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

Very Short Answer Questions (1 Mark)

Question 1. pK_a of chloroacetic acid is lower than pK_a of acetic acid. Explain.
Answer:
pK_a of chloroacetic acid is lower than pK_a of acetic acid. This means, chloroacetic acid is a stronger acid than acetic acid. This may be explained on the basis of the following two reasons :
(i) Due to –I-effect of Cl atom, the electron density in the O–H bond in chloroacetic acid is much lower than due to +I-effect of CH₃ group in acetic acid. As a result, O–H bond in chloroacetic acid is much weaker than in acetic acid and hence loses a proton more easily than acetic acid.
Cl←CH₂←CO←O←H ; CH₃→CO→O→H
(ii) Due to –I-effect of Cl, dispersal of the –ve charge occurs in chloroacetate ion but due to +I-effect of CH₃ group, intensification of –ve charge occurs in acetate ion. In other words, chloroacetate ion is much more stable than acetate ion.

Teacher's Note:
1. Electron-withdrawing groups (Cl, F, NO₂, CN) near COOH increase acidity; electron-releasing alkyl groups decrease it.

Question 2. O₂N–CH₂–COOH has lower pK_a value than CH₃COOH. Give reasons. (CBSE (Foreign) 2017 ; CBSE 2018S)
Answer:
(i) Due to +I-effect of CH₃ group, electron density increases in the O–H bond of CH₃COOH whereas due to –I-effect of the NO₂ group electron density decreases in the O–H bond of NO₂CH₂COOH
CH₃→CO→O→H ; O₂N←CH₂←CO←O←H
As a result, O–H bond in O₂NCH₂COOH is weaker than the O–H bond in CH₃COOH and hence loses a proton more easily than CH₃COOH. In other words, O₂NCH₂COOH is a stronger acid than CH₃COOH. Since a stronger acid has a lower pK_a value, therefore, O₂NCH₂COOH has lower pK_a value than CH₃COOH.
(ii) Due to –I-effect of NO₂ group dispersal of –ve charge occurs in O₂NCH₂COO⁻ but due to +I-effect of CH₃ group intensification of –ve charge occurs in CH₃COO⁻. As a result, O₂NCH₂COO⁻ is more stable than CH₃COO⁻ and hence O₂NCH₂COOH is a stronger acid than CH₃COOH. In other words, O₂NCH₂COOH has lower pK_a value than CH₃COOH.

Teacher's Note:
1. NO₂ is one of the strongest electron-withdrawing groups, so nitroacetic acid is far stronger than acetic acid.

Question 3. Arrange the following in increasing order of acid strength : ClCH₂COOH, CH₃CH₂COOH, ClCH₂CH₂COOH, (CH₃)₂CHCOOH, CH₃COOH
Answer:
Since the +I-effect of alkyl groups decreases in the order : (CH₃)₂CH– > CH₃CH₂– > CH₃–, therefore, acid strength increases in the order : (CH₃)₂CHCOOH < CH₃CH₂COOH < CH₃COOH
Further, since –I-effect decreases with distance, therefore, ClCH₂CH₂COOH is a weaker acid than ClCH₂COOH. Combining these two trends, the acid strength increases in the order :
(CH₃)₂CHCOOH < CH₃CH₂COOH < CH₃COOH < ClCH₂CH₂COOH < ClCH₂COOH

Teacher's Note:
1. The inductive effect fades quickly with distance: a Cl on the β-carbon raises acidity much less than a Cl on the α-carbon.

Question 4. How can you distinguish between an alcohol and a carboxylic acid.
Answer:
By NaHCO₃ test. Carboxylic acids react with NaHCO₃ to evolve CO₂ gas while alcohols do not.
RCOOH + NaHCO₃ → RCOONa + CO₂ + H₂O ; ROH + NaHCO₃ → No action

Teacher's Note:
1. The brisk effervescence (CO₂) with NaHCO₃ is the standard laboratory test for the COOH group.

Question 5. Complete the following reaction
Phthalic acid (benzene-1, 2-dicarboxylic acid) + 2NH₃ \( \xrightarrow{Strong\ heating} \) (CBSE Sample Paper 2018-19)
Answer:
Ammonium salts on strong heating give amides.
o-C₆H₄(COOH)₂ \( \xrightarrow{2NH_3} \) o-C₆H₄(COONH₄)₂ \( \xrightarrow[-2H_2O]{Heat} \) o-C₆H₄(CONH₂)₂ \( \xrightarrow[(-NH_3)]{Strong\ heating} \) C₆H₄(CO)₂NH
Phthalic acid → Amm. phthalate → Phthalamide → Phthalimide

Teacher's Note:
1. In phthalimide both C=O groups are joined to the same NH, forming a five-membered ring fused to the benzene ring.

Question 6. Predict the product of the following reaction
CH₃COONa \( \xrightarrow[\Delta]{NaOH/CaO} \) ? (CBSE 2015)
Answer:
Sodium salts of carboxylic acids on heating with soda-lime (NaOH/CaO) undergo decarboxylation to give an alkane with one carbon atom less than the starting carboxylic acid salts.
CH₃COONa + NaOH \( \xrightarrow[(Decarboxylation)]{CaO,\ \Delta} \) CH₄ + Na₂CO₃
Sod. acetate → Methane

Teacher's Note:
1. The COONa group is replaced by H from NaOH; CaO only keeps the NaOH dry and prevents it from attacking the glass.

