CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids Important Questions Set 05

Very Short Answer Questions (1 Mark)

Question 1. Draw the structure of the compound whose IUPAC name is 4-chloropentan-2-one.
Answer:
CH₃–CH(Cl)–CH₂–CO–CH₃

Teacher's Note:
1. Number from the end giving the C=O carbon the lowest locant (C-2); Cl then falls on C-4.

Question 2. Which type of aldehydes can undergo Cannizzaro reaction?
Answer:
Aromatic and aliphatic aldehydes which do not contain α-hydrogen.

Teacher's Note:
1. Examples to quote: HCHO, C₆H₅CHO and (CH₃)₃CCHO.

Question 3. Arrange the following in order of their increasing reactivity towards HCN:
CH₃CHO, CH₃COCH₃, HCHO, C₂H₅COCH₃
Answer:
C₂H₅COCH₃ < CH₃COCH₃ < CH₃CHO < HCHO

Teacher's Note:
1. Larger and more numerous alkyl groups reduce the positive charge on the carbonyl carbon and add steric hindrance, slowing nucleophilic addition.

Question 4. Mention industrial product obtained from HCHO.
Answer:
Bakelite

Teacher's Note:
1. Bakelite is a phenol-formaldehyde polymer; HCHO is also used to make urotropine and urea-formaldehyde resins.

Question 5. Arrange the following compounds in increasing order of their boiling point:
CH₃CHO, CH₃CH₂OH, CH₃–O–CH₃, CH₃–CH₂–CH₃
Answer:
CH₃–CH₂–CH₃ < CH₃–O–CH₃ < CH₃CHO < CH₃CH₂OH

Teacher's Note:
1. For similar molar masses: H-bonded alcohol > polar carbonyl (dipole-dipole) > weakly polar ether > non-polar alkane.

Question 6. How is acetone obtained from ethanol?
Answer:
CH₃–CH₂–OH –[Cu, 573 K]→ CH₃–CHO –[(i) CH₃MgBr, (ii) H₂O/H⁺]→ CH₃–CH(OH)–CH₃ –[Cu, 573 K]→ CH₃–CO–CH₃

Teacher's Note:
1. Ethanol has two carbons and acetone three, so a carbon must be added – the Grignard step does this.

Question 7. Why do aldehydes and ketones have lower boiling point than alcohols?
Answer:
Due to presence of associated molecules with H-bonding in alcohols.

Teacher's Note:
1. Aldehydes and ketones have no H attached to O, so they cannot form intermolecular hydrogen bonds among themselves.

Question 8. Write reaction between acetyl chloride and dimethyl cadmium.
Answer:
2CH₃COCl + Cd(CH₃)₂ –[dry ether]→ 2CH₃–CO–CH₃ + CdCl₂

Teacher's Note:
1. Dialkylcadmium is preferred over Grignard reagent because it does not react further with the ketone formed.

Question 9. What happens when CH₃CHO is treated with K₂Cr₂O₇ in presence of H₂SO₄?
Answer:
CH₃CHO + [O] –[K₂Cr₂O₇ + H₂SO₄]→ CH₃COOH

Teacher's Note:
1. The orange dichromate turns green (Cr³⁺) as the aldehyde is oxidised to the acid with the same number of carbons.

Question 10. Write the IUPAC name of
OHC–CH=C(CH₃)–CH₂–CH₂–CH=C(CH₃)–CH₃
Answer:
3,7-Dimethylocta-2,6-dienal

Teacher's Note:
1. The CHO carbon is always C-1; with two double bonds the ending is ‘-dienal’ and the ‘a’ is added before ‘dien’.

Short Answer Questions (2 Marks)

Question 11. tert-Butylbenzene does not give benzoic acid on oxidation with acidic KMnO₄. Give reasons.
Answer:
Due to absence of benzylic hydrogens, tert-butylbenzene does not undergo oxidation to give benzoic acid while other alkylbenzenes such as toluene, ethylbenzene, n-propylbenzene and isopropylbenzene which contain one or more benzylic hydrogens (on the α-carbon) undergo oxidation to form benzoic acid.
C₆H₅CH₃, C₆H₅CH₂CH₃, C₆H₅CH₂CH₂CH₃ or C₆H₅CH(CH₃)₂ →^[O] C₆H₅COOH (benzoic acid)
C₆H₅C(CH₃)₃ (tert-butylbenzene, no benzylic H) →^[O] no oxidation

Teacher's Note:
1. Whatever the length of the side chain, an alkylbenzene with at least one benzylic H is oxidised to benzoic acid.

