Class 12 Chemistry Unit 13: Aldehydes and Ketones Notes (NEB New Syllabus) | Formula, Examples, Difference | Aldehydes and Ketones PDF

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Class 12 Chemistry • Unit 13

Aldehydes and Ketones

Class 12 Chemistry Notes • NEB Syllabus 2083

Syllabus, 29 solved multiple choice questions, the full textbook exercise with solutions and diagrams, handwritten and typed PDF notes and frequently asked questions on aliphatic and aromatic aldehydes and ketones.

1. Syllabus

Unit 13: Aldehydes and Ketones

13.1 Aliphatic aldehydes and ketones

13.2 Aromatic aldehydes and ketones

These notes follow the NEB Class 12 Chemistry syllabus (+2 Chemistry, Nepal). Practise the multiple choice questions first, then work through the solved numerical and subjective questions, and finish with the PDF notes for quick revision.

2. Multiple Choice Questions

Question 1. Which one could be true about the information given in the following table?

SN	Compounds	Possible reactions/products
1.	A	Gives Tollen’s test but not Fehling’s test
2.	B	Gives Cannizzaro’s reaction
3.	C	Gives 2-hydroxy-2-methylpropanoic acid by reacting C with HCN followed by hydrolysis

a) A is acetone, B is acetaldehyde and C is formaldehyde
b) A is benzaldehyde, B is acetone and C is acetaldehyde
c) A is acetaldehyde, B is acetone and C is benzaldehyde
d) A is benzaldehyde, B is formaldehyde and C is acetone

Correct Answer: d) A is benzaldehyde, B is formaldehyde and C is acetone

Reason: A gives Tollen’s test but not Fehling’s test: an aromatic aldehyde, benzaldehyde (aliphatic aldehydes give both). B gives Cannizzaro’s reaction, so it has no alpha-hydrogen: formaldehyde. C gives 2-hydroxy-2-methylpropanoic acid after HCN and hydrolysis: \(\mathrm{CH_3COCH_3+HCN\longrightarrow(CH_3)_2C(OH)CN\xrightarrow{H_3O^+}(CH_3)_2C(OH)COOH}\), so C is acetone.

Question 2. Which of the following is true about carbonyl compounds?

a) They give a positive iodoform test
b) They give a positive Claisen condensation test
c) They give a negative iodoform test
d) They give a positive DNP test

Correct Answer: d) They give a positive DNP test

Reason: Every aldehyde and ketone has the >C=O group, which adds 2,4-dinitrophenylhydrazine to form an orange-yellow 2,4-dinitrophenylhydrazone. The iodoform test is given only by compounds with the \(\mathrm{CH_3CO-}\) group, so a and c are not true for all carbonyl compounds, and Claisen condensation is a reaction of esters, not a test.

Question 3. What would be the product when 2-methylbut-2-ene is ozonolyzed?

a) 2 molecules of ethanal
b) Ethanal and propanal
c) Ethanal and propan-2-one
d) Ethanal and methanal

Correct Answer: c) Ethanal and propan-2-one

Reason: \(\mathrm{(CH_3)_2C=CHCH_3\xrightarrow{1)\ O_3;\ 2)\ Zn/H_2O}(CH_3)_2C=O+O=CHCH_3}\). The C=C bond is broken and each carbon gets an =O: the carbon with two methyl groups gives propan-2-one and the carbon with one methyl and H gives ethanal.

Question 4. Choose the best alternative from the following table.

Reactants	Condition	Products
X	\(\mathrm{Cu(OH)_2}\) + boil	Ethanoic acid
\(\mathrm{CH_3CHO}\)	Dil. NaOH	Y
\(\mathrm{CH_3-CO-CH_3}\)	Z	cyanohydrin

a) X is ethanal, Y is sodium ethanoate and Z is potassium cyanide
b) X is ethanal, Y is 3-hydroxybutanol and Z is potassium cyanide
c) X is ethanol, Y is 3-hydroxybutanal and Z is hydrogen cyanide
d) X is ethanal, Y is 3-hydroxybutanal and Z is hydrogen cyanide

Correct Answer: d) X is ethanal, Y is 3-hydroxybutanal and Z is hydrogen cyanide

Reason: X is oxidised to ethanoic acid by \(\mathrm{Cu(OH)_2}\) (Fehling’s solution) on boiling, so X is ethanal (ethanol does not react). Ethanal with dil. NaOH gives the aldol 3-hydroxybutanal, \(\mathrm{2CH_3CHO\xrightarrow{dil.\ NaOH}CH_3CH(OH)CH_2CHO}\). Propanone adds HCN to give a cyanohydrin.

Question 5. Which of the following is correct for Z? \(\mathrm{CH_3CHO}\) with Zn-Hg/HCl gives Z

a) Z is ethene
b) Z is ethane
c) Z is ethanol
d) Z is ethanal

Correct Answer: b) Z is ethane

Reason: Zn-Hg/HCl is the Clemmensen reduction reagent, which reduces >C=O to >CH₂: \(\mathrm{CH_3CHO+4[H]\xrightarrow{Zn-Hg/HCl}CH_3CH_3+H_2O}\).

Question 6. Which reaction represents the following chemical equation? \(\mathrm{CH_3COCl+H_2}\) (Pd/\(\mathrm{BaSO_4}\)) gives \(\mathrm{CH_3CHO+HCl}\)

a) Clemmensen’s reduction
b) Wolff-Kishner’s reduction
c) Rosenmund’s reduction
d) Catalytic reduction

Correct Answer: c) Rosenmund’s reduction

Reason: Reduction of an acyl chloride to an aldehyde with hydrogen over palladium poisoned with \(\mathrm{BaSO_4}\) is Rosenmund’s reduction. The poisoned catalyst stops the reduction at the aldehyde stage and prevents further reduction to the alcohol.

Question 7. How are ketones prepared?

a) By the oxidation of primary alcohols
b) By the oxidation of secondary alcohols
c) By the oxidation of tertiary alcohols
d) By the oxidation of aldehydes

Correct Answer: b) By the oxidation of secondary alcohols

Reason: \(\mathrm{CH_3CH(OH)CH_3\xrightarrow{K_2Cr_2O_7/H^+}CH_3COCH_3}\). Primary alcohols give aldehydes (then acids), tertiary alcohols resist oxidation, and aldehydes give carboxylic acids.

Question 8. Which of the following compounds gives Cannizzaro’s reaction?

a) Formic acid
b) Acetaldehyde
c) Acetone
d) Formaldehyde

Correct Answer: d) Formaldehyde

Reason: Cannizzaro’s reaction is given by aldehydes that have no alpha-hydrogen: \(\mathrm{2HCHO+NaOH\longrightarrow CH_3OH+HCOONa}\). Acetaldehyde has alpha-hydrogens and gives aldol condensation instead; formic acid is an acid and acetone is a ketone.

Question 9. Which of the following combinations is correct for the preparation of Bakelite?

a) Phenol + formaldehyde
b) Phenol + formic acid
c) Phenol + acetaldehyde
d) Phenol + acetic acid

Correct Answer: a) Phenol + formaldehyde

Reason: Bakelite is a thermosetting polymer made by the condensation polymerisation of phenol with methanal (formaldehyde) in the presence of an acid or a base.

Question 10. Which of the following compounds gives both iodoform and Fehling’s tests?

a) Acetone
b) Ethanol
c) Acetaldehyde
d) Propan-2-ol

Correct Answer: c) Acetaldehyde

Reason: Acetaldehyde is an aliphatic aldehyde, so it reduces Fehling’s solution, and it has the \(\mathrm{CH_3CO-}\) group, so it gives iodoform. Acetone gives only iodoform; ethanol and propan-2-ol give only iodoform (they are oxidised to carbonyl compounds by the hypoiodite) and not Fehling’s test.

Question 11. What would be the product when acetone reacts with chloroform in the presence of KOH?

a) Trichloromethanol
b) Chloropicrin
c) Chloretone
d) Trichloroacetaldehyde

Correct Answer: c) Chloretone

Reason: \(\mathrm{CH_3COCH_3+CHCl_3\xrightarrow{KOH}(CH_3)_2C(OH)CCl_3}\). The product is 1,1,1-trichloro-2-methylpropan-2-ol, called chloretone, a hypnotic and preservative.

