CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids Important Questions Set 10

Very Short Answer Questions (1 Mark)

Question 1. Out of acetophenone and benzophenone, which gives iodoform test ? Write the reaction involved.
Answer:
Acetophenone (C₆H₅COCH₃) contains the group CH₃CO attached to carbon and hence gives iodoform test while benzophenone does not contain this group and hence does not give iodoform test.
C₆H₅COCH₃ + 3I₂ + 4NaOH → CHI₃ + C₆H₅COONa + 3NaI + 3H₂O
Acetophenone → Iodoform + Sod. benzoate
C₆H₅COC₆H₅ \( \xrightarrow{I_2/NaOH} \) No reaction. (Benzophenone)

Teacher's Note:
1. Balance check: 3 I₂ supply the three I atoms of CHI₃ and three I⁻ ions.

Question 2. What type of aldehydes and ketones undergo aldol condensation ?
Answer:
Aldehydes and ketones containing one or more α-hydrogen/s, i.e., acetaldehyde, acetone, etc.
2 CH₃CHO \( \xrightarrow{dil.\ NaOH} \) CH₃CH(OH)CH₂CHO \( \xrightarrow[-H_2O]{\Delta} \) CH₃CH=CHCHO
Ethanal → 3-Hydroxybutanal (Aldol) → But-2-enal

Teacher's Note:
1. The α-hydrogen is needed because dilute alkali first removes it to form the enolate (carbanion) that attacks a second molecule.

Question 3. (CH₃)₂CH–CHO undergoes aldol condensation whereas (CH₃)₃C–CHO does not. Give reasons. (CBSE Foreign 2017)
or (CH₃)₃C–CHO does not undergo aldol condensation. Comment. (CBSE Sample Paper 2017)
Answer:
Aldol condensation is characteristic of aldehydes containing α-hydrogen/s. Since (CH₃)₂CH–CHO contains α-hydrogen/s and hence it undergoes aldol condensation whereas (CH₃)₃C–CHO does not contain α-hydrogen/s and hence does not undergo aldol condensation.
(CH₃)₂C^αH–CHO : contains α-H and hence undergoes aldol condensation.
(CH₃)₃C^α–CHO : does not contain α-hydrogen and hence does not undergo aldol condensation.

Teacher's Note:
1. (CH₃)₃C–CHO has no α-H, so with concentrated alkali it undergoes the Cannizzaro reaction instead.

Question 4. Which of the following compounds will undergo cannizzaro reaction ? Benzaldehyde, Cyclohexanone, 2-methylpentanal (CBSE Sample Paper 2019-2020)
Answer:
Compounds which do not have α-hydrogen/s undergo Cannizzaro reaction.
Here, only benzaldehyde does not have α-hydrogen/s and hence undergo cannizzaro reaction.
2 C₆H₅CHO \( \xrightarrow{50\%\ NaOH} \) C₆H₅COONa + C₆H₅CH₂OH
Benzaldehyde → Sod. benzoate + Benzyl alcohol
In contrast, cyclohexanone (four α-H on C-2 and C-6) and 2-methylpentanal, CH₃CH₂CH₂–CH(CH₃)–CHO (one α-H) contain one or more α-hydrogen/s and hence do not undergo cannizzaro reaction.

Teacher's Note:
1. The carbon of the benzene ring attached to CHO carries no hydrogen, so benzaldehyde counts as having no α-H.

Question 5. Write the structures of all the possible compounds having the M.F. C₈H₈O which undergo Cannizzaro reaction.
Answer:
Since the compound with M.F. C₈H₈O, undergoes Cannizzaro reaction, therefore, it must be methyl substituted benzaldehyde. Three such aldehydes are :
CH₃–C₆H₄–CHO with CH₃ at the ortho, meta or para position to CHO, i.e., o-, m- and p-methylbenzaldehyde.

Teacher's Note:
1. C₆H₅CH₂CHO and C₆H₅COCH₃ also have this formula but have α-H, so they do not give the Cannizzaro reaction.

Question 6. How will you prepare methanol from formaldehyde without using a reducing agent ?
Answer:
Cannizzaro reaction is a disproportionation reaction in which one molecule of an aldehyde is reduced while the other is oxidised.
2 H–CHO + NaOH (50%) → CH₃OH + HCOONa
Formaldehyde → Methanol + Sod. formate
Here, a conc. solution of NaOH is used which is not a reducing agent.