Question 7. Name the reagent used to convert carboxylic acids directly to the corresponding alcohols.
Answer:
LiAlH₄/ether or B₂H₆/ether, i.e., C₆H₅COOH \( \xrightarrow{LiAlH_4\ or\ B_2H_6/ether} \) C₆H₅CH₂OH

Teacher's Note:
1. NaBH₄ does not reduce the COOH group; diborane does not reduce ester or nitro groups easily, which makes it selective.

Question 8. What is vinegar ?
Answer:
A 8-10% solution of acetic acid in water is called vinegar.

Teacher's Note:
1. Pure (100%) acetic acid freezes at about 290 K and is called glacial acetic acid; do not confuse it with vinegar.

Question 9. Mention a chemical property in which formic acid (methanoic acid) differs from acetic acid (ethanoic acid). (Tamil Nadu Board 2012)
Answer:
Formic acid contains a H-atom attached to a >C=O group and hence can be regarded as an aldehyde. Therefore, it acts as a reducing agent and hence decolourises pink-violet colour of KMnO₄ solution. Acetic acid, on the other hand, contains an alkyl (i.e., CH₃) group attached to >C=O group. Therefore, it does not act as a reducing agent and hence does not decolourise KMnO₄ solution.
5HCOOH + 2KMnO₄ + 3H₂SO₄ → 5CO₂ + K₂SO₄ + 2MnSO₄ + 8H₂O
Formic acid + Pink-violet KMnO₄ → Colourless solution
CH₃COOH \( \xrightarrow{KMnO_4/H_2SO_4} \) No change in pink-violet colour of KMnO₄ soln.
Similarly, methanoic acid reduces Tollens’ reagent and Fehling’s solution but CH₃COOH does not.
H–COOH + 2[Ag(NH₃)₂]⁺ + 2OH⁻ \( \xrightarrow{Heat} \) 2Ag↓ + 4NH₃ + CO₂ + 2H₂O (Silver mirror)
H–COOH + 2Cu²⁺ + 4OH⁻ \( \xrightarrow{Heat} \) Cu₂O↓ + CO₂ + 3H₂O (Red ppt.)
CH₃COOH \( \xrightarrow[\text{or Fehling's solution}]{\text{Tollens' reagent}} \) No action

Teacher's Note:
1. Any one of these tests (KMnO₄, Tollens’ or Fehling’s) is enough for a one-mark answer, but write its equation.

Question 10. How will you convert an acid into an ester without using an alcohol ?
Answer:
By using diazomethane.
RCOOH + CH₂N₂ \( \xrightarrow{Dry\ ether} \) RCOOCH₃ + N₂
Carboxylic acid + Diazomethane → Methyl ester

Teacher's Note:
1. This method gives only methyl esters, since diazomethane supplies the CH₃ group.

Question 11. Why is the ester hydrolysis slow in the beginning and becomes faster after some time ? (CBSE 2010)
Answer:
An ester on hydrolysis gives an alcohol and an acid.
RCOOR′ + H₂O \( \xrightarrow{H_2SO_4\ (few\ drops)} \) RCOOH + R′OH
Ester → Carboxylic acid + Alcohol
A small amount of dil. H₂SO₄ is added in the beginning to start the reaction. As the reaction proceeds, a carboxylic acid (RCOOH) is produced. As the amount of carboxylic acid produced increases, the rate of ester hydrolysis also increases.

Teacher's Note:
1. This is an example of autocatalysis: a product (the acid) catalyses the reaction that forms it.

Short Answer Questions (2 Marks)

Question 12. Write short notes on the following : Esterification reaction
Answer:
Carboxylic acids react with alcohols in presence of a small amount of conc. H₂SO₄ or dry HCl to form esters. The reaction is reversible and is acid catalysed.
RCOOH + R′OH \( \xrightarrow{H^+} \) RCOOR′ + H₂O ; e.g., CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
The equilibrium is shifted towards the ester by using excess alcohol or by removing water.

Teacher's Note:
1. Esters have fruity smells; this is a common identification test for carboxylic acids.

Question 13. Write short notes on the following : Decarboxylation reaction (CBSE 2008 ; HP Board 2011)
Answer:
Removal of CO₂ from a carboxylic acid (or its salt) is called decarboxylation. Sodium salts of carboxylic acids on heating with soda-lime give hydrocarbons with one carbon atom less.
RCOONa + NaOH \( \xrightarrow{CaO,\ \Delta} \) R–H + Na₂CO₃ ; e.g., C₆H₅COONa + NaOH → C₆H₆ + Na₂CO₃
Electrolysis of salts (Kolbe’s method) also brings about decarboxylation.

Teacher's Note:
1. Acids with an electron-withdrawing group on the α-carbon (like malonic acid) lose CO₂ on mere heating.

Question 14. Write short notes on the following : Hunsdiecker reaction.
Answer:
When the silver salt of a carboxylic acid is refluxed with bromine in CCl₄, an alkyl bromide with one carbon atom less than the acid is formed.
RCOOAg + Br₂ \( \xrightarrow{CCl_4,\ \Delta} \) R–Br + CO₂ + AgBr ; e.g., CH₃CH₂COOAg + Br₂ → CH₃CH₂Br + CO₂ + AgBr

Teacher's Note:
1. The reaction gives good yields with bromine; with iodine the products depend on the ratio of reactants.