Question 12. Which of the following reagents is used to distinguish between pentan-2-one and pentan-3-one?

a) \(\mathrm{Zn-Hg/HCl}\)
b) dil. NaOH
c) \(\mathrm{NaOH+I_2}\)
d) \(\mathrm{K_2Cr_2O_7/H^+}\)

Correct Answer: c) \(\mathrm{NaOH+I_2}\)

Reason: Pentan-2-one, \(\mathrm{CH_3COCH_2CH_2CH_3}\), is a methyl ketone and gives a yellow precipitate of iodoform. Pentan-3-one, \(\mathrm{CH_3CH_2COCH_2CH_3}\), has no \(\mathrm{CH_3CO-}\) group and gives none. Both are reduced by Zn-Hg/HCl, both give aldol-type products and both resist dichromate.

Question 13. Which of the following is correct about the information given in the table?

Reactions	Products
A. Perkin’s condensation	1. Benzoin
B. Aldol condensation	2. Beta-hydroxybutanal
C. Claisen condensation	3. Cinnamic acid
D. Benzoin condensation	4. Beta-keto ester

a) A-3, B-2, C-1, D-4
b) A-3, B-1, C-2, D-4
c) A-3, B-2, C-4, D-1
d) A-3, B-1, C-4, D-2

Correct Answer: c) A-3, B-2, C-4, D-1

Reason: Perkin’s condensation (benzaldehyde + acetic anhydride) gives cinnamic acid (3). Aldol condensation of ethanal gives beta-hydroxybutanal (2). Claisen condensation of esters gives a beta-keto ester (4). Benzoin condensation of benzaldehyde gives benzoin (1).

Question 14. The aromatic organic compound Z does not give Fehling’s test but gives Tollen’s test. What would be the compound Z?

a) Acetophenone
b) Benzophenone
c) Benzaldehyde
d) Benzoic acid

Correct Answer: c) Benzaldehyde

Reason: Aromatic aldehydes are weaker reducing agents than aliphatic ones: they reduce Tollen’s reagent to a silver mirror but not Fehling’s solution. Acetophenone and benzophenone are ketones and benzoic acid is an acid, so none of them reduces Tollen’s reagent.

Question 15. What would be the product when propyne is catalytically hydrated in the presence of dil. \(\mathrm{H_2SO_4}\) and 1% \(\mathrm{HgSO_4}\)?

a) \(\mathrm{CH_3CH_2CHO}\)
b) \(\mathrm{CH_3CH_2CH_2OH}\)
c) \(\mathrm{CH_3COCH_3}\)
d) \(\mathrm{CH_3-O-CH_2CH_3}\)

Correct Answer: c) \(\mathrm{CH_3COCH_3}\)

Reason: Addition of water follows Markovnikov’s rule, so the OH adds to the middle carbon and gives the enol \(\mathrm{CH_3C(OH)=CH_2}\), which changes (tautomerises) to the stable ketone: \(\mathrm{CH_3C\equiv CH+H_2O\xrightarrow{HgSO_4/H_2SO_4}CH_3COCH_3}\). Only ethyne gives an aldehyde (ethanal).

Question 16. Which of the following conditions is correct for Cannizzaro’s reaction?

a) Dil. NaOH, Heat, alpha-hydrogen
b) Conc. NaOH, Heat, alpha-hydrogen
c) Dil. NaOH, Heat, without alpha-hydrogen
d) Conc. NaOH, Heat, without alpha-hydrogen

Correct Answer: d) Conc. NaOH, Heat, without alpha-hydrogen

Reason: Cannizzaro’s reaction needs an aldehyde without alpha-hydrogen and a concentrated alkali. With dilute NaOH an aldehyde that has alpha-hydrogen undergoes aldol condensation instead.

Question 17. Which of the following is correct about butan-2-one?

a) It does not give the iodoform test but gives Tollen’s test
b) It does not give Tollen’s test but gives Fehling’s test
c) It does not give Fehling’s test but gives the DNP test
d) It does not give the DNP test but gives the iodoform test

Correct Answer: c) It does not give Fehling’s test but gives the DNP test

Reason: Butan-2-one, \(\mathrm{CH_3COCH_2CH_3}\), is a ketone: it gives no Tollen’s or Fehling’s test and it does give the 2,4-DNP test (all carbonyl compounds do). It also gives the iodoform test, so a and d are wrong (d says it gives no DNP test), and b is wrong. Option c is correct as stated.

Question 18. The following diagram shows the four compounds W, X, Y and Z. Each compound has hydrogen, carbon and oxygen. The histogram shows the mass of these elements.

Class 12 Chemistry Unit 13 Aldehydes and Ketones

Figure: Mass of hydrogen, carbon and oxygen in the compounds W, X, Y and Z

a) W is methanol, X is methanal, Y is ethanal and Z is propanal
b) W is methanal, X is methanol, Y is ethanal and Z is propanol
c) W is methanol, X is ethanal, Y is ethanol and Z is propanal
d) W is methanal, X is ethanal, Y is methanol and Z is propanal

Correct Answer: d) W is methanal, X is ethanal, Y is methanol and Z is propanal

Reason: Read the masses (H, C, O) from the histogram and divide by the atomic masses (1, 12, 16). W: 2, 12, 16 gives H 2 : C 1 : O 1, so \(\mathrm{CH_2O}\), methanal. X: 4, 24, 16 gives H 4 : C 2 : O 1, so \(\mathrm{C_2H_4O}\), ethanal. Y: 4, 12, 16 gives H 4 : C 1 : O 1, so \(\mathrm{CH_4O}\), methanol. Z: 6, 36, 16 gives H 6 : C 3 : O 1, so \(\mathrm{C_3H_6O}\), propanal. Only option d matches.

Question 19. What is formalin?

a) 30% formaldehyde
b) 50% formaldehyde
c) 40% formaldehyde
d) 60% formaldehyde

Correct Answer: c) 40% formaldehyde

Reason: Formalin is a colourless solution of about 40% (37 to 40%) methanal in water, with a little methanol added to prevent polymerisation. It is used to preserve biological specimens.

Question 20. Which of the following is correct about the following table?

Compounds	Products
a. \(\mathrm{CH_3CHO}\)	1. Cinnamic acid
b. \(\mathrm{CH_3COCH_3}\)	2. 4-Hydroxy-4-methylpentan-2-one
c. \(\mathrm{C_6H_5CHO}\)	3. Formalin
d. HCHO	4. Ethane

a) a-1, b-2, c-3, d-4
b) a-3, b-4, c-1, d-2
c) a-2, b-3, c-4, d-1
d) a-4, b-2, c-1, d-3

Correct Answer: d) a-4, b-2, c-1, d-3

Reason: Ethanal on Clemmensen (or Wolff-Kishner) reduction gives ethane (4). Propanone with dil. NaOH gives diacetone alcohol, 4-hydroxy-4-methylpentan-2-one (2). Benzaldehyde gives cinnamic acid by Perkin’s condensation (1). Methanal is sold as a 40% solution, formalin (3).

Question 21. Which of the following is true about benzaldehyde?

a) It gives the DNP test, iodoform test and gives benzoin condensation
b) It gives aldol condensation, iodoform test and Perkin’s condensation
c) It gives Tollen’s test, Perkin’s condensation and Clemmensen’s reduction
d) It gives Perkin’s condensation, Fehling’s test and Clemmensen’s reduction

Correct Answer: c) It gives Tollen’s test, Perkin’s condensation and Clemmensen’s reduction

Reason: Benzaldehyde reduces Tollen’s reagent, gives cinnamic acid with acetic anhydride (Perkin), and is reduced to toluene by Zn-Hg/HCl. It does not give the iodoform test (no \(\mathrm{CH_3CO-}\) group), aldol condensation (no alpha-hydrogen) or Fehling’s test.

Question 22. \(\mathrm{CH_3CHO\xrightarrow{HCN}X\xrightarrow{H_3O^+}Y}\) In the above sequence of the reaction, what could be Y?

a) Ethanoic acid
b) β-hydroxypropanoic acid
c) α-hydroxypropanoic acid
d) α-hydroxyethanoic acid

Correct Answer: c) α-hydroxypropanoic acid

Reason: \(\mathrm{CH_3CHO+HCN\longrightarrow CH_3CH(OH)CN\ (X)}\); hydrolysis of the nitrile: \(\mathrm{CH_3CH(OH)CN+2H_2O+HCl\longrightarrow CH_3CH(OH)COOH\ (Y)+NH_4Cl}\). Y is 2-hydroxypropanoic acid (lactic acid); the OH is on the carbon next to -COOH, so it is an alpha-hydroxy acid.

Question 23. Which of the following is related with \(\mathrm{(CH_2)_6N_4}\)?

