Chapter 6 of 10

Updated 6 August 2026

NCERT Solutions

Aldehydes, Ketones and Carboxylic Acids — NCERT Solutions

Q: What are the important topics in Aldehydes, Ketones and Carboxylic Acids for Madhya Pradesh Board Class 12 Chemistry?

Key topics in Aldehydes, Ketones and Carboxylic Acids include Aldehydes have the group RCHO, The carbonyl carbon is sp2, Aldehydes and ketones are prepared by, A nucleophile attacks the electrophilic carbonyl. Study these first, then practise questions on each for the Madhya Pradesh Board Class 12 board exam.

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Learn the core ideas first, then work through the 121 practice questions on Aldehydes, Ketones and Carboxylic Acids. Revise definitions regularly and use flashcards for quick recall before the exam.

Madhya Pradesh Board · Class 12 · Chemistry

NCERT Solutions for Aldehydes, Ketones and Carboxylic Acids, Madhya Pradesh Board Class 12 Chemistry: 27 textbook questions solved step by step.

121 questions80 flashcards8 formulas & key relations5 concepts

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A labeled diagram showing the selective partial reduction of nitriles and esters to aldehydes using DIBAL-H (Diisobutylaluminum hydride) at low temperatures, followed by hydrolysis. — One of 27 illustrations for Aldehydes, Ketones and Carboxylic Acids in Super Tutor — alongside flashcards, concept maps and practice questions.

27 Questions Solved · 8 Sections

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Intext Question 8.1

8.1Write the structures of the following compounds.
(i) α-Methoxypropionaldehyde
(ii) 3-Hydroxybutanal
(iii) 2-Hydroxycyclopentane carbaldehyde
(iv) 4-Oxopentanal
(v) Di-sec. butyl ketone
(vi) 4-FluoroacetophenoneShow solution

(i) α-Methoxypropionaldehyde
α-carbon is C-2 of propionaldehyde (propanal). A methoxy (–OCH₃) group is attached at C-2.
$\text{CH}_3-\underset{\displaystyle|}{{\overset{\displaystyle\text{OCH}_3}{\text{CH}}}}-\text{CHO}$
Structure: CH₃–CH(OCH₃)–CHO

(ii) 3-Hydroxybutanal
Butanal with –OH at C-3.
$\text{CH}_3-\underset{\displaystyle|}{{\overset{\displaystyle\text{OH}}{\text{CH}}}}-\text{CH}_2-\text{CHO}$
Structure: CH₃–CH(OH)–CH₂–CHO

(iii) 2-Hydroxycyclopentane carbaldehyde
A cyclopentane ring with –CHO at C-1 and –OH at C-2.
Structure: Cyclopentane ring with –CHO substituent at C-1 and –OH at C-2 (both on adjacent carbons of the ring).

(iv) 4-Oxopentanal
A five-carbon chain with an aldehyde (–CHO) at C-1 and a keto (=O) group at C-4.
$\text{OHC}-\text{CH}_2-\text{CH}_2-\underset{\displaystyle\|}{\overset{\displaystyle}{\text{C}}}-\text{CH}_3$
Structure: OHC–CH₂–CH₂–CO–CH₃

(v) Di-sec. butyl ketone
sec-Butyl group = CH₃CH₂CH(CH₃)–
Two sec-butyl groups on either side of the carbonyl.
$\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)-\overset{\displaystyle O}{\overset{\displaystyle\|}{\text{C}}}-\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3$
Structure: (CH₃CH₂CHCH₃)–CO–(CHCH₃CH₂CH₃)

(vi) 4-Fluoroacetophenone
Acetophenone (methyl phenyl ketone) with –F at the para position of the benzene ring.
$\text{F}-\text{C}_6\text{H}_4-\overset{\displaystyle O}{\overset{\displaystyle\|}{\text{C}}}-\text{CH}_3 \quad (\text{F at para position})$
Structure: p-F–C₆H₄–CO–CH₃

Intext Question 8.2

8.2Write the structures of products of the following reactions (reactions involve figures that cannot be fully reproduced from OCR, but the standard reactions referred to are given below):
(i) Oxidation of a primary alcohol to aldehyde (PCC)
(ii) Ozonolysis of an alkene
(iii) Hydration of an alkyne
(iv) Friedel-Crafts acylationShow solution

Note: The exact reagents/structures in the figures (img_3 to img_6) cannot be read from the OCR. The following are the standard products based on the context of Section 8.2 (Preparation of Aldehydes and Ketones).

(i) PCC oxidation of a primary alcohol → Aldehyde
PCC (Pyridinium chlorochromate) oxidises a primary alcohol to the corresponding aldehyde without further oxidation to carboxylic acid.
$\text{RCH}_2\text{OH} \xrightarrow{\text{PCC}} \text{RCHO}$

(ii) Ozonolysis of an alkene → Aldehydes/Ketones
Ozonolysis followed by Zn/H₂O cleaves the C=C double bond to give carbonyl compounds.
$\text{R-CH=CH-R'} \xrightarrow{\text{(i) O}_3,\;\text{(ii) Zn/H}_2\text{O}} \text{RCHO} + \text{R'CHO}$

(iii) Hydration of an alkyne → Ketone (Markovnikov addition)
Hydration of a terminal alkyne (except acetylene) in the presence of H₂SO₄/HgSO₄ gives a methyl ketone.
$\text{RC}\equiv\text{CH} \xrightarrow{\text{H}_2\text{O/H}^+/\text{HgSO}_4} \text{RCOCH}_3$

(iv) Friedel-Crafts acylation → Aryl ketone
Benzene reacts with an acyl chloride in the presence of anhydrous AlCl₃ to give an aryl ketone.
$\text{C}_6\text{H}_6 + \text{RCOCl} \xrightarrow{\text{anhy. AlCl}_3} \text{C}_6\text{H}_5\text{COR} + \text{HCl}$

Intext Question 8.4

8.4Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.
(i) Ethanal, Propanal, Propanone, Butanone.
(ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.Show solution

Concept: Reactivity in nucleophilic addition depends on:

Steric effect: More bulky groups around carbonyl carbon → less reactive.
Electronic effect: Electron-withdrawing groups increase electrophilicity of carbonyl carbon → more reactive. Electron-donating groups decrease electrophilicity → less reactive.

(i) Ethanal, Propanal, Propanone, Butanone

Ketones (Propanone, Butanone) are less reactive than aldehydes (Ethanal, Propanal) due to two alkyl groups (steric + inductive electron donation).
Among aldehydes: Ethanal (CH₃CHO) has one methyl group; Propanal (CH₃CH₂CHO) has a larger ethyl group → Propanal is slightly less reactive than Ethanal.
Among ketones: Butanone has a larger alkyl group than Propanone → Butanone is less reactive.

Increasing order of reactivity:
$\text{Butanone} < \text{Propanone} < \text{Propanal} < \text{Ethanal}$

(ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone

Acetophenone is a ketone → least reactive (steric + electronic).
p-Tolualdehyde: –CH₃ is electron-donating (EDG) → decreases electrophilicity → less reactive than benzaldehyde.
Benzaldehyde: no substituent on ring.
p-Nitrobenzaldehyde: –NO₂ is electron-withdrawing (EWG) → increases electrophilicity of carbonyl carbon → most reactive.

Increasing order of reactivity:
$\text{Acetophenone} < p\text{-Tolualdehyde} < \text{Benzaldehyde} < p\text{-Nitrobenzaldehyde}$

Intext Question 8.5

8.5Predict the products of the following reactions (reactions involve figures; standard nucleophilic addition reactions of aldehydes/ketones are addressed below).Show solution

Note: The exact structures in img_14 cannot be read from OCR. The following covers the standard nucleophilic addition reactions typically asked in this context.

(i) Reaction with HCN (Cyanohydrin formation):
$\text{RCHO} + \text{HCN} \rightarrow \text{RCH(OH)CN}$
Product: α-hydroxynitrile (cyanohydrin)

(ii) Reaction with NaHSO₃ (Sodium bisulphite addition):
$\text{RCHO} + \text{NaHSO}_3 \rightarrow \text{RCH(OH)SO}_3\text{Na}$
Product: Sodium bisulphite addition compound

(iii) Reaction with NH₂OH (Hydroxylamine → Oxime):
$\text{RCHO} + \text{NH}_2\text{OH} \rightarrow \text{RCH=NOH} + \text{H}_2\text{O}$
Product: Oxime

(iv) Reaction with RMgX (Grignard reagent) followed by H₃O⁺:
$\text{RCHO} + \text{R'MgX} \xrightarrow{\text{(i) ether, (ii) H}_3\text{O}^+} \text{RCH(OH)R'}$
Product: Secondary alcohol

Intext Question 8.6

8.6Give the IUPAC names of the following compounds:
(i) PhCH₂CH₂COOH
(ii) (CH₃)₃C=CHCOOH
(iii) CH₃COOH
(iv) and (v) [structures from figures img_17 and img_18 — cyclic/aromatic carboxylic acids]Show solution

(i) PhCH₂CH₂COOH
Parent chain: 3 carbons with –COOH at C-1, phenyl group at C-3.
IUPAC name: 3-Phenylpropanoic acid

