CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids Important Questions Set 06

Very Short Answer Questions (1 Mark)

Question 1. Give balanced equation and name of products when CH₃COOH is treated with PCl₅.
Answer:
CH₃COOH + PCl₅ → CH₃COCl + POCl₃ + HCl
Products: acetyl chloride (ethanoyl chloride), phosphorus oxychloride and hydrogen chloride.

Teacher's Note:
1. SOCl₂ is often preferred in the lab because its by-products SO₂ and HCl are gases and escape easily.

Question 2. What product is obtained when ethyl benzene is oxidized with alkaline KMnO₄?
Answer:
Benzoic acid (C₆H₅COOH) is obtained.

Teacher's Note:
1. Whatever the length of the side chain, vigorous oxidation leaves only –COOH on the ring (provided the benzylic carbon has an H).

Question 3. CH₃CHO is more reactive than CH₃COCH₃ towards reaction with HCN. Give reason.
Answer:
Because the positive charge on carbonyl carbon of CH₃CHO decreases to a lesser extent due to only one electron releasing group (–CH₃) and hence more reaction.

Teacher's Note:
1. Also mention the steric factor: two methyl groups in acetone crowd the carbonyl carbon more than one methyl in acetaldehyde.

Question 4. What is RDX?
Answer:
RDX is Research and Development Explosive which is prepared by the nitration of hexamethylenetetramine (urotropine) under controlled conditions.

Teacher's Note:
1. Urotropine itself is made from formaldehyde and ammonia: 6HCHO + 4NH₃ → (CH₂)₆N₄ + 6H₂O.

Question 5. Write IUPAC name of the following compound:
HOOC–CH=CH–COOH
Answer:
But-2-ene-1,4-dioic acid

Teacher's Note:
1. With two –COOH groups the ‘e’ of ‘-ene’ is kept because the suffix ‘dioic’ starts with a consonant.

Question 6. Write the IUPAC name of the compound shown below.

Answer:
Methyl 2-methylbenzoate.

Teacher's Note:
1. Esters are named alkyl + alkanoate; the ring carbon carrying the ester group is C-1.

Question 7. Why does benzoic acid not undergo Friedel-Crafts reaction?
Answer:
–COOH group in C₆H₅COOH is an electron withdrawing group which deactivates the benzene ring. Hence electrophilic substitution becomes difficult.

Teacher's Note:
1. The carboxyl group also forms a complex with the Lewis acid AlCl₃, which deactivates the ring further.

Question 8. Give the IUPAC names of : (i) Diacetone alcohol (ii) Crotonaldehyde (iii) Salicylic acid (CBSE 2019)
Answer:
(i) 4-Hydroxy-4-methylpentan-2-one (ii) But-2-en-1-al (iii) 2-Hydroxybenzoic acid.

Teacher's Note:
1. Diacetone alcohol is (CH₃)₂C(OH)CH₂COCH₃; the C=O gets the lower locant (2), so the OH is at C-4.

Question 9. Draw the structures of the following : (i) 3-Methylbutanal (CBSE 2013) (ii) 4-Chloropentan-2-one (CBSE 2011, 2013)
Answer:
(i) ⁴CH₃–³CH(CH₃)–²CH₂–¹CHO   (3-Methylbutanal)
(ii) ⁵CH₃–⁴CH(Cl)–³CH₂–²CO–¹CH₃   (4-Chloropentan-2-one)

Teacher's Note:
1. In aldehydes the CHO carbon is always C-1, so it never needs a locant in the name.

Question 10. Write the structural formula of (i) 3-Oxopentanal (ii) 1-Phenylpentan-1-one (CBSE 2009) (iii) p-Methylbenzaldehyde (CBSE 2013) (iv) the compound shown below (name it). (Hr. Board 2013)

Answer:
(i) ⁵CH₃⁴CH₂–³CO–²CH₂–¹CHO   (3-Oxopentanal)
(ii) C₆H₅–¹CO–²CH₂³CH₂⁴CH₂⁵CH₃   (1-Phenylpentan-1-one)
(iii) CH₃–C₆H₄–CHO (CH₃ and CHO at positions 1 and 4 of the benzene ring)   (p-Methylbenzaldehyde)
(iv) 2-Methylcyclohexanone

Teacher's Note:
1. When an aldehyde group is a substituent of a chain with a higher-priority carbonyl, it is named with the prefix ‘oxo’ for the ketone carbonyl, as in 3-oxopentanal.

