CBSE Class 12 Chemistry Aldehydes Ketones and Carboxylic Acids Important Questions Set 08

Very Short Answer Questions (1 Mark)

Question 1. Benzaldehyde reduces Tollens’ reagent but not the Fehling’s or the Benedict’s solution. Explain.
Answer:
The electron-donating resonance effect (+R-effect) of the benzene ring increases the electron density in the carbonyl group of benzaldehyde. This, in turn, increases the electron density in the C–H bond of the aldehyde group. As a result, the C–H bond becomes stronger and hence only stronger oxidising agents like Tollens’ reagent, Ag(NH₃)₂⁺ (E°_Ag⁺/Ag = +0.80 V) can oxidise C–H to C–OH to form carboxylic acids but weaker oxidising agents like Fehling’s solution (Cu²⁺ + tartrate ion + base) and Benedict’s (Cu²⁺ + citrate ion + base) solution (E°_{Cu²⁺/Cu⁺} = 0.18 V) fail to oxidise benzaldehyde to benzoic acid.
Thus, in general, all these three reagents oxidise aliphatic aldehydes but only Tollens’ reagent oxidises aromatic aldehydes.

Teacher's Note:
1. This is why Fehling’s test is used to tell an aliphatic aldehyde (positive) from an aromatic aldehyde (negative).

Question 2. Hydrazones of aldehydes and ketones are not prepared in highly acidic medium. Explain.
Answer:
In highly acidic medium, the NH₂ group of hydrazine gets protonated
NH₂–NH₂ + H⁺ → NH₂–N⁺H₃ (Protonated hydrazine, non nucleophilic)
Due to strong –I-effect of the N⁺H₃ group, the lone pair of electrons on the –NH₂ group of protonated hydrazine is not available for nucleophilic attack on the C=O group and hence hydrazone formation does not occur.

Teacher's Note:
1. These reactions are run in weakly acidic medium (pH about 3.5): enough acid to activate the C=O, not so much that the nucleophile is protonated.

Question 3. (a) Arrange the following in order of their increasing reactivity towards HCN : CH₃CHO, CH₃COCH₃, HCHO, C₂H₅COCH₃
(b) HCHO is more reactive than CH₃CHO towards addition of HCN. (CBSE 2018S)
Answer:
(a) (i) Reactivity increases as the +ve charge on carbonyl carbon increases. Since alkyl groups have +I-effect, therefore, reactivity increases as the number of alkyl groups on the carbonyl carbon decreases.
(ii) Reactivity increases as the number and size of alkyl groups decreases, i.e., steric hindrance decreases. Due to combined effect of these two factors, the reactivity increases in the order :
C₂H₅COCH₃ < CH₃COCH₃ < CH₃CHO < HCHO
(b) Due to +I-effect of CH₃ group, the +ve charge on the carbonyl carbon of CH₃CHO is less than that on the carbonyl carbon of HCHO. As a result, HCHO is more reactive than CH₃CHO towards addition of HCN.

Teacher's Note:
1. Use two words in such answers: electronic (+I effect) and steric (crowding); both point the same way here.

Question 4. Acetone is highly soluble in water but benzophenone is not. Give reasons.
Answer:
In acetone, C=O group easily forms H-bonds with water and hence acetone is highly soluble in water. However, in benzophenone, the phenyl groups are bulky and hence C=O group cannot form H-bonds with water due to steric hindrance and hence benzophenone is insoluble in water.

Teacher's Note:
1. The large hydrophobic part (two C₆H₅ groups) also outweighs the single polar C=O group in benzophenone.

Question 5. Suggest a reason for the large difference in the boiling points of butanol and butanal, although they have the same solubility in water.
Answer:
The b.p. of butanol is higher than that of butanal because butanol has strong intermolecular H-bonding while butanal has weak dipole-dipole attraction. However, both of them form H-bonds with water and hence are soluble.

Teacher's Note:
1. Butanal has no H on oxygen, so it cannot H-bond with itself, but its C=O oxygen can still accept H-bonds from water.

Question 6. Arrange the following compounds in increasing order of their boiling points : CH₃CH₂CH₂CHO, CH₃CH₂CH₂CH₂OH, CH₃CH₂COCH₃, CH₃CH₂OCH₂CH₃. (Raj. Board 2012)
Answer:
The four compounds have comparable molecular masses : CH₃CH₂CH₂CHO (72), CH₃CH₂CH₂CH₂OH (74), CH₃CH₂COCH₃ (72) and CH₃CH₂OCH₂CH₃ (74). Amongst compounds having comparable molecular masses, alcohols have the highest b.p. due to intermolecular H-bonding, i.e., CH₃CH₂CH₂CH₂OH has the highest boiling point. The boiling points of CH₃CH₂CH₂CHO, CH₃CH₂OCH₂CH₃ and CH₃CH₂COCH₃ depend upon their relative dipole moments. Since dipole moments of these three compounds decrease in the order : ketones > aldehydes > ethers, therefore, their boiling points also decrease in the same order, i.e., b.ps decrease in order : CH₃COCH₂CH₃ > CH₃CH₂CH₂CHO > CH₃CH₂OCH₂CH₃.
Thus, the boiling points of the four compounds increase in the order :
CH₃CH₂OCH₂CH₃ < CH₃CH₂CH₂CHO < CH₃COCH₂CH₃ < CH₃CH₂CH₂CH₂OH

Teacher's Note:
1. Read the question carefully: it asks for increasing order, so write the list from the lowest (ether) to the highest (alcohol).