Figure: Options A to D for Question 23 (heart, liver, brain and kidneys)

a) Heart
b) Liver
c) Brain
d) Kidney

Correct Answer: d) Kidney (the options A to D show the heart, liver, brain and kidneys)

Reason: \(\mathrm{(CH_2)_6N_4}\) is urotropine (hexamethylenetetramine, hexamine), formed from methanal and ammonia: \(\mathrm{6HCHO+4NH_3\longrightarrow(CH_2)_6N_4+6H_2O}\). It is used as a urinary antiseptic, so it is related to the kidneys and urinary system.

Question 24. Choose the best alternative from the followings.

Reaction condition	Name of the reaction
a. Dil. NaOH	1. Tollen’s test
b. Zn-Hg/HCl	2. Aldol condensation
c. Hydrazine/glycerol/KOH	3. Clemmensen’s reduction
d. Ammoniacal \(\mathrm{AgNO_3}\)	4. Wolff-Kishner’s reduction

a) a-1, b-2, c-3, d-4
b) a-3, b-4, c-1, d-2
c) a-2, b-3, c-4, d-1
d) a-4, b-2, c-1, d-3

Correct Answer: c) a-2, b-3, c-4, d-1

Reason: Dil. NaOH is the catalyst of aldol condensation (2). Zn-Hg/HCl is Clemmensen’s reduction (3). Hydrazine with KOH in glycerol is Wolff-Kishner’s reduction (4). Ammoniacal \(\mathrm{AgNO_3}\) is Tollen’s reagent (1).

Question 25. What would be the products A and B in the following reaction sequence? \(\mathrm{CH_3-CH=CH_2\xrightarrow[2)\ Zn,\ H_2O]{1)\ O_3}A+B}\)

a) A is ethanol and B is methanol
b) A is ethanol and B is methanal
c) A is ethanal and B is methanol
d) A is ethanal and B is methanal

Correct Answer: d) A is ethanal and B is methanal

Reason: Ozonolysis (the reagent in the book is printed as O₂, but ozone, O₃, is meant) cleaves the C=C bond of propene, and zinc and water prevent oxidation of the products to acids: \(\mathrm{CH_3CH=CH_2\longrightarrow CH_3CHO+HCHO}\). The two products are carbonyl compounds, not alcohols.

Question 26. The reaction: A (acetone/alkali) gives chloretone. Identify compound A.

a) Trichloromethane
b) Bromoform
c) Iodoform
d) Haloalkane

Correct Answer: a) Trichloromethane

Reason: Chloretone is made from propanone and chloroform (trichloromethane, \(\mathrm{CHCl_3}\)) with solid alkali: \(\mathrm{CH_3COCH_3+CHCl_3\xrightarrow{KOH}(CH_3)_2C(OH)CCl_3}\). Chloretone contains three chlorine atoms, so bromoform and iodoform cannot give it.

Question 27. What would be the major product when propan-2-one is treated with dil. NaOH?

a) 3-Hydroxy butan-2-one
b) 4-Hydroxypentan-2-one
c) 3-Hydroxy 3-methyl butan-2-one
d) 4-Hydroxy 4-methyl pentan-2-one

Correct Answer: d) 4-Hydroxy 4-methyl pentan-2-one

Reason: The enolate of one propanone molecule adds to the C=O of a second molecule (aldol addition): \(\mathrm{2CH_3COCH_3\xrightarrow{dil.\ NaOH}CH_3C(OH)(CH_3)CH_2COCH_3}\). The product is diacetone alcohol, 4-hydroxy-4-methylpentan-2-one.

Question 28. Which of the following compounds give iodoform test?

a) Methanal
b) Propanal
c) Butanal
d) Propan-2-one

Correct Answer: d) Propan-2-one

Reason: The iodoform test needs the \(\mathrm{CH_3CO-}\) group (or \(\mathrm{CH_3CH(OH)-}\)). Only propan-2-one has it: \(\mathrm{CH_3COCH_3+3I_2+4NaOH\longrightarrow CHI_3\downarrow+CH_3COONa+3NaI+3H_2O}\). Methanal, propanal and butanal have no methyl group on the carbonyl carbon.

Question 29. What could be W, X, Y and Z in the following reaction sequence? W gives positive test with 2,4-DNP; X gives Tollen’s test; Y gives Fehling’s test; Z is carboxylic acid.

Figure: Reference diagram for Question 29

a) W – alcoholic compounds, X – ketone, Y – aliphatic aldehyde and Z can be ethanoic acid
b) W – carbonyl compounds, X – aldehyde, Y – aliphatic aldehyde and Z can be ethanoic acid
c) W – carboxylic acid, X – aldehyde, Y – aliphatic aldehyde and Z can be benzoic acid
d) W – alcohol, X – ketone, Y – aliphatic aldehyde and Z can be benzoic acid

Correct Answer: b) W – carbonyl compounds, X – aldehyde, Y – aliphatic aldehyde and Z can be ethanoic acid

Reason: A positive 2,4-DNP test shows a carbonyl compound (aldehyde or ketone). Tollen’s test shows an aldehyde (ketones do not give it). Fehling’s test shows an aliphatic aldehyde. Z is a carboxylic acid, which can be ethanoic acid. Options a and d name an alcohol and a ketone, which do not give Tollen’s test, and option c names an acid for W, which does not give the DNP test.

3. Numerical and Subjective Questions

Short Answer Questions

Question 1. How do you confirm \(\mathrm{CH_3CHO}\) and \(\mathrm{CH_3COCH_3}\) by chemical tests? Write the molecular structure of butan-2-one and 3-hydroxybutanal.

Ethanal and propanone both contain the \(\mathrm{CH_3CO-}\) group and both give the 2,4-DNP test and the iodoform test, so those tests cannot separate them. They are separated by tests that need an aldehyde group: Tollen’s test and Fehling’s test.

Test	Ethanal, \(\mathrm{CH_3CHO}\)	Propanone, \(\mathrm{CH_3COCH_3}\)
Tollen’s reagent (ammoniacal \(\mathrm{AgNO_3}\), warm)	Bright silver mirror	No change
Fehling’s solution (boil)	Red precipitate of \(\mathrm{Cu_2O}\)	Blue colour remains
2,4-DNP	Orange-yellow precipitate	Orange-yellow precipitate
Iodine + NaOH	Yellow \(\mathrm{CHI_3}\)	Yellow \(\mathrm{CHI_3}\)

\[ \mathrm{CH_3CHO+2[Ag(NH_3)_2]^++3OH^-\xrightarrow{\Delta}CH_3COO^-+2Ag\downarrow+4NH_3+2H_2O} \]

\[ \mathrm{CH_3CHO+2Cu^{2+}+5OH^-\xrightarrow{\Delta}CH_3COO^-+Cu_2O\downarrow+3H_2O} \]

Structures.

Name	Condensed formula	Type
Butan-2-one	\(\mathrm{CH_3-CO-CH_2-CH_3}\)	Ketone (ethyl methyl ketone)
3-Hydroxybutanal	\(\mathrm{CH_3-CH(OH)-CH_2-CHO}\)	Hydroxy aldehyde (aldol)

Question 2. Study the following cyclic pattern of reactions and complete the cycle with suitable compounds in place of A, B, C, D and E with their names, where A is an aromatic aldehyde having a molecular mass of 106 amu. E gives A, A gives B, B gives C, C gives D and D gives E.

A. The aromatic aldehyde with molecular mass 106 is benzaldehyde: \(\mathrm{C_6H_5CHO}\), M = 7(12) + 6(1) + 16 = 106. The other members of the cycle follow from its reactions:

	Name	Formula	Step
E	Toluene	\(\mathrm{C_6H_5CH_3}\)	D to E: Friedel-Crafts methylation of benzene
A	Benzaldehyde	\(\mathrm{C_6H_5CHO}\)	E to A: oxidation of toluene
B	Benzoic acid	\(\mathrm{C_6H_5COOH}\)	A to B: oxidation of the aldehyde
C	Sodium benzoate	\(\mathrm{C_6H_5COONa}\)	B to C: neutralisation with NaOH
D	Benzene	\(\mathrm{C_6H_6}\)	C to D: decarboxylation with soda lime

\[ \mathrm{C_6H_5CH_3\ (E)+[O]\xrightarrow{CeO_2/H^+}C_6H_5CHO\ (A)+H_2O} \]

\[ \mathrm{C_6H_5CHO\ (A)\xrightarrow{[O]}C_6H_5COOH\ (B)} \]

\[ \mathrm{C_6H_5COOH\ (B)+NaOH\longrightarrow C_6H_5COONa\ (C)+H_2O} \]

\[ \mathrm{C_6H_5COONa\ (C)+NaOH\xrightarrow{CaO,\ heat}C_6H_6\ (D)+Na_2CO_3} \]

\[ \mathrm{C_6H_6\ (D)+CH_3Cl\xrightarrow{anhyd.\ AlCl_3}C_6H_5CH_3\ (E)+HCl} \]

Figure: Cyclic pattern of reactions, where A is benzaldehyde (molecular mass 106) and the cycle returns to toluene

Question 3. An organic compound A, \(\mathrm{C_5H_{10}O}\) gives positive 2,4-DNP test but a negative silver nitrate test. Compound A is oxidized in the presence of alkaline \(\mathrm{KMnO_4}\) in vigorous condition to carboxylic acid B, \(\mathrm{C_3H_6O_2}\). The sodium salt of B gives a hydrocarbon C on Kolbe’s electrolytic reduction. Identify A, B and C with suitable reactions.