Question 11. Name the reaction and the reagent used for the conversion of acid chlorides to the corresponding aldehydes.
Answer:
Name : Rosenmund reaction. Reagent : H₂ in presence of Pd (supported over BaSO₄) and partially poisoned by addition of sulphur or quinoline.
R–COCl + H₂ \( \xrightarrow[+S\ or\ quinoline]{Pd/BaSO_4} \) R–CHO + HCl
Acid chloride → Aldehyde

Teacher's Note:
1. The catalyst is poisoned so that the aldehyde formed is not reduced further to a primary alcohol.

Short Answer Questions (2 Marks)

Question 12. What compound is formed when ethanal is heated with dilute NaOH solution? (Assam Board 2012)
Answer:
Ethanal undergoes aldol condensation to form 3-hydroxybutanal (aldol), which on heating loses water to form but-2-enal (crotonaldehyde).
2CH₃CHO \( \xrightarrow{dil.\ NaOH} \) CH₃CH(OH)CH₂CHO \( \xrightarrow[-H_2O]{\Delta} \) CH₃CH=CHCHO

Teacher's Note:
1. The dehydration is easy because the new C=C bond is conjugated with the C=O group.

Question 13. Write giving chemical equations, a brief account of the following : Aldol condensation (HP Board 2011 ; Hr. Board 2011, 2012 ; CBSE 2014)
Answer:
Aldehydes and ketones having at least one α-hydrogen undergo self-condensation in presence of dilute alkali to give β-hydroxy aldehydes (aldols) or β-hydroxy ketones (ketols), which on heating lose water to give α, β-unsaturated carbonyl compounds.
2CH₃CHO \( \xrightarrow{dil.\ NaOH} \) CH₃CH(OH)CH₂CHO (3-Hydroxybutanal) \( \xrightarrow[-H_2O]{\Delta} \) CH₃CH=CHCHO (But-2-enal)
2CH₃COCH₃ \( \xrightarrow{Ba(OH)_2} \) (CH₃)₂C(OH)CH₂COCH₃ (4-Hydroxy-4-methylpentan-2-one) \( \xrightarrow[-H_2O]{\Delta} \) (CH₃)₂C=CHCOCH₃ (4-Methylpent-3-en-2-one)

Teacher's Note:
1. The name ‘aldol’ comes from aldehyde + alcohol, the two groups present in the product.

Question 14. Write giving chemical equations, a brief account of the following : Cannizzaro reaction (CBSE 2009, 2010, 2014 ; Hr. Board 2011, 2012 ; Maharashtra Board 2012 ; MP Assam Board 2012)
Answer:
Aldehydes which do not have an α-hydrogen undergo self oxidation and reduction (disproportionation) on heating with concentrated alkali: one molecule is reduced to an alcohol and another is oxidised to the salt of a carboxylic acid.
2HCHO + NaOH (conc.) \( \xrightarrow{\Delta} \) CH₃OH + HCOONa
2C₆H₅CHO + NaOH (conc.) \( \xrightarrow{\Delta} \) C₆H₅CH₂OH + C₆H₅COONa

Teacher's Note:
1. The key step is transfer of a hydride ion (H⁻) from one aldehyde molecule to another.

Question 15. Write giving chemical equations, a brief account of the following : Wolff-Kishner reduction (HP Board 2011 ; Hr. Board 2012 ; HP Board 2013 ; CBSE 2010, 2014, 2017)
Answer:
The carbonyl group of aldehydes and ketones is reduced to a CH₂ group by treatment with hydrazine followed by heating the hydrazone with KOH (or sodium ethoxide) in a high boiling solvent such as ethylene glycol.
>C=O \( \xrightarrow[-H_2O]{NH_2NH_2} \) >C=NNH₂ \( \xrightarrow{KOH/ethylene\ glycol,\ \Delta} \) >CH₂ + N₂
e.g., CH₃COCH₃ → CH₃CH₂CH₃

Teacher's Note:
1. The evolution of N₂ gas drives the reaction to completion.