Teacher's Note:
1. Half of the formaldehyde acts as the reducing agent for the other half; that is the point of a disproportionation.

Question 7. Write method of preparation of urotropine and give its use. (MP Board 2011)
Answer:
Hexamethylene tetramine is called urotropine. It is formed when formaldehyde reacts with ammonia.
6 HCHO + 4 NH₃ → (CH₂)₆N₄ + 6 H₂O
Formaldehyde → Urotropine
It is used as antiseptic for urinary tract infections.

Teacher's Note:
1. Remember the numbers 6 : 4 : 6 – six HCHO, four NH₃, giving C₆H₁₂N₄ and six water.

Question 8. Give the structure and the IUPAC name of the lowest molecular mass aliphatic monocarboxylic acid containing a chiral carbon. Or Write the structure of active valeric acid.
Answer:
CH₃CH₂–C*H(CH₃)–COOH (2-Methylbutanoic acid).

Teacher's Note:
1. A chiral carbon needs four different groups: here H, CH₃, C₂H₅ and COOH.

Question 9. Give the IUPAC name of the following compounds.
(i) the cyclic compound shown below
(ii) CH₃–CH(OH)–CH₂–COOH (CBSE 2014)

Answer:
(i) 2-Methylcyclopent-3-ene-1-carboxylic acid. (ii) 3-Hydroxybutan-1-oic acid.

Teacher's Note:
1. When COOH is attached to a ring, its carbon is not counted in the ring name; the ring carbon carrying it is C-1 and the suffix is ‘carboxylic acid’.

Question 10. Why are carboxylic acids called fatty acids ? (J & K Board 2011)
Answer:
Since the higher acids such as palmitic acid, stearic acid, oleic acid, etc., were first obtained by saponification of oils and fats.

Teacher's Note:
1. Palmitic acid is C₁₅H₃₁COOH and stearic acid is C₁₇H₃₅COOH.

Question 11. Name the reagent used in the following reaction :
C₆H₅–CH₂–CH₃ \( \xrightarrow{?} \) C₆H₅–COO⁻K⁺ (CBSE 2015)
Answer:
When an alkylbenzene is heated with KMnO₄ – KOH, the entire side chain is oxidised to COO⁻K⁺ irrespective of the length of alkyl side chain.
C₆H₅–CH₂–CH₃ \( \xrightarrow[\Delta]{KMnO_4-KOH} \) C₆H₅–COO⁻K⁺
Ethyl benzene → Pot. benzoate

Teacher's Note:
1. Side-chain oxidation needs at least one benzylic H; tert-butylbenzene is not oxidised this way.

Short Answer Questions (2 Marks)

Question 12. Why are carboxylic acids stronger acids than phenol? (Hr. Board 2013 ; CBSE 2014)
Answer:
Both carboxylate ion and phenoxide ion are stabilised by resonance. In carboxylate ion (R–COO⁻), the negative charge is delocalised over two equivalent, highly electronegative oxygen atoms. In phenoxide ion, the charge is delocalised over one oxygen atom and the less electronegative carbon atoms of the ring, through non-equivalent structures. Hence carboxylate ion is more stable than phenoxide ion, and carboxylic acids (pK_a about 4–5) are much stronger acids than phenol (pK_a 10).

Teacher's Note:
1. Experimental proof: carboxylic acids decompose NaHCO₃ with effervescence, phenol does not.

Question 13. Write balanced chemical equations for the following reactions :
(i) Thionyl chloride reacts with benzoic acid
(ii) Decarboxylation of malonic acid.
(iii) Acetic acid is heated with red phosphorus and HI.
Answer:
(i) C₆H₅COOH + SOCl₂ → C₆H₅COCl + SO₂↑ + HCl↑
(ii) HOOC–CH₂–COOH \( \xrightarrow{\Delta} \) CH₃COOH + CO₂
(iii) CH₃COOH + 6HI \( \xrightarrow{Red\ P,\ \Delta} \) CH₃CH₃ + 2H₂O + 3I₂

Teacher's Note:
1. SOCl₂ is preferred for making acid chlorides because both by-products are gases and escape.

