CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids NEET JEE Questions Set 01

Single Correct Answer

Question 1. The IUPAC name of the following compound is :
(JEE Main 2020)
(a) 3-Bromo-5-methylcyclopentanecarboxylic acid
(b) 4-Bromo-2-methylcyclopentanecarboxylic acid
(c) 5-Bromo-3-methylcyclopentanoic acid
(d) 3-Bromo-5-methylcyclopentanoic acid
Answer:
Ans. (b)
The ring carbon carrying –COOH is C-1. Numbering towards the CH₃ group gives the lowest locants {1, 2, 4}: CH₃ at C-2 and Br at C-4.
Since –COOH is attached to a ring, the suffix is ‘carboxylic acid’: 4-bromo-2-methylcyclopentanecarboxylic acid.

Teacher's Note:
1. When –COOH is attached to a ring, use the suffix ‘carboxylic acid’; the suffix ‘-oic acid’ is only for open chains where the COOH carbon is counted in the chain.

Question 2. The reagents to carry out the following conversion are :
Me–C≡CH → Me–CH₂–CO–Me (West Bengal JEE 2014)
(a) HgSO₄/dil. H₂SO₄
(b) BH₃ ; H₂O₂/NaOH
(c) OsO₄, HIO₄
(d) NaNH₂/CH₃I ; HgSO₄/dil. H₂SO₄
Answer:
Ans. (d)
NaNH₂ converts propyne into its sodium salt, which with CH₃I gives but-2-yne: CH₃C≡CH → CH₃C≡C⁻Na⁺ → CH₃C≡CCH₃.
Hydration of but-2-yne with HgSO₄/dil. H₂SO₄ gives butan-2-one, CH₃CH₂COCH₃.

Teacher's Note:
1. The product has one carbon more than the starting alkyne, so a C–C bond must be formed first (alkylation of the terminal alkyne) before hydration.

Question 3. Which of the following reactions will yield benzaldehyde as a product ?
(I) C₆H₅COOH (i) SOCl₂, quinoline
⟶
(ii) H₂/Pd/BaSO₄
(II) C₆H₅CH₂OH CrO₃/H₂SO₄
⟶
(III) C₆H₅COOCH₃ (i) NaBH₄
⟶
(ii) PCC
(IV) C₆H₅CH₃ (i) CrO₃, (CH₃CO)₂O
⟶
(ii) H₃O⁺, Δ (CBSE 2022)
(a) (II) and (III)
(b) (II) and (IV)
(c) (I) and (IV)
(d) (I) and (III)
Answer:
Ans. (c)
(I) SOCl₂ converts benzoic acid into benzoyl chloride, which on Rosenmund reduction (H₂/Pd–BaSO₄) gives benzaldehyde.
(IV) Toluene with CrO₃ in acetic anhydride gives benzylidene diacetate, C₆H₅CH(OCOCH₃)₂, which on acid hydrolysis gives benzaldehyde.
(II) CrO₃/H₂SO₄ oxidises benzyl alcohol right up to benzoic acid, and (III) NaBH₄ does not reduce esters, so these do not give benzaldehyde.

Teacher's Note:
1. Strong oxidants in aqueous acid (CrO₃/H₂SO₄, K₂Cr₂O₇/H₂SO₄) take a 1° alcohol to the acid; PCC or Etard-type reagents stop at the aldehyde.

Question 4. Which one of the following reactions does not represent correct combination of substrate and product under the given condition ? (JEE Main 2022)
(a) p-CH₃C₆H₄COCl H₂, Pd/BaSO₄
⟶ p-CH₃C₆H₄CHO
(b) p-CH₃C₆H₄CN (i) DIBAL–H
⟶
(ii) H₂O p-CH₃C₆H₄CHO
(c) p-CH₃C₆H₄CO₂C₂H₅ (i) AlH(iso Bu)₂
⟶
(ii) H₂O p-CH₃C₆H₄CHO
(d) p-CH₃C₆H₄CH₂OH (i) Na₂Cr₂O₇
⟶
(ii) H₂SO₄, H₂O p-CH₃C₆H₄CHO
Answer:
Ans. (d)
Alcohols on oxidation with acidified Na₂Cr₂O₇ give acids (here p-toluic acid, p-CH₃C₆H₄COOH) but not aldehydes.

