CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids Important Questions Set 11

Very Short Answer Questions (1 Mark)

Question 1. Write the structures of A and B in the following reaction
CH₃MgBr \( \xrightarrow[(ii)\ H_3O^+]{(i)\ CO_2} \) A \( \xrightarrow{PCl_5} \) B (CBSE 2016)
Answer:
CH₃MgBr \( \xrightarrow{CO_2} \) [CH₃–CO–OMgBr] \( \xrightarrow[-Mg(OH)Br]{H_3O^+} \) CH₃–COOH \( \xrightarrow[-POCl_3,\ -HCl]{PCl_5} \) CH₃–COCl
A = CH₃COOH (Acetic acid); B = CH₃COCl (Acetyl chloride)

Teacher's Note:
1. Carboxylation of a Grignard reagent adds one carbon: RMgX gives RCOOH.

Question 2. Suggest a scheme to convert a primary alcohol into an acid with one more carbon atom.
Answer:
RCH₂OH \( \xrightarrow[-SO_2,\ -HCl]{SOCl_2} \) RCH₂Cl \( \xrightarrow[-KCl]{Alc.\ KCN} \) RCH₂CN \( \xrightarrow[Hydrolysis]{H^+/H_2O} \) RCH₂COOH
1° Alcohol → Carboxylic acid

Teacher's Note:
1. The CN group is the step that adds the extra carbon; hydrolysis then turns –CN into –COOH.

Question 3. How is CH₃OH converted into CH₃COOH (in one step) ?
Answer:
CH₃OH \( \xrightarrow[-1/2H_2]{Na} \) CH₃O⁻Na⁺ \( \xrightarrow[under\ pressure]{CO} \) CH₃COONa \( \xrightarrow[-NaCl]{HCl/H_2O} \) CH₃COOH
Methanol → Ethanoic acid

Teacher's Note:
1. Industrially, methanol is carbonylated directly with CO in presence of a rhodium catalyst to give acetic acid.

Question 4. Why is benzoic acid less soluble in water than acetic acid ?
Answer:
Due to polar nature of the C=O and O–H parts of COOH group, both CH₃COOH and C₆H₅COOH form H bonds with water. But due to large hydrocarbon part the extent of H-bonding is much lower in benzoic acid than in acetic acid and hence benzoic acid is much less soluble in water than acetic acid.

Teacher's Note:
1. Solubility falls as the non-polar part grows; benzoic acid dissolves well in hot water, which is used for its recrystallisation.

Question 5. Arrange the following in decreasing order of the boiling points : CH₃CH₂CH₂CH₂OH, CH₃CH₂OCH₂CH₃, CH₃CH₂CH₂COOH.
Answer:
Due to much stronger H-bonds in carboxylic acids than in alcohols, the boiling points of carboxylic acids are much higher than the boiling points of the corresponding alcohols. However, due to absence of H-bonding in ethers, the boiling points of ethers are much lower than alcohols and carboxylic acids of comparable molecular masses. Thus, the boiling points decrease in the order :
CH₃CH₂CH₂COOH > CH₃CH₂CH₂CH₂OH > CH₃CH₂OCH₂CH₃

Teacher's Note:
1. Acid > alcohol > ether (similar masses) is a pattern worth memorising for such questions.

Question 6. What makes acetic acid a stronger acid than phenol ?
Answer:
Greater resonance stabilization of acetate ion over phenoxide ion.
In acetate ion, the negative charge is delocalised over two equivalent electronegative oxygen atoms. In phenoxide ion, the charge is delocalised over one oxygen atom and the less electronegative carbon atoms of the ring, and the resonance structures are not equivalent. Thus, acetate ion is more stable than phenoxide ion and acetic acid is a stronger acid than phenol.

Teacher's Note:
1. A quick test: carboxylic acids liberate CO₂ from NaHCO₃, but phenol does not, because phenol is a weaker acid than carbonic acid.

Question 7. Why carboxylic acids do not give the characteristic reactions of carbonyl group ? (CBSE 2022)
Answer:
Due to the presence of lone pairs of electrons on the oxygen atom of the OH group, the carboxylic acids may be regarded as a resonance hybrid of structures (I and II) :
R–C(=O)–Ö–H (I) ↔ R–C(–O⁻)=O⁺–H (II)
Similarly, a carbonyl group of aldehydes and ketones may be regarded as a resonance hybrid of structures (III and IV) :
>C=O (III) ↔ >C⁺–O⁻ (IV)
Due to contribution of structure (IV), the carbonyl carbon in aldehydes and ketones is electrophilic. However, due to contribution of structure (II), the electrophilic character of carboxyl carbon is reduced. In other words, carbonyl carbon of carboxyl group is less electrophilic than carbonyl carbon in aldehydes and ketones and hence nucleophilic addition reactions (such as formation of oximes, hydrazones, phenylhydrazones, 2, 4-dinitrophenylhydrazones and semicarbazones) of aldehydes and ketones do not take place with carboxylic acids.