Question 14. How will you obtain trichloroacetic acid from acetic acid ?
Answer:
By Hell-Volhard-Zelinsky reaction, using excess of chlorine in presence of red phosphorus :
CH₃COOH \( \xrightarrow[-HCl]{Cl_2/red\ P} \) ClCH₂COOH \( \xrightarrow[-HCl]{Cl_2/red\ P} \) Cl₂CHCOOH \( \xrightarrow[-HCl]{Cl_2/red\ P} \) Cl₃CCOOH

Teacher's Note:
1. All three α-hydrogens of acetic acid are replaced one by one; the amount of Cl₂ controls how far it goes.

Question 15. Convert (i) RCOOH → RCOCl (West Bengal Board 2012)
(ii) CH₃CH₂COOH → CH₃CH₂Br (West Bengal Board 2012)
Answer:
(i) RCOOH + SOCl₂ → RCOCl + SO₂ + HCl ; or RCOOH + PCl₅ → RCOCl + POCl₃ + HCl ; or 3RCOOH + PCl₃ → 3RCOCl + H₃PO₃
(ii) CH₃CH₂COOH \( \xrightarrow{AgNO_3\ (via\ NaOH)} \) CH₃CH₂COOAg \( \xrightarrow[(Hunsdiecker\ reaction)]{Br_2,\ CCl_4,\ \Delta} \) CH₃CH₂Br + CO₂ + AgBr

Teacher's Note:
1. The Hunsdiecker reaction gives an alkyl halide with one carbon less than the acid.

Question 16. What is ammonolysis of esters? Explain. (Karnataka Board 2012)
Answer:
The reaction of an ester with ammonia to form an amide and an alcohol is called ammonolysis of the ester. The NH₃ molecule (nucleophile) attacks the carbonyl carbon and replaces the –OR′ group by –NH₂.
RCOOR′ + NH₃ → RCONH₂ + R′OH
e.g., CH₃COOC₂H₅ + NH₃ → CH₃CONH₂ (Acetamide) + C₂H₅OH

Teacher's Note:
1. Compare with hydrolysis (water splits the ester) and alcoholysis (an alcohol exchanges the alkoxy group).

Question 17. How will you convert ? Acetic acid into ethane (J & K Board 2011)
Answer:
2CH₃COOH + 2KOH → 2CH₃COOK + 2H₂O ; 2CH₃COOK + 2H₂O \( \xrightarrow[\text{(Kolbe method)}]{Electrolysis} \) CH₃–CH₃ (at anode) + 2CO₂ + H₂ (at cathode) + 2KOH
Or CH₃COOH + 6HI \( \xrightarrow{Red\ P,\ \Delta} \) CH₃CH₃ + 2H₂O + 3I₂

Teacher's Note:
1. Kolbe electrolysis joins two CH₃ radicals, which is why a two-carbon acid gives ethane, not methane.

Question 18. How will you convert ? Acetic acid to ethyl alcohol (Jharkhand Board 2011)
Answer:
CH₃COOH \( \xrightarrow[(ii)\ H_3O^+]{(i)\ LiAlH_4/ether} \) CH₃CH₂OH

Teacher's Note:
1. B₂H₆ can also be used; NaBH₄ does not reduce carboxylic acids.

Question 19. How will you convert ? Propanoic acid to 1-propanol
Answer:
CH₃CH₂COOH \( \xrightarrow[(ii)\ H_3O^+]{(i)\ LiAlH_4/ether} \) CH₃CH₂CH₂OH

Teacher's Note:
1. Reduction of RCOOH always gives the 1° alcohol RCH₂OH.

Question 20. How will you convert ? Propanoic acid to lactic acid.
Answer:
CH₃CH₂COOH \( \xrightarrow[(HVZ\ reaction)]{Cl_2/red\ P} \) CH₃CHClCOOH \( \xrightarrow{aq.\ KOH,\ then\ H^+} \) CH₃CH(OH)COOH (Lactic acid)

Teacher's Note:
1. HVZ puts the halogen on the α-carbon, which is exactly where lactic acid has its OH group.

Question 21. How will you convert ? Benzoic acid to 3-bromobenzoic acid (CBSE 2012)
Answer:
C₆H₅COOH + Br₂ \( \xrightarrow{FeBr_3} \) m-BrC₆H₄COOH (3-Bromobenzoic acid) + HBr

Teacher's Note:
1. COOH is meta-directing, so a single bromination step gives the 3-bromo product.

