CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids Important Questions Set 09

Very Short Answer Questions (1 Mark)

Question 1. How will you convert propanone to propan-2-ol ? (CBSE 2013)
Answer:
Propanone can be reduced to propan-2-ol by a number of reagents such as catalytic hydrogenation with H₂ in presence of Ni, Pt or Pd, LiAlH₄ or NaBH₄.
(CH₃)₂C=O \( \xrightarrow[or\ LiAlH_4\ or\ NaBH_4]{H_2 + Ni/Pt\ or\ Pd} \) (CH₃)₂CH–OH
Propanone → Propan-2-ol

Teacher's Note:
1. Reduction of a ketone gives a 2° alcohol, while reduction of an aldehyde gives a 1° alcohol.

Question 2. Name two methods which are commonly used to convert a >C=O group into a >CH₂ group.
Answer:
Clemmensen reduction and Wolff-Kishner reduction.
>C=O \( \xrightarrow[(Clemmensen\ reduction)]{Zn\text{-}Hg/conc.\ HCl} \) >CH₂ + H₂O
>C=O \( \xrightarrow[-H_2O]{NH_2NH_2} \) >C=NNH₂ \( \xrightarrow[(Wolff\text{-}Kishner\ reduction)]{KOH/ethylene\ glycol,\ \Delta} \) >CH₂ + N₂

Teacher's Note:
1. Use Clemmensen for compounds stable to acid and Wolff-Kishner for compounds stable to base.

Question 3. Predict the product of the following reaction :
CH₃–CH₂–CHO \( \xrightarrow[(ii)\ KOH/Glycol,\ \Delta]{(i)\ H_2N\text{-}NH_2} \) (CBSE 2019)
Answer:
Wolff-Kishner reduction reduces >C=O to >CH₂ group
CH₃–CH₂–CHO \( \xrightarrow[(ii)\ KOH/Glycol,\ \Delta]{(i)\ H_2N\text{-}NH_2} \) CH₃CH₂–CH₃
Propanal → Propane

Teacher's Note:
1. The carbon skeleton is unchanged: a three-carbon aldehyde gives the three-carbon alkane.

Question 4. Give the chemical reaction when ethanal is heated with hydrogen iodide and red phosphorus under high pressure.
Answer:
Aldehydes are reduced to alkanes.
CH₃CHO + 4HI \( \xrightarrow{red\ P,\ pressure} \) CH₃CH₃ + 2I₂ + H₂O
Ethanal → Ethane

Teacher's Note:
1. Red phosphorus converts the I₂ formed back into HI, so a smaller amount of HI is enough.

Question 5. Write the structure of the main product of the following reaction, when the compound shown is treated with NaBH₄. (CBSE 2018)

Answer:
NaBH₄ is a mild reducing agent, it reduces ketones to corresponding alcohols but do not reduce esters. Thus,
Ethyl (2-oxocyclohexyl)acetate \( \xrightarrow{NaBH_4} \) Ethyl (2-hydroxycyclohexyl)acetate, i.e., the ring C=O becomes CH–OH while the –CH₂–CO–OC₂H₅ group stays unchanged.

Teacher's Note:
1. NaBH₄ reduces aldehydes and ketones but leaves esters, acids and C=C bonds untouched; LiAlH₄ would also reduce the ester.

Question 6. What is Fehling’s solution ?
Answer:
Alkaline solution of CuSO₄ containing Rochelle salt is called Fehling’s solution. It is obtained by adding an alkaline solution of Rochelle salt (sodium potassium tartrate) called Fehling solution ‘B’ to an aqueous solution of CuSO₄ called Fehling solution ‘A’ until blue ppt. of Cu(OH)₂ first formed redissolve to form a deep blue solution.

Teacher's Note:
1. Tartrate complexes Cu²⁺ so that it stays in solution in the alkaline medium; the positive test is a red-brown ppt. of Cu₂O.

Question 7. Give the IUPAC name of the only aldehyde which undergoes iodoform test.
Answer:
Ethanal (CH₃CHO) contains the grouping CH₃CO– and hence undergoes iodoform test and gives yellow ppt. of iodoform (CHI₃) on treatment with I₂/NaOH.
CH₃CHO \( \xrightarrow{I_2/NaOH} \) CHI₃ + HCOONa
Ethanal → Iodoform + Sod. formate

Teacher's Note:
1. Ethanol also gives the iodoform test, because NaOI first oxidises it to ethanal.

Question 8. What type of ketones undergo iodoform test ?
Answer:
Methyl ketones, i.e., ketones containing the grouping, CH₃CO give iodoform test. For example, acetone, butanone, 2-pentanone, acetophenone, etc.
CH₃COCH₃ \( \xrightarrow{I_2/NaOH} \) CHI₃ + CH₃COONa
Acetone → Iodoform + Sod. acetate

Teacher's Note:
1. Look for a CH₃ directly attached to the C=O carbon; the other group may be alkyl or aryl.

