Aldehydes, Ketones and Carboxylic Acids - MCQs with Explanations

1. The enol form of acetone, after treatment with $D_{2}O$ gives:

• (a) $CH_{3}-C(OD)=CH_{2}$
• (b) $CD_{3}-C(=O)-CD_{3}$
• (c) $CH_{2}=C(OH)-CH_{2}D$
• (d) $CD_{2}=C(OD)-CD_{3}$

Correct Answer: (b)

Explanation: Through repeated keto-enol tautomerism in the presence of $D_{2}O$, all the $\alpha$-hydrogen atoms of acetone are eventually replaced by deuterium atoms, resulting in hexadeuteroacetone.

2. Cyclohexanone on aldol condensation followed by heating gives:

• (a) [structure a]
• (b) [structure b]
• (c) [structure c]
• (d) [structure d]

Correct Answer: (c)

Explanation: Aldol condensation of cyclohexanone followed by dehydration (heating) leads to the formation of 2-(cyclohex-1-en-1-yl)cyclohexan-1-one.

3. The acid produced (A) in the sequence $C_{2}H_{5}I \xrightarrow{\text{alc. KOH}} X \xrightarrow{Br_{2}/CCl_{4}} Y \xrightarrow{KCN} Z \xrightarrow{H_{3}O^{+}} A$ is:

• (a) Succinic acid
• (b) Malonic acid
• (c) Oxalic acid
• (d) Maleic acid

Correct Answer: (a)

Explanation: Ethyl iodide eliminates to ethene (X), which brominates to ethylene dibromide (Y). Reaction with KCN yields ethylene dicyanide (Z), which upon hydrolysis forms succinic acid (butanedioic acid).

4. Acetic acid is obtained when:

• (a) methyl alcohol is oxidised with $KMnO_{4}$
• (b) calcium acetate is distilled in the presence of calcium formate
• (c) acetaldehyde is oxidised with potassium dichromate and sulphuric acid
• (d) glycerol is heated with sulphuric acid.

Correct Answer: (c)

Explanation: Oxidation of acetaldehyde ($CH_{3}CHO$) with acidic potassium dichromate ($K_{2}Cr_{2}O_{7}/H_{2}SO_{4}$) produces acetic acid ($CH_{3}COOH$).

5. $HCOONa \xrightarrow{\text{heat}} X + H_{2}$. X is:

• (a) $Na_{2}CO_{3}$
• (b) $CO_{2}$
• (c) $(COONa)_{2}$
• (d) CO

Correct Answer: (c)

Explanation: Heating sodium formate at 360-400°C results in the formation of sodium oxalate and the liberation of hydrogen gas.

6. $Ph-C \equiv N \xrightarrow[H_{2}SO_{4}]{HN_{3}} \text{Products}$. Which of the following is not a possible product?

• (a) [structure a]
• (b) [structure b]
• (c) [structure c]
• (d) [structure d]

Correct Answer: (b)

Explanation: The Schmidt reaction with nitriles involves the reaction with hydrazoic acid ($HN_{3}$) to yield substituted amides or related derivatives, but the specified "not possible" product usually refers to a mismatch in the reaction mechanism intermediates.

7. A liquid was mixed with ethanol and a drop of concentrated $H_{2}SO_{4}$ was added. A compound with a fruity smell was formed. The liquid was:

• (a) $CH_{3}OH$
• (b) HCHO
• (c) $CH_{3}COCH_{3}$
• (d) $CH_{3}COOH$

Correct Answer: (d)

Explanation: The reaction between a carboxylic acid ($CH_{3}COOH$) and an alcohol (ethanol) in the presence of an acid catalyst is esterification, which produces an ester (ethyl acetate) characterized by a fruity smell.

8. In the Cannizzaro reaction ($2 PhCHO \xrightarrow{OH^{-}} PhCH_{2}OH + PhCOO^{-}$), the slowest step is:

• (a) the attack of $OH^{-}$ at the carbonyl group
• (b) the transfer of hydride to the carbonyl group
• (c) the abstraction of proton from the carboxylic group
• (d) the deprotonation of $PhCH_{2}OH$.

Correct Answer: (b)

Explanation: The rate-determining step of the Cannizzaro reaction is the intermolecular transfer of a hydride ion from the intermediate anion to another aldehyde molecule.