(ii) (CH₃)₃C=CHCOOH
Parent chain: pent-2-enoic acid backbone.
The double bond is between C-2 and C-3; a tert-butyl-like group is present.
Actually: (CH₃)₂C=CHCOOH would be 3-methylbut-2-enoic acid.
For (CH₃)₃C=CHCOOH: The carbon bearing three methyl groups and double bond — this is 3,3-dimethylbut-2-enoic acid.
IUPAC name: 3,3-Dimethylbut-2-enoic acid

(iii) CH₃COOH
Two-carbon carboxylic acid.
IUPAC name: Ethanoic acid (Common name: Acetic acid)

(iv) [Cyclic structure — assumed to be cyclohexane carboxylic acid based on context]
IUPAC name: Cyclohexanecarboxylic acid

(v) [Aromatic structure — assumed to be benzoic acid based on context]
IUPAC name: Benzenecarboxylic acid (Common name: Benzoic acid)

Intext Question 8.7

8.7Show how each of the following compounds can be converted to benzoic acid.
(i) Ethylbenzene
(ii) Acetophenone
(iii) Bromobenzene
(iv) Phenylethene (Styrene)Show solution

(i) Ethylbenzene → Benzoic acid
Ethylbenzene is oxidised by acidic KMnO₄ (or K₂Cr₂O₇/H⁺). The alkyl side chain is oxidised to –COOH regardless of chain length.
$\text{C}_6\text{H}_5\text{CH}_2\text{CH}_3 \xrightarrow{\text{KMnO}_4/\text{H}^+,\Delta} \text{C}_6\text{H}_5\text{COOH}$

(ii) Acetophenone → Benzoic acid
Acetophenone (C₆H₅COCH₃) is oxidised by alkaline KMnO₄ followed by acidification. The –COCH₃ group is oxidised to –COOH.
$\text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{\text{(i) KMnO}_4/\text{OH}^-,\Delta\;\text{(ii) H}^+} \text{C}_6\text{H}_5\text{COOH}$

(iii) Bromobenzene → Benzoic acid
Step 1: Convert bromobenzene to Grignard reagent.
$\text{C}_6\text{H}_5\text{Br} \xrightarrow{\text{Mg/dry ether}} \text{C}_6\text{H}_5\text{MgBr}$
Step 2: React with CO₂ followed by hydrolysis.
$\text{C}_6\text{H}_5\text{MgBr} \xrightarrow{\text{(i) CO}_2\;\text{(ii) H}_3\text{O}^+} \text{C}_6\text{H}_5\text{COOH}$

(iv) Phenylethene (Styrene) → Benzoic acid
Styrene (C₆H₅CH=CH₂) is oxidised by acidic/alkaline KMnO₄. The vinyl side chain is oxidised to –COOH.
$\text{C}_6\text{H}_5\text{CH}=\text{CH}_2 \xrightarrow{\text{KMnO}_4/\text{H}^+,\Delta} \text{C}_6\text{H}_5\text{COOH} + \text{CO}_2 + \text{H}_2\text{O}$

Intext Question 8.8

8.8Which acid of each pair shown here would you expect to be stronger?
(i) CH₃CO₂H or CH₂FCO₂H
(ii) CH₂FCO₂H or CH₂ClCO₂H
(iii) CH₂FCH₂CH₂CO₂H or CH₃CHFCH₂CO₂H
(iv) F₃C–COOH or H₃C–COOHShow solution

Concept: Electron-withdrawing groups (–I effect) stabilise the carboxylate anion (conjugate base) → increase acid strength. The closer and more electronegative the substituent, the stronger the acid.

(i) CH₃CO₂H or CH₂FCO₂H
–F is strongly electron-withdrawing (–I effect), stabilises COO⁻ more than –CH₃ (which is electron-donating).
Stronger acid: CH₂FCO₂H (fluoroacetic acid)

(ii) CH₂FCO₂H or CH₂ClCO₂H
Both F and Cl are electron-withdrawing, but F is more electronegative than Cl → F withdraws electrons more effectively → better stabilisation of carboxylate.
Stronger acid: CH₂FCO₂H (fluoroacetic acid)

(iii) CH₂FCH₂CH₂CO₂H or CH₃CHFCH₂CO₂H
In CH₃CHFCH₂CO₂H, the –F is at C-3 (β-carbon, closer to –COOH) compared to CH₂FCH₂CH₂CO₂H where –F is at C-4 (γ-carbon, farther from –COOH). Closer the EWG to –COOH, stronger the inductive effect.
Stronger acid: CH₃CHFCH₂CO₂H (F at β-position, i.e., 3-fluorobutanoic acid)

(iv) F₃C–COOH or H₃C–COOH
Three fluorine atoms in –CF₃ exert a very strong combined –I effect, greatly stabilising the carboxylate anion.
Stronger acid: F₃C–COOH (trifluoroacetic acid)

Exercises

8.1What is meant by the following terms? Give an example of the reaction in each case.
(i) Cyanohydrin
(ii) Acetal
(iii) Semicarbazone
(iv) Aldol
(v) Hemiacetal
(vi) Oxime
(vii) Ketal
(viii) Imine
(ix) 2,4-DNP-derivative
(x) Schiff's baseShow solution

(i) Cyanohydrin:
A cyanohydrin is a compound containing both a hydroxyl (–OH) and a cyano (–CN) group on the same carbon atom. It is formed by the nucleophilic addition of HCN to an aldehyde or ketone.
$\text{CH}_3\text{CHO} + \text{HCN} \rightarrow \text{CH}_3\text{CH(OH)CN}$
(Acetaldehyde cyanohydrin / Lactonitrile)

(ii) Acetal:
An acetal is a compound with two alkoxy (–OR) groups on the same carbon atom. It is formed by the reaction of an aldehyde with two equivalents of a monohydric alcohol in the presence of an acid catalyst.
$\text{CH}_3\text{CHO} + 2\text{CH}_3\text{OH} \xrightarrow{\text{dry HCl}} \text{CH}_3\text{CH(OCH}_3)_2 + \text{H}_2\text{O}$
(Acetaldehyde dimethyl acetal)

(iii) Semicarbazone:
A semicarbazone is formed by the condensation of an aldehyde or ketone with semicarbazide (H₂N–NH–CO–NH₂) in the presence of a weak acid.
$\text{RCHO} + \text{H}_2\text{N–NHCONH}_2 \xrightarrow{\text{weak acid}} \text{RCH=N–NHCONH}_2 + \text{H}_2\text{O}$

(iv) Aldol:
Aldol is a β-hydroxy carbonyl compound (β-hydroxy aldehyde or β-hydroxy ketone) formed by the aldol condensation of aldehydes or ketones having α-hydrogen in the presence of a dilute base or acid.
$2\text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO}$
(3-Hydroxybutanal — the aldol)

(v) Hemiacetal:
A hemiacetal is a compound containing one –OH and one –OR group on the same carbon. It is formed by the addition of one molecule of alcohol to an aldehyde.
$\text{RCHO} + \text{R'OH} \rightarrow \text{RCH(OH)(OR')}$

(vi) Oxime:
An oxime is formed by the condensation of an aldehyde or ketone with hydroxylamine (NH₂OH) in the presence of a weak acid.
$\text{CH}_3\text{CHO} + \text{NH}_2\text{OH} \rightarrow \text{CH}_3\text{CH=NOH} + \text{H}_2\text{O}$
(Acetaldoxime)

(vii) Ketal:
A ketal is a compound with two alkoxy (–OR) groups on the same carbon derived from a ketone (analogous to acetal from aldehyde). It is formed by the reaction of a ketone with a diol or two equivalents of alcohol in the presence of an acid catalyst.
$\text{CH}_3\text{COCH}_3 + \text{HOCH}_2\text{CH}_2\text{OH} \xrightarrow{\text{H}^+} \text{Cyclic ketal} + \text{H}_2\text{O}$

(viii) Imine:
An imine (Schiff's base) is a compound containing a C=N– bond. It is formed by the condensation of an aldehyde or ketone with a primary amine.
$\text{RCHO} + \text{R'NH}_2 \rightarrow \text{RCH=NR'} + \text{H}_2\text{O}$

(ix) 2,4-DNP derivative:
The 2,4-dinitrophenylhydrazone is formed by the reaction of an aldehyde or ketone with 2,4-dinitrophenylhydrazine (Brady's reagent). It gives a yellow/orange/red precipitate and is used to identify carbonyl compounds.
$\text{RCHO} + \text{2,4-(NO}_2)_2\text{C}_6\text{H}_3\text{NHNH}_2 \rightarrow \text{RCH=N–NH–C}_6\text{H}_3(\text{NO}_2)_2 + \text{H}_2\text{O}$

(x) Schiff's base:
Schiff's base is an imine (R–CH=N–R') formed by the condensation of an aldehyde with a primary amine. (Same as imine above.)
$\text{C}_6\text{H}_5\text{CHO} + \text{C}_6\text{H}_5\text{NH}_2 \rightarrow \text{C}_6\text{H}_5\text{CH=N–C}_6\text{H}_5 + \text{H}_2\text{O}$
(Benzylideneaniline)

8.2Name the following compounds according to IUPAC system of nomenclature:
(i) CH₃CH(CH₃)CH₂CH₂CHO
(ii) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl
(iii) CH₃CH=CHCHO
(iv) CH₃COCH₂COCH₃
(v) CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
(vi) (CH₃)₄CCH₂COOH
(vii) OHCC₆H₄CHO-pShow solution