Teacher's Note:
1. The OH group of COOH donates its lone pair to the C=O, so the carbonyl carbon is no longer strongly positive.

Question 8. Fluorine is more electronegative than chlorine even then p-fluorobenzoic acid is a weaker acid than p-chlorobenzoic acid. Explain.
Answer:
Since halogens are more electronegative than carbon and also possess lone pairs of electrons, therefore, they exert both –I- and +R- effects. Now in F, the lone pairs of electrons are present in 2p-orbitals but in Cl, they are present in 3p-orbitals. Since 2p-orbitals of F and C are of almost equal size, therefore, the +R-effect is more pronounced in p-fluorobenzoic acid than in p-chlorobenzoic acid
F–C₆H₄–COOH (p) : 2p–2p overlap, stronger +R-effect; Cl–C₆H₄–COOH (p) : 3p–2p overlap, weaker +R-effect.
Thus, in p-fluorobenzoic acid, +R-effect outweighs the –I-effect but in p-chlorobenzoic acid, it is the –I-effect which outweighs the +R-effect. Consequently, p-fluorobenzoic acid is a weaker acid than p-chlorobenzoic acid.

Teacher's Note:
1. Resonance acts strongly at the para position, which is why the size match of 2p orbitals of F and C matters here.

Question 9. Why is pK_a of FCH₂COOH lower than that of chloroacetic acid ? (CBSE 2016)
Answer:
pK_a of FCH₂COOH is lower than that of ClCH₂COOH implies that FCH₂COOH is a stronger acid than ClCH₂COOH. This is due to the following two reasons :
(i) Due to stronger –I-effect of F over Cl, the electron density in the O–H bond in FCH₂COOH is much lower than in ClCH₂COOH. As a result, O–H bond in FCH₂COOH is much weaker than in ClCH₂COOH and hence FCH₂COOH loses a proton more easily than ClCH₂COOH i.e., FCH₂COOH is a stronger acid than ClCH₂COOH.
(ii) Due to stronger –I-effect of F over Cl, the dispersal of the –ve charge in FCH₂COO⁻ occurs more effectively than in ClCH₂COO⁻.
F←CH₂←COO⁻ (stronger –I-effect of F disperses the –ve charge more effectively) ; Cl←CH₂←COO⁻ (weaker –I-effect of Cl disperses the –ve charge less effectively)
In other words, FCH₂COO⁻ is more stable than ClCH₂COO⁻ and hence FCH₂COOH is a stronger acid than ClCH₂COOH, i.e., FCH₂COOH has lower pK_a than ClCH₂COOH.

Teacher's Note:
1. Here only the –I-effect matters because the halogen is on an sp³ carbon; there is no resonance as in the benzoic acid case.

Question 10. Formic acid reduces Tollens’ reagent but acetic acid does not. Explain.
Answer:
Formic acid can be easily oxidised to CO₂ and H₂O and hence it acts as a strong reducing agent. As such it reduces Tollens’ reagent to shining silver mirror.
H–COOH + 2[Ag(NH₃)₂]⁺ + 2OH⁻ \( \xrightarrow{\Delta} \) 2Ag↓ + 4NH₃ + CO₂ + 2H₂O
Formic acid → Silver mirror

Teacher's Note:
1. HCOOH has an H attached directly to the C=O carbon, so it behaves partly like an aldehyde.

Short Answer Questions (2 Marks)

Question 11. How will you convert ? Acetaldehyde to ethyl acetamide (Raj. Board 2011 ; Karnataka Board 2012)
Answer:
CH₃CHO \( \xrightarrow{K_2Cr_2O_7/H_2SO_4} \) CH₃COOH \( \xrightarrow{SOCl_2} \) CH₃COCl
CH₃CHO \( \xrightarrow{H_2/Ni} \) CH₃CH₂OH \( \xrightarrow{PBr_3} \) CH₃CH₂Br \( \xrightarrow{NH_3\ (excess)} \) CH₃CH₂NH₂
CH₃COCl + H₂NCH₂CH₃ \( \xrightarrow[-HCl]{pyridine} \) CH₃CONHCH₂CH₃ (N-Ethylacetamide)

Teacher's Note:
1. Split the target into its two halves (acetyl part and ethylamine part); acetaldehyde supplies both.