Reasoning. A positive 2,4-DNP test shows a carbonyl group, and a negative Tollen’s (silver nitrate) test shows that A is not an aldehyde. So A is a ketone of formula \(\mathrm{C_5H_{10}O}\). Vigorous oxidation breaks the ketone at the carbonyl group and gives acids; the acid with the formula \(\mathrm{C_3H_6O_2}\) is propanoic acid, \(\mathrm{CH_3CH_2COOH}\). This is formed from pentan-3-one (and also from pentan-2-one), so A is taken as the symmetrical ketone pentan-3-one.

	Name	Structure
A	Pentan-3-one	\(\mathrm{CH_3CH_2COCH_2CH_3}\)
B	Propanoic acid	\(\mathrm{CH_3CH_2COOH}\)
C	Butane	\(\mathrm{CH_3CH_2CH_2CH_3}\)

\[ \mathrm{CH_3CH_2COCH_2CH_3\xrightarrow{alk.\ KMnO_4,\ [O]}CH_3CH_2COOH+CH_3COOH} \]

\[ \mathrm{CH_3CH_2COOH+NaOH\longrightarrow CH_3CH_2COONa+H_2O} \]

Kolbe’s electrolysis: two propanoate ions lose electrons at the anode, lose CO₂, and the two ethyl radicals join:

\[ \mathrm{2CH_3CH_2COONa+2H_2O\xrightarrow{electrolysis}CH_3CH_2CH_2CH_3+2CO_2\uparrow+H_2\uparrow+2NaOH} \]

(Pentan-2-one, \(\mathrm{CH_3COCH_2CH_2CH_3}\), also oxidises to propanoic acid and ethanoic acid, so it fits the data in the same way. Butane is the answer for C either way.)

Question 4. Identify the compounds in the following reaction sequence and complete the reactions. W (zinc dust/heat) gives X; X (\(\mathrm{CH_3Cl}\)/\(\mathrm{AlCl_3}\)/heat) gives Y; Y (\(\mathrm{CeO_2}\)/\(\mathrm{H^+}\)) gives Z. X is aromatic hydrocarbon having molecular mass 78 amu.

Reasoning. An aromatic hydrocarbon with M = 78 is benzene, \(\mathrm{C_6H_6}\) (6 × 12 + 6 × 1 = 78), so X is benzene. W changes into benzene on heating with zinc dust: that is the reduction of phenol. Benzene with \(\mathrm{CH_3Cl}\) and anhydrous \(\mathrm{AlCl_3}\) (Friedel-Crafts alkylation) gives toluene (Y), and oxidation of the methyl group of toluene gives benzaldehyde (Z).

	Name	Formula
W	Phenol	\(\mathrm{C_6H_5OH}\)
X	Benzene	\(\mathrm{C_6H_6}\)
Y	Toluene	\(\mathrm{C_6H_5CH_3}\)
Z	Benzaldehyde	\(\mathrm{C_6H_5CHO}\)

\[ \mathrm{C_6H_5OH\ (W)+Zn\xrightarrow{heat}C_6H_6\ (X)+ZnO} \]

\[ \mathrm{C_6H_6\ (X)+CH_3Cl\xrightarrow{anhyd.\ AlCl_3,\ heat}C_6H_5CH_3\ (Y)+HCl} \]

\[ \mathrm{C_6H_5CH_3\ (Y)+[O]\xrightarrow{CeO_2/H^+}C_6H_5CHO\ (Z)+H_2O} \]

Figure: Reaction sequence W to Z. X is benzene (molecular mass 78)

Question 5. What happens when: a. Formaldehyde is treated with 2,4-DNP? b. Methanal is heated with ammonia? c. Acetone is reacted with hydroxyl amine? d. Ethanal is treated with HCN followed by acid hydrolysis? e. Benzaldehyde is heated with alc. KCN?

a. Formaldehyde is treated with 2,4-DNP?

Methanal adds 2,4-dinitrophenylhydrazine and loses water (condensation). A yellow-orange precipitate of methanal 2,4-dinitrophenylhydrazone is formed; it shows the presence of a carbonyl group.

\[ \mathrm{HCHO+H_2N-NH-C_6H_3(NO_2)_2\longrightarrow H_2C=N-NH-C_6H_3(NO_2)_2+H_2O} \]

Figure: Methanal gives a yellow-orange precipitate of its 2,4-dinitrophenylhydrazone with 2,4-DNP

b. Methanal is heated with ammonia?

Urotropine (hexamethylenetetramine, hexamine) is formed. It is used as a urinary antiseptic and in the manufacture of the explosive RDX.

\[ \mathrm{6HCHO+4NH_3\xrightarrow{\Delta}(CH_2)_6N_4+6H_2O} \]

c. Acetone is reacted with hydroxyl amine?

Propanone oxime (acetoxime) is formed with the loss of water:

\[ \mathrm{CH_3COCH_3+NH_2OH\xrightarrow{H^+}(CH_3)_2C=NOH+H_2O} \]

d. Ethanal is treated with HCN followed by acid hydrolysis?

Ethanal adds HCN to give the cyanohydrin, which on hydrolysis gives 2-hydroxypropanoic acid (lactic acid):

\[ \mathrm{CH_3CHO+HCN\longrightarrow CH_3CH(OH)CN\xrightarrow{H_3O^+}CH_3CH(OH)COOH} \]

e. Benzaldehyde is heated with alc. KCN?

Benzoin condensation takes place: two molecules of benzaldehyde join to give benzoin, an alpha-hydroxy ketone (the cyanide ion is the catalyst):

\[ \mathrm{2C_6H_5CHO\xrightarrow{alc.\ KCN,\ \Delta}C_6H_5CH(OH)COC_6H_5} \]

Figure: Benzoin condensation. Two molecules of benzaldehyde couple in the presence of alcoholic KCN

Question 6. Give an example of each of the followings: a. Tollen’s test b. Fehling’s test c. Aldol condensation d. Cannizzaro’s reaction e. Clemmensen’s reduction reaction.

a. Tollen’s test

Ethanal gives a silver mirror when warmed with ammoniacal silver nitrate (any aldehyde, including benzaldehyde, gives the test):

\[ \mathrm{CH_3CHO+2[Ag(NH_3)_2]^++3OH^-\xrightarrow{\Delta}CH_3COO^-+2Ag\downarrow+4NH_3+2H_2O} \]

b. Fehling’s test

Ethanal (an aliphatic aldehyde) gives a red precipitate of copper(I) oxide on boiling with Fehling’s solution:

\[ \mathrm{CH_3CHO+2Cu^{2+}+5OH^-\xrightarrow{\Delta}CH_3COO^-+Cu_2O\downarrow+3H_2O} \]

c. Aldol condensation

Ethanal, which has alpha-hydrogens, gives 3-hydroxybutanal with dilute NaOH:

\[ \mathrm{2CH_3CHO\xrightarrow{dil.\ NaOH}CH_3CH(OH)CH_2CHO} \]

d. Cannizzaro’s reaction

Methanal (no alpha-hydrogen) with concentrated NaOH gives methanol and sodium methanoate. Benzaldehyde behaves in the same way (benzyl alcohol and sodium benzoate):

\[ \mathrm{2HCHO+NaOH\xrightarrow{conc.}CH_3OH+HCOONa} \]