(i) CH₃CH(CH₃)CH₂CH₂CHO
Longest chain containing –CHO: 5 carbons → pentanal.
Methyl branch at C-4 (numbering from –CHO end).
IUPAC name: 4-Methylpentanal

(ii) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl
Longest chain containing C=O: number to give lowest locant to ketone.
Chain: CH₃CH₂–CO–CH(C₂H₅)–CH₂–CH₂Cl
Count: C1(CH₃)–C2(CH₂)–C3(CO)–C4(CH)–C5(CH₂)–C6(CH₂Cl) → hexan-3-one
Substituents: ethyl at C-4, chloro at C-6.
IUPAC name: 6-Chloro-4-ethylhexan-3-one

(iii) CH₃CH=CHCHO
Four-carbon chain with –CHO at C-1 and double bond between C-2 and C-3.
IUPAC name: But-2-enal

(iv) CH₃COCH₂COCH₃
Five-carbon chain with two keto groups at C-2 and C-4.
IUPAC name: Pentane-2,4-dione

(v) CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
Longest chain containing C=O:
CH₃–CO–C(CH₃)₂–CH₂–CH(CH₃)–CH₃
Numbering from ketone end: C1(CH₃)–C2(CO)–C3(C(CH₃)₂)–C4(CH₂)–C5(CH(CH₃))–C6(CH₃)
This gives hexan-2-one with 3,3-dimethyl and 5-methyl substituents.
IUPAC name: 2,2,4-Trimethylhexan-5-one
(Alternatively numbered: 3,3,5-trimethylhexan-2-one — numbering from the methyl end gives lower locants to substituents: C1(CH₃)–C2(CO)–C3(C(CH₃)₂)–C4(CH₂)–C5(CH(CH₃))–C6(CH₃) → 3,3,5-trimethylhexan-2-one)
IUPAC name: 3,3,5-Trimethylhexan-2-one

(vi) (CH₃)₄CCH₂COOH
Parent chain: 3 carbons → propanoic acid (–COOH at C-1, –CH₂– at C-2, C(CH₃)₃ at C-3).
Substituent: tert-butyl [C(CH₃)₃] at C-3? Actually the longest chain through –COOH:
COOH–CH₂–C(CH₃)₃: 3 carbons in main chain = propanoic acid; C(CH₃)₃ at C-3 means three methyl groups at C-3.
IUPAC name: 3,3-Dimethylbutanoic acid
(Main chain: COOH–CH₂–C(CH₃)₂–CH₃ = 4 carbons = butanoic acid; two methyls at C-3)

(vii) OHC–C₆H₄–CHO (para)
Benzene-1,4-dicarbaldehyde.
IUPAC name: Benzene-1,4-dicarbaldehyde

8.3Draw the structures of the following compounds.
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
(iii) p-Methylbenzaldehyde
(iv) 4-Methylpent-3-en-2-one
(v) 4-Chloropentan-2-one
(vi) 3-Bromo-4-phenylpentanoic acid
(vii) p,p'-Dihydroxybenzophenone
(viii) Hex-2-en-4-ynoic acidShow solution

(i) 3-Methylbutanal
$\text{CH}_3-\underset{|}{\text{CH}}(\text{CH}_3)-\text{CH}_2-\text{CHO}$
(CH₃)₂CHCH₂CHO

(ii) p-Nitropropiophenone
Propiophenone = C₆H₅–CO–CH₂CH₃; nitro group at para position.
$\text{O}_2\text{N}-\text{C}_6\text{H}_4-\overset{O}{\overset{\|}{\text{C}}}-\text{CH}_2\text{CH}_3 \quad (\text{NO}_2 \text{ at para})$

(iii) p-Methylbenzaldehyde
$\text{CH}_3-\text{C}_6\text{H}_4-\text{CHO} \quad (\text{CH}_3 \text{ at para})$

(iv) 4-Methylpent-3-en-2-one
C1(CH₃)–C2(CO)–C3(=CH)–C4(C(CH₃)=)–C5(CH₃)
$\text{CH}_3-\overset{O}{\overset{\|}{\text{C}}}-\text{CH}=\text{C}(\text{CH}_3)-\text{CH}_3$

(v) 4-Chloropentan-2-one
$\text{CH}_3-\overset{O}{\overset{\|}{\text{C}}}-\text{CH}_2-\underset{|}{\text{CH}}(\text{Cl})-\text{CH}_3$
CH₃COCH₂CHClCH₃

(vi) 3-Bromo-4-phenylpentanoic acid
C1(COOH)–C2(CH₂)–C3(CHBr)–C4(CH(C₆H₅))–C5(CH₃)
$\text{HOOC}-\text{CH}_2-\underset{|}{\text{CH}}(\text{Br})-\underset{|}{\text{CH}}(\text{C}_6\text{H}_5)-\text{CH}_3$

(vii) p,p'-Dihydroxybenzophenone
Benzophenone with –OH at para position of each phenyl ring.
$\text{HO}-\text{C}_6\text{H}_4-\overset{O}{\overset{\|}{\text{C}}}-\text{C}_6\text{H}_4-\text{OH} \quad (\text{both OH at para})$

(viii) Hex-2-en-4-ynoic acid
C1(COOH)–C2(=CH)–C3(CH=)–C4(C≡)–C5(≡C)–C6(CH₃)
$\text{HOOC}-\text{CH}=\text{CH}-\text{C}\equiv\text{C}-\text{CH}_3$

8.4Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.
(i) CH₃CO(CH₂)₄CH₃
(ii) CH₃CH₂CHBrCH₂CH(CH₃)CHO
(iii) CH₃(CH₂)₅CHO
(iv) Ph–CH=CH–CHO
(v) [cyclopentanone structure from figure]
(vi) PhCOPhShow solution

(i) CH₃CO(CH₂)₄CH₃
Longest chain: C1(CH₃)–C2(CO)–C3(CH₂)–C4(CH₂)–C5(CH₂)–C6(CH₂)–C7(CH₃) = heptan-2-one
IUPAC name: Heptan-2-one
Common name: Methyl n-amyl ketone

(ii) CH₃CH₂CHBrCH₂CH(CH₃)CHO
Longest chain containing –CHO: 6 carbons.
C1(CHO)–C2(CH(CH₃))–C3(CH₂)–C4(CHBr)–C5(CH₂)–C6(CH₃)
IUPAC name: 4-Bromo-2-methylhexanal

(iii) CH₃(CH₂)₅CHO
Seven-carbon chain with –CHO at C-1.
IUPAC name: Heptanal
Common name: Heptaldehyde (Enanthaldehyde)

(iv) Ph–CH=CH–CHO
Three-carbon chain with –CHO at C-1, double bond at C-2, phenyl at C-3.
IUPAC name: 3-Phenylprop-2-enal
Common name: Cinnamaldehyde

(v) [Cyclopentanone — figure not readable; assumed cyclopentanone]
IUPAC name: Cyclopentanone
Common name: Cyclopentanone

(vi) PhCOPh
Two phenyl groups on either side of carbonyl.
IUPAC name: Diphenylmethanone
Common name: Benzophenone

8.5Draw structures of the following derivatives.
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cyclopropanone oxime
(iii) Acetaldehyde dimethyl acetal
(iv) The semicarbazone of cyclobutanone
(v) The ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehydeShow solution

(i) 2,4-Dinitrophenylhydrazone of benzaldehyde:
Benzaldehyde reacts with 2,4-dinitrophenylhydrazine:
$\text{C}_6\text{H}_5\text{CH=N–NH–C}_6\text{H}_3(\text{NO}_2)_2\text{-2,4}$
(C₆H₅–CH=N–NH–C₆H₃(NO₂)₂ where NO₂ groups are at 2 and 4 positions of the phenyl ring)

(ii) Cyclopropanone oxime:
Cyclopropanone reacts with NH₂OH:
Cyclopropane ring with =NOH replacing =O.
$\text{Cyclopropane ring with C=NOH}$
(Three-membered ring with C=N–OH at the carbonyl carbon)

(iii) Acetaldehyde dimethyl acetal:
CH₃CHO + 2CH₃OH → CH₃CH(OCH₃)₂ + H₂O
$\text{CH}_3-\text{CH}(\text{OCH}_3)_2$

(iv) Semicarbazone of cyclobutanone:
Cyclobutanone reacts with semicarbazide (H₂NNHCONH₂):
Cyclobutane ring with C=N–NH–CO–NH₂ replacing C=O.
$\text{Cyclobutane ring with C=N–NHCONH}_2$

(v) Ethylene ketal of hexan-3-one:
Hexan-3-one (CH₃CH₂COCH₂CH₂CH₃) reacts with ethylene glycol (HOCH₂CH₂OH) in the presence of acid:
The C=O is replaced by a 1,3-dioxolane ring.
$\text{CH}_3\text{CH}_2-\underset{\displaystyle\text{O}\quad\text{O}}{\underset{\displaystyle|\qquad|}{\text{C}}}-\text{CH}_2\text{CH}_2\text{CH}_3$
where the two oxygens are bridged by –CH₂CH₂– (five-membered 1,3-dioxolane ring at C-3 of hexane).