Question 7. How will you convert acetone into 2-methyl-2-propanol ?
Or How is tert-butyl alcohol obtained from acetone ?
Answer:
By reaction of acetone with CH₃MgBr followed by hydrolysis.
(CH₃)₂C=O \( \xrightarrow[Dry\ ether]{CH_3MgBr} \) [(CH₃)₃C–OMgBr] \( \xrightarrow[-Mg(OH)Br]{H^+/H_2O} \) (CH₃)₃C–OH
Propanone or Acetone → Addition product → 2-Methyl-2-propanol or tert-Butyl alcohol

Teacher's Note:
1. Ketone + Grignard reagent always gives a 3° alcohol; the new alkyl group joins the carbonyl carbon.

Question 8. Alpha carbon of aldehydes and ketones are acidic in nature. (CBSE 2018S)
Answer:
Firstly, due to strong –I-effect of the C=O group in aldehydes and ketones, electron density in the α-C–H bonds decreases and hence α-hydrogen atoms can be abstracted by strong bases such as OH⁻, OR⁻ etc. As a result, they are weakly acidic.
Secondly, the carbanion or the enolate anion left after the removal of a proton is stabilized by resonance.
HO⁻ + H–CH₂–CHO \( \xrightarrow[-H_2O]{} \) ⁻CH₂–CH=O ↔ CH₂=CH–O⁻ (Carbanion or enolate anion)
HO⁻ + H–CH₂–COCH₃ \( \xrightarrow[-H_2O]{} \) ⁻CH₂–C(CH₃)=O ↔ CH₂=C(CH₃)–O⁻
Due to these two reasons, α-hydrogens of aldehydes and ketones are acidic in nature.

Teacher's Note:
1. In the enolate the negative charge sits mainly on the electronegative oxygen; that resonance stabilisation is the main reason for the acidity.

Question 9. Draw the structure of the semicarbazone of ethanal. (CBSE 2016)
Answer:
Reaction occurs through the more nucleophilic NH₂ (attached to NH) and not through NH₂ attached to CO group. Thus,
CH₃–CH=O + H₂NNHCONH₂ \( \xrightarrow{pH\ 3.5} \) CH₃–CH=NNHCONH₂ + H₂O
Ethanal + Semicarbazide → Ethanal semicarbazone

Teacher's Note:
1. The NH₂ next to C=O is not nucleophilic because its lone pair is delocalised into the carbonyl group of the amide part.

Question 10. Write the structures of A and B in the following reaction.
CH₃COCl \( \xrightarrow{H_2,\ Pd\text{-}BaSO_4} \) A \( \xrightarrow{H_2N\text{-}OH} \) B (CBSE 2016)
Answer:
CH₃COCl \( \xrightarrow[(Rosenmund\ reduction)]{H_2,\ Pd\text{-}BaSO_4} \) CH₃CHO \( \xrightarrow[-H_2O]{H_2N\text{-}OH} \) CH₃CH=NOH
A = CH₃CHO (Acetaldehyde); B = CH₃CH=NOH (Acetaldoxime)

Teacher's Note:
1. Hydroxylamine replaces the carbonyl oxygen by =NOH; the products are called oximes.

Question 11. Acetophenone on reaction with hydroxylamine hydrochloride can produce two isomeric oximes. Write structures of the oximes.
Answer:
Like alkenes, oximes also show geometrical isomerism. For example, acetophenone exists in the following two geometrical isomeric forms :
(C₆H₅)(CH₃)C=N–OH with the OH group on the same side as C₆H₅ : syn-Methyl phenyl ketoxime
(C₆H₅)(CH₃)C=N–OH with the OH group on the side opposite to C₆H₅ : anti-Methyl phenyl ketoxime
In both, the lone pair on N occupies the position opposite to the OH group.

Teacher's Note:
1. Geometrical isomerism is possible because rotation about C=N is restricted and the carbon carries two different groups (C₆H₅ and CH₃).

Short Answer Questions (2 Marks)

Question 12. How will you convert : methanal to propan-1-ol.
Answer:
HCHO + CH₃CH₂MgBr \( \xrightarrow{Dry\ ether} \) CH₃CH₂CH₂OMgBr \( \xrightarrow{H^+/H_2O} \) CH₃CH₂CH₂OH

Teacher's Note:
1. The Grignard reagent supplies two carbons and HCHO the third, which carries the OH.