Question 16. Write giving chemical equations, a brief account of the following : Clemmensen reduction (CBSE 2009, 2010 ; Assam Board 2012 ; Hr. Board 2012 ; HP Board 2013)
Answer:
The carbonyl group of aldehydes and ketones is reduced to a CH₂ group on treatment with zinc amalgam and concentrated hydrochloric acid.
>C=O \( \xrightarrow{Zn\text{-}Hg/conc.\ HCl} \) >CH₂ + H₂O
e.g., C₆H₅COCH₃ \( \xrightarrow{Zn\text{-}Hg/HCl} \) C₆H₅CH₂CH₃

Teacher's Note:
1. Clemmensen uses strong acid, so it is not suitable for acid-sensitive compounds; use Wolff-Kishner for those.

Question 17. Write giving chemical equations, a brief account of the following : Friedel-Crafts reaction (CBSE 2010 ; Hr. Board 2012)
Answer:
Benzene reacts with an acyl chloride (or acid anhydride) in presence of anhydrous AlCl₃ to form an aromatic ketone (Friedel-Crafts acylation).
C₆H₆ + CH₃COCl \( \xrightarrow{Anhyd.\ AlCl_3} \) C₆H₅COCH₃ (Acetophenone) + HCl
C₆H₆ + C₆H₅COCl \( \xrightarrow{Anhyd.\ AlCl_3} \) C₆H₅COC₆H₅ (Benzophenone) + HCl

Teacher's Note:
1. The electrophile is the acylium ion, R–C≡O⁺, generated by AlCl₃.

Question 18. Write giving chemical equations, a brief account of the following : Tollens’ reagent (CBSE 2010)
Answer:
Tollens’ reagent is ammoniacal silver nitrate solution, prepared by adding NH₄OH to AgNO₃ solution until the brown ppt. of Ag₂O first formed just dissolves. Aldehydes (aliphatic and aromatic) reduce it to metallic silver, which deposits as a silver mirror (Tollens’ test); ketones do not.
RCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → RCOO⁻ + 2Ag↓ + 2H₂O + 4NH₃

Teacher's Note:
1. Formic acid also gives the silver mirror, because it contains an aldehyde-like H–C=O unit.

Question 19. Write giving chemical equations, a brief account of the following : Gattermann-Koch reaction (MP Board 2012 ; Pb. Board 2012 ; Raj. Board 2012)
Answer:
When a mixture of carbon monoxide and hydrogen chloride is passed through benzene (or its derivative) in presence of anhydrous AlCl₃ and CuCl, benzaldehyde is formed.
C₆H₆ + CO + HCl \( \xrightarrow{Anhyd.\ AlCl_3/CuCl} \) C₆H₅CHO + HCl

Teacher's Note:
1. CO + HCl together act like the unstable formyl chloride, HCOCl.

Question 20. Write giving chemical equations, a brief account of the following : Decarboxylation (CBSE 2011 ; HP Board 2011)
Answer:
Removal of CO₂ from a carboxylic acid is called decarboxylation. Sodium salts of carboxylic acids on heating with soda-lime (NaOH + CaO) give alkanes with one carbon atom less.
R–COONa + NaOH \( \xrightarrow{CaO,\ \Delta} \) R–H + Na₂CO₃ ; e.g., CH₃COONa + NaOH → CH₄ + Na₂CO₃
Electrolysis of aqueous sodium or potassium salts of carboxylic acids also brings about decarboxylation (Kolbe electrolysis) : 2CH₃COOK + 2H₂O → CH₃–CH₃ + 2CO₂ + 2KOH + H₂

Teacher's Note:
1. Soda-lime decarboxylation shortens the chain by one carbon; Kolbe electrolysis doubles the alkyl part.

Question 21. Write giving chemical equations, a brief account of the following : Haloform reaction (Hr. Board)
Answer:
Aldehydes and ketones having at least one CH₃CO– group, on treatment with sodium hypohalite (X₂ + NaOH), give haloform (CHX₃) and the sodium salt of an acid with one carbon atom less.
R–COCH₃ + 3NaOX → R–COCX₃ + 3NaOH ; R–COCX₃ + NaOH → R–COONa + CHX₃
e.g., CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃↓ (yellow) + CH₃COONa + 3NaI + 3H₂O

Teacher's Note:
1. With iodine the haloform is a yellow solid with a characteristic smell, which makes it a convenient test.