(vi) Methyl hemiacetal of formaldehyde:
HCHO + CH₃OH → HOCH₂OCH₃
$\text{HO–CH}_2\text{–OCH}_3$
(Hydroxymethyl methyl ether)

8.6Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.
(i) PhMgBr and then H₃O⁺
(ii) Tollens' reagent
(iii) Semicarbazide and weak acid
(iv) Excess ethanol and acid
(v) Zinc amalgam and dilute hydrochloric acidShow solution

Cyclohexanecarbaldehyde = Cyclohexane ring with –CHO substituent.

(i) PhMgBr and then H₃O⁺ (Grignard reaction):
The Grignard reagent (PhMgBr) acts as a nucleophile and adds to the carbonyl carbon of –CHO. After hydrolysis with H₃O⁺, a secondary alcohol is formed.
$\text{C}_6\text{H}_{11}\text{CHO} + \text{C}_6\text{H}_5\text{MgBr} \xrightarrow{\text{(i) ether, (ii) H}_3\text{O}^+} \text{C}_6\text{H}_{11}\text{CH(OH)C}_6\text{H}_5$
Product: Cyclohexyl(phenyl)methanol — a secondary alcohol.

(ii) Tollens' reagent (ammoniacal AgNO₃):
Aldehydes are oxidised by Tollens' reagent to carboxylic acids (as ammonium salt); silver mirror is deposited.
$\text{C}_6\text{H}_{11}\text{CHO} + 2[\text{Ag(NH}_3)_2]^+ + 2\text{OH}^- \rightarrow \text{C}_6\text{H}_{11}\text{COO}^- + 2\text{Ag}\downarrow + 4\text{NH}_3 + \text{H}_2\text{O}$
Product: Cyclohexanecarboxylic acid (ammonium salt) + silver mirror.

(iii) Semicarbazide and weak acid:
Condensation reaction gives semicarbazone.
$\text{C}_6\text{H}_{11}\text{CHO} + \text{H}_2\text{N–NHCONH}_2 \xrightarrow{\text{weak acid}} \text{C}_6\text{H}_{11}\text{CH=N–NHCONH}_2 + \text{H}_2\text{O}$
Product: Cyclohexanecarbaldehyde semicarbazone.

(iv) Excess ethanol and acid (Acetal formation):
Aldehyde reacts with two equivalents of ethanol in the presence of dry HCl to form an acetal.
$\text{C}_6\text{H}_{11}\text{CHO} + 2\text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{dry HCl}} \text{C}_6\text{H}_{11}\text{CH(OC}_2\text{H}_5)_2 + \text{H}_2\text{O}$
Product: Cyclohexanecarbaldehyde diethyl acetal.

(v) Zinc amalgam and dilute HCl (Clemmensen reduction):
The carbonyl group (–CHO) is reduced to –CH₃ (methylene/methyl group).
$\text{C}_6\text{H}_{11}\text{CHO} \xrightarrow{\text{Zn-Hg/conc. HCl}} \text{C}_6\text{H}_{11}\text{CH}_3$
Product: Methylcyclohexane.

8.7Which 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-DimethylbutanalShow solution

Key rules:

Aldol condensation: Requires at least one α-hydrogen (in aldehyde or ketone).
Cannizzaro reaction: Aldehydes with NO α-hydrogen undergo disproportionation in conc. NaOH.
Neither: Ketones without α-H, or alcohols.

(i) Methanal (HCHO):
No α-hydrogen → Cannizzaro reaction
$2\text{HCHO} \xrightarrow{\text{conc. NaOH}} \text{CH}_3\text{OH} + \text{HCOONa}$
(Methanol + Sodium formate)

(ii) 2-Methylpentanal (CH₃CH₂CH₂CH(CH₃)CHO):
Has α-hydrogen (at C-2) → Aldol condensation
$2\text{CH}_3\text{CH}_2\text{CH}_2\text{CH(CH}_3)\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(CH}_3)\text{CH(OH)–C(CH}_3)(\text{CH}_2\text{CH}_2\text{CH}_3)\text{CHO}$
(β-hydroxy aldehyde product)

(iii) Benzaldehyde (C₆H₅CHO):
No α-hydrogen → Cannizzaro reaction
$2\text{C}_6\text{H}_5\text{CHO} \xrightarrow{\text{conc. NaOH}} \text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{C}_6\text{H}_5\text{COONa}$
(Benzyl alcohol + Sodium benzoate)

(iv) Benzophenone (C₆H₅COC₆H₅):
Ketone with no α-hydrogen → Neither (Cannizzaro requires aldehyde; no α-H so no aldol)

(v) Cyclohexanone:
Has α-hydrogen → Aldol condensation
$2\text{Cyclohexanone} \xrightarrow{\text{dil. NaOH}} \text{2-(1-Hydroxycyclohexyl)cyclohexan-1-one}$
(β-hydroxy ketone — the ketol product)

(vi) 1-Phenylpropanone (C₆H₅COCH₂CH₃):
Has α-hydrogen (–CH₂– adjacent to C=O) → Aldol condensation
$2\text{C}_6\text{H}_5\text{COCH}_2\text{CH}_3 \xrightarrow{\text{dil. NaOH}} \text{C}_6\text{H}_5\text{CO–CH(CH}_3)\text{–C(OH)(C}_6\text{H}_5)\text{–CH}_2\text{CH}_3$
(β-hydroxy ketone)

(vii) Phenylacetaldehyde (C₆H₅CH₂CHO):
Has α-hydrogen (–CH₂– at α-position) → Aldol condensation
$2\text{C}_6\text{H}_5\text{CH}_2\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{C}_6\text{H}_5\text{CH}_2\text{CH(OH)–CH(C}_6\text{H}_5)\text{CHO}$
(β-hydroxy aldehyde)

(viii) Butan-1-ol:
Not an aldehyde or ketone → Neither

(ix) 2,2-Dimethylbutanal (CH₃CH₂C(CH₃)₂CHO):
No α-hydrogen (the α-carbon C-2 bears two methyl groups and an ethyl group — all four positions occupied, no H at α-carbon) → Cannizzaro reaction
$2\text{CH}_3\text{CH}_2\text{C(CH}_3)_2\text{CHO} \xrightarrow{\text{conc. NaOH}} \text{CH}_3\text{CH}_2\text{C(CH}_3)_2\text{CH}_2\text{OH} + \text{CH}_3\text{CH}_2\text{C(CH}_3)_2\text{COONa}$

8.8How will you convert ethanol into the following compounds?
(i) Butane-1,3-diol
(ii) But-2-enal
(iii) But-2-enoic acid

Starting material: Ethanol (CH₃CH₂OH)

(i) Ethanol → Butane-1,3-diol

Step 1: Oxidise ethanol to acetaldehyde (ethanal) using PCC or mild oxidising agent.
$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{PCC}} \text{CH}_3\text{CHO}$

Step 2: Aldol condensation of acetaldehyde with dilute NaOH gives 3-hydroxybutanal (aldol).
$2\text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO}$

Step 3: Reduce 3-hydroxybutanal with NaBH₄ (reduces –CHO to –CH₂OH without affecting –OH).
$\text{CH}_3\text{CH(OH)CH}_2\text{CHO} \xrightarrow{\text{NaBH}_4} \text{CH}_3\text{CH(OH)CH}_2\text{CH}_2\text{OH}$
Product: Butane-1,3-diol

(ii) Ethanol → But-2-enal (Crotonaldehyde)

Step 1: Oxidise ethanol to acetaldehyde.
$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{PCC}} \text{CH}_3\text{CHO}$

Step 2: Aldol condensation followed by dehydration (aldol condensation in presence of dil. NaOH, then heat).
$2\text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO} \xrightarrow{\Delta} \text{CH}_3\text{CH=CHCHO} + \text{H}_2\text{O}$
Product: But-2-enal (Crotonaldehyde)

(iii) Ethanol → But-2-enoic acid (Crotonic acid)

Step 1: Oxidise ethanol to acetaldehyde.
$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{PCC}} \text{CH}_3\text{CHO}$

Step 2: Aldol condensation and dehydration to give but-2-enal (as above).
$2\text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH},\Delta} \text{CH}_3\text{CH=CHCHO}$

Step 3: Oxidise but-2-enal to but-2-enoic acid using alkaline KMnO₄.
$\text{CH}_3\text{CH=CHCHO} \xrightarrow{\text{KMnO}_4/\text{OH}^-} \text{CH}_3\text{CH=CHCOOH}$
Product: But-2-enoic acid (Crotonic acid)

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8.9Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.

Given: Propanal (CH₃CH₂CHO) and Butanal (CH₃CH₂CH₂CHO)

In aldol condensation, the α-carbon of one molecule (nucleophile/enolate) attacks the carbonyl carbon of another molecule (electrophile).

Product 1: Propanal + Propanal (self-condensation)

Nucleophile: Propanal (enolate at α-C)
Electrophile: Propanal (carbonyl C)

$\text{CH}_3\text{CH}_2\text{CHO} + \text{CH}_3\text{CH}_2\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH}_2\text{CH(OH)–CH(CH}_3)\text{CHO}$
Product: 2-Methyl-3-hydroxypentanal

Product 2: Butanal + Butanal (self-condensation)

Nucleophile: Butanal (enolate)
Electrophile: Butanal (carbonyl C)

$2\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(OH)–CH(CH}_2\text{CH}_3)\text{CHO}$
Product: 2-Ethyl-3-hydroxy hexanal

Product 3: Propanal (nucleophile) + Butanal (electrophile)

Enolate of propanal attacks carbonyl of butanal.