Question 13. How will you convert : acetaldehyde to methane (CBSE 2007)
Answer:
CH₃CHO \( \xrightarrow{K_2Cr_2O_7/H_2SO_4\ or\ [Ag(NH_3)_2]^+} \) CH₃COOH \( \xrightarrow{NaOH} \) CH₃COONa \( \xrightarrow[(Decarboxylation)]{NaOH/CaO,\ \Delta} \) CH₄ + Na₂CO₃

Teacher's Note:
1. Soda-lime decarboxylation removes one carbon, taking the two-carbon acid to methane.

Question 14. How will you convert : propene to acetone. (Assam Board 2012)
Answer:
CH₃CH=CH₂ + H₂O \( \xrightarrow{H^+} \) CH₃CH(OH)CH₃ \( \xrightarrow{Cu,\ 573\ K\ or\ K_2Cr_2O_7/H_2SO_4} \) CH₃COCH₃

Teacher's Note:
1. Acid-catalysed hydration follows Markovnikov’s rule, which is why the 2° alcohol (and then the ketone) is formed.

Question 15. How will you convert : acetaldehyde to ethyl acrylate ?
Answer:
CH₃CHO \( \xrightarrow{HCN} \) CH₃CH(OH)CN \( \xrightarrow{H^+/H_2O} \) CH₃CH(OH)COOH (Lactic acid) \( \xrightarrow[-H_2O]{conc.\ H_2SO_4,\ \Delta} \) CH₂=CH–COOH (Acrylic acid)
CH₂=CH–COOH + C₂H₅OH \( \xrightarrow[(Esterification)]{conc.\ H_2SO_4} \) CH₂=CH–COOC₂H₅ (Ethyl acrylate) + H₂O

Teacher's Note:
1. Ethyl acrylate has three carbons in the acid part, so one carbon must be added to acetaldehyde (here through HCN).

Question 16. How will you bring about the following conversions in not more than two steps ? Propanone to propene (CBSE 2010 ; Hr. Board 2012)
Answer:
CH₃COCH₃ \( \xrightarrow[(Reduction)]{NaBH_4,\ CH_3OH} \) CH₃CH(OH)CH₃ \( \xrightarrow[(Dehydration)]{Conc.\ H_2SO_4,\ 443\ K} \) CH₃CH=CH₂

Teacher's Note:
1. Reduction then dehydration is the standard way to turn a ketone into an alkene with the same skeleton.

Question 17. How will you bring about the following conversions in not more than two steps ? Propanal to butanal
Answer:
A one-carbon chain lengthening is needed. Propanal is first reduced to propan-1-ol, converted to the bromide, then to the nitrile, which is partially reduced :
CH₃CH₂CHO \( \xrightarrow{NaBH_4} \) CH₃CH₂CH₂OH \( \xrightarrow{PBr_3} \) CH₃CH₂CH₂Br \( \xrightarrow{KCN\ (alc.)} \) CH₃CH₂CH₂CN \( \xrightarrow[(ii)\ H_2O]{(i)\ DIBAL\text{-}H} \) CH₃CH₂CH₂CHO

Teacher's Note:
1. Adding a carbon atom to a chain is usually done through a nitrile (–CN), which is then converted to –CHO or –COOH.

Question 18. How will you bring about the following conversions in not more than two steps ? Ethanol to 3-hydroxybutanal (CBSE 2009 ; Hr. Board 2012)
Answer:
CH₃CH₂OH \( \xrightarrow{Cu/573\ K\ or\ PCC/CH_2Cl_2} \) CH₃CHO \( \xrightarrow[(Aldol\ condensation)]{Dil.\ NaOH} \) CH₃CH(OH)CH₂CHO

Teacher's Note:
1. Use a mild oxidant (PCC or Cu/573 K) so that ethanol stops at ethanal and is not oxidised to acetic acid.

Question 19. How will you bring about the following conversions in not more than two steps ? Benzaldehyde to benzophenone (CBSE 2009 ; Hr. Board 2012)
Answer:
C₆H₅CHO \( \xrightarrow[(ii)\ H^+/H_2O]{(i)\ C_6H_5MgBr} \) C₆H₅CH(OH)C₆H₅ (Diphenylcarbinol) \( \xrightarrow{PCC,\ CH_2Cl_2} \) C₆H₅COC₆H₅ (Benzophenone)

Teacher's Note:
1. A 2° alcohol is oxidised to a ketone and no further, so the choice of oxidant is not critical here.

Question 20. How will you bring about the following conversions in not more than two steps ? Benzoic acid to benzaldehyde
Answer:
C₆H₅COOH \( \xrightarrow[-SO_2,\ -HCl]{SOCl_2} \) C₆H₅COCl \( \xrightarrow[(Rosenmund\ reduction)]{H_2/Pd\text{-}BaSO_4\ +\ S\ or\ quinoline} \) C₆H₅CHO

Teacher's Note:
1. Direct reduction of the acid with LiAlH₄ would go to benzyl alcohol, not stop at the aldehyde.