Question 22. How will you convert ? Benzoic acid to benzaldehyde (CBSE 2008, 2010 ; Hr. Board 2012)
Answer:
C₆H₅COOH \( \xrightarrow[-SO_2,\ -HCl]{SOCl_2} \) C₆H₅COCl \( \xrightarrow[(Rosenmund\ reduction)]{H_2,\ Pd/BaSO_4} \) C₆H₅CHO

Teacher's Note:
1. Going through the acid chloride lets the reduction stop at the aldehyde stage.

Question 23. How will you convert ? Benzoic acid to benzene
Answer:
C₆H₅COOH + NaOH → C₆H₅COONa + H₂O ; C₆H₅COONa + NaOH \( \xrightarrow[(Decarboxylation)]{CaO,\ \Delta} \) C₆H₆ + Na₂CO₃

Teacher's Note:
1. Decarboxylation with soda-lime replaces –COONa by –H.

Question 24. How will you convert ? Benzoic acid to benzyl alcohol
Answer:
C₆H₅COOH \( \xrightarrow[(ii)\ H_3O^+]{(i)\ LiAlH_4/dry\ ether} \) C₆H₅CH₂OH

Teacher's Note:
1. The COOH carbon becomes the CH₂OH carbon; the ring is not affected.

Question 25. How will you convert ? Benzoic acid to benzyl ethanoate
Answer:
C₆H₅COOH \( \xrightarrow[dry\ ether]{LiAlH_4} \) C₆H₅CH₂OH (Benzyl alcohol) \( \xrightarrow[pyridine]{CH_3COCl} \) C₆H₅CH₂OCOCH₃ (Benzyl ethanoate)

Teacher's Note:
1. Pyridine removes the HCl formed and drives the acylation forward.

Question 26. How will you convert ? Benzoic acid to benzyl chloride
Answer:
C₆H₅COOH \( \xrightarrow[(ii)\ H_3O^+]{(i)\ LiAlH_4} \) C₆H₅CH₂OH \( \xrightarrow{SOCl_2\ or\ PCl_5} \) C₆H₅CH₂Cl

Teacher's Note:
1. Do not confuse benzyl chloride (C₆H₅CH₂Cl) with benzoyl chloride (C₆H₅COCl).

Question 27. How will you convert ? Benzoyl chloride to benzonitrile
Answer:
C₆H₅COCl \( \xrightarrow[-NH_4Cl]{NH_3\ (excess)} \) C₆H₅CONH₂ (Benzamide) \( \xrightarrow[(-H_2O)]{P_2O_5,\ \Delta\ or\ SOCl_2,\ \Delta} \) C₆H₅CN (Benzonitrile)

Teacher's Note:
1. Dehydration of a primary amide (loss of H₂O) gives a nitrile.

Question 28. How will you convert ? Benzoic acid to benzamide (CBSE 2007)
Answer:
C₆H₅COOH \( \xrightarrow{SOCl_2} \) C₆H₅COCl \( \xrightarrow[-NH_4Cl]{2NH_3} \) C₆H₅CONH₂
Or C₆H₅COOH + NH₃ → C₆H₅COONH₄ \( \xrightarrow[-H_2O]{\Delta} \) C₆H₅CONH₂

Teacher's Note:
1. The acid chloride route works at room temperature and gives a better yield.

Question 29. How will you convert ? Benzamide to benzoic acid (Hr. Board 2011)
Answer:
C₆H₅CONH₂ + H₂O \( \xrightarrow{H^+,\ \Delta} \) C₆H₅COOH + NH₄⁺

Teacher's Note:
1. With alkali the products are sodium benzoate and NH₃ gas; acidify to get benzoic acid.

Short Answer Questions (3 Marks)