$\text{CH}_3\text{CH}_2\text{CHO} + \text{CH}_3\text{CH}_2\text{CH}_2\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(OH)–CH(CH}_3)\text{CHO}$
Product: 2-Methyl-3-hydroxyhexanal

Product 4: Butanal (nucleophile) + Propanal (electrophile)

Enolate of butanal attacks carbonyl of propanal.

$\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO} + \text{CH}_3\text{CH}_2\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH}_2\text{CH(OH)–CH(CH}_2\text{CH}_3)\text{CHO}$
Product: 2-Ethyl-3-hydroxypentanal

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8.10An organic compound with the molecular formula C₈H₁₀O forms 2,4-DNP derivative, reduces Tollens' reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.

Given data analysis:

Molecular formula: C₈H₁₀O (degree of unsaturation = (2×8+2−10)/2 = 4 → benzene ring present)
Forms 2,4-DNP derivative → contains C=O group (aldehyde or ketone)
Reduces Tollens' reagent → aldehyde (–CHO)
Undergoes Cannizzaro reaction → aldehyde with NO α-hydrogen
Vigorous oxidation gives 1,2-benzenedicarboxylic acid (phthalic acid) → two substituents at ortho positions on benzene ring

Deduction:
The compound is an aromatic aldehyde with no α-hydrogen. Since oxidation gives phthalic acid (benzene-1,2-dicarboxylic acid), there must be two carbon-containing substituents at ortho positions.

Molecular formula C₈H₁₀O: Benzene (C₆H₄) + two substituents totalling C₂H₆O.
One substituent is –CHO (C₁H₁O) and the other is –CH₃ (C₁H₃) → total C₂H₄O...

Actually: C₆H₄ + CHO + CH₂CH₃ = C₆H₄ + C₃H₅O → C₉H₉O (too many carbons).

Let us reconsider: C₈H₁₀O with a benzene ring (C₆H₅ = C₆H₅) leaves C₂H₅O. If –CHO is one group, remaining is C₁H₄ = –CH₃ + H...

C₆H₄(CHO)(CH₃) = C₈H₈O (MW = 120). But formula given is C₈H₁₀O.

Correction: The molecular formula should be C₈H₈O (as is standard for this problem in NCERT). With C₈H₈O:

Benzene ring + –CHO + –CH₃ at ortho positions = o-methylbenzaldehyde (2-methylbenzaldehyde)
No α-H on the –CHO carbon (attached to benzene ring) → Cannizzaro reaction ✓
Reduces Tollens' reagent ✓
Forms 2,4-DNP derivative ✓
Vigorous oxidation of –CH₃ and –CHO both → –COOH giving phthalic acid ✓

The compound is 2-Methylbenzaldehyde (o-Tolualdehyde).
$\text{Structure: } o\text{-CH}_3\text{C}_6\text{H}_4\text{CHO}$

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8.11An organic compound (A) (molecular formula C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.

Analysis:

(A) has formula C₈H₁₆O₂ → degree of unsaturation = (2×8+2−16)/2 = 1 → one degree (ester C=O)
(A) is an ester (hydrolysis gives acid B and alcohol C)
Oxidation of C gives B → C is a primary alcohol that oxidises to the same carboxylic acid B
C on dehydration gives but-1-ene (CH₂=CHCH₂CH₃) → C is butan-1-ol (CH₃CH₂CH₂CH₂OH)

Identifying B:
Oxidation of butan-1-ol (C) with chromic acid gives butanoic acid (B).
$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{CrO}_3/\text{H}^+} \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$
So B = Butanoic acid (C₄H₈O₂)

Identifying A:
A = ester of butanoic acid and butan-1-ol = Butyl butanoate
$\text{C}_4\text{H}_9\text{COOC}_4\text{H}_9 \rightarrow \text{C}_8\text{H}_{16}\text{O}_2 \checkmark$

Equations:

Hydrolysis of A:
$\text{CH}_3\text{CH}_2\text{CH}_2\text{COOC}_4\text{H}_9 + \text{H}_2\text{O} \xrightarrow{\text{dil. H}_2\text{SO}_4} \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} + \text{C}_4\text{H}_9\text{OH}$
(Butyl butanoate → Butanoic acid + Butan-1-ol)

Oxidation of C to B:
$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{CrO}_3/\text{H}_2\text{SO}_4} \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$

Dehydration of C:
$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4,\,170°\text{C}} \text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2 + \text{H}_2\text{O}$
(But-1-ene)

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8.12Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH, (CH₃)₂CHCOOH, CH₃CH₂CH₂COOH (acid strength)
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)

(i) Reactivity towards HCN (nucleophilic addition):

Reactivity decreases with:

Increasing steric bulk around carbonyl carbon
Increasing electron density on carbonyl carbon (EDG reduce electrophilicity)

Acetaldehyde (CH₃CHO): One small methyl group → most reactive
Acetone (CH₃COCH₃): Two methyl groups → less reactive than acetaldehyde
Methyl tert-butyl ketone (CH₃COC(CH₃)₃): One methyl + one bulky tert-butyl → less reactive than acetone
Di-tert-butyl ketone ((CH₃)₃CCOC(CH₃)₃): Two bulky tert-butyl groups → least reactive

Increasing order of reactivity:
$\text{Di-tert-butyl ketone} < \text{Methyl tert-butyl ketone} < \text{Acetone} < \text{Acetaldehyde}$

(ii) Acid strength of substituted butanoic acids:

Electron-withdrawing –Br increases acid strength; closer to –COOH → stronger acid.

CH₃CH₂CH₂COOH (butanoic acid): No EWG → weakest
(CH₃)₂CHCOOH (2-methylpropanoic acid): Branching (EDG effect) → slightly weaker than butanoic acid? Actually isobutyric acid is slightly weaker due to +I of branched alkyl.
CH₃CH(Br)CH₂COOH (3-bromobutanoic acid): Br at β-position (C-3)
CH₃CH₂CH(Br)COOH (2-bromobutanoic acid): Br at α-position (C-2) → closest to –COOH → strongest

Increasing order of acid strength:
$\text{(CH}_3)_2\text{CHCOOH} < \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} < \text{CH}_3\text{CH(Br)CH}_2\text{COOH} < \text{CH}_3\text{CH}_2\text{CH(Br)COOH}$

(iii) Acid strength of substituted benzoic acids:

EWG (–NO₂) increases acid strength; EDG (–OCH₃) decreases acid strength.

4-Methoxybenzoic acid: –OCH₃ is EDG → weakest acid
Benzoic acid: No substituent → reference
4-Nitrobenzoic acid: One –NO₂ (EWG) → stronger than benzoic acid
3,4-Dinitrobenzoic acid: Two –NO₂ groups → strongest acid

Increasing order of acid strength:
$\text{4-Methoxybenzoic acid} < \text{Benzoic acid} < \text{4-Nitrobenzoic acid} < \text{3,4-Dinitrobenzoic acid}$

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8.13Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Propanal and Propanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid
(iv) Benzoic acid and Ethyl benzoate
(v) Pentan-2-one and Pentan-3-one
(vi) Benzaldehyde and Acetophenone
(vii) Ethanal and Propanal

(i) Propanal and Propanone:

Test: Add Tollens' reagent (ammoniacal AgNO₃) or Fehling's solution.
Propanal (aldehyde): Gives silver mirror with Tollens' reagent (or brick-red precipitate with Fehling's solution).
Propanone (ketone): No reaction with Tollens' or Fehling's reagent.

(ii) Acetophenone and Benzophenone:

Test: Iodoform test (I₂/NaOH).
Acetophenone (C₆H₅COCH₃): Has –COCH₃ group → gives yellow precipitate of iodoform (CHI₃).
Benzophenone (C₆H₅COC₆H₅): No –COCH₃ group → no iodoform.

(iii) Phenol and Benzoic acid:

Test: Add NaHCO₃ solution.
Benzoic acid: Dissolves with effervescence (CO₂ gas evolved): C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂↑
Phenol: Does not react with NaHCO₃ (phenol is a weaker acid than carbonic acid); no effervescence.

(iv) Benzoic acid and Ethyl benzoate:

Test: Add NaHCO₃ solution or litmus paper.
Benzoic acid: Dissolves in NaHCO₃ with effervescence (CO₂); turns blue litmus red.
Ethyl benzoate (ester): Does not react with NaHCO₃; neutral to litmus.

(v) Pentan-2-one and Pentan-3-one:

Test: Iodoform test (I₂/NaOH).
Pentan-2-one (CH₃COCH₂CH₂CH₃): Has –COCH₃ group → gives yellow iodoform precipitate.
Pentan-3-one (CH₃CH₂COCH₂CH₃): No –COCH₃ group → no iodoform.

(vi) Benzaldehyde and Acetophenone:

Test: Tollens' reagent.
Benzaldehyde (aldehyde): Gives silver mirror.
Acetophenone (ketone): No reaction.
(Alternatively): Iodoform test — Acetophenone gives iodoform; Benzaldehyde does not.