Teacher's Note:
1. Rosenmund reduction (acid chloride) and DIBAL-H (nitrile, ester) stop at the aldehyde; aqueous dichromate does not.

Question 5. In the following reaction,
C₆H₅CH₃ + CrO₂Cl₂ CS₂
⟶ X H₃O⁺
⟶ C₆H₅CHO
the intermediate ‘X’ is (NEET 2021)
(a) C₆H₅CH(Cl)H
(b) C₆H₅CH(OCrOHCl₂)₂
(c) C₆H₅CH(OCOCH₃)₂
(d) C₆H₅CHCl₂
Answer:
Ans. (b)
C₆H₅CH₃ + CrO₂Cl₂ CS₂
⟶ C₆H₅CH(OCrOHCl₂)₂ (X), a chromium complex which on hydrolysis gives benzaldehyde.
Thus, option (b) is correct.

Teacher's Note:
1. In the Etard reaction the brown chromium complex C₆H₅CH(OCrOHCl₂)₂ is the intermediate; C₆H₅CH(OCOCH₃)₂ belongs to the CrO₃/acetic anhydride method.

Question 6. Identify products A and B
dil. KMnO₄
⟶
273 K A CrO₃
⟶ B (JEE Main 2021)
(a) A : B :
(b) A : B :
(c) A : B :
(d) A : OHC–CH₂CH₂CH₂–CO–CH₃ B : HOOC–CH₂CH₂CH₂–CO–CH₃
Answer:
Ans. (c)
Cold dilute KMnO₄ at 273 K brings about cis-hydroxylation: 1-methylcyclopentene → 1-methylcyclopentane-1,2-diol (A).
CrO₃ oxidises the 2° OH to C=O, while the 3° OH is resistant to oxidation: B = 2-hydroxy-2-methylcyclopentanone.
Thus, option (c) is correct.

Teacher's Note:
1. Cold, dilute KMnO₄ only adds two OH groups; hot or acidic KMnO₄ would cleave the ring (option d).

Question 7. For the reaction below
(CH₃)₂CH–CN (i) PhMgBr, THF
⟶
(ii) H₃O⁺ Q
the structure of the product Q is (JEE Main 2019)
(a) (CH₃)₂CH–CH(OH)–Ph
(b) (CH₃)₂CH–CO–Ph
(c) (CH₃)₂C(Ph)–CN
(d) (CH₃)₂C=C(NH₂)–Ph
Answer:
Ans. (b)
PhMgBr adds to the C≡N group (Grignard addition): (CH₃)₂CH–C≡N + PhMgBr THF
⟶ (CH₃)₂CH–C(Ph)=NMgBr
Hydrolysis (H₃O⁺, −NH₃, −Mg(OH)Br) gives the ketone (CH₃)₂CH–CO–Ph (Q).

Teacher's Note:
1. Grignard reagents add only once to a nitrile because the imine salt formed is negatively charged; hydrolysis then gives a ketone, not an alcohol.

Question 8. The major product obtained in the following reaction is
DIBAL–H
⟶ (JEE Main 2017)
(a)
(b)
(c)
(d)
Answer:
Ans. (d)
DIBAL-H reduces cyanides, esters, lactones, carboxylic acids and amides to the corresponding aldehydes.
Here the lactone is reduced to a hydroxy aldehyde (the ring O becomes –OH and the C=O becomes –CHO) and –COOH is reduced to –CHO. Thus, option (d) is correct.

Teacher's Note:
1. On reduction a lactone opens to give two groups: the alkyl-side oxygen ends up as –OH and the acyl carbon as –CHO.

Question 9. The main product of the following reaction is
(i) DIBAL–H
⟶
(ii) H₃O⁺ (JEE Main 2019)
(a)
(b)
(c)
(d)
Answer:
Ans. (a)
DIBAL-H converts CN to CHO and the lactone gets converted into hydroxy aldehyde. Thus, option (a) is correct.