Question 21. How will you bring about the following conversions in not more than two steps ? Ethanol to acetone (CBSE 2007, 2008)
Answer:
CH₃CH₂OH \( \xrightarrow{PCC\ or\ Cu,\ 573\ K} \) CH₃CHO \( \xrightarrow[(ii)\ H_3O^+]{(i)\ CH_3MgBr} \) CH₃CH(OH)CH₃ \( \xrightarrow{K_2Cr_2O_7/H_2SO_4} \) CH₃COCH₃
Or CH₃CH₂OH \( \xrightarrow{K_2Cr_2O_7/H_2SO_4} \) CH₃COOH \( \xrightarrow{CaCO_3} \) (CH₃COO)₂Ca \( \xrightarrow{Dry\ distil.} \) CH₃COCH₃

Teacher's Note:
1. Ethanol has two carbons and acetone three, so a carbon must be added (Grignard) or two units combined (calcium salt).

Question 22. How will you bring about the following conversions in not more than two steps ? Ethanal to but-2-enal or Acetaldehyde to crotonaldehyde (Hr. Board 2012 ; CBSE 2010, 2011, 2013)
Answer:
2CH₃CHO \( \xrightarrow[(Aldol\ condensation)]{Dil.\ NaOH} \) CH₃CH(OH)CH₂CHO \( \xrightarrow[-H_2O]{\Delta} \) CH₃CH=CHCHO (But-2-enal or Crotonaldehyde)

Teacher's Note:
1. Crotonaldehyde is the common name of but-2-enal.

Question 23. How will you bring about the following conversions in not more than two steps ? Acetone to mesitylene (Jharkhand Board 2011)
Answer:
When distilled with conc. H₂SO₄, three molecules of acetone condense with loss of three molecules of water to give mesitylene, i.e., 1, 3, 5-trimethylbenzene.
3CH₃COCH₃ \( \xrightarrow[-3H_2O]{Conc.\ H_2SO_4,\ distil.} \) C₆H₃(CH₃)₃ (1, 3, 5-Trimethylbenzene)

Teacher's Note:
1. Check the formula: 3 C₃H₆O − 3 H₂O = C₉H₁₂, which is trimethylbenzene.

Question 24. Give a chemical test to distinguish between : acetophenone and benzophenone (CBSE 2008 ; Hr. Board 2012)
Write the balanced chemical equation in each case.
Answer:
Iodoform test : Acetophenone, being a methyl ketone, gives a yellow ppt. of iodoform on warming with I₂ and NaOH while benzophenone does not.
C₆H₅COCH₃ + 3I₂ + 4NaOH → CHI₃↓ + C₆H₅COONa + 3NaI + 3H₂O

Teacher's Note:
1. Both are ketones, so Tollens’ and Fehling’s tests cannot separate them.

Question 25. Give a chemical test to distinguish between : Ethanal and propanal (CBSE 2010, 2013)
Write the balanced chemical equation in each case.
Answer:
Iodoform test : Ethanal contains the CH₃CO– group and gives a yellow ppt. of iodoform on warming with I₂ and NaOH while propanal does not.
CH₃CHO + 3I₂ + 4NaOH → CHI₃↓ + HCOONa + 3NaI + 3H₂O

Teacher's Note:
1. Both are aldehydes, so Tollens’ and Fehling’s tests are positive for both and are of no use here.

Question 26. Give a chemical test to distinguish between : Propanal and diethyl ether
Write the balanced chemical equation in each case.
Answer:
Tollens’ test : Propanal gives a silver mirror on warming with Tollens’ reagent while diethyl ether does not.
CH₃CH₂CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃CH₂COO⁻ + 2Ag↓ + 2H₂O + 4NH₃

Teacher's Note:
1. Fehling’s test (red ppt. with propanal) works equally well, since propanal is aliphatic.

Question 27. Give a chemical test to distinguish between : Acetaldehyde and acetone (Pb. Board 2012)
Write the balanced chemical equation in each case.
Answer:
Tollens’ test : Acetaldehyde gives a silver mirror with Tollens’ reagent while acetone does not.
CH₃CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃COO⁻ + 2Ag↓ + 2H₂O + 4NH₃
(Acetaldehyde also gives a red ppt. with Fehling’s solution; acetone does not.)

Teacher's Note:
1. Iodoform test is useless here: both acetaldehyde and acetone have the CH₃CO– group.

Question 28. Give a chemical test to distinguish between : C₆H₅COCH₃ and C₆H₅CHO (CBSE 2016)
Write the balanced chemical equation in each case.
Answer:
Iodoform test : C₆H₅COCH₃ gives a yellow ppt. of iodoform on warming with I₂ and NaOH while C₆H₅CHO does not.
C₆H₅COCH₃ + 3I₂ + 4NaOH → CHI₃↓ + C₆H₅COONa + 3NaI + 3H₂O
Tollens’ test : C₆H₅CHO gives a silver mirror; C₆H₅COCH₃ does not.

Teacher's Note:
1. One positive result for each compound makes the answer convincing.