Question 9. Predict the product of the following reaction :
C₆H₅–CO–CH₃ \( \xrightarrow{NaOH/I_2} \) ? + ? (CBSE 2015)
Answer:
NaOH/I₂ i.e., sodium hypoiodite (NaOI) oxidises methyl ketones to –COO⁻Na⁺ + CHI₃ – iodoform reaction.
C₆H₅–CO–CH₃ \( \xrightarrow[(Iodoform\ reaction)]{NaOH/I_2} \) C₆H₅–COO⁻Na⁺ + CHI₃
Acetophenone → Sod. benzoate + Iodoform

Teacher's Note:
1. The product acid has one carbon less than the methyl ketone; the lost CH₃ leaves as CHI₃.

Question 10. Name one reagent to distinguish between 2-pentanone and 3-pentanone.
Answer:
I₂/NaOH (Iodoform test). 2-Pentanone being a methyl ketone gives iodoform test while 3-pentanone does not.
CH₃CH₂CH₂COCH₃ + 3I₂ + 4NaOH → CHI₃↓ (yellow) + CH₃CH₂CH₂COONa + 3NaI + 3H₂O
CH₃CH₂COCH₂CH₃ \( \xrightarrow{I_2/NaOH} \) No yellow ppt.

Teacher's Note:
1. Both are ketones, so Tollens’ and Fehling’s tests are useless here; only the iodoform test separates them.

Short Answer Questions (2 Marks)

Question 11. Write the names and structures of the products formed in the following reactions :
(i) Oxidation of ethylbenzene with alkaline KMnO₄.
(ii) Propene reacts with CO and steam under pressure in presence of H₃PO₄.
Answer:
(i) C₆H₅CH₂CH₃ \( \xrightarrow{KMnO_4/KOH,\ \Delta} \) C₆H₅COOK (Potassium benzoate) \( \xrightarrow{H_3O^+} \) C₆H₅COOH (Benzoic acid)
(ii) CH₃CH=CH₂ + CO + H₂O \( \xrightarrow{H_3PO_4,\ high\ P,\ T} \) (CH₃)₂CH–COOH (2-Methylpropanoic acid)

Teacher's Note:
1. In (ii) the COOH group attaches to the carbon that would carry H⁺-generated positive charge (Markovnikov), giving the branched acid.

Question 12. How are the following conversions carried out ? Butanoic acid from 1-butanol (CBSE 2011)
Answer:
CH₃CH₂CH₂CH₂OH \( \xrightarrow{K_2Cr_2O_7/H_2SO_4\ or\ alk.\ KMnO_4,\ H_3O^+} \) CH₃CH₂CH₂COOH

Teacher's Note:
1. A strong oxidant is needed; PCC would stop at butanal.

Question 13. How are the following conversions carried out ? Acetophenone to benzoic acid
Answer:
C₆H₅COCH₃ \( \xrightarrow{I_2/NaOH} \) C₆H₅COONa + CHI₃ ; C₆H₅COONa \( \xrightarrow{H_3O^+} \) C₆H₅COOH
Or C₆H₅COCH₃ \( \xrightarrow{KMnO_4/KOH,\ \Delta} \) C₆H₅COOK \( \xrightarrow{H_3O^+} \) C₆H₅COOH

Teacher's Note:
1. The iodoform route works for any methyl ketone and removes exactly one carbon.

Question 14. How are the following conversions carried out ? Ethanol to ethanoic acid
Answer:
CH₃CH₂OH \( \xrightarrow{K_2Cr_2O_7/H_2SO_4} \) CH₃COOH

Teacher's Note:
1. The colour change orange (Cr₂O₇²⁻) to green (Cr³⁺) shows the oxidation has occurred.

Question 15. How are the following conversions carried out ? Methyl bromide to acetic acid
Answer:
CH₃Br + Mg \( \xrightarrow{Dry\ ether} \) CH₃MgBr \( \xrightarrow[(ii)\ H_3O^+]{(i)\ CO_2} \) CH₃COOH
Or CH₃Br \( \xrightarrow{KCN} \) CH₃CN \( \xrightarrow{H^+/H_2O} \) CH₃COOH

Teacher's Note:
1. Both routes add one carbon atom to the alkyl group.

Question 16. How are the following conversions carried out ? Benzotrichloride to benzoic acid
Answer:
C₆H₅CCl₃ + 3H₂O \( \xrightarrow[-3HCl]{aq.\ NaOH\ or\ H_2O,\ \Delta} \) [C₆H₅C(OH)₃] \( \xrightarrow{-H_2O} \) C₆H₅COOH (with NaOH, the sodium salt is acidified)

Teacher's Note:
1. Three OH groups on one carbon are unstable; loss of water gives the COOH group.

Question 17. How are the following conversions carried out ? Ethyl chloride to propanoic acid
Answer:
CH₃CH₂Cl \( \xrightarrow[-KCl]{KCN\ (alc.)} \) CH₃CH₂CN \( \xrightarrow{H^+/H_2O} \) CH₃CH₂COOH

Teacher's Note:
1. Alternatively, C₂H₅MgCl + CO₂ followed by hydrolysis gives the same acid.

Question 18. How are the following conversions carried out ? Benzyl alcohol to benzoic acid
Answer:
C₆H₅CH₂OH \( \xrightarrow{KMnO_4/KOH,\ \Delta;\ then\ H_3O^+\ (or\ K_2Cr_2O_7/H_2SO_4)} \) C₆H₅COOH

Teacher's Note:
1. A 1° alcohol is oxidised first to the aldehyde and then to the acid.