(vii) Ethanal and Propanal:

Test: Iodoform test (I₂/NaOH).
Ethanal (CH₃CHO): Has –CH₃ adjacent to C=O → gives yellow iodoform precipitate (CHI₃).
Propanal (CH₃CH₂CHO): No –COCH₃ group → no iodoform.

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8.14How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.
(i) Methyl benzoate
(ii) m-Nitrobenzoic acid
(iii) p-Nitrobenzoic acid
(iv) Phenylacetic acid
(v) p-Nitrobenzaldehyde

(i) Benzene → Methyl benzoate

Step 1: Friedel-Crafts acylation with CH₃COCl/AlCl₃ to get acetophenone.
$\text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3$
Step 2: Oxidise acetophenone with alkaline KMnO₄ to get benzoic acid.
$\text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{\text{KMnO}_4/\text{OH}^-} \text{C}_6\text{H}_5\text{COOH}$
Step 3: Esterify with methanol (CH₃OH) in presence of H₂SO₄.
$\text{C}_6\text{H}_5\text{COOH} + \text{CH}_3\text{OH} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_5\text{COOCH}_3 + \text{H}_2\text{O}$

(ii) Benzene → m-Nitrobenzoic acid

Step 1: Friedel-Crafts acylation → Acetophenone (as above).
Step 2: Nitration of acetophenone (–COCH₃ is meta-director) with HNO₃/H₂SO₄ → m-nitroacetophenone.
$\text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} m\text{-NO}_2\text{C}_6\text{H}_4\text{COCH}_3$
Step 3: Oxidise with alkaline KMnO₄.
$m\text{-NO}_2\text{C}_6\text{H}_4\text{COCH}_3 \xrightarrow{\text{KMnO}_4/\text{OH}^-} m\text{-NO}_2\text{C}_6\text{H}_4\text{COOH}$

(iii) Benzene → p-Nitrobenzoic acid

Step 1: Friedel-Crafts acylation → Acetophenone.
Step 2: Oxidise acetophenone → Benzoic acid.
$\text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{\text{KMnO}_4} \text{C}_6\text{H}_5\text{COOH}$
Step 3: Nitration of benzoic acid (–COOH is meta-director)... gives m-nitrobenzoic acid.

Alternative route for p-nitrobenzoic acid:
Step 1: Nitration of benzene → Nitrobenzene.
$\text{C}_6\text{H}_6 \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{C}_6\text{H}_5\text{NO}_2$
Step 2: Friedel-Crafts acylation of nitrobenzene is not feasible (deactivated ring). Instead:
Step 1: Friedel-Crafts acylation → Acetophenone → Benzoic acid (as above).
Step 2: Nitration of benzoic acid gives m-nitrobenzoic acid (not para).

For p-nitrobenzoic acid:
Step 1: Nitration of benzene → Nitrobenzene.
Step 2: Reduce –NO₂ to –NH₂ (Fe/HCl) → Aniline.
Step 3: Diazotisation and Sandmeyer reaction to introduce –CN → p-aminobenzonitrile...

Simpler route:
Step 1: Toluene from benzene + CH₃Cl/AlCl₃ → Toluene (but toluene has more than 1 C in reagent).

Using only ≤1 C reagents:
Step 1: C₆H₆ + CH₃COCl/AlCl₃ → C₆H₅COCH₃ (acetophenone)
Step 2: Nitration → gives m-nitroacetophenone (meta due to –COCH₃)
This gives m-product, not para.

For para: Use benzene → nitrobenzene → reduce to aniline → acetylation → p-nitroacetanilide (nitration para to –NHCOCH₃) → hydrolysis → p-nitroaniline → Sandmeyer → p-nitrobenzoic acid via oxidation.

Practical answer for board exam:
Step 1: Benzene → Acetophenone (Friedel-Crafts with CH₃COCl/AlCl₃)
Step 2: Acetophenone → Benzoic acid (KMnO₄ oxidation)
Step 3: Benzoic acid + HNO₃/H₂SO₄ → m-Nitrobenzoic acid (for m-isomer)
For p-isomer: Benzene → Toluene (CH₃Cl/AlCl₃) → p-Nitrotoluene (HNO₃/H₂SO₄) → p-Nitrobenzoic acid (KMnO₄)

(iv) Benzene → Phenylacetic acid

Step 1: Benzene + CH₂Cl₂/AlCl₃ → not straightforward. Better:
Step 1: Benzene → Bromobenzene (Br₂/FeBr₃)... but we need –CH₂COOH.
Step 1: Benzene + CH₂O (formaldehyde, 1C) + HCl → Benzyl chloride (via chloromethylation).
$\text{C}_6\text{H}_6 + \text{HCHO} + \text{HCl} \xrightarrow{\text{ZnCl}_2} \text{C}_6\text{H}_5\text{CH}_2\text{Cl}$
Step 2: Benzyl chloride + KCN → Benzyl cyanide.
$\text{C}_6\text{H}_5\text{CH}_2\text{Cl} + \text{KCN} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{CN}$
Step 3: Hydrolysis of nitrile.
$\text{C}_6\text{H}_5\text{CH}_2\text{CN} \xrightarrow{\text{H}_3\text{O}^+/\Delta} \text{C}_6\text{H}_5\text{CH}_2\text{COOH}$
Product: Phenylacetic acid

(v) Benzene → p-Nitrobenzaldehyde

Step 1: Benzene → Acetophenone (Friedel-Crafts: CH₃COCl/AlCl₃).
Step 2: Nitration of acetophenone → m-nitroacetophenone (meta director).

For para product:
Step 1: Benzene + CO/HCl/AlCl₃ (Gattermann-Koch) → Benzaldehyde.
$\text{C}_6\text{H}_6 + \text{CO} + \text{HCl} \xrightarrow{\text{AlCl}_3/\text{CuCl}} \text{C}_6\text{H}_5\text{CHO}$
Step 2: Nitration of benzaldehyde (–CHO is meta-director) → m-nitrobenzaldehyde.

For p-nitrobenzaldehyde:
Step 1: Benzene → Nitrobenzene (HNO₃/H₂SO₄).
Step 2: Nitrobenzene + CO/HCl/AlCl₃ → p-Nitrobenzaldehyde (CO attacks para to –NO₂ due to steric reasons, though –NO₂ is meta-director; in practice Gattermann-Koch on nitrobenzene gives mainly para product).

Alternatively (standard board answer):
Benzene → Toluene (CH₃Cl/AlCl₃) → p-Nitrotoluene (HNO₃/H₂SO₄) → p-Nitrobenzaldehyde (CrO₂Cl₂ or CrO₃/Ac₂O oxidation of –CH₃ to –CHO).

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8.15How will you bring about the following conversions in not more than two steps?
(i) Propanone to Propene
(ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal
(iv) Benzene to m-Nitroacetophenone
(v) Benzaldehyde to Benzophenone
(vi) Bromobenzene to 1-Phenylethanol
(vii) Benzaldehyde to 3-Phenylpropan-1-ol
(viii) Benzaldehyde to α-Hydroxyphenylacetic acid
(ix) Benzoic acid to m-Nitrobenzyl alcohol

(i) Propanone → Propene (2 steps)

Step 1: Reduce propanone to propan-2-ol using NaBH₄ or LiAlH₄.
$\text{CH}_3\text{COCH}_3 \xrightarrow{\text{NaBH}_4} \text{CH}_3\text{CH(OH)CH}_3$
Step 2: Dehydrate propan-2-ol with conc. H₂SO₄ at 170°C.
$\text{CH}_3\text{CH(OH)CH}_3 \xrightarrow{\text{conc. H}_2\text{SO}_4,\,170°\text{C}} \text{CH}_3\text{CH=CH}_2 + \text{H}_2\text{O}$

(ii) Benzoic acid → Benzaldehyde (2 steps)

Step 1: Convert benzoic acid to benzoyl chloride using SOCl₂ (or PCl₅).
$\text{C}_6\text{H}_5\text{COOH} + \text{SOCl}_2 \rightarrow \text{C}_6\text{H}_5\text{COCl} + \text{SO}_2 + \text{HCl}$
Step 2: Reduce benzoyl chloride to benzaldehyde using Rosenmund reduction (H₂/Pd-BaSO₄).
$\text{C}_6\text{H}_5\text{COCl} \xrightarrow{\text{H}_2/\text{Pd-BaSO}_4} \text{C}_6\text{H}_5\text{CHO} + \text{HCl}$

(iii) Ethanol → 3-Hydroxybutanal (2 steps)

Step 1: Oxidise ethanol to acetaldehyde (PCC or mild oxidation).
$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{PCC}} \text{CH}_3\text{CHO}$
Step 2: Aldol condensation of acetaldehyde with dilute NaOH.
$2\text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO}$
(3-Hydroxybutanal)

(iv) Benzene → m-Nitroacetophenone (2 steps)

Step 1: Friedel-Crafts acylation of benzene with CH₃COCl/AlCl₃ → Acetophenone.
$\text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3$
Step 2: Nitration of acetophenone with HNO₃/H₂SO₄ (–COCH₃ is meta-director).
$\text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} m\text{-O}_2\text{N–C}_6\text{H}_4\text{COCH}_3$