Question 30. Give reasons for the following : (i) The boiling points of aldehydes and ketones are lower than those of the corresponding alcohols and the carboxylic acids.
(ii) Dipole moments of aldehydes and ketones are higher than those of alcohols.
(iii) Acetone is highly soluble in water but acetophenone is not.
(iv) Why do aldehydes have lower boiling points than the corresponding alcohols ?
Answer:
(i) Aldehydes and ketones do not have a H-atom attached to oxygen, so they cannot form intermolecular H-bonds; their molecules are held only by weaker dipole-dipole attractions. Alcohols and carboxylic acids are associated by strong intermolecular H-bonding (acids even form dimers). Hence aldehydes and ketones have lower boiling points.
(ii) In the C=O group, the mobile π-electrons are strongly pulled towards oxygen (>C=O ↔ >C⁺–O⁻), so the charge separation is large. In alcohols, only the σ-electrons of C–O and O–H bonds are displaced. Hence aldehydes and ketones (e.g., acetone 2.88 D) have higher dipole moments than alcohols (e.g., ethanol 1.69 D).
(iii) Acetone forms H-bonds with water through its C=O group and its hydrocarbon part is small, so it is miscible with water. In acetophenone, the large hydrophobic phenyl group decreases the extent of H-bonding with water and hence it is insoluble.
(iv) Alcohols have intermolecular H-bonding while aldehydes do not (no H attached to O); only dipole-dipole attractions are present between aldehyde molecules, which are weaker. Hence aldehydes have lower boiling points.

Teacher's Note:
1. Higher dipole moment does not mean higher boiling point: H-bonding outweighs dipole-dipole forces, so alcohols still boil higher.

Question 31. What do you mean by a nucleophilic addition reaction ? Give its mechanism.
Answer:
Nucleophilic addition reaction is an addition reaction in which the first step is the attack of a nucleophile on the electrophilic carbonyl carbon; a proton (or other electrophile) then adds to the oxygen.
Mechanism : Step 1. The nucleophile (Nu⁻) attacks the electrophilic carbon of the polar C=O group from a direction roughly perpendicular to the plane of the sp² carbon. The π-electrons shift to oxygen and a tetrahedral alkoxide intermediate is formed; the carbon changes from sp² to sp³. This step is slow and rate-determining.
>C^δ+=O^δ− + Nu⁻ \( \xrightarrow{Slow} \) >C(Nu)–O⁻ (Tetrahedral intermediate)
Step 2. The alkoxide ion picks up a proton from the medium to give the neutral addition product.
>C(Nu)–O⁻ + H⁺ \( \xrightarrow{Fast} \) >C(Nu)–OH (Addition product)
For example, with HCN : CN⁻ adds to carbon and H⁺ to oxygen to give a cyanohydrin, >C(OH)CN.

Teacher's Note:
1. The rate-determining step is the attack of the nucleophile, so anything that makes the carbonyl carbon more positive or less crowded speeds up the reaction.

Question 32. Explain the mechanism of a nucleophilic attack on the carbonyl group of an aldehyde or a ketone. (CBSE 2010)
Answer:
Mechanism : Step 1. The nucleophile (Nu⁻) attacks the electrophilic carbon of the polar C=O group from a direction roughly perpendicular to the plane of the sp² carbon. The π-electrons shift to oxygen and a tetrahedral alkoxide intermediate is formed; the carbon changes from sp² to sp³. This step is slow and rate-determining.
>C^δ+=O^δ− + Nu⁻ \( \xrightarrow{Slow} \) >C(Nu)–O⁻ (Tetrahedral intermediate)
Step 2. The alkoxide ion picks up a proton from the medium to give the neutral addition product.
>C(Nu)–O⁻ + H⁺ \( \xrightarrow{Fast} \) >C(Nu)–OH (Addition product)
For example, with HCN : CN⁻ adds to carbon and H⁺ to oxygen to give a cyanohydrin, >C(OH)CN.

Teacher's Note:
1. In the answer, show the change of hybridisation (sp² → sp³) and the tetrahedral intermediate; examiners look for both.

Question 33. (a) Explain why aldehydes undergo nucleophilic addition reactions more readily than ketones ? (Pb. Board 2012)
(b) Arrange the following in increasing order of reactivity towards nucleophilic addition : HCHO ; CH₃CHO ; CH₃COCH₃.
Answer:
(a) (i) Electronic factor : Ketones have two electron-releasing alkyl groups (+I-effect) on the carbonyl carbon while aldehydes have only one. Hence the positive charge on the carbonyl carbon, and so its electrophilic character, is less in ketones than in aldehydes.
(ii) Steric factor : Two alkyl groups in ketones crowd the carbonyl carbon more than one alkyl group and a small H-atom in aldehydes, so the approach of the nucleophile is more hindered in ketones; the tetrahedral intermediate is also more crowded.
Hence aldehydes are more reactive than ketones.
(b) CH₃COCH₃ < CH₃CHO < HCHO

Teacher's Note:
1. HCHO, with two H-atoms and no alkyl group, is the most reactive carbonyl compound of all.