Question 12. How will you convert ? Ethyl acetate to acetamide. (Raj. Board 2011)
Answer:
CH₃COOC₂H₅ + NH₃ → CH₃CONH₂ + C₂H₅OH (Ammonolysis)

Teacher's Note:
1. Esters react with NH₃ more slowly than acid chlorides, but no acid by-product is formed.

Question 13. How will you convert ? Acetamide to acetic acid. (Raj. Board 2011)
Answer:
CH₃CONH₂ + H₂O \( \xrightarrow{H^+,\ \Delta} \) CH₃COOH + NH₄⁺

Teacher's Note:
1. Amides are the least reactive acid derivatives, so hydrolysis needs heating with acid or alkali.

Question 14. Write the products of the following reactions : CH₃CH₂COOH \( \xrightarrow[(ii)\ H_2O]{(i)\ Br_2,\ Red\ P} \) (CBSE 2013 ; West Bengal Board 2013 ; CBSE Sample Paper 2019-20)
Answer:
CH₃CH₂COOH \( \xrightarrow[(ii)\ H_2O]{(i)\ Br_2/red\ P} \) CH₃CHBrCOOH (2-Bromopropanoic acid)

Teacher's Note:
1. Only the α-H is replaced in the HVZ reaction.

Question 15. Write the products of the following reactions : CH₃COOH + CH₃OH \( \xrightarrow{Conc.\ H_2SO_4} \)
Answer:
CH₃COOH + CH₃OH \( \xrightarrow{Conc.\ H_2SO_4} \) CH₃COOCH₃ (Methyl acetate) + H₂O

Teacher's Note:
1. Conc. H₂SO₄ acts as catalyst and also absorbs the water formed.

Question 16. Write the products of the following reactions : CH₃COOH + SOCl₂ →
Answer:
CH₃COOH + SOCl₂ → CH₃COCl (Acetyl chloride) + SO₂ + HCl

Teacher's Note:
1. Gaseous SO₂ and HCl leave the mixture, so the acid chloride is easily obtained pure.

Question 17. Write the products of the following reactions : C₆H₅COOH + Na₂CO₃ (aq) →
Answer:
2C₆H₅COOH + Na₂CO₃ → 2C₆H₅COONa (Sodium benzoate) + CO₂↑ + H₂O

Teacher's Note:
1. Effervescence of CO₂ shows that benzoic acid is stronger than carbonic acid.

Question 18. Write the products of the following reactions : C₆H₅COOH + PCl₅ \( \xrightarrow{heat} \)
Answer:
C₆H₅COOH + PCl₅ \( \xrightarrow{\Delta} \) C₆H₅COCl (Benzoyl chloride) + POCl₃ + HCl

Teacher's Note:
1. PCl₅ replaces the OH of COOH by Cl, just as it does with alcohols.

Question 19. What happens when : Sodium acetate is heated with soda-lime
Answer:
Methane is formed (decarboxylation) : CH₃COONa + NaOH \( \xrightarrow{CaO,\ \Delta} \) CH₄ + Na₂CO₃

Teacher's Note:
1. This is the usual laboratory method for preparing methane.

Question 20. What happens when : An aqueous solution of potassium acetate is electrolysed
Answer:
Ethane is formed at the anode and hydrogen at the cathode (Kolbe’s electrolysis) :
2CH₃COOK + 2H₂O \( \xrightarrow{Electrolysis} \) CH₃–CH₃ + 2CO₂ (anode) + H₂ + 2KOH (cathode)

Teacher's Note:
1. At the anode, CH₃COO⁻ loses an electron and CO₂; two CH₃ radicals then combine.

Question 21. What happens when : Calcium acetate alone is heated strongly (J & K Board 2012)
Answer:
Acetone is formed : (CH₃COO)₂Ca \( \xrightarrow{\Delta} \) CH₃COCH₃ + CaCO₃

Teacher's Note:
1. The calcium salt of any acid other than formic acid gives a ketone on dry distillation.

Question 22. What happens when : Calcium acetate is heated with calcium formate (J & K Board 2012)
Answer:
Acetaldehyde is formed : (CH₃COO)₂Ca + (HCOO)₂Ca \( \xrightarrow{\Delta} \) 2CH₃CHO + 2CaCO₃

Teacher's Note:
1. Some acetone and formaldehyde are also formed as by-products.

Question 23. What happens when : Malonic acid is heated. (J & K Board 2012)
Answer:
It loses CO₂ to form acetic acid : HOOC–CH₂–COOH \( \xrightarrow{\Delta} \) CH₃COOH + CO₂

Teacher's Note:
1. Two COOH groups on the same carbon make decarboxylation easy on simple heating.