Teacher's Note:
1. DIBAL-H reduces both a nitrile and a lactone to the aldehyde level; it does not take the aldehydes further to alcohols when used carefully.

Question 10. The product “F” in the following reaction is
(i) DIBAL–H, toluene, −78°C
⟶
(ii) H₃O⁺ “F” (Major product) (JEE Main 2021)
(a)
(b)
(c)
(d)
Answer:
Ans. (d)
DIBAL-H selectively reduces lactones to hydroxy aldehydes, i.e., the ring C–O bond is cleaved: the oxygen stays on the ring carbon as –OH and the acyl carbon becomes –CHO.
Thus, option (d) is correct.

Teacher's Note:
1. At −78°C DIBAL-H stops at the aldehyde stage; LiAlH₄ would reduce the lactone all the way to a diol.

Question 11. (i) DIBAL–H
⟶
(ii) H₂O
Consider the above reaction and predict the major product (JEE Main 2022)
(a)
(b)
(c)
(d)
Answer:
Ans. (a)
DIBAL-H reduces both esters and nitriles to aldehydes. Thus, option (a) is correct.

Teacher's Note:
1. DIBAL-H does not touch the ring, so any option that changes the ring (here, introduces a double bond) can be rejected at once.

Question 12. In the following reaction :
RMgX + HC(OEt)₃ Ether
⟶ H₃O⁺
⟶ P
The product ‘P’ is (West Bengal JEE 2015)
(a) RCHO
(b) R₂CHOEt
(c) R₃CH
(d) RCH(OEt)₂
Answer:
Ans. (a)
R–MgX + HC(OEt)₃ Ether
⟶ R–CH(OEt)₂ H₃O⁺
⟶ R–CH=O + 2EtOH
The Grignard reagent displaces one OEt group of ethyl orthoformate to give an acetal, which on acid hydrolysis gives the aldehyde.

Teacher's Note:
1. RCH(OEt)₂ is only the intermediate acetal; the H₃O⁺ step hydrolyses it to the aldehyde.

Question 13. In the following reaction, A is
A (i) Br₂, hν (ii) KOH (alc.) (iii) O₃
⟶
(iv) (CH₃)₂S (v) NaOH (aq.), Δ (JEE Main 2020)
(a)
(b)
(c)
(d)
Answer:
Ans. (d)
Cyclohexane Br₂/hν
⟶
free radical halogenation bromocyclohexane KOH (alc.)
⟶
dehydrobromination cyclohexene.
Cyclohexene O₃, then (CH₃)₂S
⟶ hexane-1,6-dial, OHC(CH₂)₄CHO.
With NaOH (aq.) the α-carbanion of one CHO group attacks the other CHO group (intramolecular aldol condensation) and on heating (−H₂O) gives cyclopent-1-enecarbaldehyde.
Thus, A is cyclohexane, option (d).

Teacher's Note:
1. Count carbons: the product has 6 carbons (5 in the ring + CHO), and ozonolysis of a cycloalkene keeps all carbons, so A must be a C₆ ring.

Question 14. The increasing order of reactivity of the following compounds in nucleophilic addition reaction is : Propanal, Benzaldehyde, Propanone, Butanone (JEE Main 2020)
(a) Butanone < Propanone < Benzaldehyde < Propanal
(b) Benzaldehyde < Butanone < Propanone < Propanal
(c) Propanal < Propanone < Butanone < Benzaldehyde
(d) Benzaldehyde < Propanal < Propanone < Butanone
Answer:
Ans. (a)
In general, aliphatic aldehydes are more reactive than aliphatic ketones towards nucleophilic addition reactions. Thus, propanal is more reactive than benzaldehyde.
Among aliphatic ketones, bigger the size of the alkyl groups around C=O group, less reactive is the ketone. Thus, butanone is less reactive than propanone.
Combining both the trends, the overall reactivity increases in the order : Butanone < propanone < benzaldehyde < propanal, i.e., option (a) is correct.

