Question:

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

Answer:

(i) Butane-1,3-diol: Step 1: Oxidize ethanal with a strong oxidizing agent such as potassium permanganate (KMnO₄) to form acetaldehyde.

Step 2: React acetaldehyde with a strong base such as sodium hydroxide (NaOH) in the presence of an alcohol such as butanol to form butane-1,3-diol.

(ii) But-2-enal: Step 1: Oxidize ethanal with a strong oxidizing agent such as potassium permanganate (KMnO₄) to form acetaldehyde.

Step 2: React acetaldehyde with a strong acid such as sulfuric acid (H₂SO₄) to form but-2-enal.

(iii) But-2-enoic acid: Step 1: Oxidize ethanal with a strong oxidizing agent such as potassium permanganate (KMnO₄) to form acetaldehyde.

Step 2: React acetaldehyde with an oxidizing agent such as potassium dichromate (K2Cr2O7) in the presence of sulfuric acid (H₂SO₄) to form but-2-enoic acid.

Question:

Which of the following compounds would undergo aldol condensation, which undergoe Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction. (i) Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde
(iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone (vii) Phenylacetaldehyde (viii) Butan-1-ol (ix) 2,2-Dimethylbutanal

Answer:

(i) Methanal: Aldol condensation - Aldol product: CH₃CH(OH)CH₂CHO Cannizzaro reaction - No reaction

(ii) 2-Methylpentanal: Aldol condensation - Aldol product: CH₃CH(OH)CH(CH₃)CH₂CHO Cannizzaro reaction - No reaction

(iii) Benzaldehyde: Aldol condensation - Aldol product: C₆H5CH(OH)CH₂CHO Cannizzaro reaction - Cannizzaro product: C₆H5CH₂OH + C₆H5CHO

(iv) Benzophenone: Aldol condensation - No reaction Cannizzaro reaction - No reaction

(v) Cyclohexanone: Aldol condensation - Aldol product: CH₃CH(OH)CH₂CH₂CH₂CH₂CHO Cannizzaro reaction - No reaction

(vi) 1-Phenylpropanone: Aldol condensation - Aldol product: C₆H5CH(OH)CH₂CH(CH₃)CHO Cannizzaro reaction - No reaction

(vii) Phenylacetaldehyde: Aldol condensation - Aldol product: C₆H5CH(OH)CH₂CHO Cannizzaro reaction - Cannizzaro product: C₆H5CH₂OH + C₆H5CHO

(viii) Butan-1-ol: Aldol condensation - No reaction Cannizzaro reaction - No reaction

(ix) 2,2-Dimethylbutanal: Aldol condensation - Aldol product: CH₃CH(OH)CH(CH₃)CH(CH₃)CHO Cannizzaro reaction - No reaction

Question:

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

Answer:

(i) Acetylation: Acetylation is a process that involves the addition of an acetyl group (-COCH₃) to a molecule. It is a common form of post-translational modification and is involved in many biological processes.

(ii) Cannizzaro reaction: The Cannizzaro reaction is an organic redox reaction in which an aldehyde is oxidized to form an acid and a base. The reaction is named after Italian chemist Stanislao Cannizzaro, who first observed it in 1853.

(iii) Cross aldol condensation: Cross aldol condensation is a type of aldol condensation reaction in which two different aldehydes or ketones are reacted to form a β-hydroxy aldehyde or ketone.

(iv) Decarboxylation: Decarboxylation is a chemical reaction in which a carboxyl group is removed from a molecule, releasing carbon dioxide (CO2) and resulting in a new molecule. It is a common reaction in organic chemistry and is used in the synthesis of a variety of compounds.

Question:

Give 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.

Answer:

(i) Cyclohexanone forms cyanohydrin in good yield because it is a relatively more reactive molecule compared to 2,2,6-trimethylcyclohexanone. The former molecule has a larger surface area and a higher electron density, which allows it to react more quickly and efficiently with cyanide ions. On the other hand, 2,2,6-trimethylcyclohexanone is a more sterically hindered molecule, which reduces its reactivity and thus the formation of the cyanohydrin.

(ii) The two NH₂​ groups in semicarbazide are not equally reactive. The one that is involved in the formation of semicarbazones is the one that is more exposed and accessible to reactants, whereas the other one is more sterically hindered and thus less reactive.

(iii) During the preparation of esters from a carboxylic acid and an alcohol, the water or the ester should be removed as soon as it is formed in order to prevent the reverse reaction from occurring. If the water or ester is not removed, the reaction may proceed in the reverse direction, resulting in the formation of the carboxylic acid and alcohol instead of the desired ester.

Question:

An 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 give positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acid. Write the possible structure of the compound.

Answer:

1. The empirical formula of the compound can be calculated using the percentages of carbon, hydrogen and oxygen present in it.

2. The empirical formula of the compound is C₂H₄O.

3. The molecular mass of the compound is 86, which means that the compound has two molecules of C₂H₄O.

4. The compound does not reduce Tollens reagent but forms an addition compound with sodium hydrogensulphite and gives positive iodoform test.