Question 19. How are the following conversions carried out ? Carbon monoxide into formic acid
Answer:
CO + NaOH \( \xrightarrow{473\ K,\ 6\text{-}10\ atm} \) HCOONa \( \xrightarrow{dil.\ H_2SO_4\ or\ HCl} \) HCOOH

Teacher's Note:
1. Heating under pressure is needed because CO is not very reactive.

Question 20. Starting from methylmagnesium bromide, how will you synthesize acetic acid ?
Answer:
Methylmagnesium bromide reacts with dry ice (solid CO₂) in dry ether to form an addition product which on hydrolysis with dil. acid gives acetic acid.
CH₃MgBr + CO₂ \( \xrightarrow{Dry\ ether} \) CH₃COOMgBr \( \xrightarrow{H^+/H_2O} \) CH₃COOH + Mg(OH)Br

Teacher's Note:
1. Solid CO₂ is used in excess so that the Grignard reagent does not react with the product.

Question 21. How is Grignard reagent employed to prepare an acid ?
Answer:
Grignard reagents react with carbon dioxide (dry ice) to form salts of carboxylic acids, which on acidification give carboxylic acids with one carbon atom more than the alkyl group of the Grignard reagent.
R–MgX + O=C=O \( \xrightarrow{Dry\ ether} \) R–COOMgX \( \xrightarrow{H_3O^+} \) R–COOH + Mg(OH)X
e.g., C₂H₅MgBr → C₂H₅COOH (Propanoic acid)

Teacher's Note:
1. This method and the nitrile method both lengthen the carbon chain by one carbon.

Question 22. Formic acid (methanoic acid) is a stronger acid than acetic acid (ethanoic acid). Explain. (HP Board 2011 ; Assam Board 2012)
Answer:
In acetic acid, the CH₃ group has a +I-effect which increases the electron density on the carboxylate oxygens, i.e., intensifies the negative charge and destabilises the acetate ion (CH₃→COO⁻). In formic acid, there is no alkyl group (H is attached), so the formate ion (HCOO⁻) is more stable. Also, the O–H bond in acetic acid is made slightly stronger by the +I-effect. Hence formic acid (pK_a 3.75) is a stronger acid than acetic acid (pK_a 4.76).

Teacher's Note:
1. Formic acid is the strongest of the unsubstituted aliphatic monocarboxylic acids.

Question 23. Explain why pK_a of methanoic acid is lower than that of ethanoic acid.
Answer:
Lower pK_a means a stronger acid. In ethanoic acid, the electron-releasing CH₃ group (+I-effect) intensifies the negative charge on the ethanoate ion and destabilises it, while methanoate ion (HCOO⁻) has no such group. Hence methanoate ion is more stable, methanoic acid is the stronger acid, and its pK_a (3.75) is lower than that of ethanoic acid (4.76).

Teacher's Note:
1. Alkyl groups always lower acid strength of carboxylic acids, so HCOOH > CH₃COOH > CH₃CH₂COOH.

Question 24. Out of methanoic acid and ethanoic acid, which has higher pK_a value and why ?
Answer:
Ethanoic acid has the higher pK_a value. Due to the +I-effect of the CH₃ group, the negative charge on the ethanoate ion is intensified and the ion is destabilised, so ethanoic acid is the weaker acid. Methanoate ion has no electron-releasing alkyl group and is more stable. A weaker acid has a higher pK_a (ethanoic acid 4.76, methanoic acid 3.75).

Teacher's Note:
1. Higher pK_a = weaker acid; keep this inverse relation in mind before writing the answer.

Question 25. Justify : Dichloroacetic acid is a stronger acid than acetic acid.
Answer:
In dichloroacetic acid, the two Cl atoms exert a strong –I-effect which weakens the O–H bond and disperses the negative charge of the dichloroacetate ion (Cl₂CH←COO⁻), stabilising it. In acetic acid, the +I-effect of the CH₃ group intensifies the negative charge of the acetate ion and destabilises it. Hence dichloroacetic acid (pK_a 1.26) is a much stronger acid than acetic acid (pK_a 4.76).

Teacher's Note:
1. Each extra Cl on the α-carbon raises the acid strength further: CH₃COOH < ClCH₂COOH < Cl₂CHCOOH < Cl₃CCOOH.

Question 26. Compare the acid strengths of the following : (i) Formic acid (ii) Acetic acid (iii) Benzoic acid.
Answer:
Acetic acid is the weakest because the +I-effect of CH₃ destabilises the acetate ion. In benzoic acid, the COOH group is attached to the electron-withdrawing sp² carbon of the phenyl ring, so benzoate ion is more stable than acetate ion. Formic acid has no electron-releasing group at all and is the strongest. Thus, acid strength :
Formic acid (pK_a 3.75) > Benzoic acid (pK_a 4.19) > Acetic acid (pK_a 4.76)

Teacher's Note:
1. Quoting pK_a values is a quick way to justify such an order in the exam.

Question 27. Arrange the following compounds in increasing order of their acidity : (CH₃)₂CH–COOH, H–COOH, CH₃–COOH, (CH₃)₃C–COOH. (Raj. Board 2012)
Answer:
The +I-effect of the alkyl group increases in the order CH₃– < (CH₃)₂CH– < (CH₃)₃C–. Greater the +I-effect, less stable is the carboxylate ion and weaker is the acid. H has no +I-effect. Hence acidity increases in the order :
(CH₃)₃C–COOH < (CH₃)₂CH–COOH < CH₃–COOH < H–COOH

Teacher's Note:
1. More branching at the α-carbon means more alkyl groups pushing electrons, so the acid becomes weaker.