Question 12. Me₃CCH₂COOH is more acidic than Me₃SiCH₂COOH.
Answer:
Si (E.N. = 1.8) is more electropositive than C (E.N. = 2.5), therefore, Me₃Si (trimethylsilyl group) has greater +I-effect than that of Me₃C (t-butyl group). As a result, Me₃Si intensifies the −ve charge on the carboxylate ion relative to t-butyl group and hence Me₃CCH₂COOH is a stronger acid than Me₃SiCH₂COOH.
Me₃Si→CH₂→COO⁻ (greater +I-effect)    Me₃C→CH₂→COO⁻ (smaller +I-effect)

Teacher's Note:
1. Electron-donating groups destabilise the carboxylate ion and so lower acidity; the more electropositive the atom, the stronger its +I effect.

Question 13. Explain why the carbonyl oxygen atom of a carboxylic acid is more basic than hydroxyl oxygen.
Answer:
Protonation at the hydroxyl oxygen of a carboxylic acid gives structure (I) but protonation at the carbonyl oxygen gives structure (II) which can be regarded as a resonance hybrid of two structures (II and III). Thus, structure (II) is more stable than structure (I) and hence carbonyl oxygen atom of a carboxylic acid is more basic than hydroxyl oxygen.
R–CO–OH →^H+ R–CO–O⁺H₂ (I)   or   R–C(=O⁺H)–OH (II) ↔ R–C(OH)=O⁺H (III)

Teacher's Note:
1. The more stable the conjugate acid (here resonance-stabilised), the more basic the site that was protonated.

Question 14. A compound with molecular formula, C₄H₁₀O₃ on acetylation with acetic anhydride gives a compound with molecular weight 190. Find out the number of hydroxyl groups present in the compound.
Answer:
During acetylation, one H-atom (at. mass = 1 amu) of the OH group is replaced by an acetyl group, i.e., CH₃CO (mol. mass = 43 amu).
–OH + (CH₃CO)₂O → –O–COCH₃ + CH₃COOH
In other words, acetylation of each OH group increases the mass by 43 − 1 = 42 amu. Now the mol. mass of C₄H₁₀O₃ = 106 amu while that of the acetylated product is 190 amu, therefore, the number of OH groups in the compound = \( \frac{190 - 106}{42} \) = 2.

Teacher's Note:
1. Each acetylated –OH adds 42 u; divide the total increase in molar mass by 42 to count the OH groups.

Question 15. Write structures of various possible carbonyl compounds having the molecular formula, C₄H₈O. Give their common as well as IUPAC names.
Answer:
Three carbonyl compounds are possible :
(i) CH₃CH₂CH₂CHO : n-Butyraldehyde (Butanal)
(ii) (CH₃)₂CH–CHO : Isobutyraldehyde (2-Methylpropanal)
(iii) CH₃COCH₂CH₃ : Ethyl methyl ketone (Butan-2-one or Butanone)

Teacher's Note:
1. Butan-1-one is not possible: a carbonyl at the end of the chain is an aldehyde, so a ketone needs at least three carbons.

Question 16. Write the formula of the following : (i) 4-Methylpent-3-en-2-one (ii) 2-Ethoxy-4-methoxypentan-3-one
Answer:
(i) (CH₃)₂C=CHCOCH₃ (ii) CH₃–CH(OCH₃)–CO–CH(OCH₂CH₃)CH₃

Teacher's Note:
1. Draw the parent chain first, number it, then hang the substituents on the stated carbons; this avoids misplaced groups.

Question 17. In what respects, the C=C and C=O bonds resemble and differ from each other ? (Hr. Board 2012)
Answer:
Resemblance : Both are double bonds made of one σ- and one π-bond; in both, the carbon atom is sp²-hybridised and the atoms directly attached lie in one plane with bond angles of about 120°. Both undergo addition reactions.
Difference : C=C bond is non-polar, while C=O bond is polar (C^δ+=O^δ−) because oxygen is more electronegative than carbon. Hence C=C undergoes electrophilic addition reactions, whereas C=O undergoes nucleophilic addition reactions. Also, the C=O bond (122 pm) is shorter and stronger than the C=C bond (134 pm).