Question 22. Complete the following equations and name the products formed : HC≡CH + H₂O \( \xrightarrow{40\%\ H_2SO_4,\ 1\%\ HgSO_4} \)
Answer:
HC≡CH + H₂O \( \xrightarrow{40\%\ H_2SO_4,\ 1\%\ HgSO_4} \) [CH₂=CHOH] \( \xrightarrow{Tautomerism} \) CH₃CHO (Acetaldehyde)

Teacher's Note:
1. Hydration of any alkyne other than ethyne gives a ketone, not an aldehyde.

Question 23. Complete the following equations and name the products formed : C₆H₅COCl \( \xrightarrow{H_2,\ Pd\text{-}BaSO_4} \) (CBSE Sample Paper 2019-20)
Answer:
C₆H₅COCl + H₂ \( \xrightarrow{Pd\text{-}BaSO_4} \) C₆H₅CHO (Benzaldehyde) + HCl (Rosenmund reduction)

Teacher's Note:
1. Benzoyl chloride is the acid chloride of benzoic acid; Rosenmund reduction converts –COCl into –CHO.

Question 24. Complete the following equations and name the products formed : C₆H₆ + RCOCl \( \xrightarrow{anhydrous\ AlCl_3} \)
Answer:
C₆H₆ + RCOCl \( \xrightarrow{Anhyd.\ AlCl_3} \) C₆H₅COR (Alkyl phenyl ketone) + HCl, e.g., with R = CH₃, acetophenone.

Teacher's Note:
1. This is Friedel-Crafts acylation; the product ketone does not react further.

Question 25. Complete the following equations and name the products formed : CH₃CHO \( \xrightarrow{HCN} \) (West Bengal Board 2013)
Answer:
CH₃CHO + HCN → CH₃CH(OH)CN (Acetaldehyde cyanohydrin or 2-Hydroxypropanenitrile)

Teacher's Note:
1. Hydrolysis of this cyanohydrin gives lactic acid (2-hydroxypropanoic acid).

Question 26. Complete the following equations and name the products formed : CH₃COCH₃ + CH₃MgBr → Product \( \xrightarrow{H_2O} \)
Answer:
CH₃COCH₃ + CH₃MgBr → (CH₃)₃C–OMgBr \( \xrightarrow{H_2O} \) (CH₃)₃C–OH (2-Methylpropan-2-ol) + Mg(OH)Br

Teacher's Note:
1. Ketone + RMgX gives a 3° alcohol after hydrolysis.

Question 27. Complete the following equations and name the products formed : Cyclohexanone + H₂NOH \( \xrightarrow{H^+} \) (CBSE 2014)
Answer:
Cyclohexanone + H₂NOH \( \xrightarrow{H^+} \) Cyclohexanone oxime (the ring C=O becomes C=N–OH) + H₂O

Teacher's Note:
1. Cyclohexanone oxime is the starting material for nylon-6 (via caprolactam).

Question 28. Complete the following equations and name the products formed : CH₃CHO + NH₂NH₂ (HP Board 2011)
Answer:
CH₃CHO + H₂NNH₂ → CH₃CH=NNH₂ (Acetaldehyde hydrazone) + H₂O

Teacher's Note:
1. The C=O oxygen leaves with the two H atoms of NH₂ as water.

Short Answer Questions (3 Marks)

Question 29. Write structures of compounds A, B and C in each of the following reactions:
(i) C₆H₅Br → [Mg/dry ether] → A → [(a) CO₂(g) (b) H₃O⁺] → B → [PCl₅] → C
(ii) CH₃CN → [(a) SnCl₂/HCl (b) H₃O⁺] → A → [dil. NaOH] → B → [Δ] → C [CBSE Delhi Set-1, 2, 3 2017]
Answer:
(i) A: C₆H₅MgBr, B: C₆H₅COOH, C: C₆H₅COCl
(ii) A: CH₃CHO, B: CH₃CH(OH)CH₂CHO, C: CH₃CH=CHCHO

Teacher's Note:
1. SnCl₂/HCl followed by hydrolysis is Stephen reduction (nitrile → aldehyde); the aldehyde then undergoes aldol condensation.