Short Answer Questions (3 Marks)

Question 28. An alkene A with molecular formula C₅H₁₀ on ozonolysis gives a mixture of two compounds B and C. Compound B gives positive Fehling’s test and also reacts with iodine and NaOH solution. Compound C does not give Fehling solution test but forms iodoform. Identify compounds A, B and C. (CBSE 2019)
Answer:
To determine the structure of alkene A with M.F. C₅H₁₀.
(i) Since compound B gives positive Fehling’s test, it must be an aldehyde.
(ii) Since compound B reacts with I₂ and NaOH solution, it must contain CH₃CO– group attached to H, i.e., B must be ethanal.
CH₃CH=O (B) →^{Fehling test} produces red ppt. of Cu₂O;   CH₃CHO →^I₂/NaOH CHI₃ (Iodoform)
(iii) Since compound C does not give Fehling solution test, it must be a ketone. Since it forms iodoform on treatment with I₂/NaOH, it must be a methyl ketone, i.e., acetone.
(iv) If B is ethanal and C is acetone, then the structure of the alkene (A) is 2-methylbut-2-ene.
CH₃CH=O (ethanal, B) + O=C(CH₃)₂ (acetone, C) →^{remove O atoms and join the two fragments with a d.b.} CH₃CH=C(CH₃)₂ (2-methylbut-2-ene, A)
Thus, A is 2-methylbut-2-ene, B is ethanal and C is acetone.

Teacher's Note:
1. To work back from ozonolysis products, delete the C=O oxygens and join the two carbonyl carbons by a double bond; check the carbon count (2 + 3 = 5).

Question 29. Give the IUPAC names of the following compounds :
(i) CH₃–CH(NO₂)–CO–CH(CH₃)–CH₃
(ii) (CH₃)₂C(OH)–CH₂–CO–CH₃ (Jharkhand Board 2011)
(iii) the compound shown in the figure
(iv) CH₃–CO–CO–CH₃
(v) HOOC–CH₂–COOH (Hr. Board 2013)
(vi) CH₃–CH(CHO)–CH₂–CH₃
(vii) CH₃–C≡C–CH₂–CHO
(viii) CH₃COCH₂COCH₃ (CBSE 2007)
(ix) CH₃–CH(CH₃)–CO–CH(CH₃)–CH₃ (CBSE 2007)
(x) C₆H₅–COCH₂CH₃ (CBSE 2010)
(xi) O₂N–C₆H₄–COCH₂CH₃ (NO₂ para to the side chain) (CBSE 2010 S)
(xii) CH₃–CH=CH–CHO (CBSE 2019)

Answer:
(i) 2-Methyl-4-nitropentan-3-one (ii) 4-Hydroxy-4-methylpentan-2-one
(iii) 2-(2-Bromophenyl)ethanal (iv) Butane-2, 3-dione (v) Propane-1, 3-dioic acid (vi) 2-Methylbutanal
(vii) Pent-3-yn-1-al (viii) Pentane-2, 4-dione (ix) 2, 4-Dimethylpentan-3-one (x) 1-Phenylpropan-1-one
(xi) 1-(4-Nitrophenyl)propan-1-one (xii) But-2-en-1-al

Teacher's Note:
1. In (vi) the CHO carbon must be C-1 of the longest chain containing it, so the compound is a butanal, not a propane derivative.

Question 30. Draw the structure of a carbonyl group and indicate clearly (i) hybridised state of carbon, (ii) σ- and π-bonds present and (iii) electrophilic and nucleophilic centres in it.
Answer:
In the carbonyl group, >C^δ+=O^δ− :
(i) The carbonyl carbon is sp²-hybridised. Its three sp² orbitals form three σ-bonds (one with O and two with the attached atoms) which lie in one plane at about 120°.
(ii) The C=O double bond consists of one σ-bond (sp²–p overlap) and one π-bond formed by sideways overlap of the unhybridised 2p orbital of carbon with a 2p orbital of oxygen, above and below the plane. Oxygen also carries two lone pairs.
(iii) The carbonyl carbon (C^δ+) is the electrophilic centre (attacked by nucleophiles) and the carbonyl oxygen (O^δ−) is the nucleophilic centre (attacked by electrophiles such as H⁺).

Teacher's Note:
1. A planar sp² carbon can be attacked from either face; this is why the nucleophile can approach easily from above or below.