Figure: Cannizzaro reaction of benzaldehyde, which has no α-hydrogen

e. Clemmensen’s reduction

Propanone is reduced to propane by zinc amalgam and concentrated HCl:

\[ \mathrm{CH_3COCH_3+4[H]\xrightarrow{Zn-Hg/conc.\ HCl}CH_3CH_2CH_3+H_2O} \]

Question 7. The given table shows the compounds and their molecular formula. How would you convert A to B? How A obtained from E, where E is primary alcohol. Write down the reactions involved in conversion A to C and D. (5×1=5)

Compound	Molecular formula	Identity
A	\(\mathrm{C_2H_5Cl}\)	Chloroethane
B	\(\mathrm{C_4H_{10}}\)	Butane
C	\(\mathrm{CH_2O}\)	Methanal
D	\(\mathrm{C_2H_4O}\)	Ethanal
E	\(\mathrm{C_2H_6O}\)	Ethanol (primary alcohol)

A to B (Wurtz reaction). Chloroethane is heated with sodium metal in dry ether:

\[ \mathrm{2C_2H_5Cl+2Na\xrightarrow{dry\ ether}C_4H_{10}+2NaCl} \]

E to A. Ethanol is converted into chloroethane with thionyl chloride (or PCl₅, or HCl/ZnCl₂):

\[ \mathrm{C_2H_5OH+SOCl_2\xrightarrow{pyridine}C_2H_5Cl+SO_2\uparrow+HCl\uparrow} \]

A to D (ethanal). Hydrolysis with aqueous KOH gives ethanol, and mild oxidation with PCC stops at the aldehyde:

\[ \mathrm{C_2H_5Cl+KOH(aq.)\xrightarrow{\Delta}C_2H_5OH+KCl} \]

\[ \mathrm{C_2H_5OH\xrightarrow{PCC}CH_3CHO+H_2O} \]

A to C (methanal). Dehydrohalogenation with alcoholic KOH gives ethene, and ozonolysis of ethene gives two molecules of methanal:

\[ \mathrm{C_2H_5Cl+KOH(alc.)\xrightarrow{\Delta}CH_2=CH_2+KCl+H_2O} \]

\[ \mathrm{CH_2=CH_2\xrightarrow{1)\ O_3;\ 2)\ Zn/H_2O}2HCHO} \]

Question 8. Compound A (\(\mathrm{C_4H_{10}O}\)) does not react with phenyl hydrazine but when it is oxidized in presence of acidified potassium dichromate gives B having molecular formula (\(\mathrm{C_4H_8O}\)). Compound B reacts with 2,4-DNP but does not give Tollen’s test. When the compound A is dehydrated in presence of conc. sulphuric acid it gives a compound C(\(\mathrm{C_4H_8}\)). After ozonolysis of C, two similar molecules are formed? Identify the compounds A, B, C with pertinent reactions. What would be the product when acetone is treated with HCN followed by acid hydrolysis?

Reasoning. A does not react with phenylhydrazine, so it has no C=O group; it forms a carbonyl compound B on oxidation, so it is an alcohol. B reacts with 2,4-DNP (carbonyl) but not with Tollen’s reagent (not an aldehyde), so B is a ketone, \(\mathrm{C_4H_8O}\) = butan-2-one, and A is the secondary alcohol butan-2-ol. Dehydration of A gives but-2-ene (Saytzeff’s rule), and its ozonolysis gives two identical molecules of ethanal, which agrees with the data.

	Name	Structure
A	Butan-2-ol	\(\mathrm{CH_3CH(OH)CH_2CH_3}\)
B	Butan-2-one	\(\mathrm{CH_3COCH_2CH_3}\)
C	But-2-ene	\(\mathrm{CH_3CH=CHCH_3}\)

\[ \mathrm{CH_3CH(OH)CH_2CH_3\ (A)\xrightarrow{K_2Cr_2O_7/H^+}CH_3COCH_2CH_3\ (B)+H_2O} \]

\[ \mathrm{CH_3CH(OH)CH_2CH_3\ (A)\xrightarrow{conc.\ H_2SO_4,\ 170^\circ C}CH_3CH=CHCH_3\ (C)+H_2O} \]

\[ \mathrm{CH_3CH=CHCH_3\ (C)\xrightarrow{1)\ O_3;\ 2)\ Zn/H_2O}2CH_3CHO} \]

Acetone with HCN, then acid hydrolysis. The cyanohydrin is hydrolysed to 2-hydroxy-2-methylpropanoic acid:

\[ \mathrm{CH_3COCH_3+HCN\longrightarrow(CH_3)_2C(OH)CN\xrightarrow{H_3O^+}(CH_3)_2C(OH)COOH} \]

Question 9. An alkene ‘A’ with molecular formula \(\mathrm{C_6H_{12}}\) on ozonolysis gives only one type of carbonyl compound ‘B’. The carbonyl compound does not give red precipitate with Fehling’s solution but gives yellow precipitate of compound ‘C’ with iodine solution in presence of sodium hydroxide. The compound C on heating with silver powder gives a gas ‘D’. The gas ‘D’ gives a compound ‘E’ when treated with aq. Sulphuric acid in presence of 1% \(\mathrm{HgSO_4}\). Finally, when the compound E is oxidized, ethanoic acid is formed. Identify the compounds ‘A’, ‘B’, ‘C’, ‘D’ and ‘E’ with pertinent reactions.

Reasoning (working backwards). E gives ethanoic acid on oxidation, so E is ethanal; it is formed by hydration of ethyne with \(\mathrm{HgSO_4}\)/aq. \(\mathrm{H_2SO_4}\), so D is ethyne. Ethyne is formed when iodoform is heated with silver powder, so C is iodoform, and the yellow precipitate shows that B has the \(\mathrm{CH_3CO-}\) group. B gives no Fehling’s test and is the only product of ozonolysis, so it is propanone. The symmetrical alkene which gives two molecules of propanone is 2,3-dimethylbut-2-ene (3 + 3 = 6 carbon atoms).

	Name	Formula
A	2,3-Dimethylbut-2-ene	\(\mathrm{(CH_3)_2C=C(CH_3)_2}\)
B	Propanone	\(\mathrm{CH_3COCH_3}\)
C	Iodoform (triiodomethane)	\(\mathrm{CHI_3}\)
D	Ethyne	\(\mathrm{HC\equiv CH}\)
E	Ethanal	\(\mathrm{CH_3CHO}\)

\[ \mathrm{(CH_3)_2C=C(CH_3)_2\xrightarrow{1)\ O_3;\ 2)\ Zn/H_2O}2CH_3COCH_3\ (B)} \]

\[ \mathrm{CH_3COCH_3+3I_2+4NaOH\longrightarrow CHI_3\downarrow\ (C)+CH_3COONa+3NaI+3H_2O} \]

\[ \mathrm{2CHI_3\ (C)+6Ag\xrightarrow{\Delta}HC\equiv CH\ (D)+6AgI} \]

\[ \mathrm{HC\equiv CH\ (D)+H_2O\xrightarrow{1\%\ HgSO_4/H_2SO_4}CH_3CHO\ (E)} \]

\[ \mathrm{CH_3CHO\ (E)\xrightarrow{[O]}CH_3COOH} \]

Long Answer Questions

Question 1. The following diagram shows the four compounds W, X, Y and Z. Each compound has hydrogen, carbon and oxygen. The histogram shows the mass of these elements. a. Identify the compounds W, X, Y and Z based on the above diagram. b. What happens when compound W is heated with ammonia? c. Write the isomers of compound Z. d. Write difference between compounds X and Y with a chemical test of each.

a. Identify the compounds W, X, Y and Z.

Divide each mass in the histogram by the atomic mass (H = 1, C = 12, O = 16) to get the mole ratio:

Compound	Mass H	Mass C	Mass O	Mole ratio H : C : O	Formula	Compound
W	2	12	16	2 : 1 : 1	\(\mathrm{CH_2O}\)	Methanal
X	4	24	16	4 : 2 : 1	\(\mathrm{C_2H_4O}\)	Ethanal
Y	4	12	16	4 : 1 : 1	\(\mathrm{CH_4O}\)	Methanol
Z	6	36	16	6 : 3 : 1	\(\mathrm{C_3H_6O}\)	Propanal

Figure: Mass of hydrogen, carbon and oxygen in the compounds W, X, Y and Z

b. What happens when compound W is heated with ammonia?

Methanal and ammonia condense to form urotropine (hexamethylenetetramine), a white crystalline solid used as a urinary antiseptic:

\[ \mathrm{6HCHO+4NH_3\xrightarrow{\Delta}(CH_2)_6N_4+6H_2O} \]

c. Write the isomers of compound Z.