Question 15. Aromatic carboxylic acids do not undergo Friedel-Crafts reaction. (CBSE 2018)
Answer:
The carboxyl group is an electron-withdrawing (deactivating) group and reduces the electron density of the benzene ring. Moreover, the Lewis acid catalyst AlCl₃ gets bonded to the carboxyl group (through the lone pairs of its oxygen), which deactivates the ring further. Hence aromatic carboxylic acids do not undergo Friedel-Crafts alkylation or acylation.

Teacher's Note:
1. Benzoic acid can still be nitrated or brominated (at the meta position), because those reactions use stronger electrophiles.

Question 16. How can a phenol be distinguished from a carboxylic acid ? (Raj. Board 2012 ; Hr. Board 2012)
Answer:
Sodium bicarbonate test : Carboxylic acids decompose NaHCO₃ with brisk effervescence of CO₂, while phenols do not.
RCOOH + NaHCO₃ → RCOONa + CO₂↑ + H₂O ; C₆H₅OH + NaHCO₃ → No reaction
Ferric chloride test : Phenol gives a violet colour with neutral FeCl₃ solution while aliphatic carboxylic acids do not give a violet colour.

Teacher's Note:
1. Phenol (pK_a 10) is weaker than carbonic acid, which is why it cannot release CO₂ from bicarbonate.

Question 17. Give at least two chemical tests to distinguish between the following pairs of compounds : Phenol and benzoic acid (CBSE 2008, 2009, 2010 ; Manipur Board 2012 ; Hr. Board 2012)
Answer:
(1) Benzoic acid gives brisk effervescence of CO₂ with NaHCO₃ solution while phenol does not : C₆H₅COOH + NaHCO₃ → C₆H₅COONa + CO₂↑ + H₂O
(2) Phenol gives a violet colour with neutral FeCl₃ solution while benzoic acid (as its neutral salt) gives a buff-coloured ppt. of ferric benzoate.

Teacher's Note:
1. The NaHCO₃ test is the quickest and most reliable for any COOH versus OH pair.

Question 18. Give at least two chemical tests to distinguish between the following pairs of compounds : Acetic acid and formic acid or CH₃COOH and HCOOH (Bihar Board 2011, CBSE 2016)
Answer:
(1) Formic acid gives a silver mirror with Tollens’ reagent while acetic acid does not : HCOOH + 2[Ag(NH₃)₂]⁺ + 2OH⁻ → 2Ag↓ + CO₂ + 4NH₃ + 2H₂O
(2) Formic acid reduces Fehling’s solution to a red ppt. of Cu₂O and decolourises acidified KMnO₄; acetic acid does neither.

Teacher's Note:
1. Formic acid is the only carboxylic acid that behaves as a reducing agent.

Question 19. Give at least two chemical tests to distinguish between the following pairs of compounds : Acetic acid and acetone.
Answer:
(1) Acetic acid gives brisk effervescence of CO₂ with NaHCO₃ solution while acetone does not.
(2) Acetone gives a yellow ppt. of iodoform with I₂/NaOH and an orange ppt. with 2, 4-DNP reagent; acetic acid does not : CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃↓ + CH₃COONa + 3NaI + 3H₂O

Teacher's Note:
1. Acetic acid has a CH₃CO– unit but does not give the iodoform test, because its C=O is part of the carboxyl group.

Question 20. Give at least two chemical tests to distinguish between the following pairs of compounds : Methyl acetate and ethyl acetate. (CBSE 2007, 2014)
Answer:
On hydrolysis with NaOH, methyl acetate gives methanol while ethyl acetate gives ethanol. The alcohol is then tested :
(1) Iodoform test : Ethanol (from ethyl acetate) gives a yellow ppt. of iodoform with I₂/NaOH while methanol (from methyl acetate) does not : C₂H₅OH + 4I₂ + 6NaOH → CHI₃↓ + HCOONa + 5NaI + 5H₂O
(2) The alcohol from methyl acetate on oxidation gives formaldehyde/formic acid, which reduces Tollens’ reagent strongly; ethanol gives acetaldehyde/acetic acid (acetic acid does not reduce it).

Teacher's Note:
1. Esters themselves do not give these tests; the test is done on the alcohol obtained after hydrolysis.

Question 21. Give two uses each of methanoic acid and ethanoic acid.
Answer:
Methanoic acid : (i) in the coagulation of rubber latex, (ii) in dyeing and tanning of leather, and as a preservative for fruits.
Ethanoic acid : (i) as vinegar (6–8% solution) for flavouring and preserving food, (ii) as a solvent and in the manufacture of cellulose acetate (rayon), plastics, dyes and drugs.

Teacher's Note:
1. Two clear uses per acid are enough for full marks; avoid vague answers like ‘in industry’.

Question 22. Identify the products A, B and C in the following reactions : CH₃CHO \( \xrightarrow{K_2Cr_2O_7,\ H^+} \) (A) \( \xrightarrow{SOCl_2} \) (B) (Manipur Board 2012)
Answer:
A = CH₃COOH (Acetic acid); B = CH₃COCl (Acetyl chloride)

Teacher's Note:
1. SOCl₂ converts the COOH group into COCl.