Teacher's Note:
1. Benzaldehyde is less reactive than aliphatic aldehydes because of resonance of C=O with the ring, but it is still more reactive than simple ketones.

Question 15. The increasing order of the following compounds towards HCN addition is
(i) m-methoxybenzaldehyde (ii) o-nitrobenzaldehyde (iii) o-methoxybenzaldehyde (iv) m-nitrobenzaldehyde
(JEE Main 2020)
(a) (i) < (iii) < (iv) < (ii)
(b) (iv) < (iii) < (i) < (ii)
(c) (iii) < (i) < (iv) < (ii)
(d) (iii) < (iv) < (ii) < (i)
Answer:
Ans. (c)
Electron-withdrawing groups at o- and p-positions increase the electrophilicity (or magnitude of +ve charge on the carbon atom of the carbonyl group) more than at m-position. Therefore, o-nitrobenzaldehyde (ii) is the most reactive and hence is more reactive than m-nitrobenzaldehyde (iv) towards addition of HCN.
Conversely, electron-donating groups at o- and p-position decrease the reactivity towards nucleophilic addition more than at m-position. Since CH₃O group can exert its +R-effect only at o- and p-positions but not at m-position. At m-position, it can exert only its −I-effect. As a result, m-methoxybenzaldehyde (i) is more reactive than o-methoxybenzaldehyde (iii).
Thus, the overall reactivity increases in the order : (iii) < (i) < (iv) < (ii), i.e., option (c) is correct.

Teacher's Note:
1. A –OCH₃ group at the meta position cannot donate by resonance, so it acts only through its −I effect and slightly activates the CHO group.

Question 16. Which of the following will react with water
(a) CHCl₃
(b) Cl₃CCHO
(c) CCl₄
(d) ClCH₂CH₂Cl
Answer:
Ans. (b)
Chloral (CCl₃CHO) readily reacts with H₂O to form chloral hydrate, CCl₃CH(OH)₂.
This is due to the reason that the three electron-withdrawing Cl atoms in CCl₃CHO increase the positive charge on the carbonyl carbon which holds the two OH groups firmly and thus favours the formation of stable hydrate.

Teacher's Note:
1. Chloral hydrate is one of the few stable gem-diols; strong −I groups next to C=O favour hydrate formation.

Question 17. The major product formed in the following reaction is
CH₃CH₂CH₂CH₂OH + H⁺
⟶ (JEE Main 2022)
(a)
(b) CH₃CH₂CH₂CH₂–O–CH₂CH₂CH₂CH=CH–OH
(c) CH₃CH₂CH₂CH₂–O–CH₂CH₂CH₂CH₂CHO
(d)
Answer:
Ans. (d)
H⁺ adds to the C=C of dihydropyran to give the carbocation next to the ring oxygen (stabilised by the lone pair of O).
The alcohol (nucleophile) attacks this carbocation and loss of H⁺ gives the tetrahydropyranyl (THP) ether, 2-butoxytetrahydropyran. Thus, option (d) is correct.

Teacher's Note:
1. Protonation of an enol ether gives an oxocarbenium ion on the carbon bonded to O, so the alcohol always ends up on that carbon (an acetal).

Question 18. The product “A” in the following reaction is
CH₃COCH₂COOC₂H₅ Ethylene glycol
⟶
H⁺ A (Major Product) (JEE Main 2021)
(a)
(b)
(c)
(d)
Answer:
Ans. (b)
The ketonic C=O group reacts with ethylene glycol (HO–CH₂–CH₂–OH) in presence of H⁺ with loss of H₂O to form a cyclic ketal; the ester group does not react.
Thus, option (b) is correct.

Teacher's Note:
1. Ketones form cyclic ketals with ethylene glycol; the ester C=O is much less electrophilic and survives (this is how a keto group is protected).

Teacher's Note:
1. Acetal formation puts two OR groups and an H on one carbon, so bulky alcohols such as t-BuOH react very poorly.