9. Identify 'Z' in: $\text{Benzaldehyde} + 2 C_{2}H_{5}OH \xrightarrow{HCl} Z$.

• (a) $C_{6}H_{5}CH_{2}-OC_{2}H_{5}$
• (b) $C_{6}H_{5}CH(OC_{2}H_{5})_{2}$
• (c) $C_{6}H_{5}COC_{2}H_{5}$
• (d) $C_{6}H_{5}COOC_{2}H_{5}$

Correct Answer: (b)

Explanation: Aldehydes react with two equivalents of alcohol in the presence of dry HCl gas to form acetals. Here, benzaldehyde forms benzaldehyde diethyl acetal.

10. Benzoic acid on reaction with thionyl chloride gives:

• (a) o-chlorobenzoic acid
• (b) benzaldehyde
• (c) acetophenone
• (d) benzoyl chloride.

Correct Answer: (d)

Explanation: Thionyl chloride ($SOCl_{2}$) converts carboxylic acids into acid chlorides. Benzoic acid thus yields benzoyl chloride.

11. Which of the following will neither undergo aldol condensation nor Cannizzaro reaction?

• (a) Formaldehyde
• (b) Ethanal
• (c) Benzophenone
• (d) Acetophenone

Correct Answer: (c)

Explanation: Aldol condensation requires $\alpha$-hydrogens (ethanal/acetophenone have them). Cannizzaro requires an aldehyde with no $\alpha$-hydrogens (formaldehyde has none). Benzophenone is a ketone with no $\alpha$-hydrogens, so it undergoes neither.

12. Friedel-Craft acylation of benzene with benzoyl chloride gives:

• (a) $C_{6}H_{5}COCH_{3}$
• (b) $CH_{3}COCH_{3}$
• (c) $C_{6}H_{5}COC_{6}H_{5}$
• (d) None of these.

Correct Answer: (c)

Explanation: Benzene reacts with benzoyl chloride ($C_{6}H_{5}COCl$) in the presence of anhydrous $AlCl_{3}$ to produce benzophenone ($C_{6}H_{5}COC_{6}H_{5}$).

13. The correct order of acidic strength of (I) phenol, (II) p-nitrophenol, (III) m-nitrophenol, and (IV) o-nitrophenol is:

• (a) $I > III > IV$
• (b) $III < IV < II < I$
• (c) $II > III > IV > I$
• (d) $II < III < IV < I$

Correct Answer: (c)

Explanation: Nitrophenols are more acidic than phenol. The para-isomer (II) is strongest due to the -R and -I effects, followed by meta (III) and ortho (IV) isomers based on the positioning of the electron-withdrawing nitro group.

14. Identify 'X' in: $\text{Phthalic acid} \xrightarrow[ (ii) \Delta ]{(i) NH_{3}} X$.

• (a) Benzamide
• (b) Phthalamide
• (c) Phthalimide
• (d) Benzoyl chloride

Correct Answer: (c)

Explanation: Phthalic acid reacts with ammonia to form ammonium phthalate, which upon heating loses water to form phthalamide, and then loses ammonia to form phthalimide.

15. Correct IUPAC name of OHC-CH2-CH(OH)-CH2-C(=O)-CH2-COOH:

• (a) 6-Formyl-5-hydroxy-3-oxohexan-1-oic acid
• (b) 3-Hydroxy-5-oxo-1-formylheptan-7-oic acid
• (c) 1-Formyl-2-hydroxy-4-oxohexan-5-oic acid
• (d) 5-Hydroxy-3-oxo-6-formylhexan-1-oic acid

Correct Answer: (a)

Explanation: The carboxyl group (-COOH) is the principal functional group (C1). The aldehyde is named as a "formyl" substituent at C6, and the ketone and hydroxyl groups are substituents at C3 and C5.

16. Product of $\text{Cyclohexene} \xrightarrow[ \text{heat} ]{KMnO_{4}-H_{2}SO_{4}}$:

• (a) $HOOC(CH_{2})_{4}COOH$
• (b) [structure b]
• (c) $CH_{3}(CH_{2})_{4}CH_{3}$
• (d) $OHC(CH_{2})_{4}CHO$

Correct Answer: (a)

Explanation: Vigorous oxidation of cyclohexene with acidic potassium permanganate breaks the double bond and oxidizes the terminal carbons to carboxyl groups, forming adipic acid (hexane-1,6-dioic acid).

17. Acetic acid on heating with $P_{2}O_{5}$ gives:

• (a) phthalic acid
• (b) terephthalic acid
• (c) acetic anhydride
• (d) none of these.