Question 23. Identify the products A, B and C in the following reactions : CH₃COOH \( \xrightarrow{Cl_2/P} \) (A) \( \xrightarrow{Aq.\ NaOH} \) (B)
Answer:
A = ClCH₂COOH (Chloroacetic acid); B = HOCH₂COONa (Sodium hydroxyacetate or sodium glycollate)

Teacher's Note:
1. Aqueous NaOH both replaces Cl by OH and neutralises the COOH group.

Question 24. Identify the products A, B and C in the following reactions : C₆H₅Br \( \xrightarrow[Dry\ ether]{Mg} \) A \( \xrightarrow[(ii)\ H_3O^+]{(i)\ CO_2} \) B \( \xrightarrow{PCl_5} \) C (CBSE 2016, 2017)
Answer:
A = C₆H₅MgBr; B = C₆H₅COOH (Benzoic acid); C = C₆H₅COCl (Benzoyl chloride)

Teacher's Note:
1. PCl₅ gives POCl₃ and HCl as by-products.

Question 25. Identify the products A, B and C in the following reactions : CH₃CH₂Br \( \xrightarrow{Aq.\ NaOH} \) (A) \( \xrightarrow{K_2Cr_2O_7,\ H^+} \) (B)
Answer:
A = CH₃CH₂OH (Ethanol); B = CH₃COOH (Acetic acid)

Teacher's Note:
1. Aqueous NaOH substitutes; alcoholic KOH would instead eliminate to give ethene.

Question 26. Identify the products A, B and C in the following reactions : CH₃CHO \( \xrightarrow{Oxidation} \) (A) \( \xrightarrow{PCl_5} \) (B) \( \xrightarrow{CH_3COONa} \) (CH₃CO)₂O
Answer:
A = CH₃COOH (Acetic acid); B = CH₃COCl (Acetyl chloride); CH₃COCl + CH₃COONa → (CH₃CO)₂O + NaCl

Teacher's Note:
1. Acid chloride + carboxylate salt is a standard method for making acid anhydrides.

Question 27. Identify the products A, B and C in the following reactions : CH₃COCl \( \xrightarrow[(ii)\ H_2O,\ Ag_2O,\ Heat]{(i)\ CH_2N_2\ (excess)} \) (West Bengal Board 2013)
Answer:
CH₃COCl \( \xrightarrow{CH_2N_2} \) CH₃COCHN₂ (Diazoketone) \( \xrightarrow{H_2O,\ Ag_2O,\ \Delta} \) CH₃CH₂COOH (Propanoic acid) + N₂
This is the Arndt-Eistert synthesis : the acid chain is lengthened by one CH₂ group.

Teacher's Note:
1. Acetyl chloride (2 C) gives propanoic acid (3 C): the extra carbon comes from diazomethane.

Question 28. Identify the products formed in the following reaction : CH₃COOC₂H₅ \( \xrightarrow{NH_3,\ Heat} \) (West Bengal Board 2013)
Answer:
CH₃COOC₂H₅ + NH₃ \( \xrightarrow{\Delta} \) CH₃CONH₂ (Acetamide) + C₂H₅OH (Ammonolysis)

Teacher's Note:
1. The OC₂H₅ group of the ester leaves as ethanol.

Question 29. Identify the products A, B and C in the following reactions : C₆H₅COOH \( \xrightarrow{Br_2/FeBr_3} \) (CBSE 2013)
Answer:
C₆H₅COOH + Br₂ \( \xrightarrow{FeBr_3} \) m-BrC₆H₄COOH (3-Bromobenzoic acid) + HBr

Teacher's Note:
1. COOH deactivates the ring and directs the incoming Br to the meta position.

Short Answer Questions (3 Marks)

Question 30. How is benzoic acid prepared from :
(i) Ethylbenzene (CBSE 2009, 2011) (ii) Toluene (CBSE 2013, 2018)
(iii) Bromobenzene (CBSE 2010) (iv) Benzonitrile (v) Benzaldehyde ?
Answer:
(i) C₆H₅CH₂CH₃ \( \xrightarrow{KMnO_4/KOH,\ \Delta} \) C₆H₅COOK \( \xrightarrow{H_3O^+} \) C₆H₅COOH
(ii) C₆H₅CH₃ \( \xrightarrow{KMnO_4/KOH,\ \Delta} \) C₆H₅COOK \( \xrightarrow{H_3O^+} \) C₆H₅COOH
(iii) C₆H₅Br + Mg \( \xrightarrow{Dry\ ether} \) C₆H₅MgBr \( \xrightarrow[(ii)\ H_3O^+]{(i)\ CO_2} \) C₆H₅COOH
(iv) C₆H₅CN + 2H₂O \( \xrightarrow{H^+\ or\ OH^-,\ \Delta} \) C₆H₅COOH + NH₃
(v) C₆H₅CHO \( \xrightarrow{KMnO_4/KOH\ or\ Tollens\ reagent;\ then\ H_3O^+} \) C₆H₅COOH

Teacher's Note:
1. With KMnO₄/KOH, the product first is the potassium salt; always finish with acidification to get the free acid.