Question 34. Account for the following :
(i) Ethanal is more reactive towards nucleophilic addition reactions than propanone. or Propanone is less reactive than ethanal towards nucleophilic addition reactions. (CBSE (Foreign) 2017) or CH₃CHO is more reactive than CH₃COCH₃. (CBSE 2014)
(ii) Di-tert-butyl ketone does not give a NaHSO₃ adduct but acetone does.
(iii) Aldehydes are more reactive than ketones towards nucleophilic addition reactions.
(iv) HCHO reacts with HCN faster than CH₃CHO.
(v) Benzaldehyde is less reactive than acetaldehyde towards nucleophilic addition reactions.
Answer:
(i) In propanone, two CH₃ groups (+I-effect) reduce the positive charge on the carbonyl carbon more than the single CH₃ group of ethanal; they also hinder the approach of the nucleophile more. Hence ethanal is more reactive than propanone.
(ii) Due to steric hindrance caused by the bulky tert-butyl groups, the bisulphite ion cannot easily approach the carbonyl group for addition. In acetone, the small CH₃ groups allow the addition.
(iii) Aldehydes have only one alkyl group on the carbonyl carbon while ketones have two. Due to smaller +I-effect and less steric hindrance, the carbonyl carbon of aldehydes is more electrophilic and more easily approached by nucleophiles. Hence aldehydes are more reactive.
(iv) In CH₃CHO, the +I-effect of CH₃ decreases the positive charge on the carbonyl carbon and the CH₃ group also causes some steric hindrance. HCHO has no alkyl group, so its carbonyl carbon is more electrophilic and less hindered. Hence HCHO reacts with HCN faster.
(v) In benzaldehyde, the +R-effect of the benzene ring (C₆H₅–CH=O ↔ ⁺C₆H₅=CH–O⁻) reduces the positive charge on the carbonyl carbon much more than the +I-effect of CH₃ in acetaldehyde. Hence benzaldehyde is less reactive.

Teacher's Note:
1. For every reactivity comparison of carbonyl compounds, give the electronic reason and the steric reason in separate sentences.

Question 35. Write chemical equations when :
(i) Acetone reacts with ethanol.
(ii) Ethanal reacts with HCN.
(iii) Acetone reacts with HCN.
(iv) Acetone reacts with NaHSO₃.
(v) Ethanal reacts with NaHSO₃.
(vi) Acetaldehyde reacts with C₂H₅MgBr.
(vii) Cyclohexanone + HCN → (CBSE 2017)
(viii) Benzaldehyde \( \xrightarrow{NaCN/HCl} \) (CBSE 2019)
Answer:
(i) (CH₃)₂C=O + 2C₂H₅OH \( \xrightarrow{Dry\ HCl} \) (CH₃)₂C(OC₂H₅)₂ + H₂O (Acetone diethyl ketal)
(ii) CH₃CHO + HCN → CH₃CH(OH)CN (Ethanal cyanohydrin or 2-Hydroxypropanenitrile)
(iii) (CH₃)₂C=O + HCN → (CH₃)₂C(OH)CN (Acetone cyanohydrin or 2-Hydroxy-2-methylpropanenitrile)
(iv) (CH₃)₂C=O + NaHSO₃ → (CH₃)₂C(OH)SO₃Na (Acetone sodium bisulphite adduct)
(v) CH₃CHO + NaHSO₃ → CH₃CH(OH)SO₃Na (Acetaldehyde sodium bisulphite adduct)
(vi) CH₃CHO + C₂H₅MgBr \( \xrightarrow{Dry\ ether} \) CH₃CH(OMgBr)C₂H₅ \( \xrightarrow{H_3O^+} \) CH₃CH(OH)CH₂CH₃ (Butan-2-ol) + Mg(OH)Br
(vii) Cyclohexanone + HCN → Cyclohexanone cyanohydrin (1-Hydroxycyclohexane-1-carbonitrile), the ring carbon carrying both OH and CN.
(viii) C₆H₅CHO + HCN (from NaCN/HCl) → C₆H₅CH(OH)CN (Benzaldehyde cyanohydrin or 2-Hydroxy-2-phenylethanenitrile)

Teacher's Note:
1. Addition of HCN is catalysed by a base: CN⁻ is the actual nucleophile, which is why NaCN with a little acid is used.