Z is propanal, \(\mathrm{C_3H_6O}\). Its carbonyl isomers are:

Isomer	Structure	Type
Propanal	\(\mathrm{CH_3CH_2CHO}\)	Aldehyde (Z)
Propanone	\(\mathrm{CH_3COCH_3}\)	Ketone (functional isomer)

Other compounds of the same formula are the unsaturated alcohol prop-2-en-1-ol, \(\mathrm{CH_2=CHCH_2OH}\), the ether methoxyethene, \(\mathrm{CH_2=CHOCH_3}\), and the cyclic compounds methyloxirane and oxetane.

d. Write difference between compounds X and Y with a chemical test of each.

	X: Ethanal, \(\mathrm{CH_3CHO}\)	Y: Methanol, \(\mathrm{CH_3OH}\)
Functional group	Aldehyde, -CHO	Alcohol, -OH
Chemical test	Tollen’s test: silver mirror with ammoniacal \(\mathrm{AgNO_3}\); Fehling’s test: red \(\mathrm{Cu_2O}\); 2,4-DNP: orange precipitate	Sodium metal: bubbles of \(\mathrm{H_2}\); no reaction with Tollen’s, Fehling’s or 2,4-DNP
Reaction with Na	No reaction	Gives \(\mathrm{CH_3ONa}\) and hydrogen
Oxidation product	Ethanoic acid	Methanal, then methanoic acid

\[ \mathrm{2CH_3OH+2Na\longrightarrow2CH_3ONa+H_2\uparrow} \]

Question 2. A. V (Zn/heat) gives W; W (\(\mathrm{CH_3Cl}\)/\(\mathrm{AlCl_3}\)) gives X; X (\(\mathrm{CeO_2}\)/\(\mathrm{H^+}\)) gives Y; Y (\(\mathrm{CH_3COONa}\)/heat) gives Z. W is also obtained by passing acetylene gas into red hot iron or copper tube at about 15-25 atmospheric pressure. [1+1+1+1+1] B. Compound M \(\mathrm{C_6H_{12}}\) decolorizes bromine, but gives no reaction with sodium metal or phenyl hydrazine. Ozonolysis of compound M gives two compounds N and O, both react with phenyl hydrazine. Compound N gives positive Tollen’s and iodoform tests. Compound O has molecular mass 72 amu. Suggest the structures for A, B and C. Give chemical equations for the reaction. [1+1+1]

A. Identify V, W, X, Y and Z

W is formed when acetylene is passed through a red-hot iron tube (cyclic polymerisation), so W is benzene: \(\mathrm{3HC\equiv CH\xrightarrow{red\ hot\ Fe\ tube}C_6H_6}\). V gives benzene on heating with zinc dust, so V is phenol. Benzene with \(\mathrm{CH_3Cl}\)/anhydrous \(\mathrm{AlCl_3}\) gives toluene (X), toluene is oxidised to benzaldehyde (Y), and benzaldehyde heated with sodium ethanoate (with ethanoic anhydride) gives cinnamic acid (Z) by Perkin’s condensation.

	Name	Formula
V	Phenol	\(\mathrm{C_6H_5OH}\)
W	Benzene	\(\mathrm{C_6H_6}\)
X	Toluene	\(\mathrm{C_6H_5CH_3}\)
Y	Benzaldehyde	\(\mathrm{C_6H_5CHO}\)
Z	Cinnamic acid (3-phenylprop-2-enoic acid)	\(\mathrm{C_6H_5CH=CHCOOH}\)

\[ \mathrm{C_6H_5OH\ (V)+Zn\xrightarrow{heat}C_6H_6\ (W)+ZnO} \]

\[ \mathrm{C_6H_6\ (W)+CH_3Cl\xrightarrow{AlCl_3}C_6H_5CH_3\ (X)+HCl} \]

\[ \mathrm{C_6H_5CH_3\ (X)+[O]\xrightarrow{CeO_2/H^+}C_6H_5CHO\ (Y)+H_2O} \]

\[ \mathrm{C_6H_5CHO\ (Y)+(CH_3CO)_2O\xrightarrow{CH_3COONa,\ heat}C_6H_5CH=CHCOOH\ (Z)+CH_3COOH} \]

Figure: Reaction sequence V to Z. W (benzene) is also formed from acetylene in a red-hot iron tube

B. Identify M, N and O

(The compounds in this part are M, N and O.) M decolorises bromine, so it is an alkene. It does not react with sodium (no -OH) or with phenylhydrazine (no C=O). Ozonolysis gives two carbonyl compounds. N gives Tollen’s and iodoform tests, so it is ethanal, \(\mathrm{CH_3CHO}\). O has M = 72 and is a carbonyl compound: \(\mathrm{C_4H_8O}\) (4 × 12 + 8 + 16 = 72). Since M has 6 carbons and N supplies 2, O has 4. Taking O as butan-2-one, the two carbonyl carbons join through a double bond:

	Name	Structure
M	3-Methylpent-2-ene	\(\mathrm{CH_3CH=C(CH_3)CH_2CH_3}\)
N	Ethanal	\(\mathrm{CH_3CHO}\)
O	Butan-2-one	\(\mathrm{CH_3COCH_2CH_3}\)

\[ \mathrm{CH_3CH=C(CH_3)CH_2CH_3\ (M)\xrightarrow{1)\ O_3;\ 2)\ Zn/H_2O}CH_3CHO\ (N)+CH_3COCH_2CH_3\ (O)} \]

The data also fit M = hex-2-ene if O is butanal, or M = 4-methylpent-2-ene if O is 2-methylpropanal, because each of these carbonyl compounds has the molecular mass 72 and reacts with phenylhydrazine. The answer commonly given is 3-methylpent-2-ene with butan-2-one.

Question 3. A. Identify P, Q, R, S and T from the following information: Compound P is a compound having same functional group as of ethanol. Compound Q gives cannizzaro’s reaction but is not an aliphatic compound. Compound R is obtained when compound Q is oxidized. Compound S is a compound of ammonia. Compound T is a compound which is obtained by nitration of fundamental aromatic compound. B. W (acidified \(\mathrm{K_2Cr_2O_7}\)) gives X; X (2,4-DNP) gives Y. W gives positive iodoform test and is primary alcohol. Identify W, X and Y.

A. Identify P, Q, R, S and T

	Clue	Compound
P	Same functional group (-OH) as ethanol: an alcohol	Methanol, \(\mathrm{CH_3OH}\) (any alcohol fits; methanol is the simplest)
Q	Not aliphatic and gives Cannizzaro’s reaction (aldehyde, no alpha-hydrogen)	Benzaldehyde, \(\mathrm{C_6H_5CHO}\)
R	Oxidation product of Q	Benzoic acid, \(\mathrm{C_6H_5COOH}\)
S	A compound of ammonia, formed from an aldehyde (this unit)	Urotropine, \(\mathrm{(CH_2)_6N_4}\) (from methanal and ammonia)
T	Nitration of benzene, the fundamental aromatic compound	Nitrobenzene, \(\mathrm{C_6H_5NO_2}\)

\[ \mathrm{2C_6H_5CHO\xrightarrow{conc.\ NaOH}C_6H_5CH_2OH+C_6H_5COONa} \]

\[ \mathrm{C_6H_5CHO\ (Q)\xrightarrow{[O]}C_6H_5COOH\ (R)} \]

\[ \mathrm{6HCHO+4NH_3\longrightarrow(CH_2)_6N_4\ (S)+6H_2O} \]

\[ \mathrm{C_6H_6+HNO_3\xrightarrow{conc.\ H_2SO_4}C_6H_5NO_2\ (T)+H_2O} \]

Figure: Cannizzaro reaction of benzaldehyde, which has no α-hydrogen

B. Identify W, X and Y

A primary alcohol that gives the iodoform test must contain the \(\mathrm{CH_3CH(OH)-}\) group, and the only primary alcohol that does is ethanol. So W is ethanol. Acidified dichromate oxidises it to ethanal (X), and ethanal gives an orange-yellow precipitate with 2,4-DNP (Y).