Question 30. Give reasons:
(i) Propanone is less reactive than ethanal towards nucleophilic addition reactions.
(ii) O₂N–CH₂–COOH has lower pK_a value than CH₃COOH.
(iii) (CH₃)₂CH–CHO undergoes aldol condensation whereas (CH₃)₃C–CHO does not. [CBSE Foreign Set-1, 2, 3 2017]
Answer:
(i) Due to steric hindrance and +I effect caused by two alkyl groups in propanone.
(ii) Due to electron withdrawing nature of –NO₂ group which increases the acidic strength and decreases the pK_a value.
(iii) (CH₃)₂CH–CHO has one α-H atom whereas α-H atom is absent in (CH₃)₃C–CHO.
Detailed Answer:
(i) As alkyl group is an electron releasing group (+I inductive effect), electron releasing power of two alkyl groups in propanone is higher than that of one alkyl group in ethanal. The reduced positive charge on carbon in propanone discourages the attack of nucleophiles making propanone less reactive than ethanal. Also, presence of two alkyl groups increases the steric hindrance to the attack of nucleophile reducing the reactivity of carbonyl. Thus, propanone is less reactive than ethanal towards nucleophilic addition reactions.
(iii) Aldol condensation occurs due to presence of α-hydrogen. As (CH₃)₂CH–CHO has one α-H atom whereas (CH₃)₃C–CHO does not have α-H atom. Thus, (CH₃)₂CH–CHO undergoes aldol condensation whereas (CH₃)₃C–CHO does not.

Teacher's Note:
1. For ‘reactivity towards nucleophiles’ questions, always give both reasons – electronic (+I effect) and steric – for full marks.

Teacher's Note:
1. Name reactions are scored by the reagent: Pd-BaSO₄ = Rosenmund, SnCl₂/HCl on a nitrile = Stephen, X₂/red P = HVZ.

Question 32. Illustrate the following name reactions:
(i) Hell-Volhard-Zelinsky reaction
(ii) Wolff-Kishner reduction
(iii) Etard reaction
Answer:
(i) RCH₂COOH –[(i) X₂/Red P, (ii) H₂O]→ RCH(X)COOH   (X = Cl, Br)
(ii) >C=O –[(i) NH₂NH₂, (ii) KOH/ethylene glycol, heat]→ >CH₂ + N₂
(iii) C₆H₅CH₃ (toluene) –[(1) CrO₂Cl₂, CS₂; (2) H₃O⁺]→ C₆H₅CHO

Teacher's Note:
1. In Etard reaction the intermediate is a chromium complex; hydrolysis in step (2) is essential to get benzaldehyde.

Question 33. Predict the organic products of the following reactions:
(i) C₆H₅CH₂CH₃ –[KMnO₄, KOH, Δ]→
(ii) o-C₆H₄(COOH)₂ (phthalic acid) –[SOCl₂, Δ]→
(iii) C₆H₅CHO –[H₂NCONHNH₂]→
(iv) C₆H₆ + ........ → C₆H₅COC₆H₅
(v) 4-Oxocyclohexane-1-carbaldehyde –[[Ag(NH₃)₂]⁺]→
(vi) 2-Formylbenzoic acid (o-OHC–C₆H₄–COOH) –[NaCN/HCl]→
Answer:
(i) C₆H₅–COOK
(ii) o-C₆H₄(COCl)₂ (phthaloyl chloride)
(iii) C₆H₅CH=N–NHCONH₂
(iv) C₆H₅COCl/AlCl₃
(v) 4-Oxocyclohexane-1-carboxylate ion (the ring C=O is untouched; only –CHO is oxidised to –COO⁻)
(vi) o-HOOC–C₆H₄–CH(OH)CN (cyanohydrin)

Teacher's Note:
1. Tollens' reagent is selective: it oxidises an aldehyde group but leaves a ketone group in the same molecule unchanged.