Question 31. Give reasons for the following : (i) The melting points and boiling points of aromatic acids are usually higher than those of aliphatic acids of comparable molecular masses.
(ii) The melting point of an aliphatic carboxylic acid containing an even number of carbon atoms is higher than the next lower and the next higher homologues containing odd number of carbon atoms.
(iii) Carboxylic acids have higher boiling points than alcohols of comparable molecular masses.
(iv) Ethanoic acid is highly soluble in water but hexanoic acid is only slightly soluble.
(v) Most aromatic acids are solids but acids of acetic acid group are mostly liquids.
Answer:
(i) Aromatic acids have planar, rigid molecules which pack closely in the crystal lattice and have stronger van der Waals forces between their rings. Hence they have higher m.p. and b.p. than aliphatic acids of comparable molecular masses.
(ii) In acids with an even number of carbon atoms, the terminal CH₃ and COOH groups lie on opposite sides of the zig-zag carbon chain, so the molecules fit more closely in the crystal lattice than those of odd-carbon acids, where these groups lie on the same side. The stronger intermolecular forces give higher melting points.
(iii) Carboxylic acids form stronger and more extensive intermolecular H-bonds than alcohols, since the O–H bond is more polar; they exist as cyclic dimers held by two H-bonds even in the vapour state. Hence more energy is needed to boil them.
(iv) Ethanoic acid forms H-bonds with water and its hydrocarbon part is small, so it is miscible with water. In hexanoic acid, the large non-polar (hydrophobic) C₅H₁₁ chain hinders H-bonding with water, so its solubility is low.
(v) Aromatic acids have higher molecular masses and planar molecules that pack well in a crystal lattice, so they are solids. Lower aliphatic acids of the acetic acid group (up to about C₉) have flexible chains and weaker lattice forces, so they are liquids.

Teacher's Note:
1. Boiling point depends on intermolecular forces in the liquid; melting point also depends on how well molecules pack in the solid.

Question 32. Explain the following about acetic acid :
(i) Its boiling point is higher than that of n- propanol.
(ii) It is a weaker acid than chloroacetic acid and formic acid.
(iii) Acetic acid is a stronger acid than phenol. (CBSE 2013)
Answer:
(i) Both have the same molecular mass (60), but acetic acid forms stronger intermolecular H-bonds and exists as a cyclic dimer held by two H-bonds. Hence its b.p. (391 K) is higher than that of n-propanol (370 K).
(ii) In chloroacetic acid, the –I-effect of Cl disperses the negative charge of the chloroacetate ion and stabilises it. In acetic acid, the +I-effect of CH₃ intensifies the negative charge on the acetate ion and destabilises it. Formic acid has no electron-releasing alkyl group, so formate ion is more stable than acetate ion. Hence acetic acid is weaker than both.
(iii) In acetate ion, the negative charge is delocalised over two equivalent oxygen atoms, while in phenoxide ion it is delocalised over one oxygen and the less electronegative carbons of the ring through non-equivalent structures. Acetate ion is thus more stable, and acetic acid is a stronger acid than phenol.

Teacher's Note:
1. pK_a values to remember: ClCH₂COOH 2.86, HCOOH 3.75, CH₃COOH 4.76, phenol 10.0.

Question 33. Account for the following :
(i) Chloroacetic acid is a stronger acid than acetic acid. (HP Board 2011 ; Kerala Board 2012)
(ii) Trichloroacetic acid is a stronger acid than dichloroacetic acid.
(iii) Fluoroacetic acid is a stronger acid than chloroacetic acid.
(iv) Dichloroacetic acid is a stronger acid than chloroacetic acid.
(v) 2-Fluorobutanoic acid is a stronger acid than 3-fluorobutanoic acid. (CBSE Sample Paper 2017)
Answer:
Electron-withdrawing groups (–I-effect) disperse the negative charge of the carboxylate ion and stabilise it, thereby increasing the acid strength.
(i) Due to the –I-effect of Cl, chloroacetate ion (ClCH₂COO⁻) is more stable than acetate ion, in which the +I-effect of CH₃ intensifies the charge. Hence ClCH₂COOH is stronger.
(ii) Three Cl atoms exert a greater –I-effect than two; Cl₃CCOO⁻ is more stable than Cl₂CHCOO⁻. Hence Cl₃CCOOH is stronger.
(iii) F is more electronegative than Cl and exerts a stronger –I-effect; FCH₂COO⁻ is more stable than ClCH₂COO⁻. Hence FCH₂COOH is stronger.
(iv) Two Cl atoms exert a greater –I-effect than one, so Cl₂CHCOO⁻ is more stable than ClCH₂COO⁻. Hence Cl₂CHCOOH is stronger.
(v) The –I-effect decreases with distance. In 2-fluorobutanoic acid, F is on the α-carbon, nearer to COOH, than in 3-fluorobutanoic acid (F on β-carbon). Hence the anion of the 2-fluoro acid is more stabilised and it is the stronger acid.

Teacher's Note:
1. Three factors raise the –I-effect on acidity: more halogen atoms, more electronegative halogen, and halogen closer to COOH.