Teacher's Note:
1. The polarity of C=O is the single fact that explains why its additions start with a nucleophile, unlike alkenes.

Question 18. Why do aldehydes behave like polar compounds ?
Answer:
Oxygen is more electronegative than carbon. Therefore, the electrons of the C=O bond, especially the mobile π-electrons, are pulled towards oxygen. The carbonyl group is thus a resonance hybrid of >C=O ↔ >C⁺–O⁻, with a partial positive charge on carbon and a partial negative charge on oxygen. Due to this, aldehydes have large dipole moments (about 2.3–2.8 D) and behave like polar compounds.

Teacher's Note:
1. The polarity gives aldehydes higher boiling points than hydrocarbons and ethers of similar mass.

Question 19. How is benzophenone prepared from benzene ? Give only equations. (Hr. Board 2011)
Answer:
C₆H₆ + C₆H₅COCl \( \xrightarrow{Anhyd.\ AlCl_3} \) C₆H₅COC₆H₅ + HCl
Or 2C₆H₆ + COCl₂ \( \xrightarrow{Anhyd.\ AlCl_3} \) C₆H₅COC₆H₅ + 2HCl

Teacher's Note:
1. Friedel-Crafts acylation stops after one substitution because the C=O group deactivates the ring.

Question 20. How will you obtain 3-pentanone from propionic acid ?
Answer:
2CH₃CH₂COOH \( \xrightarrow{MnO,\ 573\ K} \) CH₃CH₂COCH₂CH₃ + CO₂ + H₂O
Or 2CH₃CH₂COOH + CaCO₃ → (CH₃CH₂COO)₂Ca \( \xrightarrow{Dry\ distil.} \) CH₃CH₂COCH₂CH₃ + CaCO₃

Teacher's Note:
1. Two molecules of the acid combine: the ketone has 2n − 1 carbons (here 2 × 3 − 1 = 5).

Question 21. Show with chemical equations, how will you convert (i) acetaldehyde into acetone and (ii) benzaldehyde into acetophenone ?
Answer:
(i) CH₃CHO + CH₃MgBr \( \xrightarrow{Dry\ ether} \) CH₃CH(OMgBr)CH₃ \( \xrightarrow{H_3O^+} \) CH₃CH(OH)CH₃ \( \xrightarrow{K_2Cr_2O_7/H_2SO_4\ or\ Cu,\ 573\ K} \) CH₃COCH₃
(ii) C₆H₅CHO + CH₃MgBr \( \xrightarrow{Dry\ ether} \) C₆H₅CH(OMgBr)CH₃ \( \xrightarrow{H_3O^+} \) C₆H₅CH(OH)CH₃ \( \xrightarrow{K_2Cr_2O_7/H_2SO_4} \) C₆H₅COCH₃

Teacher's Note:
1. A useful pattern: aldehyde + RMgX gives a 2° alcohol, whose oxidation gives a ketone with one more alkyl group.

Question 22. Aldehydes and ketones undergo nucleophilic addition reactions. Justify.
Answer:
Due to the greater electronegativity of oxygen, the C=O bond is polar: the carbonyl carbon carries a partial positive charge (>C^δ+=O^δ−) and is electrophilic. Moreover, the carbonyl carbon is sp²-hybridised and the group is planar, so it is open to attack from above or below the plane. Hence the reaction is initiated by a nucleophile attacking this carbon, followed by addition of H⁺ to the negatively charged oxygen, i.e., aldehydes and ketones undergo nucleophilic addition reactions.
e.g., (CH₃)₂C=O + HCN → (CH₃)₂C(OH)CN

Teacher's Note:
1. Contrast with alkenes: a non-polar C=C is electron-rich and is attacked by electrophiles instead.

Question 23. How is Grignard reagent used to prepare tert-butyl alcohol.
Answer:
Acetone reacts with methylmagnesium bromide in dry ether to form an addition product which on hydrolysis gives tert-butyl alcohol.
(CH₃)₂C=O + CH₃MgBr \( \xrightarrow{Dry\ ether} \) (CH₃)₃C–OMgBr \( \xrightarrow[-Mg(OH)Br]{H^+/H_2O} \) (CH₃)₃C–OH (2-Methylpropan-2-ol)

Teacher's Note:
1. All three alkyl groups on the 3° carbon are CH₃: two come from acetone and one from the Grignard reagent.