Question 24. What happens when : Ethanoic acid (acetic acid) is treated with LiAlH₄.
Answer:
Ethanol is formed : CH₃COOH \( \xrightarrow[(ii)\ H_3O^+]{(i)\ LiAlH_4/ether} \) CH₃CH₂OH

Teacher's Note:
1. LiAlH₄ is one of the few reagents strong enough to reduce a COOH group.

Question 25. What happens when : Acetic acid is treated with Cl₂ in presence of red phosphorus.
Answer:
α-Hydrogen atoms are replaced by Cl (HVZ reaction) giving chloroacetic acid and, with excess Cl₂, dichloro- and trichloroacetic acids :
CH₃COOH \( \xrightarrow[-HCl]{Cl_2/red\ P} \) ClCH₂COOH \( \xrightarrow{Cl_2/red\ P} \) Cl₂CHCOOH \( \xrightarrow{Cl_2/red\ P} \) Cl₃CCOOH

Teacher's Note:
1. Each chlorination step makes the acid stronger.

Question 26. Write short notes on the following : Kolbe’s electrolytic reaction (Hr. Board 2011 ; HP Board 2011)
Answer:
When a concentrated aqueous solution of the sodium or potassium salt of a carboxylic acid is electrolysed, an alkane containing twice the number of carbon atoms in the alkyl group of the acid is formed at the anode.
2RCOONa + 2H₂O \( \xrightarrow{Electrolysis} \) R–R + 2CO₂ (anode) + H₂ + 2NaOH (cathode)
e.g., 2CH₃COONa + 2H₂O → CH₃CH₃ + 2CO₂ + H₂ + 2NaOH

Teacher's Note:
1. Only alkanes with an even number of carbons are obtained cleanly by this method; methane cannot be made.

Question 27. Write short notes on the following : Hell-Volhard-Zelinsky reaction (CBSE 2010, 2012 ; AP Board 2012 ; Hr. Board 2012 ; HP Board 2013)
Answer:
Carboxylic acids having α-hydrogen atoms react with Cl₂ or Br₂ in presence of a small amount of red phosphorus to give α-halocarboxylic acids.
R–CH₂–COOH \( \xrightarrow[(ii)\ H_2O]{(i)\ X_2/red\ P} \) R–CH(X)–COOH (X = Cl, Br)
e.g., CH₃CH₂COOH + Br₂ → CH₃CHBrCOOH + HBr

Teacher's Note:
1. The α-halo acids formed are useful for making α-hydroxy acids, α-amino acids and unsaturated acids.

Short Answer Questions (3 Marks)

Question 28. Write the products of oxidation of (i) 2-pentanone and (ii) 3-pentanone. Are the major products same or different ? Explain.
Answer:
Ketones are oxidised only by strong oxidising agents (e.g., K₂Cr₂O₇/H₂SO₄, hot) with cleavage of a C–C bond on either side of the carbonyl group, giving a mixture of acids with fewer carbon atoms. According to Popoff’s rule, the carbonyl group stays with the smaller alkyl group.
(i) CH₃COCH₂CH₂CH₃ \( \xrightarrow{[O]} \) CH₃COOH + CH₃CH₂COOH (major) ; CH₃CH₂CH₂COOH + CO₂ + H₂O (minor)
(ii) CH₃CH₂COCH₂CH₃ \( \xrightarrow{[O]} \) CH₃COOH + CH₃CH₂COOH
Thus, the major products (ethanoic acid and propanoic acid) are the same in both cases, although 2-pentanone also gives some butanoic acid by cleavage on the other side of the C=O group.

Teacher's Note:
1. Count the carbons as a check: 5 carbons of the ketone split into 2 + 3 for both major products.

Question 29. How are aldehydes distinguished from ketones using Tollens’ and Fehling’s reagents ? Give complete chemical reactions.
Answer:
Aldehydes are easily oxidised and hence reduce Tollens’ reagent and Fehling’s solution; ketones do not.
Tollens’ test : On warming an aldehyde with Tollens’ reagent, a shining silver mirror is formed.
RCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → RCOO⁻ + 2Ag↓ + 2H₂O + 4NH₃
Fehling’s test : On heating an aliphatic aldehyde with Fehling’s solution, a reddish-brown ppt. of Cu₂O is formed.
RCHO + 2Cu²⁺ + 5OH⁻ → RCOO⁻ + Cu₂O↓ + 3H₂O
Ketones (e.g., CH₃COCH₃) give neither the silver mirror nor the red ppt.

Teacher's Note:
1. Aromatic aldehydes give Tollens’ test but not Fehling’s test, so for benzaldehyde use Tollens’ reagent.