Question 34. (i) Explain why pK_a value of chloroacetic acid is lower than pK_a value of acetic acid.
(ii) Why pK_a of FCH₂COOH is lower than that of ClCH₂COOH ? (CBSE 2016)
(iii) O₂NCH₂COOH has lower pK_a value than CH₃COOH. (CBSE Foreign 2017)
Answer:
A lower pK_a means a stronger acid.
(i) Due to the –I-effect of Cl, the O–H bond of ClCH₂COOH is weakened and the chloroacetate ion is stabilised by dispersal of its negative charge, whereas the +I-effect of CH₃ in acetic acid does the opposite. Hence ClCH₂COOH is a stronger acid and has a lower pK_a.
(ii) F is more electronegative than Cl and exerts a stronger –I-effect, so FCH₂COO⁻ is more stable than ClCH₂COO⁻. Hence FCH₂COOH is a stronger acid and has a lower pK_a.
(iii) The NO₂ group has a strong –I-effect which weakens the O–H bond and stabilises O₂NCH₂COO⁻, while the +I-effect of CH₃ destabilises CH₃COO⁻. Hence O₂NCH₂COOH is a stronger acid and has a lower pK_a.

Teacher's Note:
1. Always convert a pK_a statement into an acid-strength statement first; then explain with the stability of the anion.

Question 35. State reasons for the following :
(i) Monochloroethanoic acid has a higher pK_a value than dichloroethanoic acid.
(ii) pK_a of 4-nitrobenzoic acid is lower than that of benzoic acid. (CBSE 2018)
(iii) Ethanoic acid is a weaker acid than benzoic acid. (CBSE 2008 ; HP Board 2013)
Answer:
(i) Two Cl atoms exert a greater –I-effect than one Cl atom. Hence dichloroethanoate ion is more stabilised than monochloroethanoate ion, and dichloroethanoic acid is the stronger acid. A weaker acid has a higher pK_a, so monochloroethanoic acid has the higher pK_a.
(ii) The NO₂ group is strongly electron-withdrawing (–I and –R effects). At the para position it withdraws electron density from the COO⁻ group and stabilises the 4-nitrobenzoate ion. Hence 4-nitrobenzoic acid is a stronger acid than benzoic acid and has a lower pK_a.
(iii) In ethanoic acid, the CH₃ group has a +I-effect which destabilises the ethanoate ion. In benzoic acid, the COOH group is attached to an sp²-hybridised carbon of the phenyl group, which is more electronegative than the sp³ carbon of CH₃ and withdraws electrons; the benzoate ion is therefore more stable. Hence ethanoic acid (pK_a 4.76) is a weaker acid than benzoic acid (pK_a 4.19).

Teacher's Note:
1. Electron-withdrawing groups (NO₂, CN, halogens) on the ring increase the acidity of benzoic acid; electron-donating groups (CH₃, OCH₃) decrease it.

Question 36. What happens when acetic acid is treated with :
(i) NaHCO₃ (ii) Zn metal (iii) LiAlH₄ (iv) ethanol and a drop of conc. H₂SO₄ (v) NH₃, then heat (vi) PCl₅ (vii) Soda-lime ?
Answer:
(i) CH₃COOH + NaHCO₃ → CH₃COONa + CO₂↑ + H₂O (brisk effervescence)
(ii) 2CH₃COOH + Zn → (CH₃COO)₂Zn + H₂↑
(iii) CH₃COOH \( \xrightarrow[(ii)\ H_3O^+]{(i)\ LiAlH_4/ether} \) CH₃CH₂OH (Ethanol)
(iv) CH₃COOH + C₂H₅OH \( \xrightarrow{conc.\ H_2SO_4} \) CH₃COOC₂H₅ (Ethyl acetate) + H₂O
(v) CH₃COOH + NH₃ → CH₃COONH₄ \( \xrightarrow[-H_2O]{\Delta} \) CH₃CONH₂ (Acetamide)
(vi) CH₃COOH + PCl₅ → CH₃COCl (Acetyl chloride) + POCl₃ + HCl
(vii) Acetic acid is first neutralised to sodium acetate, which on heating with soda-lime gives methane : CH₃COONa + NaOH \( \xrightarrow{CaO,\ \Delta} \) CH₄ + Na₂CO₃

Teacher's Note:
1. Learn this list as a set: it covers acidity, reduction, esterification, amide and acid chloride formation, and decarboxylation.

Question 37. What is esterification? Explain the mechanism of esterification of carboxylic acids in detail. (Uttarakhand Board 2012)
Answer:
The reaction of a carboxylic acid with an alcohol in presence of a mineral acid (conc. H₂SO₄ or dry HCl) as catalyst to form an ester and water is called esterification. It is reversible.
RCOOH + R′OH \( \xrightarrow{H^+} \) RCOOR′ + H₂O
Mechanism : Step 1. Protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic : R–C(=O)–OH + H⁺ ⇌ R–C(=O⁺H)–OH
Step 2. Nucleophilic attack of the alcohol oxygen on the carbonyl carbon gives a tetrahedral intermediate : R–C(OH)₂–O⁺(H)R′
Step 3. A proton is transferred from the alcohol oxygen to one OH group, converting it into a good leaving group (–O⁺H₂) : R–C(OH)(OR′)–O⁺H₂
Step 4. Loss of a water molecule gives the protonated ester : R–C(=O⁺H)–OR′ + H₂O
Step 5. Loss of a proton gives the ester and regenerates the catalyst : R–C(=O)–OR′ + H⁺
Since the oxygen of water comes from the acid (the C–OH bond of the acid breaks), this is an acyl–oxygen cleavage.

Teacher's Note:
1. To push the reversible reaction forward, use excess alcohol or remove water as it forms.