Question 20. The correct sequence of reagents for the following conversion will be
→ (JEE Main 2017)
(a) CH₃MgBr, [Ag(NH₃)₂]⁺OH⁻, H⁺/CH₃OH
(b) [Ag(NH₃)₂]⁺OH⁻, CH₃MgBr, H⁺/CH₃OH
(c) [Ag(NH₃)₂]⁺OH⁻, H⁺/CH₃OH, CH₃MgBr
(d) CH₃MgBr, H⁺/CH₃OH, [Ag(NH₃)₂]⁺OH⁻
Answer:
Ans. (c)
Tollens’ reagent, [Ag(NH₃)₂]⁺OH⁻, oxidises only the –CHO group to –CO₂H (the ketone is not affected).
H⁺/CH₃OH converts –CO₂H into the ester –CO₂CH₃ (esterification).
3 CH₃MgBr : addition of Grignard reagent to ketones gives 3° alcohols while to esters, it gives 3° alcohols (two CH₃ groups are added). This gives the product. Thus, option (c) is correct.

Teacher's Note:
1. Oxidise the aldehyde first: if CH₃MgBr were used first it would also attack the CHO group and give a 2° alcohol.

Question 21. Which of the following reagents would distinguish cis-cyclopentane-1,2-diol from the trans-isomer ? (NEET Phase-I, 2016)
(a) MnO₂
(b) Aluminium isopropoxide
(c) Acetone
(d) Ozone
Answer:
Ans. (c)
A cyclic ketal contains a five-membered ring. Since two fused five membered rings are more stable if the fusion is cis, therefore, cis-cyclopentane-1,2-diol forms a stable ketal with acetone (HCl(g), −H₂O) but trans-isomer does not react.

Teacher's Note:
1. Only a cis 1,2-diol on a five-membered ring can bridge both OH groups to one carbon without strain.

Question 22. The reagent used for separation of acetaldehyde and acetophenone is
(a) NaHSO₃
(b) C₆H₅NHNH₂
(c) NH₂OH
(d) NaOH–I₂
Answer:
Ans. (a)
Acetaldehyde forms addition product with NaHSO₃ but acetophenone does not.
The crystalline bisulphite adduct can be filtered off and the aldehyde regenerated from it.

Teacher's Note:
1. Phenylhydrazine and NH₂OH react with both compounds, and both give the iodoform test, so only NaHSO₃ can separate them.

Question 23. Which one of the following is reduced with zinc and hydrochloric acid to give the corresponding hydrocarbon ?
(a) ethyl acetate
(b) acetic acid
(c) acetamide
(d) butan-2-one
Answer:
Ans. (d)
Zn/HCl reduces only ketones (and not esters, acids or amides) to the corresponding hydrocarbons.
CH₃COCH₂CH₃ Zn–Hg/HCl
⟶ CH₃CH₂CH₂CH₃

Teacher's Note:
1. The C=O of esters, acids and amides is stabilised by resonance with the attached O or N, so Clemmensen conditions do not reduce them.

Question 24. During reduction of aldehydes with hydrazine and potassium hydroxide, the first is formation of
(a) R–C≡N
(b) R–CO–NH₂
(c) R–CH=NH
(d) R–CH=N–NH₂
Answer:
Ans. (d)
First hydrazones are formed: R–CHO + H₂N–NH₂ → R–CH=N–NH₂ + H₂O.
The hydrazone is then decomposed by KOH on heating to give R–CH₃ and N₂ (Wolff–Kishner reduction).

Teacher's Note:
1. In Wolff–Kishner reduction the nitrogen leaves as N₂ gas; this drives the reaction.

Question 25. The appropriate reagent for the following transformation is
→ (Karnataka CET 2018)
(a) Zn-Hg/HCl
(b) H₂N-NH₂, KOH/ethylene glycol
(c) Ni/H₂
(d) NaBH₄
Answer:
Ans. (b)
Ni/H₂ and NaBH₄ will reduce >C=O to >CHOH.
Out of reagents (a) and (b), HCl in reagent (a) will either bring about conversion of –OH to Cl or bring about dehydration to form a double bond in the ring. Therefore, only option (b) is correct.

Teacher's Note:
1. Choose Wolff–Kishner (basic) when the molecule has acid-sensitive groups such as an alcohol, and Clemmensen (acidic) when it has base-sensitive groups.