Correct Answer: (c)

Explanation: Dehydration of two molecules of acetic acid using phosphorus pentoxide ($P_{2}O_{5}$) yields acetic anhydride.

18. Order of reactivity towards nucleophilic addition (Diethylketone I, Benzaldehyde II, Propanal III, Acetaldehyde IV):

• (a) $I > II > III > IV$
• (b) $IV > III > II > I$
• (c) $II > III > IV$
• (d) $IV > III > II$

Correct Answer: (b)

Explanation: Aldehydes are more reactive than ketones due to steric and electronic (+I) factors. Small aliphatic aldehydes are most reactive, while steric bulk and resonance (in aromatic aldehydes) decrease reactivity: Acetaldehyde > Propanal > Benzaldehyde > Diethylketone.

19. Reaction product of $(CH_{3})_{2}CO + NH_{2}NHCONH_{2} \rightarrow$:

• (a) $(CH_{3})_{2}C=NCONHNH_{2}$
• (b) [intermediate]
• (c) $(CH_{3})_{2}C=NNHCONH_{2}$
• (d) Both (b) and (c)

Correct Answer: (c)

Explanation: Acetone reacts with semicarbazide to form acetone semicarbazone. The hydrazine nitrogen attacks because the amide nitrogen is involved in resonance.

20. Carbonyl compounds undergo nucleophilic addition because of:

• (a) non-polar carbonyl group
• (b) electromeric effect
• (c) more stable anion with negative charge on oxygen
• (d) none of these.

Correct Answer: (c)

Explanation: Nucleophilic addition occurs because the attack on the carbonyl carbon creates an alkoxide intermediate where the negative charge resides on the electronegative oxygen atom, which is relatively stable.

21. Match Acids (P. Phthalic, Q. Oxalic, R. Succinic, S. Adipic) with IUPAC names:

• (a) 1 2 3 4
• (b) 2 4 1 3
• (c) 2 3 4 1
• (d) 4 3 2 1

Correct Answer: (c)

Explanation: Phthalic is Benzene-1,2-dicarboxylic (2); Oxalic is Ethane-1,2-dioic (3); Succinic is Butane-1,4-dioic (4); Adipic is Hexane-1,6-dioic (1).

22. Match Pairs (P. CH3CHO/HCHO, Q. HCOOH/HCHO, R. CH3CHO/HCHO, S. CH3COCH3/Steric Ketone) with Tests:

• (a) 1 3 2 4
• (b) 3 1 2 4
• (c) 4 3 2 1
• (d) 3 1 4 2

Correct Answer: (d)

Explanation: (P) Benzaldehyde/HCHO by Fehling's (3); (Q) HCOOH/HCHO by $NaHCO_{3}$ (1); (R) CH3CHO/HCHO by Iodoform (4); (S) Acetone/Bulky ketone by $NaHSO_{3}$ (2).

23. Aldol condensation giving methyl vinyl ketone occurs between:

• (a) HCHO and $CH_{3}COCH_{3}$
• (b) HCHO and $CH_{3}CHO$
• (c) Two molecules of $CH_{3}CHO$
• (d) Two molecules of $CH_{3}COCH_{3}$

Correct Answer: (a)

Explanation: The reaction between formaldehyde and acetone forms 4-hydroxybutan-2-one, which dehydrates to form methyl vinyl ketone.

24. Identify X in the synthesis of cinnamic acid derivative:

• (a) $CH_{3}COOH$
• (b) $BrCH_{2}COOH$
• (c) $(CH_{3}CO)_{2}O$
• (d) OHC-COOH

Correct Answer: (c)

Explanation: The Perkin reaction requires an aromatic aldehyde to react with an acid anhydride (X) in the presence of its sodium salt to form an $\alpha,\beta$-unsaturated acid.

25. $CH_{3}MgBr \xrightarrow[ (ii) H_{2}O/H^{+} ]{(i) CO_{2}} X$. X is:

• (a) Acetaldehyde
• (b) Acetic acid
• (c) Formic acid
• (d) Formaldehyde

Correct Answer: (b)

Explanation: Grignard reagents react with dry ice ($CO_{2}$) to form a salt, which upon acidic hydrolysis yields the corresponding carboxylic acid (acetic acid here).

26. Reaction of HCOOH with conc. $H_{2}SO_{4}$ gives:

• (a) $CO_{2}$
• (b) CO
• (c) oxalic acid
• (d) acetic acid

Correct Answer: (b)

Explanation: Formic acid is dehydrated by concentrated sulphuric acid to yield carbon monoxide and water.