Long Answer Questions (5 Marks)

Question 38. How does acetaldehyde react with
(a) dilute aqueous caustic soda
(b) hydrazine
(c) ammoniacal silver nitrate
(d) phosphorus petachloride (Maharashtra Board 2012)
Answer:
(a) Aldol condensation occurs to give 3-hydroxybutanal, which on heating gives but-2-enal : 2CH₃CHO \( \xrightarrow{dil.\ NaOH} \) CH₃CH(OH)CH₂CHO \( \xrightarrow[-H_2O]{\Delta} \) CH₃CH=CHCHO
(b) Acetaldehyde hydrazone is formed : CH₃CHO + H₂NNH₂ → CH₃CH=NNH₂ + H₂O
(c) It reduces Tollens’ reagent to a silver mirror and is itself oxidised to acetate ion : CH₃CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃COO⁻ + 2Ag↓ + 2H₂O + 4NH₃
(d) The carbonyl oxygen is replaced by two chlorine atoms to give 1, 1-dichloroethane (ethylidene chloride) : CH₃CHO + PCl₅ → CH₃CHCl₂ + POCl₃

Teacher's Note:
1. PCl₅ gives a gem-dihalide with aldehydes and ketones, unlike alcohols, which give monohalides.

Question 39. Discuss briefly the oxidation and reduction reactions of aldehydes and ketones.
Answer:
Oxidation : (i) Aldehydes are easily oxidised to carboxylic acids even by mild oxidising agents such as Tollens’ reagent, Fehling’s solution and Benedict’s solution, as well as by HNO₃, KMnO₄ and K₂Cr₂O₇ : RCHO \( \xrightarrow{[O]} \) RCOOH
RCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → RCOO⁻ + 2Ag↓ + 2H₂O + 4NH₃ ; RCHO + 2Cu²⁺ + 5OH⁻ → RCOO⁻ + Cu₂O↓ + 3H₂O
(ii) Ketones are oxidised only by strong oxidising agents at high temperature, with cleavage of a C–C bond to give a mixture of carboxylic acids with fewer carbon atoms : CH₃COCH₂CH₃ \( \xrightarrow{[O]} \) 2CH₃COOH
(iii) Haloform reaction : Methyl ketones (and CH₃CHO) are oxidised by sodium hypohalite to carboxylates with one carbon atom less and a haloform : RCOCH₃ + 3NaOX → RCOONa + CHX₃ + 2NaOH
Reduction : (i) To alcohols : With NaBH₄, LiAlH₄ or H₂/Ni, Pt or Pd, aldehydes give 1° alcohols and ketones give 2° alcohols : RCHO \( \xrightarrow{[H]} \) RCH₂OH ; R₂CO \( \xrightarrow{[H]} \) R₂CHOH
(ii) To hydrocarbons : The >C=O group is reduced to >CH₂ by Clemmensen reduction (Zn-Hg/conc. HCl) or Wolff-Kishner reduction (NH₂NH₂, then KOH/ethylene glycol, heat) : >C=O → >CH₂
(iii) To pinacols : Ketones on reduction with Mg-Hg followed by hydrolysis give pinacols : 2(CH₃)₂CO \( \xrightarrow[(ii)\ H_2O]{(i)\ Mg\text{-}Hg} \) (CH₃)₂C(OH)C(OH)(CH₃)₂

Teacher's Note:
1. The easy oxidation of aldehydes (because of the H on the carbonyl carbon) is the basis of all the tests that separate aldehydes from ketones.

Question 40. What is α-hydrogen in a carbonyl compound ? Outline, with relevant chemical equations wherever possible, the similarities and differences on the reactivity of the compounds CH₃CHO and C₆H₅CHO with the following reagents :
(i) Tollens’ reagent, (ii) NaOH. (Manipur Board 2011)
Answer:
A hydrogen atom attached to the carbon next to the carbonyl group (the α-carbon) is called an α-hydrogen. Due to the –I-effect of C=O and resonance stabilisation of the carbanion formed on its removal, α-hydrogens are acidic. CH₃CHO has three α-H atoms; C₆H₅CHO has none.
(i) Tollens’ reagent (similarity) : Both aldehydes reduce Tollens’ reagent to a silver mirror :
CH₃CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃COO⁻ + 2Ag↓ + 2H₂O + 4NH₃
C₆H₅CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → C₆H₅COO⁻ + 2Ag↓ + 2H₂O + 4NH₃
(ii) NaOH (difference) : CH₃CHO, having α-H atoms, undergoes aldol condensation with dilute NaOH : 2CH₃CHO \( \xrightarrow{dil.\ NaOH} \) CH₃CH(OH)CH₂CHO
C₆H₅CHO, having no α-H, does not undergo aldol condensation; with concentrated NaOH it undergoes Cannizzaro reaction : 2C₆H₅CHO + NaOH (conc.) → C₆H₅CH₂OH + C₆H₅COONa

Teacher's Note:
1. The presence or absence of α-H decides how an aldehyde behaves with alkali: aldol or Cannizzaro.