Question 26. Which of the following reagent is suitable for the preparation of the product in the following reaction
?
⟶ (JEE Main 2021)
(a) Red P + Cl₂
(b) NH₂NH₂/C₂H₅O⁻Na⁺
(c) Ni/H₂
(d) NaBH₄
Answer:
Ans. (b)
Same explanation as given in ans. 26: Ni/H₂ would also reduce the C=C bonds and NaBH₄ gives only an alcohol.
>C=O + NH₂NH₂ −H₂O
⟶ >C=N–NH₂ C₂H₅ONa, −N₂
⟶ >CH₂, the C=C bonds remaining unaffected. Thus, option (b) is correct.

Teacher's Note:
1. Wolff–Kishner reduction converts C=O to CH₂ without touching C=C double bonds.

Question 27. Which of the following ketone will NOT give enamine on treatment with secondary amines ? (JEE Main 2021)
(a) C₂H₅COC₂H₅
(b) C₂H₅COCH₃
(c) (CH₃)₃C–CO–C(CH₃)₃ (t-Bu–CO–t-Bu)
(d) Cyclohexanone
Answer:
Ans. (c)
t-Butyl groups, (CH₃)₃C do not have α-hydrogen so enolization cannot take place and hence enamine formation is prevented.

Teacher's Note:
1. Enamine formation needs an α-H; a ketone with two tert-butyl groups has none.

One or More Correct Answers

Question 28. Tautomerism is exhibited by
(a) C₆H₅–CH=CH–OH
(b)
(c)
(d)
Answer:
Ans. (a, c, d)
(a) C₆H₅–CH=CH–OH ⇌ C₆H₅–CH₂–CHO (Aldehydes are more stable than vinyl alcohols)
(b) does not show tautomerism since its α-hydrogen is not acidic enough for migration.
(c) Cyclohexane-1,3-dione ⇌ resorcinol (enol form, i.e., resorcinol is stabilized by resonance energy of the benzene ring)
(d) Cyclohexane-1,2-dione ⇌ its enol (Enol form is stabilized by H-bonding)

Teacher's Note:
1. Keto–enol tautomerism needs an H on an sp³ carbon next to C=O; in p-benzoquinone every ring C–H is on an sp² carbon.

Passage-Based Questions

Question 29. Comprehension 1: In chemical analysis, the presence of a methyl ketonic group is tested with I₂ in presence of an alkali like NaOH or KOH. In another chemical analysis, aldehyde group is tested either by Tollens’ reagent (AgNO₃ + NH₄OH) which gives silver mirror or by Fehling’s solution (CuSO₄ + NaOH + Roschelle salt) which gives red ppt. of copper (I) oxide.
Consider the following compounds :
I. CH₃–CO–CH₂–CO–CH₃   II. CH₃–CO–CO–CH₃   III. CH₃–CO–OH   IV. CH₃–CO–Cl
Which will give iodoform test ?
(a) Only I
(b) Both I and II
(c) Only II
(d) All
Answer:
Ans. (c)
Although CH₃COCH₂COCH₃ contains a CH₃CO– group linked to carbon, yet it does not undergo iodoform test. The reason being that the iodination occurs at the more reactive CH₂ group rather than at the terminal CH₃ which is essential for iodoform test to occur.
CH₃CO–CH₂–COCH₃ I₂/NaOH
⟶ CH₃CO–CI₂–COCH₃, which does not give CHI₃ with NaOH.
Only CH₃CO–COCH₃ gives iodoform test.

Teacher's Note:
1. In CH₃COOH and CH₃COCl the CH₃CO group is attached to O or Cl, not to C or H, so they do not give the iodoform test.

Matrix Match Questions

Teacher's Note:
1. Schiff’s bases (imines, >C=N–R) come from primary amines; oximes (>C=N–OH) come from hydroxylamine.