	Name	Formula
W	Ethanol	\(\mathrm{CH_3CH_2OH}\)
X	Ethanal	\(\mathrm{CH_3CHO}\)
Y	Ethanal 2,4-dinitrophenylhydrazone	\(\mathrm{CH_3CH=N-NH-C_6H_3(NO_2)_2}\)

\[ \mathrm{CH_3CH_2OH\ (W)\xrightarrow{K_2Cr_2O_7/H^+}CH_3CHO\ (X)+H_2O} \]

\[ \mathrm{CH_3CHO\ (X)+H_2N-NH-C_6H_3(NO_2)_2\longrightarrow CH_3CH=N-NH-C_6H_3(NO_2)_2\ (Y)+H_2O} \]

Question 4. A. Benzene (\(\mathrm{CH_3Cl}\)/\(\mathrm{AlCl_3}\)) gives A; A (\(\mathrm{CeO_2}\)/\(\mathrm{H^+}\)) gives B; B (\(\mathrm{CH_3COONa}\)/heat) gives C. Identify in the given reaction sequence. [1+1+1] B. Compound P, \(\mathrm{C_6H_{12}O}\) gives the following results. i. P does not react with Tollen’s reagent. ii. P on catalytic hydrogenation gives Q, \(\mathrm{C_6H_{14}O}\). iii. Q on treatment with conc. \(\mathrm{H_2SO_4}\) gives R, \(\mathrm{C_6H_{12}}\). iv. R on ozonolysis gives two compounds, S (\(\mathrm{C_3H_6O}\)) and T(\(\mathrm{C_3H_6O}\)). v. S gives positive iodoform test and negative Tollen’s test. vi. T gives positive Tollen’s test and negative iodoform test.

A. Identify A, B and C

	Name	Formula
A	Toluene	\(\mathrm{C_6H_5CH_3}\)
B	Benzaldehyde	\(\mathrm{C_6H_5CHO}\)
C	Cinnamic acid	\(\mathrm{C_6H_5CH=CHCOOH}\)

\[ \mathrm{C_6H_6+CH_3Cl\xrightarrow{anhyd.\ AlCl_3}C_6H_5CH_3\ (A)+HCl} \]

\[ \mathrm{C_6H_5CH_3\ (A)+[O]\xrightarrow{CeO_2/H^+}C_6H_5CHO\ (B)+H_2O} \]

\[ \mathrm{C_6H_5CHO\ (B)+(CH_3CO)_2O\xrightarrow{CH_3COONa,\ heat}C_6H_5CH=CHCOOH\ (C)+CH_3COOH} \]

The last step is Perkin’s condensation; sodium ethanoate acts as the base and ethanoic anhydride supplies the -CH₂COOH unit.

Figure: Reaction sequence from benzene to A, B and C

B. Identify P, Q, R, S and T

S and T. S (\(\mathrm{C_3H_6O}\)) gives iodoform and no Tollen’s test: propanone. T gives Tollen’s test and no iodoform: propanal. R is the alkene that gives these two on ozonolysis: \(\mathrm{(CH_3)_2C=CHCH_2CH_3}\), 2-methylpent-2-ene. Q is an alcohol that loses water to give R and is formed by hydrogenation of the carbonyl compound P. P does not react with Tollen’s reagent, so it is a ketone and Q is a secondary alcohol: 2-methylpentan-3-ol, whose dehydration gives the more substituted alkene R (Saytzeff’s rule). So P is 2-methylpentan-3-one.

	Name	Structure
P	2-Methylpentan-3-one	\(\mathrm{(CH_3)_2CHCOCH_2CH_3}\)
Q	2-Methylpentan-3-ol	\(\mathrm{(CH_3)_2CHCH(OH)CH_2CH_3}\)
R	2-Methylpent-2-ene	\(\mathrm{(CH_3)_2C=CHCH_2CH_3}\)
S	Propanone	\(\mathrm{CH_3COCH_3}\)
T	Propanal	\(\mathrm{CH_3CH_2CHO}\)

\[ \mathrm{(CH_3)_2CHCOCH_2CH_3\ (P)+H_2\xrightarrow{Ni}(CH_3)_2CHCH(OH)CH_2CH_3\ (Q)} \]

\[ \mathrm{(CH_3)_2CHCH(OH)CH_2CH_3\ (Q)\xrightarrow{conc.\ H_2SO_4,\ \Delta}(CH_3)_2C=CHCH_2CH_3\ (R)+H_2O} \]

\[ \mathrm{(CH_3)_2C=CHCH_2CH_3\ (R)\xrightarrow{1)\ O_3;\ 2)\ Zn/H_2O}CH_3COCH_3\ (S)+CH_3CH_2CHO\ (T)} \]

Question 5. a. Compound ‘B’ reacts with dil NaOH and gives 4-hydroxy 4-methyl pentan-2-one. ‘A’ reacts with \(\mathrm{Cu(OH)_2}\) to give a carboxylic acid and forms antiseptic reagent by the reduction of ‘A’. ‘C’ reacts with ammoniacal silver nitrate solution but does not react with \(\mathrm{Cu(OH)_2}\). It reacts with acetic anhydride and gives aromatic unsaturated carboxylic acid. Identify A, B, and C with all possible reactions involved. [5] The molecular weights are A 44 amu, B 58 amu and C 106 amu. b. An aliphatic aldehyde is treated with hydrogen cyanide and gives 2-hydroxy propanoic acid after hydrolysis. Identify the compound and write the reactions followed. Also write a reaction to prepare the aldehyde. [2+1]

a. Identify A, B and C

B gives 4-hydroxy-4-methylpentan-2-one with dilute NaOH: this is the aldol product of propanone, \(\mathrm{CH_3COCH_3}\) (M = 58). A (M = 44) is oxidised by Fehling’s solution to a carboxylic acid, so it is an aliphatic aldehyde: ethanal, \(\mathrm{CH_3CHO}\) (2 × 12 + 4 + 16 = 44). Its reduction gives ethanol, which is the antiseptic. C (M = 106) reduces Tollen’s reagent but not Fehling’s solution, so it is an aromatic aldehyde, benzaldehyde, which forms cinnamic acid (an aromatic unsaturated acid) with ethanoic anhydride.

	Name	Formula	M
A	Ethanal	\(\mathrm{CH_3CHO}\)	44
B	Propanone	\(\mathrm{CH_3COCH_3}\)	58
C	Benzaldehyde	\(\mathrm{C_6H_5CHO}\)	106

Reactions of A.

\[ \mathrm{CH_3CHO+2Cu(OH)_2\xrightarrow{\Delta}CH_3COOH+Cu_2O\downarrow+2H_2O} \]

\[ \mathrm{CH_3CHO+H_2\xrightarrow{Ni}CH_3CH_2OH\ (antiseptic)} \]

Reaction of B.

\[ \mathrm{2CH_3COCH_3\xrightarrow{dil.\ NaOH}CH_3C(OH)(CH_3)CH_2COCH_3} \]

Reactions of C.

\[ \mathrm{C_6H_5CHO+2[Ag(NH_3)_2]^++3OH^-\longrightarrow C_6H_5COO^-+2Ag\downarrow+4NH_3+2H_2O} \]

\[ \mathrm{C_6H_5CHO+(CH_3CO)_2O\xrightarrow{CH_3COONa,\ heat}C_6H_5CH=CHCOOH+CH_3COOH} \]

Figure: Perkin condensation. Benzaldehyde and acetic anhydride give cinnamic acid

b. Identify the aldehyde, write the reactions and its preparation

The product 2-hydroxypropanoic acid, \(\mathrm{CH_3CH(OH)COOH}\), has three carbon atoms, and the -CN group of the cyanohydrin supplies one. So the aldehyde has two carbons: ethanal, \(\mathrm{CH_3CHO}\).

\[ \mathrm{CH_3CHO+HCN\longrightarrow CH_3CH(OH)CN} \]

\[ \mathrm{CH_3CH(OH)CN+2H_2O+HCl\longrightarrow CH_3CH(OH)COOH+NH_4Cl} \]

Preparation of ethanal. Mild oxidation of ethanol with PCC (or by passing ethanol over hot copper); ethyne with water and \(\mathrm{HgSO_4}\)/\(\mathrm{H_2SO_4}\) also gives ethanal:

\[ \mathrm{CH_3CH_2OH\xrightarrow{PCC}CH_3CHO+H_2O} \]

\[ \mathrm{HC\equiv CH+H_2O\xrightarrow{HgSO_4/H_2SO_4}CH_3CHO} \]

4. View PDF Notes

i. Handwritten Notes

ii. Typed PDF Notes

5. Frequently Asked Questions

What are aldehydes and ketones? What is their general formula and functional group? (Aldehydes and ketones formula)

Aldehydes and ketones are carbonyl compounds, which contain the carbonyl group >C=O. In an aldehyde the carbonyl carbon is bonded to at least one hydrogen atom, R-CHO, and the functional group is the formyl group -CHO. In a ketone the carbonyl carbon is bonded to two carbon groups, R-CO-R’, and the functional group is the keto group >C=O. The general formula of saturated aliphatic aldehydes and ketones is CnH2n+O, written CnH2nO, so methanal is CH2O, ethanal is C2H4O and propanone is C3H6O.