Question 29. Give two reactions in which aliphatic aldehydes differ from aromatic aldehydes ?
Answer:
(i) Aliphatic aldehydes being more reactive reduce Fehling’s solution to red ppt. of Cu₂O while aromatic aldehydes do not.
RCHO + 2Cu²⁺ + 5OH⁻ → RCOO⁻ + Cu₂O↓ + 3H₂O
(ii) Aromatic aldehydes on treatment with an alcoholic solution of KCN undergo benzoin condensation but aliphatic aldehydes do not.
2C₆H₅CHO \( \xrightarrow{alc.\ KCN,\ \Delta} \) C₆H₅CH(OH)COC₆H₅ (Benzoin)

Teacher's Note:
1. Another difference: aromatic aldehydes (no α-H) give Cannizzaro reaction, while most aliphatic aldehydes give aldol condensation.

Short Answer Questions (3 Marks)

Question 30. Although both >C=C< and >C=O have a double bond, they exhibit different type of addition reactions. Explain.
Answer:
The >C=C< undergoes electrophilic addition reactions while >C=O shows nucleophilic addition reactions. This difference in behaviour is due to the different shapes of their π-electron clouds.
The π-electron cloud of >C=C< bond due to similar electronegatives of the two carbon atoms is almost symmetrical and surrounds both the carbon atoms equally. Consequently, if a reagent is to attack one of the carbon atoms of the >C=C< bond, it has to pass through this electron cloud. Since the π-electron cloud consists of loosely held electrons, therefore, it can readily allow electrophiles to react. As a result, the typical reactions of >C=C< are electrophilic addition reactions.
In contrast, the π-electron cloud of the >C=O is unsymmetrical i.e., shifts towards oxygen due to greater electronegativity of O than C. As a result, the C-atom of the >C=O bond acquires a partial +ve charge and hence is readily attacked by nucleophiles. Thus, the typical reactions of >C=O are nucleophilic addition reactions.

Teacher's Note:
1. The key word is polarity: the non-polar C=C attracts electrophiles, the polar C^δ+=O^δ− attracts nucleophiles to carbon.

Question 31. (a) Why are aldehydes more reactive than ketones? (Hr. Board 2013)
(b) Oxidation of aldehydes is easier than ketones. Give reason. (CBSE 2022)
Answer:
(a) Aldehydes are more reactive than ketones due to the following two reasons:
(i) Due to smaller +I-effect of one alkyl group in aldehydes as compared to larger +I-effect of two alkyl groups, the magnitude of positive charge on the carbonyl carbon is more in aldehydes than in ketones. As a result, nucleophilic addition reactions occur more readily in aldehydes than in ketones.
(ii) Due to presence of a H-atom on the carbonyl group, aldehydes can be more easily oxidised than ketones. As a result, aldehydes act as reducing agents and thus reduce Tollens’ reagent, Fehling’s solution, etc.
(b) Aldehydes contain a hydrogen atom attached to >C=O group but ketones contain a carbon atom. Since it is easier to break a C–H bond than a C–C bond. Therefore, oxidation of aldehydes is easier than ketones.

Teacher's Note:
1. Also mention steric hindrance: two alkyl groups in ketones crowd the carbonyl carbon more than one alkyl group and a small H atom in aldehydes.

Question 32. Suggest a suitable oxidising agent for the conversion,
(CH₃)₂C=CHCOCH₃ → (CH₃)₂C=CHCO₂H
Answer:
Alkaline KMnO₄, acidified K₂Cr₂O₇ or HNO₃ cannot be used since all of these will cleave the molecule at the site of the double bond giving a mixture of ketones/acids. The most suitable reagent for this oxidation is NaOI (I₂/NaOH) since methyl ketones on treatment with NaOI undergo iodoform reaction to give iodoform along with the Na salt of a carboxylic acid having one carbon atom less than the starting methyl ketone.
(CH₃)₂C=CH–CO–CH₃ (4-methylpent-3-en-2-one, mesityl oxide) + 3NaOI → (CH₃)₂C=CHCOONa + CHI₃ + 2NaOH
(CH₃)₂C=CHCOONa →^H+/H₂O (CH₃)₂C=CHCOOH (3-methylbut-2-en-1-oic acid, β,β-dimethylacrylic acid)

Teacher's Note:
1. Hypohalite oxidises only the CH₃CO– group and leaves C=C untouched; strong oxidants would cleave the double bond.

Question 33. Addition of Grignard reagents to dry ice followed by hydrolysis gives carboxylic acids while that of organolithium compounds under similar conditions gives ketones. Explain
Answer:
Since electronegativity of Li (E.N. = 1.0) is lower than that of Mg (E.N. = 1.2), therefore, organolithium compounds are more nucleophilic than Grignard reagents. As a result, organolithium compounds not only add to the more reactive CO₂ but also to the less reactive resonance stabilized lithium salt of carboxylic acid thus formed to yield ketones.
R–Li + O=C=O →^{dry ether} R–COO⁻Li⁺ →^R–Li R₂C(O⁻Li⁺)₂ →^H₃O+ [R₂C(OH)₂] (unstable) → R₂C=O (ketone) + H₂O
Grignard reagents, on the other hand, being less nucleophilic add only to the more reactive CO₂ but not to the less reactive resonance stabilized magnesium salt of the carboxylic acid from which carboxylic acid can be generated by hydrolysis with mineral acids.
R–MgX + O=C=O →^{dry ether} [R–COOMgX] →^H₃O+ R–COOH (carboxylic acid) + Mg(OH)X

Teacher's Note:
1. The more ionic the C–metal bond (lower E.N. of the metal), the more reactive the organometallic compound.