Question 36. (a) Sodium bisulphite is used for the purification of aldehydes and ketones. Explain. (Jharkhand Board 2011)
(b) Benzophenone does not react with NaHSO₃.
Answer:
(a) Aldehydes and ketones (methyl ketones) react with saturated NaHSO₃ solution to form crystalline bisulphite addition compounds, which separate out and can be filtered off from the non-carbonyl impurities. On treatment with dilute HCl or Na₂CO₃ solution, the adduct decomposes and the pure aldehyde or ketone is regenerated.
>C=O + NaHSO₃ ⇌ >C(OH)SO₃Na (crystalline); >C(OH)SO₃Na \( \xrightarrow{dil.\ HCl\ or\ Na_2CO_3} \) >C=O
(b) In benzophenone, the two bulky phenyl groups cause steric hindrance to the approach of the large bisulphite ion; also, the +R-effect of the two phenyl groups reduces the positive charge on the carbonyl carbon. Hence benzophenone does not react with NaHSO₃.

Teacher's Note:
1. The reaction is reversible, which is exactly what makes bisulphite useful for purification.

Question 37. Write the names and structures of the products formed in the following reactions.
(i) Reaction of ethylmagnesium bromide with 2-butanone
(ii) Acetaldehyde is treated with methylmagnesium halide and then hydrolysed.
(iii) Propanone is treated with methylmagnesium iodide and then hydrolysed.
(iv) Ethanal is treated with ethanol in the presence of dry HCl gas.
Answer:
(i) CH₃COCH₂CH₃ + C₂H₅MgBr \( \xrightarrow{Dry\ ether} \) CH₃C(OMgBr)(C₂H₅)₂ \( \xrightarrow{H^+/H_2O} \) CH₃C(OH)(CH₂CH₃)₂ : 3-Methylpentan-3-ol
(ii) CH₃CHO + CH₃MgX \( \xrightarrow{Dry\ ether} \) CH₃CH(OMgX)CH₃ \( \xrightarrow{H^+/H_2O} \) CH₃CH(OH)CH₃ : Propan-2-ol
(iii) CH₃COCH₃ + CH₃MgI \( \xrightarrow{Dry\ ether} \) (CH₃)₃COMgI \( \xrightarrow{H^+/H_2O} \) (CH₃)₃COH : 2-Methylpropan-2-ol
(iv) CH₃CHO + C₂H₅OH \( \xrightarrow{Dry\ HCl} \) CH₃CH(OH)OC₂H₅ (Hemiacetal) \( \xrightarrow[-H_2O]{C_2H_5OH,\ dry\ HCl} \) CH₃CH(OC₂H₅)₂ : 1, 1-Diethoxyethane (Acetaldehyde diethyl acetal)

Teacher's Note:
1. Dry HCl both protonates the carbonyl oxygen and removes the water formed, pushing the acetal equilibrium forward.

Long Answer Questions (5 Marks)

Question 38. (i) Write structure of the product(s) formed:
(a) CH₃–CH₂–COOH → [Cl₂, red phosphorus] →
(b) C₆H₅COCl → [H₂, Pd–BaSO₄] →
(c) 2HCHO → [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 → [NaBH₄] → CH₃CH(OH)CH₃ → [conc. H₂SO₄, 443 K] → CH₃–CH=CH₂
(b) C₆H₅CH₂Cl → [KCN] → C₆H₅CH₂CN → [H₃O⁺] → C₆H₅CH₂COOH

Teacher's Note:
1. Cyanide substitution followed by hydrolysis adds one carbon as –COOH, turning benzyl chloride (C₇) into phenylethanoic acid (C₈).

Question 39. 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 → [NH₃/heat] → C₆H₅CONH₂ → [Br₂ + NaOH] → B → [(CH₃CO)₂O] → C
C₆H₅CONH₂ → [LiAlH₄/ether] → D
B → [Br₂(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 (Br at both ortho positions and the para position to NH₂)

Teacher's Note:
1. Br₂ + NaOH on an amide is Hoffmann bromamide degradation (loses one carbon), while LiAlH₄ reduces the amide to an amine keeping all carbons.