Question 31. How will you obtain the following ?
(i) Acetone from acetic acid
(ii) acetaldehyde from acetic acid
(iii) Acetaldehyde from acetyl chloride (MP Board 2011)
(iv) Acetone from calcium acetate (MP Board 2011)
(v) Formaldehyde from methylene chloride. (Maharashtra Board 2012)
Answer:
(i) 2CH₃COOH + CaCO₃ → (CH₃COO)₂Ca + CO₂ + H₂O; (CH₃COO)₂Ca \( \xrightarrow{Dry\ distil.} \) CH₃COCH₃ + CaCO₃
(ii) CH₃COOH \( \xrightarrow[-SO_2,\ -HCl]{SOCl_2} \) CH₃COCl \( \xrightarrow{H_2,\ Pd/BaSO_4} \) CH₃CHO + HCl
(iii) CH₃COCl + H₂ \( \xrightarrow[(Rosenmund\ reduction)]{Pd/BaSO_4,\ S} \) CH₃CHO + HCl
(iv) (CH₃COO)₂Ca \( \xrightarrow{Dry\ distil.} \) CH₃COCH₃ + CaCO₃
(v) CH₂Cl₂ + 2NaOH (aq) \( \xrightarrow[-2NaCl]{\Delta} \) [CH₂(OH)₂] \( \xrightarrow{-H_2O} \) HCHO

Teacher's Note:
1. Two OH groups on one carbon (a gem-diol) are unstable and lose water to give the carbonyl compound; this is the basis of (v).

Question 32. How will you obtain
(i) Benzaldehyde from benzoyl chloride (CBSE 2013)
(ii) Acetophenone from benzene
(iii) Butanone from 2-butanol
(iv) Ethanal from 2-butene (CBSE 2009)
(v) Benzaldehyde from toluene
(vi) Acetaldehyde from acetylene? (MP Board 2013)
Answer:
(i) C₆H₅COCl + H₂ \( \xrightarrow[(Rosenmund\ reduction)]{Pd/BaSO_4} \) C₆H₅CHO + HCl
(ii) C₆H₆ + CH₃COCl \( \xrightarrow[(Friedel\text{-}Crafts\ acylation)]{Anhyd.\ AlCl_3} \) C₆H₅COCH₃ + HCl
(iii) CH₃CH(OH)CH₂CH₃ \( \xrightarrow{Cu,\ 573\ K\ or\ K_2Cr_2O_7/H_2SO_4} \) CH₃COCH₂CH₃
(iv) CH₃CH=CHCH₃ + O₃ → Ozonide \( \xrightarrow{Zn/H_2O} \) 2CH₃CHO + ZnO (Ozonolysis)
(v) C₆H₅CH₃ + CrO₂Cl₂ \( \xrightarrow{CS_2} \) Chromium complex \( \xrightarrow{H_3O^+} \) C₆H₅CHO (Étard reaction)
(vi) HC≡CH + H₂O \( \xrightarrow[333\ K]{40\%\ H_2SO_4,\ 1\%\ HgSO_4} \) [CH₂=CHOH] \( \xrightarrow{Tautomerism} \) CH₃CHO

Teacher's Note:
1. Zn dust in ozonolysis destroys the H₂O₂ formed, so the aldehyde is not oxidised further to an acid.

Question 33. How are the following conversions carried out ? (Write equations).
(i) Propyne into propanone
(ii) Acetyl chloride into acetaldehyde
(iii) Acetyl chloride to acetone (Raj. Board 2011)
(iv) Acetylene to acetaldehyde
(v) Ethanol to ethanal. (CBSE 2019)
(vi) Ethanol to propanone. (AIPMT Mains 2009)
(vii) PhCN to PhCOPh
(viii) R–CN to RCHO (West Bengal Board 2012)
(ix) acetophenone to 2-phenyl-2-butanol
(x) Propene to acetone
(xi) Cyclohexanol to cyclohexan-1-one. (CBSE 2010)
Answer:
(i) CH₃C≡CH + H₂O \( \xrightarrow{dil.\ H_2SO_4,\ HgSO_4} \) [CH₃C(OH)=CH₂] \( \xrightarrow{Tautomerism} \) CH₃COCH₃
(ii) CH₃COCl + H₂ \( \xrightarrow{Pd/BaSO_4} \) CH₃CHO + HCl
(iii) 2CH₃COCl + (CH₃)₂Cd \( \xrightarrow{Dry\ ether} \) 2CH₃COCH₃ + CdCl₂
(iv) HC≡CH + H₂O \( \xrightarrow{40\%\ H_2SO_4,\ 1\%\ HgSO_4} \) CH₃CHO
(v) CH₃CH₂OH \( \xrightarrow{PCC,\ CH_2Cl_2\ (or\ Cu,\ 573\ K)} \) CH₃CHO
(vi) CH₃CH₂OH \( \xrightarrow{K_2Cr_2O_7/H_2SO_4} \) CH₃COOH \( \xrightarrow{CaCO_3} \) (CH₃COO)₂Ca \( \xrightarrow{Dry\ distil.} \) CH₃COCH₃
(vii) C₆H₅CN + C₆H₅MgBr \( \xrightarrow{Dry\ ether} \) (C₆H₅)₂C=NMgBr \( \xrightarrow{H_3O^+} \) C₆H₅COC₆H₅ + NH₃ + Mg(OH)Br
(viii) R–CN + SnCl₂ + HCl → R–CH=NH \( \xrightarrow{H_3O^+} \) R–CHO + NH₃ (Stephen reaction); or R–CN \( \xrightarrow[(ii)\ H_2O]{(i)\ DIBAL\text{-}H} \) R–CHO
(ix) C₆H₅COCH₃ + C₂H₅MgBr \( \xrightarrow{Dry\ ether} \) C₆H₅C(CH₃)(C₂H₅)OMgBr \( \xrightarrow{H_3O^+} \) C₆H₅C(OH)(CH₃)CH₂CH₃ (2-Phenyl-2-butanol)
(x) CH₃CH=CH₂ + H₂O \( \xrightarrow{H^+} \) CH₃CH(OH)CH₃ \( \xrightarrow{Cu,\ 573\ K\ or\ CrO_3} \) CH₃COCH₃
(xi) Cyclohexanol \( \xrightarrow{PCC\ or\ K_2Cr_2O_7/H_2SO_4\ or\ Cu,\ 573\ K} \) Cyclohexan-1-one, i.e., the ring CH–OH becomes C=O.