Question 30. How will you distinguish benzaldehyde from (i) acetaldehyde, (ii) acetone ? Give reactions.
Answer:
(i) Fehling’s test : Acetaldehyde gives a red ppt. of Cu₂O with Fehling’s solution while benzaldehyde does not.
CH₃CHO + 2Cu²⁺ + 5OH⁻ → CH₃COO⁻ + Cu₂O↓ + 3H₂O ; C₆H₅CHO + Fehling’s solution → No red ppt.
(Alternatively, acetaldehyde gives a yellow ppt. of CHI₃ with I₂/NaOH while benzaldehyde does not.)
(ii) Tollens’ test : Benzaldehyde gives a silver mirror with Tollens’ reagent while acetone does not.
C₆H₅CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → C₆H₅COO⁻ + 2Ag↓ + 2H₂O + 4NH₃

Teacher's Note:
1. Name the test, state what each compound shows, and write the equation for the compound that reacts.

Question 31. What is haloform reaction ? Discuss its utility in organic analysis.
Answer:
Compounds containing the CH₃CO– group (methyl ketones and acetaldehyde) or the CH₃CH(OH)– group (which is oxidised to CH₃CO–) react with a halogen in presence of an alkali (NaOX) to give a haloform (CHX₃) and the sodium salt of an acid with one carbon atom less. This is called the haloform reaction.
R–COCH₃ \( \xrightarrow{NaOX\ (X_2 + NaOH)} \) R–COONa + CHX₃ ; e.g., CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃↓ + CH₃COONa + 3NaI + 3H₂O
Utility : With I₂/NaOH, a yellow ppt. of iodoform (iodoform test) is formed. It is used to detect the CH₃CO– or CH₃CH(OH)– group and hence to distinguish, e.g., ethanal from propanal, 2-pentanone from 3-pentanone, ethanol from methanol, and acetophenone from benzophenone. It is also used to prepare carboxylic acids having one carbon atom less than the methyl ketone.

Teacher's Note:
1. The C=C double bond, if present, is not attacked by NaOX, so haloform reaction can convert an unsaturated methyl ketone into an unsaturated acid.

Question 32. Identify A, B, C and D in the following reactions.
(i) CH₃Br \( \xrightarrow{Aq.\ KOH} \) A \( \xrightarrow{Cu,\ 573} \) B \( \xrightarrow{Conc.\ NaOH} \) C + D
(ii) CH₃CH₂OH \( \xrightarrow{K_2Cr_2O_7,\ H^+} \) A \( \xrightarrow[Calcium\ salt]{Distil\ the} \) B \( \xrightarrow{Cl_2,\ NaOH} \) C
(iii) CH₃COOH \( \xrightarrow{Ca(OH)_2} \) A \( \xrightarrow{Dry\ distillation} \) B \( \xrightarrow{LiAlH_4} \) C \( \xrightarrow{SOCl_2} \) D (Assam Board 2012)
(iv) CH₃CHO \( \xrightarrow{NH_2NH_2} \) A \( \xrightarrow[453\text{-}473\ K]{KOH,\ ethylene\ glycol} \) B (West Bengal Board 2012)
(v) CH≡CH \( \xrightarrow[HgSO_4]{Dil.\ H_2SO_4} \) (A) \( \xrightarrow{Dil.\ NaOH} \) (B) \( \xrightarrow{Heat} \) (C)
(vi) CH₃CN \( \xrightarrow[(b)\ H_3O^+]{(a)\ SnCl_2/HCl} \) A \( \xrightarrow{dil.\ NaOH} \) B \( \xrightarrow{\Delta} \) C (CBSE 2017)
Answer:
(i) A = CH₃OH (Methanol), B = HCHO (Formaldehyde), C + D = CH₃OH + HCOONa (Cannizzaro reaction : 2HCHO + NaOH → CH₃OH + HCOONa)
(ii) A = CH₃COOH (Acetic acid), B = CH₃COCH₃ (Acetone, by dry distillation of (CH₃COO)₂Ca), C = CHCl₃ (Chloroform, haloform reaction; CH₃COONa is the other product)
(iii) A = (CH₃COO)₂Ca (Calcium acetate), B = CH₃COCH₃ (Acetone), C = CH₃CH(OH)CH₃ (Propan-2-ol), D = CH₃CHClCH₃ (2-Chloropropane)
(iv) A = CH₃CH=NNH₂ (Acetaldehyde hydrazone), B = CH₃CH₃ (Ethane) (Wolff-Kishner reduction)
(v) A = CH₃CHO (Acetaldehyde), B = CH₃CH(OH)CH₂CHO (3-Hydroxybutanal), C = CH₃CH=CHCHO (But-2-enal)
(vi) A = CH₃CHO (Ethanal, Stephen reduction), B = CH₃CH(OH)CH₂CHO (3-Hydroxybutanal), C = CH₃CH=CHCHO (But-2-enal)

Teacher's Note:
1. In such chains, write the structure after every arrow and name the reaction; one wrong step makes every later letter wrong.