Question 40. Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction:
(i) Methanal   (ii) 2-Methylpentanal   (iii) Benzaldehyde
(iv) Benzophenone   (v) Cyclohexanone   (vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde   (viii) Butan-1-ol   (ix) 2,2-Dimethylbutanal
Answer:
(i), (iii) and (ix) will give Cannizzaro reaction due to absence of α-hydrogen.
(ii), (v), (vi) and (vii) will give aldol condensation due to presence of α-hydrogen.
(iv) and (viii) will neither undergo Cannizzaro reaction nor aldol condensation, because benzophenone is a ketone without α-hydrogen and butan-1-ol is an alcohol, not an aldehyde.
Cannizzaro products:
(i) 2HCHO –[conc. NaOH]→ CH₃OH + HCOONa
(iii) 2C₆H₅CHO –[conc. NaOH]→ C₆H₅CH₂OH + C₆H₅COONa
(ix) 2CH₃CH₂C(CH₃)₂CHO –[conc. NaOH]→ CH₃CH₂C(CH₃)₂CH₂OH + CH₃CH₂C(CH₃)₂COONa
Aldol products (dil. NaOH):
(ii) 2CH₃CH₂CH₂CH(CH₃)CHO → CH₃CH₂CH₂CH(CH₃)–CH(OH)–C(CH₃)(CHO)–CH₂CH₂CH₃
(v) 2 Cyclohexanone → 2-(1-hydroxycyclohexyl)cyclohexan-1-one, which on heating loses water to give 2-cyclohexylidenecyclohexan-1-one
(vi) 2C₆H₅COCH₂CH₃ → C₆H₅–C(OH)(CH₂CH₃)–CH(CH₃)–CO–C₆H₅
(vii) 2C₆H₅CH₂CHO → C₆H₅CH₂–CH(OH)–CH(C₆H₅)–CHO, which on heating gives C₆H₅CH₂–CH=C(C₆H₅)–CHO

Teacher's Note:
1. In an aldol product, the α-carbon of one molecule bonds to the carbonyl carbon of the other; draw this new C–C bond first, then put –OH on the old carbonyl carbon.

Question 41. An organic compound ‘A’ (C₃H₆O) is resistant to oxidation but forms compound ‘B’ (C₃H₈O) on reduction. ‘B’ reacts with HBr to form the compound ‘C’. ‘C’ with Mg forms Grignard's reagent ‘D’ which reacts with ‘A’ to form a product which on hydrolysis gives ‘E’. Identify ‘A’ to ‘E’.
Answer:
‘A’ must be a ketone (resistant to oxidation).
CH₃–CO–CH₃ (A) –[Red.]→ CH₃–CH(OH)–CH₃ (B) –[HBr]→ CH₃–CH(Br)–CH₃ (C) –[Mg, dry ether]→ CH₃–CH(MgBr)–CH₃ (D)
CH₃–CO–CH₃ (A) + (CH₃)₂CH–MgBr (D) → (CH₃)₂CH–C(CH₃)₂–OMgBr –[H₂O/H⁺]→ (CH₃)₂CH–C(CH₃)₂–OH (E)
A = propanone, B = propan-2-ol, C = 2-bromopropane, D = isopropylmagnesium bromide, E = 2,3-dimethylbutan-2-ol

Teacher's Note:
1. Ketone + Grignard reagent always gives a 3° alcohol after hydrolysis – a quick check on E.

Question 42. (a) Write the products of the following reactions:
(i) Cyclohexanone + H₂N–OH –[H⁺]→
(ii) 2C₆H₅CHO + conc. NaOH →
(iii) CH₃COOH –[Cl₂/P]→
(b) Give simple tests to distinguish between the following pairs of compounds.
(i) Benzaldehyde and Benzoic acid
(ii) Propanal and Propanone
Answer:
(a) (i) Cyclohexanone oxime, C₆H₁₀=N–OH
(ii) C₆H₅CH₂OH + C₆H₅COO⁻Na⁺
(iii) Cl–CH₂COOH
(b) (i) Add NaHCO₃; benzoic acid will give brisk effervescence (CO₂) whereas benzaldehyde will not give this test.
(ii) Silver mirror is formed on adding ammoniacal silver nitrate (Tollens' reagent) to propanal and not with propanone.

Teacher's Note:
1. Propanone also gives the iodoform test while propanal does not – a second valid way to distinguish them.