27. Sequence from Acetic acid to product D:

• (a) [structure a]
• (b) [structure b]
• (c) [structure c]
• (d) [structure d]

Correct Answer: (a)

Explanation: Acetic acid $\rightarrow$ Acetyl chloride (A) $\xrightarrow{\text{Benzene}/AlCl_{3}}$ Acetophenone (B) $\xrightarrow{HCN}$ Acetophenone cyanohydrin (C) $\xrightarrow{H_{2}O}$ Atrolactic acid (D).

28. Isomers $C_{3}H_{6}Cl_{2}$ A (gives propionaldehyde) and B (gives acetone) are:

• (a) $CH_{3}-CCl_{2}-CH_{3}$ and $CH_{3}-CH_{2}-CHCl_{2}$
• (b) $CH_{3}-CHCl-CHCl_{2}$ and $CH_{3}-CH_{2}-CHCl_{2}$
• (c) $CH_{3}-CH_{2}-CHCl_{2}$ and $CH_{3}-CCl_{2}-CH_{3}$
• (d) $CH_{3}-CHCl-CHCl_{2}$ and $CH_{3}-CCl_{2}-CH_{3}$

Correct Answer: (c)

Explanation: Terminal gem-dihalides yield aldehydes upon hydrolysis, while internal gem-dihalides yield ketones.

29. End product of the sequence involving ethyne and Grignard:

• (a) [option a]
• (b) [option b]
• (c) [option c]
• (d) [option d]

Correct Answer: (b)

Explanation: Ethyne reacts with Grignard to form ethynyl magnesium bromide, which then undergoes carboxylation/hydrolysis to propiolic acid. Further hydration with $HgSO_{4}/H_{2}SO_{4}$ followed by mild oxidation leads to malonic acid.

30. $(CH_{3})_{2}C=CH-CH_{2}-CH_{3} + \text{alk. } KMnO_{4}$ gives:

• (a) [diol product]
• (b) [acid mixture]
• (c) [alcohol mixture]
• (d) Acetone and Propanoic acid

Correct Answer: (d)

Explanation: Alkaline $KMnO_{4}$ cleaves the double bond. The trisubstituted alkene yields a ketone (acetone) and an aldehyde (propanal), but since aldehydes oxidize easily under these conditions, it forms propanoic acid.

31. Product of $(CH_{3})_{2}CHCOOH$ sequence ($LiAlH_{4}, PBr_{3}, KCN, H_{2}O$):

• (a) [bromide]
• (b) [nitrile]
• (c) [amine]
• (d) 3-methylbutanoic acid

Correct Answer: (d)

Explanation: The acid is reduced to an alcohol, converted to a bromide, then to a nitrile, and finally hydrolyzed to an acid with one additional carbon atom.

32. Reagents for converting benzyl bromide to benzyl methyl ether:

• (a) $CH_{3}\overline{O}Na^{+}$ in THF
• (b) $CH_{3}\overline{O}Na^{+}$ in DMSO
• (c) $CH_{3}I$ then $H_{3}O^{+}$
• (d) $NaOH$ then $CH_{3}I$

Correct Answer: (a)

Explanation: Direct nucleophilic substitution ($S_{N}2$) of benzyl bromide with sodium methoxide in a suitable solvent yields the ether.

33. Strongest acid among the following:

• (a) $CH_{3}COOH$
• (b) HCOOH
• (c) $CH_{3}CH_{2}CH(Cl)COOH$
• (d) $ClCH_{2}CH_{2}CH_{2}COOH$

Correct Answer: (c)

Explanation: Electron-withdrawing groups increase acidity. Chlorine at the $\alpha$-position (C) has a much stronger inductive effect than at the distal $\gamma$-position (D), making (C) the strongest.

34. Silver mirror test is given by:

• (a) Acetaldehyde
• (b) Acetone
• (c) Butanoic acid
• (d) Benzophenone

Correct Answer: (a)

Explanation: Tollens' reagent (silver mirror test) is used to detect aldehydes. Ketones and simple carboxylic acids do not react.

35. Products of $CH_{3}CH_{2}COOC_{2}H_{5} \xrightarrow[ H_{3}O^{+} ]{NaOH}$ contain:

• (a) [anion]
• (b) [neutral]
• (c) [aldehyde]
• (d) Both (a) and (b)

Correct Answer: (d)

Explanation: Saponification of ethyl propanoate yields sodium propanoate (anion) and ethanol (neutral alcohol).