Teacher's Note:
1. Markovnikov addition of water to propene puts the OH on C-2, so its oxidation gives a ketone (acetone), not an aldehyde.

Question 34. Write reactions stating conditions for the following conversions :
(i) Toluene to benzaldehyde.
(ii) Benzene to acetophonone. (Assam Board 2012 ; HP Board 2013)
(iii) Ethyl cyanide or propanenitrile to 1-phenylpropanone.
(iv) Benzene to benzaldehyde (CBSE 2011 ; Hr. Board 2011)
Answer:
(i) C₆H₅CH₃ \( \xrightarrow{CrO_2Cl_2,\ CS_2} \) C₆H₅CH(OCrOHCl₂)₂ \( \xrightarrow{H_3O^+} \) C₆H₅CHO (Étard reaction)
Or C₆H₅CH₃ \( \xrightarrow{Cl_2,\ h\nu} \) C₆H₅CHCl₂ \( \xrightarrow{H_2O,\ 373\ K} \) C₆H₅CHO
(ii) C₆H₆ + CH₃COCl \( \xrightarrow{Anhyd.\ AlCl_3} \) C₆H₅COCH₃ + HCl
(iii) CH₃CH₂CN + C₆H₅MgBr \( \xrightarrow{Ether} \) C₆H₅C(=NMgBr)CH₂CH₃ \( \xrightarrow{H_3O^+} \) C₆H₅COCH₂CH₃
(iv) C₆H₆ + CO + HCl \( \xrightarrow{Anhyd.\ AlCl_3/CuCl} \) C₆H₅CHO (Gattermann-Koch reaction)

Teacher's Note:
1. In (iii), the Grignard reagent supplies the phenyl group and the nitrile carbon becomes the ketone carbon.

Question 35. Complete the following equations giving the names of the reactants and the products :
(i) (CH₃COO)₂Ca \( \xrightarrow{heat} \) (Uttarakhand Board 2012 ; J & K Board 2012)
(ii) (HCOO)₂Ca \( \xrightarrow{Distil.} \) (Hr. Board 2011 ; Maharashtra Board 2012 ; West Bengal Board 2013)
(iii) C₆H₅CHCl₂ + NaOH (aq) \( \xrightarrow{heat} \)
(iv) CH₃COCl + H₂ \( \xrightarrow{Pd/BaSO_4} \) (CBSE 2013)
(v) C₆H₅COCl + (CH₃)₂Cd →
(vi) (C₆H₅CH₂)₂Cd + CH₃COCl → (CBSE 2019)
(vii) C₆H₅COCl \( \xrightarrow{H_2/Pd/BaSO_4} \) (CBSE 2009)
(viii) C₆H₆ + .......... \( \xrightarrow{Anhyd.\ AlCl_3} \) C₆H₅COCH₃ (CBSE 2009)
(ix) C₆H₆ + .......... → C₆H₅COC₆H₅ (CBSE 2009)
(x) C₆H₅CH₂CH₃ \( \xrightarrow[heat]{KMnO_4,\ KOH} \) (CBSE 2011, 2015)
(xi) C₆H₆ \( \xrightarrow[Anhyd.\ AlCl_3,\ CuCl]{CO,\ HCl} \) (Uttarakhand Board 2012 ; West Bengal 2012)
(xii) C₆H₆ + COCl₂ \( \xrightarrow{Anhyd.\ AlCl_3} \) (West Bengal Board 2012)
(xiii) CH₃CH=CHCH₂CN \( \xrightarrow[(ii)\ H_2O]{(i)\ (iBu)_2AlH\ or\ DIBAL\text{-}H} \) (CBSE 2016, 2017)
Answer:
(i) (CH₃COO)₂Ca (Calcium acetate) \( \xrightarrow{heat} \) CH₃COCH₃ (Acetone) + CaCO₃
(ii) (HCOO)₂Ca (Calcium formate) \( \xrightarrow{Distil.} \) HCHO (Formaldehyde) + CaCO₃
(iii) C₆H₅CHCl₂ (Benzal chloride) + 2NaOH (aq) \( \xrightarrow{heat} \) [C₆H₅CH(OH)₂] \( \xrightarrow{-H_2O} \) C₆H₅CHO (Benzaldehyde) + 2NaCl
(iv) CH₃COCl (Acetyl chloride) + H₂ \( \xrightarrow{Pd/BaSO_4} \) CH₃CHO (Acetaldehyde) + HCl
(v) 2C₆H₅COCl (Benzoyl chloride) + (CH₃)₂Cd (Dimethylcadmium) → 2C₆H₅COCH₃ (Acetophenone) + CdCl₂
(vi) (C₆H₅CH₂)₂Cd (Dibenzylcadmium) + 2CH₃COCl (Acetyl chloride) → 2C₆H₅CH₂COCH₃ (1-Phenylpropan-2-one) + CdCl₂
(vii) C₆H₅COCl (Benzoyl chloride) + H₂ \( \xrightarrow{Pd/BaSO_4} \) C₆H₅CHO (Benzaldehyde) + HCl
(viii) C₆H₆ (Benzene) + CH₃COCl (Acetyl chloride) \( \xrightarrow{Anhyd.\ AlCl_3} \) C₆H₅COCH₃ (Acetophenone) + HCl
(ix) C₆H₆ (Benzene) + C₆H₅COCl (Benzoyl chloride) \( \xrightarrow{Anhyd.\ AlCl_3} \) C₆H₅COC₆H₅ (Benzophenone) + HCl
(x) C₆H₅CH₂CH₃ (Ethylbenzene) \( \xrightarrow[heat]{KMnO_4,\ KOH} \) C₆H₅COO⁻K⁺ (Potassium benzoate) \( \xrightarrow{H_3O^+} \) C₆H₅COOH (Benzoic acid)
(xi) C₆H₆ (Benzene) + CO + HCl \( \xrightarrow{Anhyd.\ AlCl_3,\ CuCl} \) C₆H₅CHO (Benzaldehyde) (Gattermann-Koch reaction)
(xii) C₆H₆ (Benzene) + COCl₂ (Phosgene) \( \xrightarrow{Anhyd.\ AlCl_3} \) C₆H₅COCl (Benzoyl chloride) + HCl; with excess benzene : C₆H₅COCl + C₆H₆ → C₆H₅COC₆H₅ (Benzophenone) + HCl
(xiii) CH₃CH=CHCH₂CN (Pent-3-enenitrile) \( \xrightarrow[(ii)\ H_2O]{(i)\ DIBAL\text{-}H} \) CH₃CH=CHCH₂CHO (Pent-3-enal)