Question 33. Give the IUPAC names of the following :
(i) CH₃–CH(CO₂H)–CH₂–CH₃ (ii) CH₃–CH(OH)–CH₂–COOH (UP Board 2012)
(iii) CH₃COCH₂COOH (iv) CHO.COOH (v) CH₃COCOOH (vi) C₆H₅–CH=CH–COOH (vii) C₆H₅CH₂CH₂COOH
(viii) CH₃CH₂CH(CHO)CH₂COOH (ix) CH₃CH(Br)CH₂CONHCH₃
Answer:
(i) 2-Methylbutanoic acid, (ii) 3-Hydroxybutanoic acid, (iii) 3-Oxobutanoic acid, (iv) 2-Oxoethanoic acid, (v) 2-Oxopropanoic acid, (vi) 3-Phenylprop-2-enoic acid, (vii) 3-Phenylpropanoic acid, (viii) 3-Ethyl-4-oxobutanoic acid (ix) 3-Bromo-N-methylbutanamide

Teacher's Note:
1. The COOH carbon is always C-1, so its locant is not written; the other groups are numbered from it.

Question 34. How will you account for the following ?
(i) A carboxylic acid does not form an oxime or a hydrazone although there is present a carbonyl group in carboxylic functional group.
(ii) Why does not propanoic acid give test for carbonyl group ?
(iii) A >C=O group behaves differently in aldehydes and acids.
(iv) Carboxylic acids do not exhibit the properties of carbonyl compounds.
(v) The bond length of C=O in carboxylic acids is slightly longer than that in aldehydes and ketones.
Answer:
(i) to (iv) The lone pairs on the oxygen of the OH group are delocalised into the C=O group, so a carboxylic acid is a resonance hybrid :
R–C(=O)–ÖH ↔ R–C(–O⁻)=O⁺H
Due to this resonance, the positive charge on the carbonyl carbon is reduced, i.e., the carboxyl carbon is much less electrophilic than the carbonyl carbon of aldehydes and ketones. Hence nucleophiles such as NH₂OH, NH₂NH₂, 2, 4-DNP do not add to it, and carboxylic acids (e.g., propanoic acid) do not form oximes or hydrazones and do not give the usual tests of the carbonyl group. Thus, the >C=O group behaves differently in aldehydes and in acids.
(v) Because of the same resonance, the C=O bond in carboxylic acids acquires some single-bond character. Hence its bond length is slightly longer than the C=O bond in aldehydes and ketones.

Teacher's Note:
1. One resonance idea answers all five parts; state it once clearly and then apply it.

Question 35. How is acetic acid prepared from :
(i) Acetylene (ii) Ethyl alcohol (iii) Methyl iodide (iv) Acetonitrile ?
Answer:
(i) HC≡CH + H₂O \( \xrightarrow{40\%\ H_2SO_4,\ 1\%\ HgSO_4} \) CH₃CHO \( \xrightarrow{K_2Cr_2O_7/H_2SO_4} \) CH₃COOH
(ii) CH₃CH₂OH \( \xrightarrow{K_2Cr_2O_7/H_2SO_4\ or\ KMnO_4/H^+} \) CH₃COOH
(iii) CH₃I + Mg \( \xrightarrow{Dry\ ether} \) CH₃MgI \( \xrightarrow[(ii)\ H_3O^+]{(i)\ CO_2} \) CH₃COOH + Mg(OH)I
(iv) CH₃CN + 2H₂O \( \xrightarrow{H^+} \) CH₃COOH + NH₃ (as NH₄⁺)

Teacher's Note:
1. In (iii), CH₃I + KCN followed by hydrolysis is another route; both add one carbon.

Long Answer Questions (5 Marks)

Question 36. (i) How will you prepare (a) acetic anhydride and (b) acetyl chloride from CH₃COOH? Write the equation involved in each case.
(ii) Why is the boiling point of acid anhydride higher than the acid from which it is obtained?
Answer:
(i) (a) 2CH₃COOH –[P₂O₅, Δ]→ (CH₃CO)₂O + H₂O
(b) CH₃COOH + PCl₅ → CH₃COCl + POCl₃ + HCl
(ii) Acid anhydride has bigger size (higher molecular mass) than the corresponding acid and hence stronger van der Waals forces of attraction than the corresponding acid.

Teacher's Note:
1. Two acid molecules lose one water molecule to form an anhydride – do not forget the coefficient 2.

