Download CBSE MCQs for Class 12 Chemistry: Unit 8 Aldehydes Ketones and Carboxylic Acids
Review structured MCQ sets for Class 12 Chemistry Unit 8 Aldehydes Ketones and Carboxylic Acids. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.
Chapter-wise Objective Questions: Unit 8 Aldehydes Ketones and Carboxylic Acids
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Multiple Choice Questions
(a) hydrolysis of esters
(b) oxidation of primary alcohols
(c) oxidation of secondary alcohols
(d) reaction of alkyl halides with alcohols.
Show Answer & Explanation
Answer: (c) oxidation of secondary alcohols
Explanation:
1. Ketones are formed by oxidation of secondary alcohols.
2. CH3–CH(OH)–CH3 [O]
⟶ CH3–CO–CH3
Teacher's Note:
1. Primary alcohols oxidise to aldehydes (and then acids); only secondary alcohols stop at the ketone stage.
(a) tertiary alcohols
(b) secondary alcohols
(c) primary alcohols
(d) carboxylic acids.
Show Answer & Explanation
Answer: (b) secondary alcohols
Explanation:
1. Formaldehyde forms primary alcohol while all other aldehydes form secondary alcohols on reaction with Grignard’s reagent followed by hydrolysis.
2. CH3CHO + CH3MgI → (CH3)(H)C(OMgI)CH3 H+/H2O
⟶ CH3–CH(OH)–CH3 + Mg(OH)I
Teacher's Note:
1. Remember the pattern: HCHO gives 1° alcohol, other aldehydes give 2° alcohols and ketones give 3° alcohols with RMgX.
(a) CH3CH2CH2CH2CH2CHO
(b) CH3CH2CH(OH)CH2CH2CHO
(c) CH3CH2CH(OH)CH(CH3)CHO
(d) CH3CH2COOH
Show Answer & Explanation
Answer: (c) CH3CH2CH(OH)CH(CH3)CHO
Explanation:
1. CH3CH2CHO + CH3CH2CHO dil. NaOH
⟶ CH3–CH2–CH(OH)–CH(CH3)–CHO
2. The α-carbon (CH2) of one propanal molecule attacks the carbonyl carbon of the other, so the CH3 ends up as a branch on the α-carbon.
Teacher's Note:
1. In an aldol product the new C–C bond is always between the α-carbon of one molecule and the carbonyl carbon of the other.
(i) CH3–CH(CH3)–COOH
(ii) CH3–CH(CH3)–CH2–COOH
(iii) CH3COOH
(iv) HCOOH
(a) (i) and (iii)
(b) (i), (ii) and (iii)
(c) (i), (ii), (iii) and (iv)
(d) (iii) and (iv)
Show Answer & Explanation
Answer: (c) (i), (ii), (iii) and (iv)
Explanation:
1. (CH3)2CH–CO–CH2–CH(CH3)2 [O]
⟶ CH3–CH(CH3)–COOH + CH3–CH(CH3)–CH2–COOH + CH3–CO–CH3
2. The CH3–CO–CH3 formed is further oxidised: CH3COCH3 → HCOOH + CH3COOH
3. So all four products (i), (ii), (iii) and (iv) are formed.
Teacher's Note:
1. Strong oxidation of a ketone breaks the C–CO bond on either side, so check both possible cleavages before choosing.
| X | Y | Z | |
|---|---|---|---|
| (a) | C6H5COCH3 | CH3CHO | CH3COCH3 |
| (b) | CH3–CH=C(CH3)–CH3 | CH3CHO | CH3COCH3 |
| (c) | CH3CH2CH=CH2 | CH3CH2CHO | HCHO |
| (d) | CH3–CH=CH–CH3 | CH3CHO | CH3CHO |
(a) (a)
(b) (b)
(c) (c)
(d) (d)
Show Answer & Explanation
Answer: (b) (b)
Explanation:
1. CH3–CH=C(CH3)–CH3 (X) (i) O3
⟶
(ii) Zn/H2O CH3CHO (Y) + CH3–CO–CH3 (Z)
2. CH3CHO gives positive Fehling’s test and iodoform test; CH3COCH3 gives only the iodoform test.
Teacher's Note:
1. Work backwards from ozonolysis: join the two carbonyl carbons by a C=C to rebuild the alkene.
(a) pentan-3-one, propanoic acid, butane
(b) pentanal, pentanoic acid, octane
(c) 2-methylbutanone, butanoic acid, hexane
(d) 2, 2-dimethylpropanone, propanoic acid, hexane
Show Answer & Explanation
Answer: (a) pentan-3-one, propanoic acid, butane
Explanation:
1. Since the compound gives positive 2,4-DNP test and negative Tollen’s test, it is a ketone.
2. CH3CH2–CO–CH2CH3 (pentan-3-one, X) [O]
⟶ 2CH3CH2COOH (propanoic acid, Y)
3. CH3CH2COOH KOH
⟶ CH3CH2COOK + H2O; 2CH3CH2COO− Electrolysis
⟶ CH3CH2CH2CH3 (Z) + 2CO2
Teacher's Note:
1. In Kolbe electrolysis two alkyl groups join, so the alkane has twice the carbons of R: C2H5 + C2H5 = butane.
(a) FeCl3 is used in the detection of phenols.
(b) Fehling solution is used in the detection of glucose.
(c) Tollens’ reagent is used in the detection of unsaturation.
(d) NaHSO3 is used in the detection of carbonyl compounds.
Show Answer & Explanation
Answer: (c) Tollens’ reagent is used in the detection of unsaturation.
Explanation:
1. Tollens’ reagent is used to detect aldehyde group (not unsaturation).
Teacher's Note:
1. Unsaturation is tested with bromine water or Baeyer’s reagent; Tollens’ reagent is specific for the –CHO group.
(a) CH3COCl
(b) CH3COOC2H5
(c) CH3COCH3
(d) CH3CONH2
Show Answer & Explanation
Answer: (c) CH3COCH3
Explanation:
1. (CH3)2C=O + H2NNH–C6H3(NO2)2 → (CH3)2C=N–NH–C6H3(NO2)2 + H2O
2. Acetone forms the orange-red 2,4-dinitrophenylhydrazone; acid chlorides, esters and amides do not.
Teacher's Note:
1. The C=O of acid derivatives is stabilised by resonance with Cl, OR or NH2, so they do not give the 2,4-DNP test.
(a) CH3CH=CHCHO → CH3CH=CHCOOH : Ammonical AgNO3
(b) CH3CH=CHCHO → CH3CH=CHCH2OH : H2/Pt
(c) R–COOH → R–CH2OH : LiAlH4
(d) CH3CH2COCl → CH3CH2CHO : H2, Pd/BaSO4
Show Answer & Explanation
Answer: (b) CH3CH=CHCHO → CH3CH=CHCH2OH : H2/Pt
Explanation:
1. H2/Pt reduces the C=C bond as well as the –CHO group, so it cannot give CH3CH=CHCH2OH.
2. CH3CH=CHCHO LiAlH4
⟶ CH3CH=CHCH2OH (LiAlH4 or NaBH4 reduces only the carbonyl group).
Teacher's Note:
1. Catalytic hydrogenation is not selective; use LiAlH4 or NaBH4 when the C=C must survive.
R–CO–Cl H2
⟶
Pd/BaSO4 P
(a) RCHO
(b) RCH3
(c) RCOOH
(d) RCH2OH
Show Answer & Explanation
Answer: (a) RCHO
Explanation:
1. Acid chlorides are reduced to aldehydes on reaction with BaSO4 and Pd. The reaction is called Rosenmund reduction.
2. RCOCl H2
⟶
Pd/BaSO4 RCHO + HCl
Teacher's Note:
1. BaSO4 poisons the Pd catalyst so that the aldehyde is not reduced further to an alcohol.
(a) Phenyl acetaldehyde
(b) 2-Methylpentanal
(c) Benzaldehyde
(d) 1-Phenylpropanone
Show Answer & Explanation
Answer: (c) Benzaldehyde
Explanation:
1. Benzaldehyde will not give aldol condensation due to absence of α-H atom.
Teacher's Note:
1. Check the carbon next to C=O: in benzaldehyde it is a ring carbon with no hydrogen.
(a) It is a reducing agent.
(b) It is a weaker acid than acetic acid.
(c) It is an oxidising agent.
(d) When its calcium salt is heated, it forms acetone.
Show Answer & Explanation
Answer: (a) It is a reducing agent.