Question 34. Write chemical reactions to effect the following transformations:
(i) Butan-1-ol → Butanoic acid
(ii) Benzyl alcohol to phenylethanoic acid
(iii) 3-Nitrobromobenzene to 3-nitrobenzoic acid
(iv) 4-Methylacetophenone to terephthalic acid
(v) Cyclohexene → Hexane-1,6-dioic acid
(vi) Butanal → Butanoic acid
Answer:
(i) CH₃CH₂CH₂CH₂OH –[CrO₃–H₂SO₄ (Jones reagent)]→ CH₃CH₂CH₂COOH
(ii) C₆H₅CH₂OH –[HBr]→ C₆H₅CH₂Br –[KCN]→ C₆H₅CH₂CN –[H₃O⁺, Δ]→ C₆H₅CH₂COOH
(iii) m-O₂N–C₆H₄–Br –[Mg, ether]→ m-O₂N–C₆H₄–MgBr –[CO₂ (dry ice)]→ m-O₂N–C₆H₄–COOMgBr –[H₃O⁺]→ m-O₂N–C₆H₄–COOH + Mg(Br)OH
(iv) p-CH₃–C₆H₄–COCH₃ –[KMnO₄/KOH]→ p-KOOC–C₆H₄–COOK (dipotassium benzene-1,4-dicarboxylate) –[dil. H₂SO₄]→ p-HOOC–C₆H₄–COOH (terephthalic acid)
(v) Cyclohexene –[KMnO₄, dil. H₂SO₄, Δ]→ HOOC–(CH₂)₄–COOH (hexane-1,6-dioic acid, adipic acid)
(vi) CH₃CH₂CH₂CHO –[ammoniacal AgNO₃ (Tollens' reagent)]→ CH₃CH₂CH₂COOH

Teacher's Note:
1. Count carbons first: (ii) needs one extra carbon, which is why the –CN step is used; (v) keeps all six carbons by breaking the ring.

Question 35. Draw the structure of the following derivatives:
(i) 2,4-Dinitrophenylhydrazone of C₆H₅CHO
(ii) Cyclopropanone oxime
(iii) Acetaldehyde dimethylacetal
(iv) Semicarbazone of cyclobutanone
(v) Ethylene ketal of hexan-3-one
(vi) Methyl hemiacetal of formaldehyde
Answer:
(i) C₆H₅–CH=N–NH–C₆H₃(NO₂)₂ (NO₂ groups at 2,4-positions of the ring)
(ii) (CH₂)₂C=N–OH (three-membered ring carbon bearing =NOH)
(iii) CH₃–CH(OCH₃)₂
(iv) (CH₂)₃C=N–NHCONH₂ (four-membered ring carbon bearing =N–NHCONH₂)
(v) CH₃–CH₂–C(–O–CH₂–CH₂–O–)–CH₂–CH₂–CH₃ (C-3 carries both O atoms of a five-membered dioxolane ring)
(vi) H–CH(OH)–OCH₃, i.e. HOCH₂OCH₃

Teacher's Note:
1. Hemiacetal = one –OH and one –OR on the same carbon; acetal/ketal = two –OR groups on the same carbon.

Question 36. Draw the structure of a carbonyl group and indicate:
(i) hybridized state of carbon
(ii) the σ and π bonds
(iii) the electrophilic and nucleophilic centres
Answer:
(i) >C^δ+=O^δ−; hybridized state of ‘C’ is sp².
(ii) The carbonyl carbon forms three σ bonds (two to the attached groups and one to O); the C=O double bond consists of one σ bond and one π bond.
(iii) The carbonyl carbon (δ+) is the electrophilic centre and the oxygen (δ−) is the nucleophilic centre.

Teacher's Note:
1. Nucleophiles attack the carbon and electrophiles (such as H⁺) attack the oxygen – this explains acid catalysis of nucleophilic addition.

Long Answer Questions (5 Marks)