Give examples of aldehydes and ketones with their IUPAC and common names. (Aldehydes and ketones examples)

Aldehydes: HCHO methanal (formaldehyde); CH3CHO ethanal (acetaldehyde); CH3CH2CHO propanal (propionaldehyde); C6H5CHO benzaldehyde. Ketones: CH3COCH3 propanone (acetone); CH3COCH2CH3 butan-2-one (ethyl methyl ketone); CH3CH2COCH2CH3 pentan-3-one (diethyl ketone); C6H5COCH3 acetophenone (1-phenylethanone); C6H5COC6H5 benzophenone (diphenylmethanone). Aldehydes are named by changing the final -e of the alkane to -al, and ketones by changing it to -one with the position of the carbonyl carbon.

What is the difference between aldehydes and ketones?

In an aldehyde the carbonyl group is at the end of the chain and carries a hydrogen atom (R-CHO); in a ketone it is inside the chain between two carbon groups (R-CO-R’). Aldehydes are easily oxidised to carboxylic acids with the same number of carbon atoms, whereas ketones resist oxidation and break into smaller acids only with strong oxidising agents. Aldehydes give Tollen’s test (silver mirror) and aliphatic aldehydes give Fehling’s test (red precipitate of Cu2O), but ketones give neither. Both give the 2,4-DNP test and add HCN. Methanal and benzaldehyde (no alpha-hydrogen) give Cannizzaro’s reaction.

How are aldehydes and ketones prepared?

Aldehydes: by mild oxidation of primary alcohols with PCC or by dehydrogenation over copper; by Rosenmund’s reduction of acyl chlorides (RCOCl + H2 over Pd/BaSO4); by ozonolysis of alkenes; by hydration of ethyne (ethanal); and, for benzaldehyde, by oxidation of toluene (Etard reaction or CeO2/H+). Ketones: by oxidation of secondary alcohols, by hydration of alkynes other than ethyne (propyne gives propanone with dil. H2SO4 and HgSO4), by ozonolysis of alkenes and by Friedel-Crafts acylation (benzene + CH3COCl/AlCl3 gives acetophenone).

How do you distinguish between aldehydes and ketones? Which chemical tests are used?

Tollen’s test: ammoniacal silver nitrate is reduced to a silver mirror by all aldehydes and not by ketones. Fehling’s test: a red precipitate of copper(I) oxide is given by aliphatic aldehydes only (not by benzaldehyde or ketones). Iodoform test: a yellow precipitate of CHI3 with I2 and NaOH is given by ethanal and methyl ketones (CH3CO- group), for example propanone and pentan-2-one but not pentan-3-one. 2,4-DNP test: an orange-yellow precipitate shows any carbonyl compound. Sodium bisulphite adds to aldehydes and methyl ketones.

What are the reactions of aldehydes and ketones with HCN, hydroxylamine, ammonia and 2,4-DNP?

They undergo nucleophilic addition at the carbonyl carbon. With HCN they give cyanohydrins (ethanal gives CH3CH(OH)CN, which hydrolyses to lactic acid, 2-hydroxypropanoic acid; propanone gives the cyanohydrin that hydrolyses to 2-hydroxy-2-methylpropanoic acid). With hydroxylamine they give oximes, with 2,4-dinitrophenylhydrazine they give coloured 2,4-dinitrophenylhydrazones, and methanal with ammonia gives urotropine (hexamethylenetetramine), (CH2)6N4, a urinary antiseptic.

What is the aldol condensation? Which compounds give it?

Aldehydes and ketones with an alpha-hydrogen, when treated with dilute NaOH, form beta-hydroxy aldehydes or ketones. Two molecules of ethanal give 3-hydroxybutanal (aldol), and two molecules of propanone give 4-hydroxy-4-methylpentan-2-one (diacetone alcohol). Methanal and benzaldehyde have no alpha-hydrogen and do not give it.

What is Cannizzaro’s reaction?

It is the disproportionation of aldehydes without alpha-hydrogen in concentrated NaOH: one molecule is oxidised to the carboxylate salt and the other is reduced to the alcohol. 2HCHO + NaOH gives CH3OH + HCOONa, and 2C6H5CHO + NaOH gives C6H5CH2OH + C6H5COONa.

What are Clemmensen’s, Wolff-Kishner’s and Rosenmund’s reductions?

Clemmensen’s reduction converts C=O into CH2 with zinc amalgam and concentrated HCl (propanone gives propane). Wolff-Kishner’s reduction does the same with hydrazine and KOH in glycol or glycerol at high temperature, and is used when the compound is acid-sensitive. Rosenmund’s reduction is the partial reduction of an acyl chloride to an aldehyde with H2 over palladium on barium sulphate (a poisoned catalyst), for example CH3COCl to CH3CHO.

What are the reactions of benzaldehyde? (Aromatic aldehydes and ketones)

Benzaldehyde gives Tollen’s test but not Fehling’s test. With acetic anhydride and sodium acetate it undergoes Perkin’s condensation to give cinnamic acid; with alcoholic KCN it gives benzoin (benzoin condensation); with concentrated NaOH it gives Cannizzaro’s reaction (benzyl alcohol and sodium benzoate). The -CHO group withdraws electrons, so electrophilic substitution (nitration, sulphonation, halogenation) occurs at the meta position, giving m-nitrobenzaldehyde with HNO3 and H2SO4. Acetophenone is prepared by Friedel-Crafts acylation of benzene.

What is formalin? What is Bakelite? What are the uses of aldehydes and ketones?

Formalin is a 37 to 40 percent aqueous solution of methanal (with a little methanol), used to preserve biological specimens and as a disinfectant. Bakelite is a thermosetting polymer made by condensing phenol with methanal. Urotropine, from methanal and ammonia, is a urinary antiseptic. Propanone is a solvent for paints, nail polish and plastics, and is used to make chloretone (a hypnotic). Benzaldehyde is used in perfumes and flavours, and cinnamic acid esters in perfumery.

What is chloretone and how is it prepared?

Chloretone (1,1,1-trichloro-2-methylpropan-2-ol) is a hypnotic and preservative. It is made by adding chloroform to propanone in the presence of a solid alkali such as KOH: CH3COCH3 + CHCl3 gives (CH3)2C(OH)CCl3.

Where can I get Aldehydes and ketones Class 12 notes, notes PDF and handwritten notes?

This page includes the handwritten notes and the typed notes PDF of Unit 13 Aldehydes and Ketones, which you can view online free of cost, along with 29 solved multiple choice questions and the complete textbook exercise with diagrams. The notes follow the NEB Class 12 Chemistry syllabus 2083 and work as short notes for quick revision before the exam.

What topics are included in Unit 13 Aldehydes and Ketones of NEB Class 12 Chemistry?

Unit 13 has two parts: 13.1 aliphatic aldehydes and ketones, and 13.2 aromatic aldehydes and ketones. It covers nomenclature, preparation, tests (Tollen’s, Fehling’s, iodoform, 2,4-DNP), addition reactions, aldol and Cannizzaro reactions, reductions, and the preparation and reactions of benzaldehyde and acetophenone.

Where can I get Class 12 Chemistry notes of NEB (handwritten, PDF, Nepal)?

The Class 12 Chemistry notes of the NEB syllabus 2083 are available on this website unit by unit, in handwritten and typed PDF form. Students who look for notes in the style of Note Mandir will find this unit in the same way: syllabus, solved exercise, handwritten notes and typed notes. These are NEB Class 12 Chemistry notes for Chapter 13, Aldehydes and Ketones, and the syllabus items are listed at the top of the page.

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Aldehydes and Ketones

1. Syllabus

2. Multiple Choice Questions

3. Numerical and Subjective Questions

Short Answer Questions

Long Answer Questions

4. View PDF Notes

5. Frequently Asked Questions

Also Read: NEB Class 12 Notes

NEB (Grade 12)

1. Syllabus

2. Multiple Choice Questions

3. Numerical and Subjective Questions

Short Answer Questions

Long Answer Questions

4. View PDF Notes

5. Frequently Asked Questions

Also Read: NEB Class 12 Notes

NEB (Grade 12)

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