Teacher's Note:
1. In (x) the whole ethyl side chain is oxidised, giving benzoate, not phenylacetate.

Long Answer Questions (5 Marks)

Question 36. (i) Account for the following:
(a) CH₃CHO is more reactive than CH₃COCH₃ towards reaction with HCN.
(b) Carboxylic acid is stronger acid than phenol.
(ii) Write chemical reactions to illustrate the following name reactions:
(a) Wolff Kishner reduction
(b) Aldol condensation
(c) Cannizzaro reaction. [CBSE Delhi 2014]
Answer:
(i) (a) Because the positive charge on carbonyl carbon of CH₃CHO decreases to a lesser extent due to one electron releasing (+I effect) CH₃ group as compared to CH₃COCH₃ (two electron releasing CH₃ group) and hence more reactive.
(b) Because carboxylate ion (conjugate base) is more resonance stabilized than phenoxide ion.
(ii) (a) >C=O → [NH₂NH₂, −H₂O] → >C=NNH₂ → [KOH/ethylene glycol, heat] → >CH₂ + N₂
(b) CH₃–CHO + H–CH₂–CHO ⇌ [dil. NaOH] ⇌ CH₃–CH(OH)–CH₂–CHO (3-Hydroxybutanal, Aldol) → [Δ, −H₂O] → CH₃–CH=CH–CHO (But-2-enal)
(c) HCHO + HCHO + Conc. KOH → CH₃OH (Methyl alcohol) + HCOOK (Potassium formate)

Teacher's Note:
1. Wolff–Kishner works in strongly basic conditions, so it is the choice for carbonyl compounds that would be damaged by the hot HCl of Clemmensen reduction.

Question 37. (i) Illustrate the following name reactions giving suitable example in each case:
(a) Clemmensen reduction
(b) Hell Volhard-Zelinsky reaction.
(ii) How are the following conversions carried out?
(a) Ethylcyanide to ethanoic acid,
(b) Butan-1-ol to butanoic acid,
(c) Benzoic acid to m-bromobenzoic acid. [CBSE OD 2012]
Answer:
(i) (a) Clemmensen reduction: The carbonyl group of aldehydes and ketones is reduced to –CH₂ group on treatment with zinc amalgam and concentrated HCl.
>C=O → [Zn(Hg), Conc. HCl] → >CH₂
(b) Hell Volhard-Zelinsky reaction: Carboxylic acids having an α-hydrogen are halogenated at the α-position on treatment with chlorine or bromine in the presence of red phosphorus to give α-halocarboxylic acids.
R–CH₂–COOH → [X₂/P (X = Cl, Br)] → R–CH(X)–COOH (Or by any other method)
(ii) (a) CH₃CH₂CN → [H₃O⁺] → CH₃CH₂COOH → [NH₃, Δ] → CH₃CH₂CONH₂ → [Br₂/KOH] → CH₃CH₂NH₂ → [HNO₂] → CH₃CH₂OH → [KMnO₄] → CH₃COOH
(b) CH₃–CH₂–CH₂–CH₂–OH → [KMnO₄] → CH₃CH₂CH₂COOH
(c) C₆H₅COOH → [Br₂/FeBr₃] → m-BrC₆H₄COOH
(Or any other suitable method can be used for each conversion)

Teacher's Note:
1. Ethyl cyanide has three carbons and ethanoic acid two, so the route must lose one carbon – Hoffmann bromamide degradation does exactly that.