Question 24. Convert ketones into tertiary alcohols.
Answer:
Ketones react with Grignard reagents to form addition products which on hydrolysis with dil. acid give tertiary alcohols.
R₂C=O + R′MgX \( \xrightarrow{Dry\ ether} \) R₂C(R′)–OMgX \( \xrightarrow[-Mg(OH)X]{H^+/H_2O} \) R₂C(R′)–OH (3° Alcohol)
e.g., CH₃COCH₃ + C₂H₅MgBr → then H₃O⁺ → (CH₃)₂C(OH)C₂H₅ (2-Methylbutan-2-ol)

Teacher's Note:
1. The reaction must be done in dry ether; even traces of water destroy the Grignard reagent, giving an alkane.

Question 25. Explain why during preparation of ammonia derivatives from aldehydes and ketones, pH of the reaction is carefully controlled.
Answer:
The reaction of aldehydes and ketones with ammonia derivatives (H₂N–Z) is acid catalysed: protonation of the carbonyl oxygen increases the electrophilic character of the carbonyl carbon.
In strongly acidic medium, however, the ammonia derivative itself gets protonated (H₃N⁺–Z) and loses its lone pair, so it no longer acts as a nucleophile. In alkaline medium, the carbonyl group is not protonated and the reaction is slow.
Hence the pH is carefully controlled in weakly acidic range (about 3.5) where the carbonyl group is activated and enough free nucleophile is present.
>C=O + H₂N–Z \( \xrightarrow{H^+\ (pH\ 3.5)} \) >C=N–Z + H₂O

Teacher's Note:
1. Remember the balance: too much acid kills the nucleophile, too little fails to activate the carbonyl group.

Question 26. Write the structure of the product formed when semicarbazide reacts with formaldehyde.
Answer:
H₂C=O + H₂N–NHCONH₂ \( \xrightarrow{H^+} \) H₂C=N–NHCONH₂ + H₂O
Formaldehyde + Semicarbazide → Formaldehyde semicarbazone

Teacher's Note:
1. Only the terminal NH₂ of the NH–NH₂ end reacts; the amide NH₂ stays unchanged.

Question 27. Formaldehyde and benzaldehyde give Cannizzaro reaction but acetaldehyde does not. Explain. (Tamil Nadu Board 2012)
Answer:
Cannizzaro reaction is given by aldehydes which do not contain α-hydrogen atoms. Formaldehyde (HCHO) and benzaldehyde (C₆H₅CHO) have no α-H, so with conc. alkali they undergo Cannizzaro reaction.
2HCHO + NaOH (conc.) → CH₃OH + HCOONa ; 2C₆H₅CHO + NaOH (conc.) → C₆H₅CH₂OH + C₆H₅COONa
Acetaldehyde contains three α-H atoms; with alkali, an α-H is removed to form a carbanion (enolate), so it undergoes aldol condensation instead of Cannizzaro reaction.

Teacher's Note:
1. Rule of thumb: α-H present → aldol; α-H absent → Cannizzaro.

Question 28. Why acetaldehyde gives aldol condensation while formaldehyde does not. Explain. (Manipur Board 2012)
or (CH₃)₂CH–CHO undergoes aldol condensation whereas (CH₃)₃C–CHO does not. Give reasons. (CBSE (Foreign) 2017)
Answer:
Aldol condensation requires at least one α-hydrogen, which is removed by the base to form the carbanion (enolate ion) that attacks the carbonyl group of a second molecule.
Acetaldehyde (CH₃CHO) has three α-H atoms and so undergoes aldol condensation : 2CH₃CHO \( \xrightarrow{dil.\ NaOH} \) CH₃CH(OH)CH₂CHO. Formaldehyde (HCHO) has no α-carbon and hence no α-H, so it does not.
Similarly, (CH₃)₂CH–CHO has one α-H and undergoes aldol condensation, while (CH₃)₃C–CHO has no α-H and does not.

Teacher's Note:
1. The α-carbon is the carbon directly attached to the C=O group; count the H atoms only on that carbon.