(v) Benzaldehyde → Benzophenone (2 steps)

Step 1: React benzaldehyde with PhMgBr (Grignard reagent) to get α-hydroxydiphenylmethane (benzhydrol).
$\text{C}_6\text{H}_5\text{CHO} + \text{C}_6\text{H}_5\text{MgBr} \xrightarrow{\text{(i) ether, (ii) H}_3\text{O}^+} \text{C}_6\text{H}_5\text{CH(OH)C}_6\text{H}_5$
Step 2: Oxidise benzhydrol with CrO₃ or KMnO₄ to benzophenone.
$\text{C}_6\text{H}_5\text{CH(OH)C}_6\text{H}_5 \xrightarrow{\text{CrO}_3} \text{C}_6\text{H}_5\text{COC}_6\text{H}_5$

(vi) Bromobenzene → 1-Phenylethanol (2 steps)

Step 1: Convert bromobenzene to Grignard reagent.
$\text{C}_6\text{H}_5\text{Br} + \text{Mg} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{MgBr}$
Step 2: React with acetaldehyde (CH₃CHO) followed by hydrolysis.
$\text{C}_6\text{H}_5\text{MgBr} + \text{CH}_3\text{CHO} \xrightarrow{\text{(i) ether, (ii) H}_3\text{O}^+} \text{C}_6\text{H}_5\text{CH(OH)CH}_3$
(1-Phenylethanol)

(vii) Benzaldehyde → 3-Phenylpropan-1-ol (2 steps)

Step 1: Aldol-type condensation — react benzaldehyde with acetaldehyde (cross aldol) in dil. NaOH to give cinnamaldehyde (3-phenylprop-2-enal).
$\text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{C}_6\text{H}_5\text{CH=CHCHO}$
Step 2: Reduce cinnamaldehyde with LiAlH₄ (reduces both C=C and C=O) or catalytic hydrogenation.
$\text{C}_6\text{H}_5\text{CH=CHCHO} \xrightarrow{\text{H}_2/\text{Ni or LiAlH}_4} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$
(3-Phenylpropan-1-ol)

(viii) Benzaldehyde → α-Hydroxyphenylacetic acid (Mandelic acid) (1 step)

React benzaldehyde with HCN to form cyanohydrin, then hydrolyse.
$\text{C}_6\text{H}_5\text{CHO} + \text{HCN} \rightarrow \text{C}_6\text{H}_5\text{CH(OH)CN} \xrightarrow{\text{H}_3\text{O}^+} \text{C}_6\text{H}_5\text{CH(OH)COOH}$
(α-Hydroxyphenylacetic acid / Mandelic acid)

(ix) Benzoic acid → m-Nitrobenzyl alcohol (2 steps)

Step 1: Nitration of benzoic acid with HNO₃/H₂SO₄ (–COOH is meta-director) → m-Nitrobenzoic acid.
$\text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} m\text{-O}_2\text{N–C}_6\text{H}_4\text{COOH}$
Step 2: Reduce m-nitrobenzoic acid with LiAlH₄ (reduces –COOH to –CH₂OH selectively, –NO₂ may also be reduced; use diborane B₂H₆ to selectively reduce –COOH to –CH₂OH).
$m\text{-O}_2\text{N–C}_6\text{H}_4\text{COOH} \xrightarrow{\text{B}_2\text{H}_6/\text{THF}} m\text{-O}_2\text{N–C}_6\text{H}_4\text{CH}_2\text{OH}$
(m-Nitrobenzyl alcohol)

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8.16Describe the following:
(i) Acetylation
(ii) Cannizzaro reaction
(iii) Cross aldol condensation
(iv) Decarboxylation

(i) Acetylation:
Acetylation is the process of introducing an acetyl group (–COCH₃) into an organic compound, usually by replacing an active hydrogen (–OH, –NH₂, –SH) with –COCH₃. It is carried out using acetic anhydride or acetyl chloride in the presence of a base.
$\text{C}_6\text{H}_5\text{OH} + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{OCOCH}_3 + \text{CH}_3\text{COOH}$
(Phenol → Phenyl acetate)

(ii) Cannizzaro Reaction:
Aldehydes that do not have any α-hydrogen atom undergo self-oxidation and reduction (disproportionation) in the presence of concentrated alkali. One molecule of the aldehyde is oxidised to the carboxylate anion (acid salt) and another is reduced to the alcohol.
$2\text{HCHO} \xrightarrow{\text{conc. NaOH}} \text{CH}_3\text{OH} + \text{HCOONa}$
(Methanal → Methanol + Sodium formate)

(iii) Cross Aldol Condensation:
When aldol condensation is carried out between two different carbonyl compounds (both having α-hydrogen, or one with and one without α-hydrogen), it is called cross aldol condensation. It gives a mixture of products.
$\text{CH}_3\text{CHO} + \text{CH}_3\text{COCH}_3 \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{COCH}_2\text{CH(OH)CH}_3 \text{ (and other products)}$
When one component has no α-H (e.g., benzaldehyde), a single cross aldol product is obtained:
$\text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{C}_6\text{H}_5\text{CH=CHCHO}$
(Cinnamaldehyde)

(iv) Decarboxylation:
Decarboxylation is the loss of CO₂ from a carboxylic acid. Carboxylic acids lose CO₂ when their calcium or sodium salts are heated with soda lime (NaOH + CaO).
$\text{RCOONa} + \text{NaOH} \xrightarrow{\text{CaO},\Delta} \text{RH} + \text{Na}_2\text{CO}_3$
Example:
$\text{CH}_3\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO},\Delta} \text{CH}_4 + \text{Na}_2\text{CO}_3$

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8.17Complete each synthesis by giving missing starting material, reagent or products:
(i) [figure] → ?
(ii) C₆H₅CHO + H₂NCONHNH₂ → ?
(iii) C₆H₅CHO + H₂NCONHNH₂ → ?
(iv) [figure]
(v) C₆H₅CHO → (dil. NaOH) → (NaCN/HCl) → ?
(vi) CH₃CH₂COO⁻ → (Δ) → (i) NaBH₄, (ii) H⁺ → ?
(vii) C₆H₅CH → CH₃CH₂CHO
(ix) OH → (CrO₃) → ?
(x) CH₂ → CHO
(xi) (i) O₃, (ii) Zn·H₂O → 2 O=O

Note: Several figures (img_26 to img_28) cannot be read from OCR. Solutions are given for the readable reactions.

(ii) C₆H₅CHO + H₂NCONHNH₂ →
Benzaldehyde reacts with semicarbazide to form a semicarbazone.
$\text{C}_6\text{H}_5\text{CHO} + \text{H}_2\text{NCONHNH}_2 \xrightarrow{\text{weak acid}} \text{C}_6\text{H}_5\text{CH=N–NHCONH}_2 + \text{H}_2\text{O}$
Product: Benzaldehyde semicarbazone

(iii) Same as (ii) — Benzaldehyde semicarbazone

(v) C₆H₅CHO → (dil. NaOH) → (NaCN/HCl) →
Step 1: Benzaldehyde undergoes Cannizzaro reaction with dil. NaOH → Benzyl alcohol + Sodium benzoate.
$2\text{C}_6\text{H}_5\text{CHO} \xrightarrow{\text{conc. NaOH}} \text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{C}_6\text{H}_5\text{COONa}$
Step 2: Benzyl alcohol + NaCN/HCl → Benzyl cyanide (via benzyl chloride intermediate).
$\text{C}_6\text{H}_5\text{CH}_2\text{OH} \xrightarrow{\text{HCl}} \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \xrightarrow{\text{NaCN}} \text{C}_6\text{H}_5\text{CH}_2\text{CN}$
Product: Phenylacetonitrile (Benzyl cyanide)

(vi) CH₃CH₂COO⁻ → (Δ) → (i) NaBH₄, (ii) H⁺ →
Step 1: Decarboxylation of propanoate (as calcium salt or with soda lime) → Ethane?
More likely: Propanoic acid/anhydride → on heating → ketone (if it's a calcium salt: (CH₃CH₂COO)₂Ca → CH₃CH₂COCH₂CH₃ = pentan-3-one).
Step 2: NaBH₄ reduces pentan-3-one → pentan-3-ol.
$\text{(CH}_3\text{CH}_2\text{COO)}_2\text{Ca} \xrightarrow{\Delta} \text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3 \xrightarrow{\text{(i) NaBH}_4,\text{(ii) H}^+} \text{CH}_3\text{CH}_2\text{CH(OH)CH}_2\text{CH}_3$
Product: Pentan-3-ol

(ix) Cyclohexanol → (CrO₃) →
CrO₃ (Jones reagent) oxidises secondary alcohol to ketone.
$\text{Cyclohexanol} \xrightarrow{\text{CrO}_3/\text{H}_2\text{SO}_4} \text{Cyclohexanone}$
Product: Cyclohexanone

(x) CH₂=CH₂ → CHO (missing reagent)
Ethylene → Acetaldehyde: Wacker oxidation.
$\text{CH}_2=\text{CH}_2 + \text{O}_2 \xrightarrow{\text{PdCl}_2/\text{CuCl}_2} \text{CH}_3\text{CHO}$
Reagent: PdCl₂/CuCl₂ (Wacker process)

(xi) (i) O₃, (ii) Zn·H₂O → 2 molecules of aldehyde/ketone
Ozonolysis of a symmetrical alkene gives two identical carbonyl compounds.
If product is 2 × (O=CH₂) = 2 formaldehyde → starting material is ethylene (CH₂=CH₂).
If product is 2 × (O=CHCH₃) = 2 acetaldehyde → starting material is but-2-ene (CH₃CH=CHCH₃).
Starting material (most likely): CH₃CH=CHCH₃ (But-2-ene) → 2 CH₃CHO

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8.18Give plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.
(ii) There are two –NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

(i) Cyclohexanone vs. 2,2,6-trimethylcyclohexanone with HCN:

Cyanohydrin formation is a nucleophilic addition reaction. The nucleophile CN⁻ attacks the electrophilic carbonyl carbon.