36. Catalyst for the reduction of methanal to methanol:

• (a) Pyridine
• (b) Pd
• (c) $V_{2}O_{5}$
• (d) $K_{2}Cr_{2}O_{7}$

Correct Answer: (b)

Explanation: Catalytic hydrogenation of aldehydes to alcohols uses metal catalysts like Pd, Pt, or Ni.

37. Liquid acidic to litmus, oxidised by $KMnO_{4}$, reduces $HgCl_{2}$, and decomposes with $H_{2}SO_{4}$ to CO + $H_{2}O$:

• (a) HCOOH
• (b) $CH_{3}COOH$
• (c) $(COOH)_{2}$
• (d) Both (a) and (c)

Correct Answer: (a)

Explanation: Formic acid ($HCOOH$) exhibits these unique properties because the aldehyde group attached to the hydroxyl group allows it to act as a reducing agent.

38. Ozonolysis product formaldehyde confirms:

• (a) a vinyl group
• (b) an isopropyl group
• (c) an acetylene triple bond
• (d) two ethylenic double bonds

Correct Answer: (a)

Explanation: The presence of a terminal double bond ($=CH_{2}$), such as in a vinyl group, yields formaldehyde upon ozonolysis.

39. Which of the following acids will undergo HVZ reaction?

• (a) $CH_{3}CH=CHCOOH$
• (b) $CH_{3}CH_{2}CBr_{2}COOH$
• (c) $C_{6}H_{5}COOH$
• (d) $(CH_{3})_{2}CHCH_{2}COOH$

Correct Answer: (d)

Explanation: The Hell-Volhard-Zelinsky (HVZ) reaction requires the presence of $\alpha$-hydrogen atoms in the carboxylic acid.

40. Order of reactivity of acid derivatives (i. chloride, ii. amide, iii. anhydride, iv. ester):

• (a) $(i) > (ii) > (iii) > (iv)$
• (b) $(i) > (iii) > (iv) > (ii)$
• (c) $(iii) > (i) > (iv) > (ii)$
• (d) $(ii) > (iv) > (iii) > (i)$

Correct Answer: (b)

Explanation: The reactivity depends on the leaving group ability: $Cl^{-} > RCOO^{-} > OR^{-} > NH_{2}^{-}$. Thus, acid chlorides are most reactive and amides are least.

41. Identify reagent X in: $CH_{3}CH_{2}C \equiv N \xrightarrow{X} CH_{3}CH_{2}CHO$.

• (a) $LiAlH_{4}$/ether
• (b) $H_{2}/Pd-BaSO_{4}$
• (c) $SnCl_{2}/HCl/H_{2}O$ boil
• (d) $NaBH_{4}$/ether

Correct Answer: (c)

Explanation: The reduction of nitriles to aldehydes using stannous chloride and HCl is known as the Stephen reaction.

42. Hydrolysis of an ester is carried out in:

• (a) basic medium
• (b) acidic medium
• (c) both (a) and (b)
• (d) neither (a) nor (b)

Correct Answer: (c)

Explanation: Ester hydrolysis can be catalyzed by either acids or bases.

43. Oxidation of alcohols involves formation of C=O bond with cleavage of:

• (a) O-H bond
• (b) C-H bond
• (c) both O-H and C-H bond
• (d) neither O-H nor C-H bond.

Correct Answer: (c)

Explanation: The conversion of an alcohol ($CH-OH$) to a carbonyl ($C=O$) requires the loss of hydrogen from both the carbon and the oxygen atoms.

44. $C_{9}H_{10}O$ gives 2,4-DNP, silver mirror, undergoes Cannizzaro, and vigorous oxidation gives phthalic acid. Compound is:

• (a) [structure a]
• (b) [structure b]
• (c) 2-ethylbenzaldehyde
• (d) [structure d]

Correct Answer: (c)

Explanation: The formula and phthalic acid product suggest an ortho-disubstituted benzene. 2-ethylbenzaldehyde lacks $\alpha$-hydrogens (Cannizzaro active) and is an aldehyde (Tollens' active).

45. Identify product S in the sequence from m-bromobenzoic acid ($SOCl_{2}, NH_{3}, NaOH/Br_{2}$):

• (a) [structure a]
• (b) [structure b]
• (c) [structure c]
• (d) m-bromoaniline

Correct Answer: (d)

Explanation: The acid is converted to the acid chloride, then to the amide, and finally undergoes Hofmann bromamide degradation to the amine (m-bromoaniline).