Question 34. Why is ethylbenzene not prepared by simple alkylation of benzene? Devise a convenient route for its synthesis.
Answer:
During Friedel-Crafts alkylation, the first product of the reaction is an alkylbenzene. Since an alkyl group is electron-donating, therefore, electron density in the benzene ring of alkylbenzene increases. In other words, alkylbenzene is more reactive than benzene towards further electrophilic substitution reaction and thus alkylbenzene competes with benzene and hence would preferentially undergo further F.C. reaction to give first di- and then polysubstituted benzene. It is because of this disadvantage of polysubstitution that F.C. alkylation reaction is not used for preparation of alkylbenzenes. Instead F.C. acylation is used.
This is due to the reason that acyl group (RCO–) being electron-withdrawing decreases the electron density on the benzene ring. In other words, acyl group once introduced makes acylbenzene less reactive than benzene towards further reaction. Therefore, the reaction stops at the monosubstitution stage and hence complications due to polysubstitution are avoided.
In the light of the above discussion, a convenient route for the synthesis of ethylbenzene involves the following two steps.
C₆H₆ (benzene) + CH₃COCl (acetyl chloride) →^{anhyd. AlCl₃ (F.C. acylation)} C₆H₅–COCH₃ (acetophenone) + HCl
C₆H₅COCH₃ →^{NH₂NH₂.H₂O, KOH, glycol, 453–473 K (Wolff-Kishner reduction) or Zn–Hg + HCl (Clemmensen reduction)} C₆H₅CH₂CH₃ (ethylbenzene)

Teacher's Note:
1. Acylation followed by Clemmensen or Wolff-Kishner reduction is the standard route to a pure mono-alkylbenzene.

Question 35. (a) p-Nitrobenzoic acid has higher K_a value than benzoic acid. Give reasons.
Or 4-Nitrobenzoic acid is a stronger acid than benzoic acid. Give reasons. (CBSE 2022)
(b) pK_a of 4-nitrobenzoic acid is lower than that of benzoic acid. (CBSE 2018)
Answer:
(a) Higher the K_a, stronger is the acid. Thus, p-nitrobenzoic acid is a stronger acid than benzoic acid. This is due to the following two reasons:
(i) Due to −I and −R-effect of the –NO₂ group, the electron density in the O–H bond decreases. As a result, O–H bond becomes weak and hence p-nitrobenzoic acid more easily loses a proton than benzoic acid.
(ii) Due to −I and −R-effect of the NO₂ group, dispersal of the −ve charge occurs and hence p-nitrobenzoate ion becomes more stable than benzoate ion.
(b) Since higher the K_a lower the pK_a value, therefore, 4-nitrobenzoic acid has lower pK_a value than benzoic acid.

Teacher's Note:
1. pK_a = −log K_a: a stronger acid has a larger K_a but a smaller pK_a.

Question 36. Arrange the following in decreasing ease of acid-catalysed esterification:
2,4-dimethylbenzoic acid, benzoic acid, 2,4,6-trimethylbenzoic acid
Answer:
During acid catalysed esterification, the nucleophilic attack of the alcohol (R′OH) molecule occurs on the protonated carbonyl group of the acid (ArCOOH) to give the tetrahedral intermediate as shown below:
Ar–COOH →^H+ Ar–C⁺(OH)₂ (protonated carboxylic acid) →^{R′OH, −H+} Ar–C(OH)₂–OR′ (tetrahedral intermediate) →^−H₂O Ar–CO–OR′ (ester)
Evidently, as the number of methyl groups at the o-position of the benzene ring increases, the carbonyl group becomes more and more sterically hindered. As a result, the rate of nucleophilic attack by ROH on the carbonyl group decreases and hence the rate of esterification decreases accordingly. Thus, the rate of esterification decreases in the order: benzoic acid > 2, 4-dimethylbenzoic acid > 2, 4, 6-trimethyl benzoic acid.

Teacher's Note:
1. Ortho substituents hinder the approach of the alcohol to the carbonyl carbon; 2,6-disubstituted benzoic acids esterify very slowly.

Question 37. Predict the product of the following reaction
C₆H₅–O–CO–C₆H₅ →^NO₂+
Answer:
Due to the presence of lone pairs of electrons on oxygen, it has +R-effect while C=O group has −R-effect.
As a result, electron density is more in the ring attached to O- atom than in the ring attached to C=O group. In other words, NO₂⁺ will attack the ring attached to oxygen. Further, due to +R-effect of the O-atom, electron density increases at o- and p-positions. But, due to steric hindrance at o-position, electrophilic substitution mainly occurs at the p-position. Thus,
C₆H₅–O–CO–C₆H₅ →^{NO₂+, −H+} p-O₂N–C₆H₄–O–CO–C₆H₅ (4-nitrophenyl benzoate)

Teacher's Note:
1. In phenyl benzoate the –O–CO– link activates the ring on the oxygen side and deactivates the ring on the carbonyl side.