Question 37. (i) Write the chemical reaction involved in Etard reaction.
(ii) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction: CH₃–CHO, C₆H₅COCH₃, HCHO
(iii) Why pK_a of Cl–CH₂–COOH is lower than the pK_a of CH₃COOH?
(iv) Write the product in the following reaction.
CH₃CH₂CH=CH–CH₂CN \( \xrightarrow[\text{2. H}_{2}\text{O}]{\text{1. (i-Bu)}_{2}\text{AlH}} \)
(v) A and B are two functional isomers of compound C₃H₆O. On heating with NaOH and I₂, isomer A forms yellow precipitate of iodoform whereas isomer B does not form any precipitate. Write the formulae of A and B. [CBSE OD Set-2 2016]
Answer:
(i) Etard reaction: C₆H₅CH₃ (Toluene) + CrO₂Cl₂ \( \xrightarrow{\text{CS}_{2}} \) C₆H₅CH(OCrOHCl₂)₂ (Chromium complex) \( \xrightarrow{\text{H}_{3}\text{O}^{+}} \) C₆H₅CHO (Benzaldehyde)
(ii) C₆H₅COCH₃ < CH₃CHO < HCHO
The reactivity of the compound towards nucleophilic addition reaction is directly proportional to electrophilic character of carbonyl carbon. In ketone, the +I group lowers the electrophilicity. Whereas, +I of methyl group in ethanal is less than that of –C₆H₅. Hence, ethanal is most reactive than acetophenone.
(iii) –Cl being electron withdrawing group stabilizes the ClCH₂COO⁻ anion and increases the acidic strength. Therefore, chloroacetic acid has lower pK_a value than acetic acid.
(iv) CH₃–CH₂–CH=CH–CH₂–CHO
(v) A = CH₃–CO–CH₃ (Propanone); B = C₂H₅CHO (Propanal)
CH₃–CO–CH₃ (A) + NaOH + I₂ → CH₃–COO⁻Na⁺ + CHI₃ (Yellow ppt.)
C₂H₅CHO (B) + NaOH + I₂ → No ppt.

Teacher's Note:
1. C₃H₆O carbonyl isomers are only propanal and propanone; the methyl ketone (propanone) is the one that gives iodoform, so read carefully which letter forms the precipitate.

Question 38. (i) Give reasons:
(a) HCHO is more reactive than CH₃–CHO towards addition of HCN.
(b) pKa of O₂N–CH₂–COOH is lower than that of CH₃–COOH.
(c) Alpha hydrogen of aldehydes and ketones is acidic in nature.
(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Ethanal and Propanal
(b) Pentan-2-one and Pentan-3-one
Answer:
(i) (a) Due to +I effect of methyl group in CH₃CHO.
CH₃CHO has a comparatively bulky group attached to carbonyl group than HCHO which hinders the attack of nucleophile to some extent. Also, CH₃ group in CH₃CHO decreases the positive charge on carbonyl carbon by +I effect to some extent which doesn’t take place in HCHO. Since, Nu attack is favourable with more positive charge and less hindrance at carbonyl carbon, hence HCHO is more reactive than CH₃CHO.
(b) Due to −I effect of nitro group in nitroacetic acid.
(c) Due to the strong electron withdrawing effect of the carbonyl group and resonance stabilisation of the conjugate base.
(ii) (a) Add NaOH and I₂ to both the compounds and heat, ethanal gives yellow ppt of iodoform.
(b) Add NaOH and I₂ to both the compounds and heat, pentan-2-one gives yellow ppt of iodoform.

Teacher's Note:
1. Pentan-2-one (CH₃COC₃H₇) is a methyl ketone, while pentan-3-one (C₂H₅COC₂H₅) is not; that is the basis of the test.

Question 39. (i) Write structure of the product(s) formed:
(a) CH₃–CH₂–COOH \( \xrightarrow{\text{Cl}_{2}\text{, red phosphorus}} \)
(b) C₆H₅COCl \( \xrightarrow{\text{H}_{2}\text{, Pd–BaSO}_{4}} \)
(c) 2HCHO \( \xrightarrow{\text{Conc. KOH}} \)
(ii) How will you bring the following conversions in not more than two steps:
(a) Propanone to propene
(b) Benzyl chloride to phenyl ethanoic acid [CBSE Comptt. Delhi/OD 2018]
Answer:
(i) (a) CH₃–CH(Cl)–COOH
(b) C₆H₅CHO
(c) CH₃OH + HCOOK
(ii) (a) CH₃COCH₃ \( \xrightarrow{\text{NaBH}_{4}} \) CH₃CH(OH)CH₃ \( \xrightarrow{\text{conc. H}_{2}\text{SO}_{4}\text{, 443K}} \) CH₃–CH=CH₂
(b) C₆H₅CH₂Cl \( \xrightarrow{\text{KCN}} \) C₆H₅CH₂CN \( \xrightarrow{\text{H}_{3}\text{O}^{+}} \) C₆H₅CH₂COOH

Teacher's Note:
1. HVZ halogenation keeps the –COOH group; only the α-hydrogen is replaced by halogen, so the product is still an acid.