Short Answer Questions (3 Marks)

Question 29. Complete each synthesis by giving missing reagents or products in the following:
(i) Benzene-1,2-dicarboxylic acid (phthalic acid) \( \xrightarrow[\text{heat}]{SOCl_2} \) ?
(ii) C₆H₅CHO \( \xrightarrow{H_2NCONHNH_2} \) ?
(iii) Methylenecyclohexane (C₆H₁₀=CH₂) → Cyclohexanecarbaldehyde (C₆H₁₁–CHO) [CBSE (AI) 2011]
Answer:
(i) C₆H₄(COOH)₂ (1,2-) + 2SOCl₂ \( \xrightarrow{\text{heat}} \) C₆H₄(COCl)₂ (benzene-1,2-dicarbonyl chloride) + 2SO₂ + 2HCl
(ii) C₆H₅CHO \( \xrightarrow{H_2NCONHNH_2} \) C₆H₅CH=NNHCONH₂ + H₂O
(iii) C₆H₁₀=CH₂ \( \xrightarrow[(ii)\,PCC]{(i)\,B_2H_6,\,H_2O_2/\bar{O}H} \) C₆H₁₁–CHO

Teacher's Note:
1. Hydroboration-oxidation of the exocyclic C=CH₂ puts OH on the terminal CH₂ (anti-Markovnikov); PCC then stops at the aldehyde.

Question 30. How do you convert the following?
(i) Benzoic acid to Benzaldehyde
(ii) Ethyne to Ethanal
(iii) Acetic acid to Methane [CBSE (F) 2015]
Answer:
(i) C₆H₅COOH (Benzoic acid) \( \xrightarrow{+SOCl_2} \) C₆H₅COCl \( \xrightarrow[BaSO_4]{H_2/Pd} \) C₆H₅CHO (Benzaldehyde)
(ii) CH≡CH (Ethyne) \( \xrightarrow[HgSO_4/H_2SO_4]{+H_2O} \) [CH₂=CH–OH] → CH₃–CHO (Ethanal)
(iii) CH₃COOH (Acetic acid) \( \xrightarrow{+NaOH} \) CH₃COO⁻Na⁺ \( \xrightarrow[\Delta]{NaOH/CaO} \) CH₄ (Methane)

Teacher's Note:
1. Vinyl alcohol (the enol) is unstable and tautomerises at once to ethanal – show it in brackets as an intermediate.

Question 31. An organic compound (A) has characteristic odour. On treatment with NaOH, it forms compounds (B) and (C). Compound (B) has molecular formula C₇H₈O which on oxidation gives back (A). The compound (C) is a sodium salt of an acid. When (C) is treated with soda-lime, it yields an aromatic compound (D). Deduce the structures of (A), (B), (C) and (D). Write the sequence of reactions involved. [CBSE Sample Paper 2015]
Answer:
A = C₆H₅CHO (Benzaldehyde); B = C₆H₅CH₂OH (Benzyl alcohol); C = C₆H₅COO⁻Na⁺ (Sodium benzoate); D = C₆H₆ (Benzene)
Reaction involved are:
2C₆H₅CHO (A) \( \xrightarrow{\text{NaOH/Conc.}} \) C₆H₅CH₂OH (B) + C₆H₅COO⁻Na⁺ (C)
C₆H₅COO⁻Na⁺ (C) \( \xrightarrow[\Delta]{NaOH/CaO} \) C₆H₆ (D) + Na₂CO₃

Teacher's Note:
1. An aldehyde whose reaction with NaOH gives an alcohol and an acid salt is undergoing Cannizzaro reaction, so it has no α-hydrogen.