Long Answer Questions (5 Marks)

Question 38. (i) Write the structures of A, B, C and D in the following reactions:
CH₃CN \( \xrightarrow[(ii)\,H_2O]{(i)\,SnCl_2{-}HCl} \) A \( \xrightarrow{\text{dil. NaOH}} \) B \( \xrightarrow{\text{Heat}} \) C;   A \( \xrightarrow{HCN} \) D
(ii) Distinguish between:
(a) C₆H₅–CH=CH–COCH₃ and C₆H₅–CH=CH–COCH₂CH₃
(b) CH₃CH₂COOH and HCOOH
(iii) Arrange the following in the increasing order of their boiling points:
CH₃CH₂OH, CH₃COCH₃, CH₃COOH [CBSE North 2016]
Answer:
(i) A = CH₃–CHO (Ethanal); B = CH₃–CH(OH)–CH₂–CHO (3-Hydroxybutanal)
C = CH₃–CH=CH–CHO (But-2-enal); D = CH₃–CH(OH)–CN (2-Hydroxypropanenitrile)
(ii) (a) C₆H₅CH=CH–COCH₃ on warming with NaOI (I₂/NaOH) gives yellow precipitate of iodoform while C₆H₅CH=CH–CO–CH₂–CH₃ does not.
C₆H₅CH=CHCOCH₃ + 3NaOI → CHI₃↓ (Iodoform, Yellow ppt.) + C₆H₅CH=CH–COO⁻Na⁺ + 2NaOH
(b) Formic acid reduces Tollens’ reagent to metallic silver while propionic acid does not.
HCOOH (Formic acid) + 2[Ag(NH₃)₂]⁺ + 2OH⁻ → 2Ag↓ (Silver mirror) + CO₂ + 2H₂O + 4NH₃
(iii) CH₃–CO–CH₃ < CH₃–CH₂–OH < CH₃–COOH

Teacher's Note:
1. Formic acid contains an aldehyde-like H–C=O unit, which is why it alone among carboxylic acids gives the Tollens’ test.

Question 39. (i) How will you convert the following:
(a) Propanone to propan-2-ol
(b) Ethanal to 2-hydroxy propanoic acid
(c) Toluene to benzoic acid
(ii) Distinguish the following pairs of compounds:
(a) Pentan-2-one and pentan-3-one
(b) Ethanal and propanal [CBSE OD 2013]
Answer:
(i) (a) CH₃COCH₃ (Propanone) → [LiAlH₄, Reduction] → CH₃–CH(OH)–CH₃ (Propan-2-ol)
(b) CH₃CHO (Ethanal) + HCN → CH₃–CH(OH)–CN → [H₂O, HCl] → CH₃–CH(OH)–COOH (2-Hydroxy propanoic acid)
(c) C₆H₅CH₃ (Toluene) → [K₂Cr₂O₇ + H₂SO₄, Oxidation] → C₆H₅COOH (Benzoic acid)
(ii) (a) Pentan-2-one gives yellow ppt of iodoform in Iodoform test, while pentan-3-one does not give it.
CH₃CH₂CH₂COCH₃ + 3NaOI → CH₃CH₂CH₂COONa + CHI₃↓ (Yellow ppt) + 2NaOH
(b) Ethanal will give iodoform test, while propanal will not give it.
CH₃CHO + 3NaOI → CHI₃↓ (yellow ppt) + HCOONa + 2NaOH

Teacher's Note:
1. Ethanal → cyanohydrin → hydrolysis adds one carbon as –COOH: a standard way to make α-hydroxy acids.

Question 40. (i) Write the structures of A, B, C, D and E in the following reactions:
C₆H₆ → [CH₃COCl, Anhyd. AlCl₃] → A → [Zn–Hg/Conc. HCl] → B → [(i) KMnO₄–KOH, Δ (ii) H₃O⁺] → C
A → [NaOH] → D + E [CBSE Delhi 2016]
Answer:
A = C₆H₅COCH₃ (Acetophenone)
B = C₆H₅C₂H₅ (Ethylbenzene)
C = C₆H₅COOH (Benzoic acid)
D = C₆H₅COO⁻Na⁺ (Sodium benzoate)
E = CHI₃↓ (Iodoform, yellow ppt.)

Teacher's Note:
1. The step A → D + E is the haloform reaction (NaOH with I₂); a methyl ketone always splits into a carboxylate with one carbon less plus CHI₃.