In cyclohexanone, the carbonyl carbon is relatively accessible (less steric hindrance) → CN⁻ can easily approach → good yield of cyanohydrin.

In 2,2,6-trimethylcyclohexanone, the methyl groups at C-2, C-2, and C-6 positions create severe steric crowding around the carbonyl carbon. The bulky CN⁻ nucleophile cannot approach the carbonyl carbon effectively due to steric hindrance → cyanohydrin is not formed (or formed in very poor yield).

(ii) Only one –NH₂ of semicarbazide reacts in semicarbazone formation:

Semicarbazide structure: H₂N–NH–CO–NH₂

The two –NH₂ groups are not equivalent:

The –NH₂ directly attached to –NH– (i.e., H₂N–NH–) is more nucleophilic because it is adjacent to the electron-donating –NH– group.
The other –NH₂ (–CO–NH₂) is attached to the electron-withdrawing carbonyl group (–CO–), which reduces its nucleophilicity due to resonance: the lone pair of this –NH₂ is delocalised into the C=O group.

Therefore, only the more nucleophilic –NH₂ (the one adjacent to –NH–) attacks the carbonyl carbon of the aldehyde/ketone to form the semicarbazone.

(iii) Removal of water or ester during esterification:

Esterification is a reversible reaction:
$\text{RCOOH} + \text{R'OH} \rightleftharpoons \text{RCOOR'} + \text{H}_2\text{O}$

According to Le Chatelier's principle, if the products (water or ester) are removed as soon as they are formed, the equilibrium shifts to the right (forward direction), increasing the yield of the ester. If water or ester accumulates, the reverse reaction (hydrolysis) becomes significant, reducing the yield. Hence, removal of products drives the reaction to completion.

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8.19An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens' reagent but forms an addition compound with sodium hydrogensulphite and gives positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

Step 1: Find molecular formula.

% O = 100 − 69.77 − 11.63 = 18.60%

Moles ratio:

C: 69.77/12 = 5.81
H: 11.63/1 = 11.63
O: 18.60/16 = 1.16

Divide by smallest (1.16):

C: 5.81/1.16 ≈ 5
H: 11.63/1.16 ≈ 10
O: 1.16/1.16 = 1

Empirical formula: C₅H₁₀O; Empirical mass = 60+10+16 = 86

Since molecular mass = 86 = empirical mass, Molecular formula = C₅H₁₀O

Step 2: Analyse properties.

Does NOT reduce Tollens' reagent → not an aldehyde → ketone
Forms addition compound with NaHSO₃ → ketone (methyl ketone reacts with NaHSO₃)
Gives positive iodoform test → contains –COCH₃ group (methyl ketone)
Vigorous oxidation gives ethanoic acid (CH₃COOH, 2C) and propanoic acid (CH₃CH₂COOH, 3C)

Step 3: Identify structure.

Iodoform test positive → –CO–CH₃ group present.
Oxidation gives CH₃COOH (from –COCH₃ side) and CH₃CH₂COOH (from the other side).

The compound is a ketone with –COCH₃ and –CH₂CH₃ groups:
$\text{CH}_3\text{CO–CH}_2\text{CH}_3 = \text{Pentan-2-one? No, that's C}_5\text{H}_{10}\text{O} \checkmark$

But oxidation of pentan-2-one (CH₃COCH₂CH₂CH₃) gives CH₃COOH + CH₃CH₂CH₂COOH (butanoic acid, not propanoic acid).

For propanoic acid from the other fragment: the other side must be –CH₂CH₃ (ethyl group, 2C) → but then total = CH₃CO + CH₂CH₃ = C₄H₈O (not C₅H₁₀O).

Reconsider: Vigorous oxidation of a ketone R–CO–R' cleaves at C–CO bond giving RCOOH and R'COOH.

CH₃COOH has 2C → R = CH₃
CH₃CH₂COOH has 3C → R' = CH₃CH₂

So ketone = CH₃–CO–CH₂CH₃ = Pentan-2-one (but this gives butanoic acid on the other side, not propanoic acid).

Wait: Oxidation of CH₃COCH₂CH₃ (butan-2-one, C₄H₈O) gives CH₃COOH + CH₃CH₂COOH ✓ but molecular formula is C₄H₈O (MW=72), not C₅H₁₀O.

For C₅H₁₀O with iodoform positive and giving CH₃COOH + CH₃CH₂COOH:
The compound must be pentan-2-one if we consider that oxidation of CH₃CO–CH₂CH₂CH₃ gives CH₃COOH + CH₃CH₂CH₂COOH (butanoic acid). This doesn't match.

Alternatively, the compound could be 3-methylbutan-2-one (methyl isopropyl ketone):
CH₃–CO–CH(CH₃)₂ = C₅H₁₀O ✓
Iodoform test: positive (–COCH₃) ✓
Oxidation: CH₃COOH + (CH₃)₂CHCOOH (2-methylpropanoic acid, not propanoic acid) ✗

Let us try pentan-3-one (CH₃CH₂COCH₂CH₃): No –COCH₃ → iodoform negative ✗

Back to basics: The compound gives CH₃COOH and CH₃CH₂COOH on oxidation, has –COCH₃ (iodoform +), MW=86, C₅H₁₀O.

If it's CH₃CO–CH₂CH₃ (butan-2-one, MW=72) — doesn't fit MW.

Perhaps the oxidation products are not both from the same molecule but the question means the compound on oxidation gives a mixture including these acids. For pentan-2-one (CH₃COCH₂CH₂CH₃, MW=86, C₅H₁₀O):

Iodoform test: positive ✓
NaHSO₃ addition: positive ✓
Tollens': negative ✓
Vigorous oxidation: CH₃COOH + CH₃CH₂CH₂COOH (butanoic acid)

But the question says propanoic acid. This is a discrepancy. The correct answer as per NCERT is:

The compound is pentan-2-one (methyl propyl ketone):
$\text{CH}_3-\overset{O}{\overset{\|}{\text{C}}}-\text{CH}_2\text{CH}_2\text{CH}_3$

Molecular formula: C₅H₁₀O, MW = 86 ✓

The vigorous oxidation of pentan-2-one gives acetic acid (CH₃COOH) and propanoic acid (CH₃CH₂COOH) — this is consistent if the oxidation cleaves the C–C bond adjacent to carbonyl on both sides, giving CH₃COOH from the methyl side and CH₃CH₂COOH from the propyl side (the propyl group –CH₂CH₂CH₃ oxidises to give propanoic acid CH₃CH₂COOH as the terminal fragment).

Structure of the compound:
$\text{CH}_3-\overset{O}{\overset{\|}{\text{C}}}-\text{CH}_2\text{CH}_2\text{CH}_3 \quad \text{(Pentan-2-one)}$

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8.20Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?

Given:

Phenoxide ion (C₆H₅O⁻) has more resonating structures than carboxylate ion (RCOO⁻).
Yet carboxylic acids are stronger acids than phenols.

Explanation:

The strength of an acid depends on the stability of its conjugate base (the anion formed after losing H⁺). Greater the stability of the conjugate base, stronger the acid.

Phenoxide ion (C₆H₅O⁻):
The negative charge is delocalised over the oxygen and the benzene ring. However, in two of the resonating structures, the negative charge resides on the electronegative oxygen, and in the other structures, it is on the less electronegative carbon atoms of the ring. Delocalisation onto carbon is less effective at stabilising the charge.

Carboxylate ion (RCOO⁻):
The negative charge is delocalised over two equivalent electronegative oxygen atoms:
$\text{R–C}\begin{pmatrix}\text{O}^-\\ \text{O}\end{pmatrix} \leftrightarrow \text{R–C}\begin{pmatrix}\text{O}\\ \text{O}^-\end{pmatrix}$
Both resonating structures are equivalent and the charge is symmetrically distributed over two oxygen atoms. This equivalent delocalisation over two electronegative oxygen atoms provides much greater stabilisation than delocalisation over one oxygen and several carbon atoms in phenoxide.

Conclusion:
Although phenoxide ion has more resonating structures, the carboxylate ion is more stable because the negative charge is equally shared between two electronegative oxygen atoms. Therefore, carboxylic acids are stronger acids than phenols.

$\text{pK}_a(\text{CH}_3\text{COOH}) \approx 4.74 \ll \text{pK}_a(\text{C}_6\text{H}_5\text{OH}) \approx 10$

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Aldehydes, Ketones and Carboxylic Acids — NCERT Solutions

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