Question 32. How would you bring about the following conversions? Write the complete equation in each case.
(i) Ethanal to 3-hydroxybutanal
(ii) Benzoic acid to m-nitrobenzyl alcohol
(iii) Benzaldehyde to benzophenone [CBSE (AI) 2011]
Answer:
(i) 2CH₃CHO (Ethanal) \( \xrightarrow{\text{dil. NaOH}} \) CH₃–CH(OH)–CH₂–CHO (3-Hydroxybutanal)
(ii) C₆H₅COOH (Benzoic acid) \( \xrightarrow[H_2SO_4\,(conc.),\,\Delta]{HNO_3\,(conc.)} \) m-O₂N–C₆H₄–COOH \( \xrightarrow[\Delta]{+NH_3} \) m-O₂N–C₆H₄–CONH₂ \( \xrightarrow{LiAlH_4} \) m-O₂N–C₆H₄–CH₂–NH₂ \( \xrightarrow{HNO_2} \) m-O₂N–C₆H₄–CH₂OH (m-Nitrobenzyl alcohol)
(iii) C₆H₅CHO (Benzaldehyde) \( \xrightarrow[(O)]{K_2Cr_2O_7/H_2SO_4} \) C₆H₅COOH (Benzoic acid) \( \xrightarrow{SOCl_2} \) C₆H₅COCl \( \xrightarrow[AlCl_3\,(anhyd.)]{+C_6H_6} \) C₆H₅COC₆H₅ (Benzophenone)

Teacher's Note:
1. Nitrate first, while –COOH is still on the ring, because –COOH is meta-directing; this fixes the m-position of the nitro group.

Question 33. Give reasons:
(i) The α-hydrogen of aldehydes and ketones are acidic in nature. [CBSE 2020]
(ii) Propanone is less reactive than ethanal towards addition of HCN.
(iii) Benzoic acid does not give Friedel-Crafts reaction. [CBSE (F) 2016]
Answer:
(i) The acidity of α-hydrogen atom of carbonyl carbon is due to the strong withdrawing effect of the carbonyl group and resonance stabilisation of the conjugate base.
–CO–C(H)< + B⁻ → [–C(=O)–C⁻< ↔ –C(–O⁻)=C<] + BH
(ii) This is due to steric and electronic reasons. Sterically, the presence of two methyl groups in propanone hinders more the approach of nucleophile to carbonyl carbon than in ethanal having one methyl group. Electronically two methyl groups reduce the positivity of the carbonyl carbon more effectively in propanone than in ethanal.
(iii) Benzoic acid does not give Friedel Craft reaction because:
(a) the carboxyl group is strongly deactivating.
(b) the catalyst AlCl₃ which is a lewis acid gets bonded to the carboxyl group strongly.

Teacher's Note:
1. In the enolate ion the negative charge is shared between carbon and the more electronegative oxygen – that sharing is what makes the α-H acidic.

Question 34. Predict the products of the following reactions:
(i) CH₃–CO–CH₃ → [(i) H₂N–NH₂ (ii) KOH/Glycol, Δ] → ?
(ii) C₆H₅–CO–CH₃ → [NaOH/I₂] → ? + ?
(iii) CH₃COONa → [NaOH/CaO, Δ] → ? [CBSE Delhi 2015]
Answer:
(i) CH₃CH₂CH₃
(ii) C₆H₅COONa + CHI₃
(iii) CH₄

Teacher's Note:
1. Decarboxylation with soda lime removes one carbon: sodium ethanoate gives methane, not ethane.

Question 35. Write the structure of the main products of following reactions:
(i) C₆H₆ + C₆H₅COCl → [Anhy. AlCl₃, CS₂] →
(ii) H₃C–C≡C–H → [Hg²⁺, H₂SO₄] →
(iii) p-Nitrotoluene (p-O₂N–C₆H₄–CH₃) → [(i) CrO₂Cl₂ (ii) H₃O⁺] → [CBSE Delhi 2012]
Answer:
(i) C₆H₅COC₆H₅ (Benzophenone) + HCl
(ii) CH₃–CO–CH₃ (Acetone)
(iii) p-O₂N–C₆H₄–CH₃ → [CrO₂Cl₂] → p-O₂N–C₆H₄–CH(OCrOHCl₂)₂ → [H₃O⁺] → p-O₂N–C₆H₄–CHO (p-nitrobenzaldehyde)

Teacher's Note:
1. Hydration of propyne follows Markovnikov’s rule, so the enol tautomerises to a ketone (acetone), not to propanal.

Question 36. Complete the following reactions:
(i) Cyclopentanone + H₂N–OH →
(ii) Cyclohexene → [KMnO₄, H₂SO₄, Δ] →
(iii) Benzene-1,2-dicarboxylic acid (phthalic acid) + NH₃ → [Strong heating] → [CBSE SQP 2018-2019]
Answer:
(i) Cyclopentanone oxime, (CH₂)₄C=N–OH
(ii) HOOC–CH₂–CH₂–CH₂–CH₂–COOH
(iii) Phthalimide, C₆H₄(CO)₂NH (the two C=O groups and NH form a five-membered ring fused to the benzene ring)

Teacher's Note:
1. Hot acidic KMnO₄ breaks the C=C of a cyclic alkene and turns both carbons into –COOH, so cyclohexene opens up to adipic acid.