Question 37. Complete the following reactions and write main products:
(i) CH₃CONH₂ + HNO₂ →   (ii) CH₃COCl + H₂O (steam) →
(iii) CH₃MgBr –[(i) CO₂, (ii) H₂O/H⁺]→   (iv) CH₃COOH + NH₃ –[Δ]→
(v) CH₃COOH + C₂H₅OH –[conc. H₂SO₄]→   (vi) HCHO + NH₃ →
(vii) CH₃CHO –[LiAlH₄]→   (viii) CH₃CHO + HOCH₂CH₂OH –[H⁺]→
(ix) CH₃COR + NaOI →   (x) C₆H₅CHO + NH₂–NH₂ →
Answer:
(i) CH₃COOH
(ii) CH₃COOH
(iii) CH₃COOH
(iv) CH₃CONH₂
(v) CH₃COOC₂H₅
(vi) (CH₂)₆N₄ (urotropine)
(vii) CH₃CH₂OH
(viii) 2-Methyl-1,3-dioxolane: a five-membered ring –O–CH₂–CH₂–O– closed by the CH–CH₃ carbon (ethylene acetal of acetaldehyde)
(ix) CHI₃ (and RCOONa)
(x) C₆H₅CH=N–NH₂ (benzaldehyde hydrazone)

Teacher's Note:
1. Several different reactions here give CH₃COOH; in each case identify which bond breaks (C–N, C–Cl, or new C–C with CO₂).

Question 38. Give reasons for the following:
(i) C₆H₅COOH is weaker than formic acid.
(ii) HCOOH and CH₃COOH are differentiated by Tollens' reagent.
(iii) R–COOH do not give characteristic reactions of >C=O.
(iv) Carboxylic acids are stronger acids than phenols.
(v) Acid amides are weakly basic in nature.
Answer:
(i) Due to lower stability of the benzoate anion, as the phenyl group releases electrons to the carboxylate group by conjugation (resonance); formate has no such group.
(ii) Presence of H–CO– (aldehydic) group in HCOOH, so it reduces Tollens' reagent; CH₃COOH does not.
(iii) Due to resonance in the carboxyl group, R–C(=O)–OH ↔ R–C(–O⁻)=O⁺H, the electrophilic character of the carbonyl carbon is reduced, so >C=O does not behave as a true carbonyl group.
(iv) Dispersal of −ve charge is better on carboxylate ions (over two equivalent electronegative O atoms) than on phenoxide ion.
(v) Acid amides are basic due to lone pair of electrons on nitrogen atom, but only weakly basic because this lone pair is delocalised over the C=O group by resonance.

Teacher's Note:
1. The same resonance idea answers (iii) and (v): delocalisation into C=O removes electron density from O or N.

Question 39. How are aldehydes and ketones prepared from :
(i) alcohols (ii) carboxylic acid chlorides (iii) alkenes and (iv) fatty acids ?
Answer:
(i) From alcohols : (a) Oxidation : Primary alcohols are oxidised to aldehydes and secondary alcohols to ketones.
RCH₂OH \( \xrightarrow{PCC,\ CH_2Cl_2} \) RCHO ; R₂CHOH \( \xrightarrow{K_2Cr_2O_7/H_2SO_4} \) R₂C=O
(b) Dehydrogenation : Vapours of 1° and 2° alcohols passed over heated copper at 573 K give aldehydes and ketones respectively : CH₃CH₂OH \( \xrightarrow{Cu,\ 573\ K} \) CH₃CHO + H₂ ; (CH₃)₂CHOH \( \xrightarrow{Cu,\ 573\ K} \) (CH₃)₂C=O + H₂
(ii) From acid chlorides : Aldehydes by Rosenmund reduction and ketones by reaction with dialkylcadmium :
RCOCl + H₂ \( \xrightarrow{Pd/BaSO_4,\ S} \) RCHO + HCl ; 2RCOCl + R′₂Cd → 2RCOR′ + CdCl₂
(iii) From alkenes : Ozonolysis of alkenes followed by reductive cleavage with Zn/H₂O gives aldehydes and/or ketones depending on the substitution of the alkene :
R₂C=CHR′ \( \xrightarrow[(ii)\ Zn/H_2O]{(i)\ O_3} \) R₂C=O + R′CHO ; e.g., (CH₃)₂C=CHCH₃ → CH₃COCH₃ + CH₃CHO
Ethylene also gives ethanal by Wacker process : CH₂=CH₂ + H₂O + PdCl₂ \( \xrightarrow{CuCl_2,\ air} \) CH₃CHO + Pd + 2HCl
(iv) From fatty acids : (a) Dry distillation of calcium salts of fatty acids gives ketones; with calcium formate, aldehydes are obtained :
(RCOO)₂Ca \( \xrightarrow{\Delta} \) R₂CO + CaCO₃ ; (RCOO)₂Ca + (HCOO)₂Ca \( \xrightarrow{\Delta} \) 2RCHO + 2CaCO₃
(b) Vapours of fatty acids passed over MnO at 573 K give ketones (with formic acid, aldehydes) : 2CH₃COOH \( \xrightarrow{MnO,\ 573\ K} \) CH₃COCH₃ + CO₂ + H₂O ; CH₃COOH + HCOOH \( \xrightarrow{MnO,\ 573\ K} \) CH₃CHO + CO₂ + H₂O