Integer and Numerical Answer Questions

Question 31. Total number of isomeric aldehydes and ketones having the molecular formula C₅H₁₀O are.
Answer:
Ans. 7
Seven isomers are possible. These are :
Aldehydes : CH₃CH₂CH₂CH₂CHO, CH₃CH₂CH(CH₃)CHO, (CH₃)₂CHCH₂CHO, (CH₃)₃CCHO.
Ketones : CH₃COCH₂CH₂CH₃, CH₃COCH(CH₃)₂, CH₃CH₂COCH₂CH₃.

Teacher's Note:
1. Count aldehydes as C₄H₉–CHO (four butyl groups) and ketones by placing C=O inside the chain; do not forget the branched ones.

Question 32. In the scheme given below, the total number of intramolecular aldol condensation products formed from ‘Y’ is
1. O₃
⟶
2. Zn, H₂O Y 1. NaOH (aq)
⟶
2. Heat
Answer:
Ans. 1
Bicyclo[4.4.0]dec-1(6)-ene 1. O₃, 2. Zn/H₂O
⟶
(Reductive ozonolysis) cyclodecane-1,6-dione (Y)
Y NaOH (aq)
⟶
(Aldol condensation) bicyclic aldol Heat
⟶
−H₂O bicyclo[5.3.0]dec-1(7)-en-2-one
Since, diketone (Y) is symmetrical, therefore, upon intramolecular aldol condensation, it gives only one product.

Teacher's Note:
1. Because Y is symmetrical, all four α-CH₂ groups are equivalent and every intramolecular attack gives the same five-membered ring product.

Question 33. Among the following, the number of reaction(s) that produce(s) benzaldehyde is
I. C₆H₆ CO, HCl
⟶
Anhydrous AlCl₃/CuCl   II. C₆H₅CHCl₂ H₂O
⟶
100°C
III. C₆H₅COCl H₂
⟶
Pd–BaSO₄   IV. C₆H₅CO₂Me DIBAL-H
⟶
Toluene, −78°C, H₂O (JEE Advanced 2015)
Answer:
Ans. 4
All the four reactions give benzaldehyde.
I is Gattermann–Koch formylation, II is hydrolysis of benzal chloride (gem-dichloride), III is Rosenmund reduction and IV is partial reduction of an ester by DIBAL-H at −78°C.

Teacher's Note:
1. A gem-dihalide on hydrolysis gives a carbonyl compound: a terminal CHCl₂ gives an aldehyde.

Question 34. How many of the following compounds undergo aldol condensation ?
Methanal, 2-methylpentanal, benzaldehyde, benzophenone, cyclohexanone, 1-phenyl-propanone, phenylacetaldehyde, butan-1-ol and 2, 2-dimethylbutanal.
Answer:
Ans. 4
Four aldehydes/ketones are : 2-methylpentanal, cyclohexanone, 1-phenylpropanone and phenylacetaldehyde contain α-hydrogens and hence undergo aldol condensation.

Teacher's Note:
1. Butan-1-ol is not a carbonyl compound; methanal, benzaldehyde, benzophenone and 2,2-dimethylbutanal have no α-H.

Assertion–Reason Questions

Question 35. Statement-1: Fehling solution oxidises acetaldehyde to acetic acid but not benzaldehyde to benzoic acid.
Statement-2: The C–H bond of CHO group in benzaldehyde is stronger than in acetaldehyde.
(a) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1
(b) Statement-1 is True, Statement-2 is True ; Statement-2 is not a correct explanation for Statement-1
(c) Statement-1 is True, Statement-2 is False
(d) Statement-1 is False, Statement-2 is True
Answer:
Ans. (a)
Both statements are true. Due to resonance of the CHO group with the benzene ring, the C–H bond of the CHO group in benzaldehyde acquires some double-bond character and becomes stronger; hence the weak oxidant Fehling’s solution cannot oxidise it, though it oxidises acetaldehyde. Statement-2 explains Statement-1.

Teacher's Note:
1. Tollens’ reagent oxidises both aliphatic and aromatic aldehydes; Fehling’s and Benedict’s solutions oxidise only aliphatic ones.