Explanation:
1. Formic acid acts as a reducing agent: it reduces Fehling’s and Tollen’s reagent, etc.
Teacher's Note:
1. HCOOH contains an aldehyde-like H–C=O unit, which is why it alone among simple acids reduces Tollens’ reagent.
(a) C6H5CH=C(CH3)2
(b) C6H5CH2–CO–CH=CH2
(c) C6H5–CO–CH=CH–CH3
(d) C6H5–CH=CH–CO–CH3
Show Answer & Explanation
Answer: (d) C6H5–CH=CH–CO–CH3
Explanation:
1. C6H5CHO + CH3COCH3 → C6H5–CH(OH)–CH2–CO–CH3 Δ
⟶
–H2O C6H5–CH=CH–CO–CH3
Teacher's Note:
1. In this cross aldol (Claisen–Schmidt) reaction only acetone has α-hydrogens, so acetone forms the carbanion and benzaldehyde is attacked.
(a) FCH2CHO
(b) ClCH2CHO
(c) BrCH2CHO
(d) ICH2CHO
Show Answer & Explanation
Answer: (a) FCH2CHO
Explanation:
1. FCH2CHO is most reactive towards nucleophilic addition since presence of most electronegative F withdraws electrons from carbon of carbonyl group making it more polar.
Teacher's Note:
1. Electron-withdrawing groups increase the positive charge on the carbonyl carbon; the stronger the –I effect, the faster the nucleophilic addition.
(a) : 3-Cyclopentylpropanoic acid
(b) (CH3)2C=CHCOOH : 3-Methylbut-2-enoic acid
(c) PhCH2CH2COOH : 3-Phenylpropanoic acid
(d) : 2, 4, 6-Trinitrobenzoic acid
Show Answer & Explanation
Answer: (a) : 3-Cyclopentylpropanoic acid
Explanation:
1. The structure in (a) is 2-methylcyclopentanecarboxylic acid (the –COOH and –CH3 are on adjacent ring carbons), not 3-cyclopentylpropanoic acid.
Teacher's Note:
1. When –COOH is attached directly to a ring, use the suffix ‘carboxylic acid’ and number the ring from that carbon.
(a) CH3CH2CH2CH2–CO–H
(b) CH3CH2CH2–CO–CH3
(c) CH3CH2–CO–CH2CH3
(d) CH3–CO–CH(CH3)–CH3
Show Answer & Explanation
Answer: (a) CH3CH2CH2CH2–CO–H
Explanation:
1. Aldehydes are more reactive than ketones.
Teacher's Note:
1. Aldehydes have one small H on the carbonyl carbon: less steric hindrance and less +I effect than in ketones.
(a) Phenol < Ethanol < Chloroacetic acid < Acetic acid
(b) Ethanol < Phenol < Chloroacetic acid < Acetic acid
(c) Ethanol < Phenol < Acetic acid < Chloroacetic acid
(d) Chloroacetic acid < Acetic acid < Phenol < Ethanol
Show Answer & Explanation
Answer: (c) Ethanol < Phenol < Acetic acid < Chloroacetic acid
Explanation:
1. Due to –I effect of Cl, chloroacetic acid is a stronger acid than acetic acid.
2. Due to stabilization of phenoxide ion by resonance, phenol is a stronger acid than ethanol.
Teacher's Note:
1. Carboxylic acids are stronger than phenols because the negative charge of the carboxylate ion is shared equally by two oxygen atoms.
(a) Pentan-2-one will give silver mirror test
(b) Pentan-2-one will give iodoform test.
(c) Pentan-3-one will give iodoform test
(d) None of these.
Show Answer & Explanation
Answer: (b) Pentan-2-one will give iodoform test.
Explanation:
1. Pentan-2-one will give positive iodoform test while pentan-3-one will not give this test.
2. CH3COCH2CH2CH3 + I2 + NaOH → CHI3 (Iodoform)
Teacher's Note:
1. Only methyl ketones (CH3CO–) give iodoform; pentan-3-one has ethyl groups on both sides.
(a) Aldehyde is formed.
(b) Primary alcohol is formed.
(c) Ketone is formed.
(d) Grignard reagent is formed.
Show Answer & Explanation
Answer: (b) Primary alcohol is formed.
Explanation:
1. RCOOH + 4[H] LiAlH4
⟶ RCH2OH + H2O
Teacher's Note:
1. The aldehyde intermediate is reduced faster than the acid, so LiAlH4 always takes the acid all the way to the 1° alcohol.
(a) Propanoic acid
(b) Ethanoic acid
(c) 2-Methylpropanic acid
(d) 2,2-Dimethylpropanoic acid
Show Answer & Explanation
Answer: (d) 2,2-Dimethylpropanoic acid
Explanation:
1. 2,2-Dimethylpropanoic acid, (CH3)3C–COOH, will not undergo HVZ reaction due to absence of an α-H atom.