Teacher's Note:
1. For each method, state the product class clearly: 1° alcohol → aldehyde, 2° alcohol → ketone; Rosenmund → aldehyde, R₂Cd → ketone.

Question 40. How can aromatic aldehydes and ketones be prepared by the use of following reactions ?
(i) Etard reaction (ii) Reimer-Tiemann reaction and (iii) Friedel-Crafts reaction.
Illustrate your answer by giving one example in each case.
Answer:
(i) Étard reaction : Toluene (or a substituted toluene) is oxidised by chromyl chloride in CS₂ to a brown chromium complex which on hydrolysis gives benzaldehyde.
C₆H₅CH₃ + 2CrO₂Cl₂ \( \xrightarrow{CS_2} \) C₆H₅CH(OCrOHCl₂)₂ \( \xrightarrow{H_3O^+} \) C₆H₅CHO
(ii) Reimer-Tiemann reaction : Phenol on heating with chloroform and aqueous NaOH at about 340 K, followed by acidification, gives salicylaldehyde (2-hydroxybenzaldehyde). A CHO group is introduced at the ortho position.
C₆H₅OH + CHCl₃ + 3NaOH \( \xrightarrow{340\ K} \) o-NaOC₆H₄CHO + 3NaCl + 2H₂O ; o-NaOC₆H₄CHO \( \xrightarrow{H^+} \) o-HOC₆H₄CHO (Salicylaldehyde)
(iii) Friedel-Crafts reaction : (a) Aromatic ketones are obtained by Friedel-Crafts acylation of benzene with acid chlorides in presence of anhyd. AlCl₃ : C₆H₆ + CH₃COCl \( \xrightarrow{Anhyd.\ AlCl_3} \) C₆H₅COCH₃ + HCl
(b) Benzaldehyde is obtained by the Gattermann-Koch reaction (a Friedel-Crafts type reaction with CO and HCl) : C₆H₆ + CO + HCl \( \xrightarrow{Anhyd.\ AlCl_3/CuCl} \) C₆H₅CHO

Teacher's Note:
1. The Reimer-Tiemann product carries the CHO group ortho to OH; the intermediate is dichlorocarbene, :CCl₂.

Question 41. Discuss briefly the mechanism of nucleophilic addition reactions. Explain the relative reactivities of aldehydes and ketones towards these reactions. (J & K Board 2012)
Answer:
Mechanism : The carbonyl carbon is electrophilic because of the polarity of the C=O bond (>C^δ+=O^δ−).
Step 1 (slow) : The nucleophile attacks the carbonyl carbon from a direction approximately perpendicular to the plane of the sp²-hybridised orbitals. The π-electrons move to oxygen, the carbon becomes sp³-hybridised and a tetrahedral alkoxide intermediate is formed : >C=O + Nu⁻ \( \xrightarrow{Slow} \) >C(Nu)–O⁻
Step 2 (fast) : The intermediate captures a proton from the reaction medium to give the electrically neutral product : >C(Nu)–O⁻ + H⁺ \( \xrightarrow{Fast} \) >C(Nu)–OH
e.g., addition of HCN : >C=O + CN⁻ → >C(CN)O⁻ \( \xrightarrow{H^+} \) >C(OH)CN (cyanohydrin)
Relative reactivities : Aldehydes are generally more reactive than ketones for two reasons :
(i) Electronic factor : Alkyl groups are electron-releasing (+I). Ketones have two alkyl groups which reduce the positive charge on the carbonyl carbon more than one alkyl group of aldehydes; HCHO has none.
(ii) Steric factor : Two alkyl groups in ketones hinder the approach of the nucleophile and crowd the tetrahedral intermediate more than one alkyl group and a H-atom in aldehydes.
Aromatic aldehydes and ketones are less reactive than aliphatic ones because the +R-effect of the benzene ring reduces the positive charge on the carbonyl carbon.
Thus, reactivity : HCHO > CH₃CHO > CH₃COCH₃ and CH₃CHO > C₆H₅CHO > C₆H₅COCH₃ > C₆H₅COC₆H₅

Teacher's Note:
1. A neat labelled two-step mechanism and one reactivity order with both reasons usually covers all the marks.