Long Answer Questions (5 Marks)

Question 37. (i) Write the product(s) in the following reactions:
(a) Salicylic acid (2-hydroxybenzoic acid) \( \xrightarrow[\text{H}^{+}]{\text{(CH}_{3}\text{CO)}_{2}\text{O}} \) ?
(b) CH₃–CH(CH₃)–O–CH₂–CH₃ \( \xrightarrow{\text{HI}} \) ? + ?
(c) CH₃–CH=CH–CH₂–OH \( \xrightarrow{\text{PCC}} \) ?
(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Ethanol and Phenol
(b) Propanol and 2-methylpropan-2-ol [CBSE Delhi Set-1, 2, 3 2017]
Answer:
(i) (a) 2-Acetoxybenzoic acid (Aspirin), o-CH₃COO–C₆H₄–COOH (+ CH₃COOH)
(b) (CH₃)₂CHOH and CH₃CH₂I
(c) CH₃CH=CHCHO (But-2-enal)
(ii) (a) Add neutral FeCl₃ to both the compounds, phenol gives violet complex.
(b) Add anhy ZnCl₂ and conc. HCl to both the compounds, 2-methyl propan-2-ol gives turbidity immediately. (or any other correct test)

Teacher's Note:
1. PCC is a mild oxidant: it stops at the aldehyde and does not attack the C=C, so the allylic alcohol gives but-2-enal.

Question 38. (i) Account for the following:
(a) Propanal is more reactive than propanone towards nucleophilic reagents.
(b) Electrophilic substitution in benzoic acid takes place at meta-position.
(c) Carboxylic acids do not give characteristic reactions of carbonyl group.
(ii) Give simple chemical test to distinguish between the following pairs of compounds:
(a) Acetophenone and benzaldehyde
(b) Benzoic acid and ethylbenzoate
Answer:
(i) (a) Due to steric and +I effect of two methyl groups in propanone.
(b) Because it is a deactivating group/due to electron withdrawing carboxylic group resulting in decreased electron density at o- and p- position.
(c) Due to resonance, electrophilicity of carbonyl carbon is reduced.
(ii) (a) Add NaOH and I₂ to both the compounds and heat, acetophenone forms yellow ppt of iodoform.
(b) Add NaHCO₃ solution to both the compounds, benzoic acid will give effervescence and liberates CO₂. (Or any other suitable test)

Teacher's Note:
1. Propanone has two electron-releasing CH₃ groups (+I effect), not −I; quote the correct effect when comparing reactivity.

Question 39. (i) Write structures of A, B, C and D in the following reaction sequence:
CH₃COCl \( \xrightarrow{\text{H}_{2}\text{/Pd-BaSO}_{4}} \) A \( \xrightarrow{\text{dil. NaOH}} \) B \( \xrightarrow{\Delta} \) C;   A \( \xrightarrow{\text{CH}_{3}\text{MgBr/H}_{3}\text{O}^{+}} \) D
(ii) Arrange the following compounds in the increasing order of their boiling points:
CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃COOH [CBSE Comptt. OD Set-1, 2, 3 2017]
Answer:
(i) A: CH₃CHO; B: CH₃–CH(OH)–CH₂–CHO; C: CH₃–CH=CH–CHO; D: CH₃–CH(CH₃)–OH
(ii) CH₃–O–CH₃ < CH₃CHO < CH₃–CH₂–OH < CH₃–COOH
Acetic acid possesses higher boiling point than ethanol due to more extensive association of acetic acid molecules through intermolecular hydrogen bonding. So, acetic acid has higher boiling point than ethanol. Moreover, the polar C=O double bond causes acetaldehyde to have higher boiling point than dimethyl ether.

Teacher's Note:
1. Carboxylic acids exist as hydrogen-bonded dimers, which is why their boiling points are higher than those of alcohols of similar mass.