Teacher's Note:
1. HVZ needs at least one α-hydrogen; draw the carbon next to –COOH and count its H atoms.
(a) aldol condensation
(b) cross aldol condensation
(c) Cannizzaro reaction
(d) HVZ condensation.
Show Answer & Explanation
Answer: (b) cross aldol condensation
Explanation:
1. Aldol condensation between two different aldehydes, each having α-hydrogen, is called cross aldol condensation.
2. CH3CH2CHO + (CH3)2CHCHO NaOH
⟶ CH3CH2CH(OH)C(CH3)2CHO + CH3CH2CH=C(CH3)CHO + (CH3)2CHCH(OH)C(CH3)2CHO + (CH3)2CHCH=C(CH3)CHO
Teacher's Note:
1. When both carbonyl compounds have α-hydrogens, cross aldol gives a mixture of four products (two self and two cross).
(a) passing a mixture of ethyne and steam over a catalyst, magnesium at 420°C
(b) passing a mixture of ethyne and ethanol over a catalyst zinc chromite
(c) boiling ethyne with water and H2SO4
(d) treating ethyne with iodine and NaOH.
Show Answer & Explanation
Answer: (a) passing a mixture of ethyne and steam over a catalyst, magnesium at 420°C
Explanation:
1. 2CH≡CH + 3H2O 420°C
⟶
Mg or Zinc vanadate CH3COCH3 + CO2 + 2H2
Teacher's Note:
1. Hydration of ethyne with dil. H2SO4/HgSO4 gives ethanal, not propanone – a common trap.
| Column I | Column II |
|---|---|
| (A) RCOCH3 Zn-Hg ⟶ HCl RCH2CH3 | (i) Kolbe’s reaction |
| (B) 2C6H5CHO NaOH ⟶ C6H5COONa + C6H5CH2OH | (ii) Clemmensen reduction |
| (C) C6H6 + CH3COCl Anh. AlCl3 ⟶ C6H5COCH3 | (iii) Friedel–Crafts reaction |
| (D) C6H5OH + CO2 + NaOH → HOC6H4COONa | (iv) Cannizzaro reaction |
(a) (A) → (ii), (B) → (iv), (C) → (iii), (D) → (i)
(b) (A) → (i), (B) → (iii), (C) → (ii), (D) → (iv)
(c) (A) → (iii), (B) → (ii), (C) → (i), (D) → (iv)
(d) (A) → (iv), (B) → (i), (C) → (ii), (D) → (iii)
Show Answer & Explanation
Answer: (a) (A) → (ii), (B) → (iv), (C) → (iii), (D) → (i)
Explanation:
1. (A) C=O to CH2 with Zn-Hg/HCl is Clemmensen reduction; (B) disproportionation of benzaldehyde with NaOH is Cannizzaro reaction.
2. (C) acylation of benzene with CH3COCl/AlCl3 is Friedel–Crafts reaction; (D) phenol + CO2 + NaOH giving sodium salicylate is Kolbe’s reaction.
Teacher's Note:
1. Kolbe’s reaction (phenol + CO2) is different from Kolbe electrolysis (carboxylate salts giving alkanes) – read the reaction, not just the name.
Ethanal, Propanal, Propanone, Butanone
(a) Butanone < Propanone < Propanal < Ethanal
(b) Propanone < Butanone < Ethanal < Propanal
(c) Propanal < Ethanal < Propanone < Butanone
(d) Ethanal < Propanal < Propanone < Butanone
Show Answer & Explanation
Answer: (a) Butanone < Propanone < Propanal < Ethanal
Explanation:
1. Ketones are less reactive than aldehydes.
2. Larger alkyl groups increase the +I effect and steric hindrance, so butanone < propanone and propanal < ethanal.
Teacher's Note:
1. Both size and number of alkyl groups on the carbonyl carbon lower reactivity – rank ketones first, then aldehydes.
(i) Benzoyl chloride + Benzene + AlCl3
(ii) Benzoyl chloride + Phenylmagnesium bromide
(iii) Benzoyl chloride + Diphenyl cadmium
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (i), (ii) and (iii)
Show Answer & Explanation
Answer: (c) (i) and (iii)
Explanation:
1. (i) C6H5COCl + C6H6 AlCl3
⟶ C6H5–CO–C6H5 + HCl
2. (iii) 2C6H5COCl + (C6H5)2Cd → 2C6H5–CO–C6H5 + CdCl2
3. With phenylmagnesium bromide the ketone formed reacts further to give a tertiary alcohol.
Teacher's Note:
1. Dialkyl/diaryl cadmium is used instead of Grignard reagent because it is less reactive and stops at the ketone.
(a) CH3MgBr, H2O, Oxidation
(b) Oxidation, Ca(OH)2, Heat
(c) Reduction, KCN, Hydrolysis
(d) Oxidation, C2H5ONa, Heat
Show Answer & Explanation
Answer: (b) Oxidation, Ca(OH)2, Heat
Explanation:
1. CH3CHO K2Cr2O7/H+
⟶
[O] CH3COOH Ca(OH)2
⟶ (CH3COO)2Ca heat
⟶ CH3COCH3
Teacher's Note:
1. Option (a) also gives acetone (via propan-2-ol), but it adds a carbon through the Grignard step; dry distillation of calcium acetate is the expected route here.
(a) Cannizzaro reaction
(b) Clemmensen reduction
(c) Rosenmund reduction
(d) Wolff-Kishner reduction.
Show Answer & Explanation
Answer: (b) Clemmensen reduction
Explanation:
1. >C=O Zn/Hg
⟶
HCl >CH2 + H2O is called Clemmensen reduction.
Teacher's Note:
1. Clemmensen uses acid (Zn-Hg/HCl); Wolff–Kishner uses base (NH2NH2/KOH) – choose based on what else the molecule can tolerate.
(a) nucleophilic addition
(b) electrophilic addition
(c) free radical addition
(d) electromeric addition.
Show Answer & Explanation
Answer: (a) nucleophilic addition
Explanation:
1. >C=O <→ >C+–O−
2. Nucleophile (CN−) attacks at the positive C centre of carbonyl group hence the addition is nucleophilic addition.
Teacher's Note:
1. The type of addition is named after the species that attacks first – here CN−, a nucleophile.
CH3COCH3 + X → (CH3)3C–OMgCl H2O
⟶ (CH3)3C–OH + Mg(OH)Cl
(a) CH3MgCl
(b) CH3COCl + Mg
(c) MgCl2
(d) CH3CH2MgCl
Show Answer & Explanation
Answer: (a) CH3MgCl
Explanation:
1. CH3–CO–CH3 + CH3MgCl (X) → (CH3)3C–OMgCl HOH, H+
⟶ (CH3)3C–OH + Mg(OH)Cl
Teacher's Note:
1. Count carbons: acetone (3 C) + the alkyl of RMgX must give the 4-carbon tert-butyl alcohol, so R = CH3.
(a) ClCH2COOH > BrCH2COOH > FCH2COOH
(b) BrCH2COOH > ClCH2COOH > FCH2COOH
(c) FCH2COOH > ClCH2COOH > BrCH2COOH
(d) ClCH2COOH > FCH2COOH > BrCH2COOH
Show Answer & Explanation
Answer: (c) FCH2COOH > ClCH2COOH > BrCH2COOH
Explanation:
1. The electron withdrawing strength of halogen groups is in the order of F > Cl > Br. Hence, the strength of acids is FCH2COOH > ClCH2COOH > BrCH2COOH.
Teacher's Note:
1. Acid strength of haloacids follows the electronegativity of the halogen: F > Cl > Br > I.
Assertion–Reason Questions
Assertion (A): Aromatic aldehydes and formaldehyde undergo Cannizzaro reaction.
Reason (R): Aromatic aldehydes are almost as reactive as formaldehyde.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Aromatic aldehydes and formaldehyde do not contain α-hydrogen and thus undergo Cannizzaro reaction. Formaldehyde is more reactive than aromatic aldehydes.
Teacher's Note:
1. The condition for Cannizzaro reaction is the absence of α-hydrogen, not high reactivity.
Assertion (A): α-Hydrogen atoms in aldehydes and ketones are acidic.
Reason (R): The anion left after the removal of α-hydrogen is stabilised by inductive effect.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. The anion left after the removal of α-hydrogen is stabilized by resonance effect.
Teacher's Note:
1. The carbanion is delocalised onto the carbonyl oxygen (enolate ion); it is this resonance that makes α-H acidic.
Assertion (A): Hydrogen bonding in carboxylic acids is stronger than alcohols.
Reason (R): Highly branched carboxylic acids are more acidic than unbranched acids.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Highly branched carboxylic acids are less acidic than unbranched acids. The +I effect of alkyl groups in branched acid increases the magnitude of negative charge. Thus, –COOH group is shielded from solvent molecules and cannot be stabilized by solvation as effectively as in unbranched carboxylic acids.