Question 41. (i) Write the structures of A and B in the following reactions:
(a) CH₃COCl → [H₂, Pd–BaSO₄] → A → [H₂N–OH] → B.
(b) CH₃MgBr → [1. CO₂ 2. H₃O⁺] → A → [PCl₅] → B.
(ii) Distinguish between:
(a) C₆H₅–COCH₃ and C₆H₅–CHO,
(b) CH₃COOH and HCOOH.
(iii) Arrange the following in the increasing order of their boiling points:
CH₃CHO, CH₃COOH, CH₃CH₂OH. [CBSE OD 2016]
Answer:
(i) (a) CH₃COCl → [H₂, Pd–BaSO₄] → CH₃CHO [A] (Acetaldehyde) + HCl → [H₂N–OH] → CH₃–CH=N–OH [B] (Acetaldoxime)
(b) CH₃MgBr → [1. CO₂ 2. H₃O⁺] → CH₃COOH [A] (Acetic acid) → [PCl₅] → CH₃COCl [B] (Acetyl chloride) + HCl + POCl₃
(ii) (a) C₆H₅–CHO being an aldehyde reduces Tollens’ reagent to give shining silver mirror whereas C₆H₅COCH₃ being a ketone does not.
C₆H₅CHO (Benzaldehyde) + 2[Ag(NH₃)₂]⁺ (Tollens’ reagent) + 3OH⁻ → [Δ] → C₆H₅COO⁻ (Benzoate ion) + 2Ag↓ (Silver mirror) + 4NH₃ + 2H₂O
C₆H₅COCH₃ → [Tollens’ reagent] → No silver mirror
(b) HCOOH gives silver mirror test with Tollens’ reagent whereas ethanoic acid does not.
HCOOH + 2[Ag(NH₃)₂]⁺ + 2OH⁻ → 2Ag↓ (Silver mirror) + 2H₂O + CO₂ + 4NH₃
CH₃COOH → [Tollens’ reagent] → No silver mirror (or any other correct test)
(iii) CH₃CHO < CH₃CH₂OH < CH₃COOH

Teacher's Note:
1. Formic acid contains an aldehyde-like H–C=O unit, which is why it alone among carboxylic acids reduces Tollens’ reagent.

Question 42. (i) Complete the following equations:
(a) C₆H₅CHO → [conc. NaOH] →
(b) CH₃COCH₃ → [LiAlH₄] →
(c) CH₃–COOH → [(a) Cl₂/P (b) H₂O] →
(ii) Distinguish between:
(a) CH₃COOH and C₆H₅OH
(b) CH₃–CHO and CH₃–CO–CH₃ [CBSE Comptt. Delhi 2016]
Answer:
(i) (a) 2C₆H₅CHO → [Conc. NaOH] → C₆H₅CH₂OH + C₆H₅COONa
(b) 2[H] + CH₃COCH₃ → [LiAlH₄] → CH₃CH(OH)CH₃
(c) CH₃–COOH → [(i) Cl₂/Red phosphorus (ii) H₂O] → CH₂(Cl)–COOH
(ii) (a) When CH₃COOH is added to an aqueous solution of sodium carbonate, brisk effervescence of CO₂ is evolved.
RCOOH (Carboxylic acid) + NaHCO₃ → RCOONa (Sodium carboxylate) + H₂O + CO₂
Phenol do not give this test.
(b) When CH₃CHO is heated with Tollens’ reagent, they form silver mirror while CH₃COCH₃ does not.
CH₃CHO + 2[Ag(NH₃)₂]⁺ (Tollens’ reagent) + 3OH⁻ → CH₃–COO⁻ + 2Ag↓ (Silver mirror) + 2H₂O + 4NH₃ (or any other suitable test)

Teacher's Note:
1. Phenol is too weak an acid to liberate CO₂ from NaHCO₃/Na₂CO₃, so effervescence cleanly separates carboxylic acids from phenols.

Question 43. (i) Describe the following reactions:
(a) Acetylation (b) Aldol condensation
(ii) Write the main product in the following equations:
(a) CH₃–CO–CH₃ → [LiAlH₄] → ?
(b) C₆H₅CHO → [HNO₃/H₂SO₄, 273–283 K] → ?
(c) CH₃–COOH → [PCl₅] → ? [CBSE Comptt. Delhi 2015]
Answer:
(i) (a) When the acyl groups are introduced at ortho and para positions by reaction with acyl halide in the presence of anhydrous aluminium chloride (a Lewis acid) as catalyst.
C₆H₅Cl + CH₃COCl → [Anhyd. AlCl₃] → o-ClC₆H₄COCH₃ + p-ClC₆H₄COCH₃
(b) Aldehydes and ketones having at least one α-hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form β-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol), respectively.
2CH₃–CHO ⇌ [dil. NaOH] ⇌ CH₃–CH(OH)–CH₂–CHO
(ii) (a) CH₃–CO–CH₃ → [LiAlH₄] → CH₃–CH(OH)–CH₃
(b) C₆H₅CHO → [HNO₃/H₂SO₄, 273-283 K] → m-O₂N–C₆H₄–CHO
(c) CH₃COOH → [PCl₅] → CH₃COCl

Teacher's Note:
1. Aldol needs dilute alkali and an α-H; with concentrated alkali and no α-H you get the Cannizzaro reaction instead – do not mix them up.