Question 36. Statement-1: Aromatic aldehydes can be distinguished from aliphatic aldehydes by Fehling solution.
Statement-2: Fehling solution is an alkaline solution of CuSO₄ containing Rochelle salt.
(a) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1
(b) Statement-1 is True, Statement-2 is True ; Statement-2 is not a correct explanation for Statement-1
(c) Statement-1 is True, Statement-2 is False
(d) Statement-1 is False, Statement-2 is True
Answer:
Ans. (b)
Correct explanation. Aliphatic aldehydes reduce Fehling’s solution but aromatic aldehydes do not.
Statement-2 only describes the composition of Fehling’s solution, so it does not explain Statement-1.

Teacher's Note:
1. Rochelle salt (sodium potassium tartrate) only keeps Cu²⁺ in solution as a complex in alkaline medium.

Question 37. Statement-1: Acetoacetic ester (CH₃COCH₂COOC₂H₅) contains CH₃CO group but does not give iodoform test.
Statement-2: The H-atoms of the CH₃ group are more acidic than those of CH₂ group.
(a) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1
(b) Statement-1 is True, Statement-2 is True ; Statement-2 is not a correct explanation for Statement-1
(c) Statement-1 is True, Statement-2 is False
(d) Statement-1 is False, Statement-2 is True
Answer:
Ans. (c)
Correct statement-2. The H-atoms of the CH₂ group are more acidic than those of CH₃ group.
Hence iodination takes place at the CH₂ group (flanked by two C=O groups) and not at the CH₃ group, so the ester does not give iodoform.

Teacher's Note:
1. A CH₂ between two C=O groups is far more acidic than a CH₃ next to one C=O group.

Question 38. Statement-1: Acrylic acid (CH₂=CHCOOH) is a weaker acid than benzoic acid (C₆H₅COOH).
Statement-2: Ethylenic double bond is less electron-donating than benzene ring.
(a) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1
(b) Statement-1 is True, Statement-2 is True ; Statement-2 is not a correct explanation for Statement-1
(c) Statement-1 is True, Statement-2 is False
(d) Statement-1 is False, Statement-2 is True
Answer:
Ans. (c)
Correct statement-2. Ethylenic double bond is more electron-donating than a benzene ring.
That is why acrylic acid (pK_a 4.25) is slightly weaker than benzoic acid (pK_a 4.20).

Teacher's Note:
1. The more the group attached to COOH donates electrons, the less stable the carboxylate ion and the weaker the acid.

Question 39. Statement-1: p-Hydroxybenzoic acid has a lower boiling point than o-hydroxybenzoic acid.
Statement-2: o-Hydroxybenzoic acid has intramolecular hydrogen bonding. (IIT Paper I, 2007)
(a) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1
(b) Statement-1 is True, Statement-2 is True ; Statement-2 is not a correct explanation for Statement-1
(c) Statement-1 is True, Statement-2 is False
(d) Statement-1 is False, Statement-2 is True
Answer:
Ans. (d)
Correct statement-1. o-Hydroxybenzoic acid has lower boiling point than p-hydroxybenzoic acid.
Intramolecular H-bonding in the ortho isomer reduces intermolecular association, while the para isomer is associated by intermolecular H-bonds.

Teacher's Note:
1. Intramolecular H-bonding lowers boiling points and water solubility; intermolecular H-bonding raises them.

Question 40. Statement-1: Peracids are stronger acids than corresponding carboxylic acids.
Statement-2: The anion of carboxylic acids is stabilized by resonance but not that of peracids.
(a) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1
(b) Statement-1 is True, Statement-2 is True ; Statement-2 is not a correct explanation for Statement-1
(c) Statement-1 is True, Statement-2 is False
(d) Statement-1 is False, Statement-2 is True
Answer:
Ans. (d)
Correct statement-1. Peracids are weaker acids than corresponding carboxylic acids.
In RCO–O–O⁻ the negative charge is on an oxygen that is not attached to the C=O group, so it is not delocalised by resonance, whereas RCOO⁻ is resonance stabilised.

Teacher's Note:
1. Peracetic acid (pK_a about 8.2) is much weaker than acetic acid (pK_a 4.76).