Question 41. What is common in both aldehydes and ketones ? In what respects, acetaldehyde resembles and differs from acetone? (Jharkhand Board 2011)
Answer:
Both aldehydes and ketones contain the carbonyl group (>C=O), and so show many similar (nucleophilic addition) reactions.
Resemblances of acetaldehyde and acetone :
(i) Both add HCN to form cyanohydrins : CH₃CHO + HCN → CH₃CH(OH)CN ; (CH₃)₂CO + HCN → (CH₃)₂C(OH)CN
(ii) Both form bisulphite adducts with NaHSO₃ and react with Grignard reagents.
(iii) Both react with ammonia derivatives (NH₂OH, NH₂NH₂, semicarbazide, 2, 4-DNP) : CH₃CHO + NH₂OH → CH₃CH=NOH ; (CH₃)₂CO + NH₂OH → (CH₃)₂C=NOH
(iv) Both have α-H atoms and undergo aldol condensation; both contain CH₃CO– and give the iodoform test.
(v) Both are reduced by Zn-Hg/HCl or Wolff-Kishner reduction to alkanes.
Differences :
(i) Acetaldehyde reduces Tollens’ reagent (silver mirror) and Fehling’s solution (red ppt.); acetone does not.
(ii) Acetaldehyde restores the pink colour of Schiff’s reagent; acetone does not.
(iii) On reduction, acetaldehyde gives a primary alcohol (ethanol) while acetone gives a secondary alcohol (propan-2-ol).
(iv) With Grignard reagents (CH₃MgBr), acetaldehyde gives a 2° alcohol while acetone gives a 3° alcohol.
(v) Acetaldehyde is easily oxidised to acetic acid (same number of C atoms); acetone is oxidised only by strong oxidants, with breaking of a C–C bond.

Teacher's Note:
1. Organise this answer in two clear lists (resemblances, differences) with one equation for each point where possible.

Question 42. Describe briefly the general methods of preparation of carboxylic acids. (J & K Board 2012)
Answer:
(i) From primary alcohols and aldehydes : Oxidation with KMnO₄ (neutral, acidic or alkaline) or K₂Cr₂O₇/H₂SO₄ : RCH₂OH \( \xrightarrow{[O]} \) RCHO \( \xrightarrow{[O]} \) RCOOH
(ii) From alkylbenzenes : Vigorous oxidation with alkaline KMnO₄ or chromic acid oxidises the entire side chain to COOH : C₆H₅CH₂R \( \xrightarrow{KMnO_4/KOH,\ \Delta;\ H_3O^+} \) C₆H₅COOH
(iii) From nitriles and amides : Hydrolysis with dilute acid or alkali : RCN \( \xrightarrow{H^+/H_2O\ or\ OH^-/H_2O} \) RCONH₂ \( \xrightarrow{H^+/H_2O} \) RCOOH
(iv) From Grignard reagents : RMgX + CO₂ \( \xrightarrow{Dry\ ether} \) RCOOMgX \( \xrightarrow{H_3O^+} \) RCOOH
(v) From acyl halides and anhydrides : Hydrolysis with water (or alkali followed by acid) : RCOCl + H₂O → RCOOH + HCl ; (RCO)₂O + H₂O → 2RCOOH
(vi) From esters : Acid hydrolysis gives the acid directly; basic hydrolysis (saponification) gives the salt which on acidification gives the acid : RCOOR′ + H₂O \( \xrightarrow{H^+} \) RCOOH + R′OH
(vii) From alkenes (industrial) : Alkenes react with CO and steam under pressure in presence of H₃PO₄ to give acids : CH₂=CH₂ + CO + H₂O \( \xrightarrow{H_3PO_4} \) CH₃CH₂COOH

Teacher's Note:
1. Note which methods add a carbon (nitrile, Grignard + CO₂) and which keep the carbon count (oxidation, hydrolysis).

Question 43. Explain why carboxylic acids behave as acids ? Discuss briefly the effect of electron donating and electron-withdrawing substituents on the acidity of aliphatic carboxylic acids.
Answer:
Carboxylic acids ionise in water to give a proton and a carboxylate ion : RCOOH + H₂O ⇌ RCOO⁻ + H₃O⁺
They behave as acids because (i) the O–H bond is weakened by the electron-withdrawing C=O group, and (ii) the carboxylate ion formed is stabilised by resonance, the negative charge being delocalised equally over the two electronegative oxygen atoms (R–C(=O)O⁻ ↔ R–C(O⁻)=O). They turn blue litmus red and liberate CO₂ from NaHCO₃.
Effect of substituents : Any factor that stabilises the carboxylate ion (by dispersing its negative charge) increases the acid strength, and any factor that destabilises it decreases the acid strength.
(i) Electron-donating groups (alkyl groups, +I-effect) intensify the negative charge on the carboxylate ion and decrease acidity : HCOOH > CH₃COOH > CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH
(ii) Electron-withdrawing groups (–I-effect : halogens, NO₂, CN, C₆H₅, etc.) disperse the negative charge and increase acidity. The effect increases with the number and electronegativity of the groups and decreases with their distance from the COOH group :
CF₃COOH > CCl₃COOH > CHCl₂COOH > NO₂CH₂COOH > NC–CH₂COOH > FCH₂COOH > ClCH₂COOH > BrCH₂COOH > HCOOH > ClCH₂CH₂COOH > C₆H₅COOH > C₆H₅CH₂COOH > CH₃COOH > CH₃CH₂COOH

Teacher's Note:
1. Carboxylic acids are weaker than mineral acids but stronger than alcohols and most phenols.