Question 40. (i) How will you convert:
(a) Benzene to acetophenone
(b) Propanone to 2-Methylpropan-2-ol
(ii) Give reasons:
(a) Electrophilic substitution in benzoic acid take place at meta-position.
(b) Carboxylic acids are higher boiling liquids than aldehydes, ketones and alcohols of comparable molecular masses.
(c) Propanal is more reactive than propanone in nucleophilic addition reactions.
Answer:
(i) (a) C₆H₆ + CH₃–CO–Cl \( \xrightarrow{\text{Anhyd. AlCl}_{3}} \) C₆H₅COCH₃
(b) CH₃COCH₃ + CH₃MgBr → (CH₃)₃C–OMgBr \( \xrightarrow{\text{H}_{2}\text{O}} \) (CH₃)₃C–OH
(ii) (a) Because it is a deactivating group/due to electron withdrawing carboxylic group resulting in decreased electron density at o- and p- position.
(b) Due to extensive association of carboxylic acid molecules through intermolecular hydrogen bonding.
(c) Due to steric and +I effect of two methyl groups in propanone.

Teacher's Note:
1. A ketone + Grignard reagent always gives a tertiary alcohol after hydrolysis.

Question 41. (i) Write the products of the following reaction:
(a) Cyclohexanone + NH₂–NH–CO–NH₂ \( \xrightarrow{\text{H}^{+}} \)
(b) CH₃MgBr + CO₂ \( \xrightarrow[\text{H}_{3}\text{O}^{+}]{\text{Dry ether}} \)
(c) CH₃CH₂COOH + Br₂ \( \xrightarrow{\text{Red Phosphorus}} \)
(ii) Write simple chemical tests to distinguish between the following pairs of compounds?
(a) Propanal and propanone
(b) Benzaldehyde and Benzoic acid [CBSE Comptt. Delhi Set-1, 2, 3 2017]
Answer:
(i) (a) Cyclohexanone semicarbazone, (CH₂)₅C=N–NH–CO–NH₂
(b) CH₃COOH
(c) CH₃–CH(Br)–COOH
(ii) (a) Add ammonical solution of silver nitrate / Tollen’s reagent to both the compounds, propanal will give silver mirror while propanone does not.
(b) Add NaHCO₃ solution to both the compounds, benzoic acid will give effervescence and liberate CO₂ while benzaldehyde will not. (Or any other suitable test)

Teacher's Note:
1. Grignard reagent + CO₂ followed by hydrolysis adds one carbon and gives a carboxylic acid: CH₃MgBr gives ethanoic acid.

Question 42. (i) Write the products of the following reactions:
(a) Cyclohexanone + H₂N–OH \( \xrightarrow{\text{H}^{+}} \)
(b) 2C₆H₅CHO + conc. NaOH →
(c) CH₃COOH \( \xrightarrow{\text{Cl}_{2}\text{/P}} \)
(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Benzaldehyde and Benzoic acid
(b) Propanal and Propanone [CBSE Delhi 2014, SQP 2017]
Answer:
(i) (a) Cyclohexanone + H₂N–OH \( \xrightarrow{\text{H}^{+}} \) (CH₂)₅C=N–OH (Cyclohexanone oxime) + H₂O
(b) 2C₆H₅CHO + Conc. NaOH → C₆H₅CH₂OH (Benzyl alcohol) + C₆H₅COONa (Sodium benzoate)
(c) CH₃COOH \( \xrightarrow{\text{Cl}_{2}\text{/P}} \) Cl–CH₂–COOH (2-Chloro-ethanoic acid) + HCl
(ii) (a) NaHCO₃ test: Benzoic acid reacts with NaHCO₃ to give brisk effervescence of CO₂ while benzaldehyde does not.
C₆H₅COOH (Benzoic acid) + NaHCO₃ → C₆H₅COONa (Sodium benzoate) + CO₂↑ + H₂O
(b) Iodoform test / Fehling’s Test / Tollen’s Test: Propanal being aldehyde when heated with Tollen’s reagent to give silver mirror but propanone being a ketone does not.
CH₃CH₂CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃CH₂COO⁻ + 2Ag↓ (Silver mirror) + 4NH₃ + 2H₂O

Teacher's Note:
1. Oxime formation is acid-catalysed but needs only weakly acidic conditions; strong acid would protonate NH₂OH and stop it acting as a nucleophile.