Teacher's Note:
1. Alkyl groups are electron-releasing: more alkyl branching means a less stable carboxylate ion and a weaker acid.
Assertion (A): m-Chlorobenzoic acid is a stronger acid than p-chlorobenzoic acid.
Reason (R): In m-chlorobenzoic acid both –I-effect and +R-effect of Cl operate but in p-chlorobenzoic acid only +R-effect of Cl operates.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. In p-chlorobenzoic acid, both +R and –I effect operate together but in m-chlorobenzoic acid only –I effect operates. Therefore, m-chlorobenzoic acid is a stronger acid than p-chlorobenzoic acid.
Teacher's Note:
1. Resonance (+R) effects act only at ortho and para positions; at the meta position only the inductive effect works.
Assertion (A): Ketones can be converted into acids by haloform reaction.
Reason (R): Addition of Grignard reagents to dry ice followed by hydrolysis gives ketones.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Addition of Grignard reagents to dry ice followed by hydrolysis gives carboxylic acid not ketone.
2. RMgX + O=C=O Dry
⟶
ether R–CO–OMgX H+/H2O
⟶ R–CO–OH
Teacher's Note:
1. Grignard reagent + CO2 is a standard way to make an acid with one carbon more than the alkyl halide.
Assertion (A): Acetic acid in vapour state shows a molecular mass of 120.
Reason (R): It undergoes intermolecular hydrogen bonding.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Acetic acid (molar mass 60) exists as a cyclic dimer held by two intermolecular hydrogen bonds even in the vapour state, so its observed molecular mass is 2 × 60 = 120.
Teacher's Note:
1. Carboxylic acid dimers survive in the vapour phase and in non-polar solvents like benzene, doubling the apparent molar mass.
Assertion (A): Nitration of benzoic acid gives m-nitrobenzoic acid.
Reason (R): Carboxyl group increases the electron density at the meta-position.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Carboxyl group only marginally decreases the electron density at m-position relative to o- and p-positions.
Teacher's Note:
1. –COOH deactivates the whole ring; the meta position is attacked only because it is the least deactivated.
Assertion (A): Boiling point of aldehydes lie in between parent alkanes and corresponding alcohols.
Reason (R): Aldehydes cannot form intermolecular hydrogen bonds like alcohols.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Aldehydes have higher molecular weight than parent alkanes as well as polarity in aldehydes shows higher boiling point than parent alkanes. Aldehydes do not have any hydrogen atom attached directly to the oxygen so they cannot form hydrogen bond with each other.
Teacher's Note:
1. Aldehydes boil above alkanes because of dipole–dipole attraction and below alcohols because they lack O–H hydrogen bonding.
Assertion (A): Carboxylic acids are stabilised by resonance.
Reason (R): Chloroacetic acid is weaker than acetic acid.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Chlorine atom has –I effect which increases the ionisation of chloroacetic acid and stabilizes the chloroacetate ion by dispersal of negative charge. In acetic acid, methyl group due to +I effect destabilizes the acetate ion by intensification of negative charge. Hence, chloroacetic acid dissociates to a greater extent than acetic acid.
Teacher's Note:
1. Any electron-withdrawing group on the α-carbon strengthens a carboxylic acid; electron-releasing groups weaken it.
Assertion (A): Benzaldehyde undergoes aldol condensation.
Reason (R): Aldehydes having α-hydrogen atom undergo aldol condensation.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. Aldehydes having a methyl or methylene group in the α-position or more correctly having atleast one hydrogen atom in the α-position undergo dimerisation in presence of a base at low temperature to form β-hydroxy aldehydes called aldols. As benzaldehyde does not have any α-hydrogen hence it does not undergoes aldol condensation.
Teacher's Note:
1. Benzaldehyde can take part in a cross aldol (as the acceptor), but it cannot undergo self aldol condensation.
Free study material for Chemistry
Unit 8 Aldehydes Ketones and Carboxylic Acids Objective Questions & Solutions for Class 12 Chemistry
Class 12 Chemistry Unit 8 Aldehydes Ketones and Carboxylic Acids Objective Test Questions
Explore reliable practice questions for Unit 8 Aldehydes Ketones and Carboxylic Acids tailored for Class 12 Chemistry learners. Use these multiple-choice formats to evaluate preparedness and strengthen problem-solving skills.
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FAQs
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