Question 38. (i) Write a suitable chemical equation to complete each of the following transformations:
(a) Butan-1-ol to butanoic acid
(b) 4-Methyl acetophenone to benzene-1,4-dicarboxylic acid
(ii) An organic compound with molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives 1,2-benzene dicarboxylic acid. Identify the compound. [CBSE Delhi 2012; NCERT]
Answer:
(i) (a) CH₃–CH₂–CH₂–CH₂OH (Butan-1-ol) → [(O), KMnO₄] → CH₃CH₂CH₂CHO → [(O), KMnO₄] → CH₃CH₂CH₂COOH (Butanoic acid)
(b) p-CH₃C₆H₄COCH₃ (4-methyl acetophenone) → [(O) (i) KMnO₄, OH⁻ (ii) solution of (NaOCl, NaOBr)] → p-HOOC–C₆H₄–COOH (Benzene-1,4-dicarboxylic acid)
(ii) (a) It is an aldehyde or ketone as it forms 2,4-DNP derivative.
(b) As the compound reduces Tollens’ reagent and undergoes Cannizzaro reaction, it is an aldehyde and not a ketone.
(c) On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. So, it must have an alkyl group at ortho position with respect to CHO group on the benzene ring.
(d) Molecular formula suggests that it should be 2-ethyl benzaldehyde, o-C₂H₅–C₆H₄–CHO.
Reactions: 2-ethylbenzaldehyde → [Tollens’ reagent] → 2-ethylbenzoate + Ag (silver mirror); 2-ethylbenzaldehyde → [(O)] → 1,2-benzenedicarboxylic acid; 2-ethylbenzaldehyde + 2,4-DNP → o-C₂H₅C₆H₄CH=N–NH–C₆H₃(NO₂)₂ (2,4-DNP derivative) + H₂O

Teacher's Note:
1. Check the formula: C₆H₄ + CHO + C₂H₅ = C₉H₁₀O, and the two groups must be ortho because oxidation gives phthalic (1,2-) acid.

Question 39. (i) Write the product(s) in the following reactions:
(a) Cyclohexanone + HCN → ?
(b) C₆H₅COONa + NaOH → [CaO, Δ] → ?
(c) CH₃–CH=CH–CN → [(a) DIBAL-H (b) H₂O] → ?
(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Butanal and Butan-2-one
(b) Benzoic acid and Phenol [CBSE OD Set-1, 2, 3 2017]
Answer:
(i) (a) 1-Hydroxycyclohexane-1-carbonitrile (cyclohexanone cyanohydrin): OH and CN on the same ring carbon
(b) C₆H₆ (Benzene)
(c) CH₃–CH=CH–CHO
(ii) (a) Tollens’ reagent test: Add ammoniacal solution of silver nitrate (Tollens’ Reagent) in both the solutions. Butanal gives silver mirror whereas Butan-2-one does not.
(b) Add neutral FeCl₃ in both the solution, phenol forms violet colour but benzoic acid does not. (or any other correct test)

Teacher's Note:
1. Butan-2-one is a methyl ketone, so the iodoform test (positive for the ketone, negative for butanal) is an equally good answer for (ii)(a).

Question 40. (i) Write structures of A, B, C and D in the following reaction sequence:
CH₃COCl → [H₂/Pd-BaSO₄] → A → [dil. NaOH] → B → [Δ] → C
A → [CH₃MgBr/H₃O⁺] → D
(ii) Arrange the following compounds in the increasing order of their boiling points:
CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃COOH [CBSE Comptt. OD Set-1, 2, 3 2017]
Answer:
(i) A: CH₃CHO; B: CH₃–CH(OH)–CH₂–CHO; C: CH₃–CH=CH–CHO; D: CH₃–CH(CH₃)–OH
(ii) CH₃–O–CH₃ < CH₃CHO < CH₃–CH₂–OH < CH₃–COOH

Teacher's Note:
1. Carboxylic acids boil highest because they form strongly hydrogen-bonded dimers; the ether, with no O–H and little polarity, boils lowest.

Question 41. (i) Write the products of the following reaction:
(a) Cyclohexanone + NH₂–NH–CO–NH₂ → [H⁺] →
(b) CH₃MgBr + CO₂ → [Dry ether, H₃O⁺] →
(c) CH₃CH₂COOH + Br₂ → [Red Phosphorus] →
(ii) Write simple chemical tests to distinguish between the following pairs of compounds.
(a) Propanal and propanone
(b) Benzaldehyde and Benzoic acid [CBSE Comptt. Delhi Set-1, 2, 3 2017]
Answer:
(i) (a) (CH₂)₅C=N–NH–CO–NH₂ (cyclohexanone semicarbazone)
(b) CH₃COOH
(c) CH₃–CH(Br)–COOH
(ii) (a) Add ammonical solution of silver nitrate / Tollens’ reagent to both the compounds, propanal will give silver mirror while propanone does not.
(b) Add NaHCO₃ solution to both the compounds, benzoic acid will give effervescence and liberate CO₂ while benzaldehyde will not. (Or any other suitable test)

Teacher's Note:
1. Carboxylation of a Grignard reagent adds one carbon: CH₃MgBr gives ethanoic acid, not methanoic acid.