Question 40. An aromatic compound ‘A’ of molecular formula C₇H₆O₂ undergoes a series of reactions as shown below. Write the structures of A, B, C, D and E in the following reactions:
(C₇H₆O₂) A \( \xrightarrow{\text{NH}_{3}\text{/heat}} \) C₆H₅CONH₂ \( \xrightarrow{\text{Br}_{2}\text{ + NaOH}} \) B \( \xrightarrow{\text{(CH}_{3}\text{CO)}_{2}\text{O}} \) C
C₆H₅CONH₂ \( \xrightarrow{\text{LiAlH}_{4}\text{/Ether}} \) D;   B \( \xrightarrow{\text{Br}_{2}\text{(aq)}} \) E [CBSE OD 2015]
Answer:
A = C₆H₅COOH (Benzoic acid)
B = C₆H₅NH₂ (Aniline)
C = C₆H₅–NH–CO–CH₃ (Acetanilide)
D = C₆H₅CH₂NH₂ (Benzylamine)
E = 2,4,6-Tribromoaniline

Teacher's Note:
1. Hoffmann bromamide degradation (Br₂ + NaOH) removes one carbon, whereas LiAlH₄ reduction of the amide keeps all carbons: compare B and D.

Question 41. Give names of the reagents that bring about the following transformations:
(i) Hexan-1-ol to hexanal
(ii) p-Fluorotoluene to p-fluorobenzaldehyde
(iii) Ethanenitrile to ethanal
(iv) Allyl alcohol to propenal
(v) Ethanoic acid to ethanol
Answer:
(i) [C₅H₅NH]⁺[CrO₃Cl]⁻ (PCC)
(ii) CrO₃ in the presence of acetic anhydride / 1. CrO₂Cl₂ 2. H₃O⁺
(iii) Diisobutyl aluminium hydride (DIBAL-H)
(iv) PCC
(v) LiAlH₄

Teacher's Note:
1. PCC is the reagent of choice whenever an alcohol must stop at the aldehyde stage, and it does not attack C=C double bonds (part iv).

Question 42. (i) A ketone A which undergoes haloform reaction gives compound B on reduction. B on heating with sulphuric acid gives compound C, which forms mono-ozonide D. The compound D on hydrolysis in presence of zinc dust gives only acetaldehyde. Write the structures and IUPAC names of A, B and C. Write down the reactions involved.
(ii) Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.
(a) PhMgBr and then H₃O⁺
(b) Tollens’ reagent. [CBSE Sample Paper 2017]
Answer:
(i) A = CH₃–CO–CH₂–CH₃ (Butan-2-one), B = CH₃–CH(OH)–CH₂–CH₃ (Butan-2-ol), C = CH₃–CH=CH–CH₃ (But-2-ene)
The reactions are as follows:
CH₃–CO–CH₂–CH₃ (A) \( \xrightarrow{LiAlH_4} \) CH₃–CH(OH)–CH₂–CH₃ (B) \( \xrightarrow{H_2SO_4\,(conc.)} \) CH₃–CH=CH–CH₃ (C) \( \xrightarrow{+O_3} \) ozonide (D) \( \xrightarrow{H_2O/Zn} \) 2CH₃–CHO (Acetaldehyde)
(ii) (a) C₆H₁₁–CHO (Cyclohexanecarbaldehyde) \( \xrightarrow[\text{Dry ether}]{Ph{-}MgBr} \) C₆H₁₁–CH(OMgBr)–Ph \( \xrightarrow{H_3O^+} \) C₆H₁₁–CH(OH)–Ph (Cyclohexylphenylcarbinol)
(b) C₆H₁₁–CHO + 2[Ag(NH₃)₂]⁺ (Tollens’ reagent) + 3OH⁻ → C₆H₁₁–COO⁻ (Cyclohexanecarboxylate ion) + 2Ag↓ (Silver mirror) + 4NH₃ + 2H₂O

Teacher's Note:
1. Only acetaldehyde on ozonolysis means the alkene is symmetrical, CH₃CH=CHCH₃; work backwards from there to the ketone.