.

Question 49

Two statements are given below:
Statement I: The melting point of monocarboxylic acid with even number of carbon atoms is higher than that of with odd number of carbon atoms immediately below and above it in the series.
Statement II: The solubility of monocarboxylic acids in water decrease

Accepted Answer

Both Statement I and Statement II are correct.. Statement I is accurate because even-numbered monocarboxylic acids have stronger intermolecular forces due to better packing, leading to higher melting points compared to their odd-numbered counterparts. Statement II is also correct, as the increased molar mass of monocarboxylic acids generally results in a decrease in solubility in water due to the larger hydrophobic hydrocarbon chain.

Question 50

The products formed in the following reaction, A and B are

A compound containing a cyclohexanone ring with a formyl substituent (–CHO) on the bottom carbon is treated with [Ag(NH3)2]+ / OH− to give A, and then A is treated with NaBH4 to give B.

Accepted Answer

A is cyclohexanone with the –CHO group converted to –COOH; B is the corresponding alcohol with the ring carbonyl reduced to –OH and the side chain remaining –COOH. Tollens reagent [Ag(NH3)2]+/OH- selectively oxidizes the aldehyde (-CHO) to -COOH, leaving the ring ketone unchanged (A = ring ketone + -COOH). NaBH4 then reduces only the ketone to -OH and does NOT reduce the carboxylic acid, so B has the ring -OH and the side chain still -COOH.

JEE Main Chemistry: Aldehydes, Ketones and Carboxylic Acids questions with solutions

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