5. On vigorous oxidation, it gives ethanoic and propanoic acid.

6. The possible structure of the compound is ethylene glycol (C₂H₆O2).

Question:

An organic compound with the molecular formula C9​H10​O forms 2,4-DNP derivative, reduces Tollens reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.

Answer:

1. The compound has a molecular formula of C9H10O.

2. The compound forms a 2,4-DNP derivative, which is a derivative of 2,4-dinitrophenol.

3. The compound reduces Tollens reagent, which is a silver-ammonia complex that is used to test for aldehydes.

4. The compound undergoes Cannizzaro reaction, which is a reaction between an aldehyde and a base that produces an alcohol and a carboxylic acid.

5. On vigorous oxidation, the compound gives 1,2-benzenedicarboxylic acid.

Therefore, the compound is an aldehyde with the molecular formula C9H10O, likely an aldehyde derivative of benzene, such as benzaldehyde.

Question:

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

Answer:

(i) 2,4-Dinitrophenylhydrazone of benzaldehyde

O \ / C=O /
H–C–H \ / N=N /
H–N–H | | H–C–H | | H–O–H

(ii) Cyclopropanone oxime

O \ / C=O /
H–C–H \ / N=N /
H–N–H | | H–C–H | | H–O–H

(iii) Acetaldehyde dimethylacetal

O \ / C=O /
H–C–H \ / O–C–O /
H–C–H | | H–O–H

(iv) The semicarbazone of cyclobutanone

O \ / C=O /
H–C–H \ / N=N /
H–N–H | | H–C–H | | H–C–H | | H–O–H

(v) The ethylene ketal of hexan-3-one

O \ / C=O /
H–C–H \ / O–C–O /
H–C–H | | H–C–H | | H–C–H | | H–C–H | | H–O–H

(vi) The methyl hemiacetal of formaldehyde

O \ / C=O /
H–C–H \ / O–C–H /
H–C–H | | H–O–H

Question:

An organic compound (A) (molecular formula C8​H16​O2​) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidationof (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene.Write equations for the reactions involved.

Answer:

1. Hydrolysis of A with dilute sulphuric acid: C8H16O2 + H₂SO₄ → CH₁₄O₄ + C₂H₆O

2. Oxidation of C with chromic acid: C₂H₆O + CrO₃ + H₂SO₄ → CH₁₄O₄ + H₂O

3. Dehydration of C to form but-1-ene: C₂H₆O → C₄H₈ (but-1-ene)

Question:

What 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 (vii) Imine (ix) 2,4-DNP-derivative (x) Schiffs base

Answer:

(i) Cyanohydrin: Cyanohydrin is an organic compound containing both a carbonyl group and a cyanide group. An example of a reaction involving a cyanohydrin is the hydrolysis of 2-hydroxypropyl cyanide, which produces glycerol and hydrogen cyanide.

(ii) Acetal: An acetal is an organic compound containing an ether group and two alkyl groups. An example of a reaction involving an acetal is the formation of 1,1-diethoxyethane from ethylene glycol and acetic acid.

(iii) Semicarbazone: A semicarbazone is an organic compound containing both a carbonyl group and a semicarbazone group. An example of a reaction involving a semicarbazone is the formation of a semicarbazone from an aldehyde and a primary amine.

(iv) Aldol: An aldol is an organic compound containing both a carbonyl group and an alcohol group. An example of a reaction involving an aldol is the aldol reaction, which involves the condensation of two aldehydes or ketones to form an aldol product.

(v) Hemiacetal: A hemiacetal is an organic compound containing an ether group and one alkyl group. An example of a reaction involving a hemiacetal is the formation of a hemiacetal from an aldehyde and an alcohol.

(vi) Oxime: An oxime is an organic compound containing both a carbonyl group and an oxime group. An example of a reaction involving an oxime is the formation of an oxime from an aldehyde and a hydroxylamine.

(vii) Ketal: A ketal is an organic compound containing an ether group and two ketone groups. An example of a reaction involving a ketal is the formation of a ketal from an aldehyde and a ketone.

(viii) Imine: An imine is an organic compound containing both a carbonyl group and an amine group. An example of a reaction involving an imine is the formation of an imine from an aldehyde and an amine.

(ix) 2,4-DNP-derivative: A 2,4-DNP-derivative is an organic compound containing both a carbonyl group and a 2,4-dinitrophenylhydrazone group. An example of a reaction involving a 2,4-DNP-derivative is the formation of a 2,4-DNP-derivative from an aldehyde and a 2,4-dinitrophenylhydrazine.

(x) Schiffs base: A Schiffs base is an organic compound containing both a carbonyl group and an amine group. An example of a reaction involving a Schiffs base is the formation of a Schiffs base from an aldehyde and a primary amine.

Question:

How 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) Benazaldehyde to α-Hydroxyphenylacetic acid (ix) Benzoic acid to m- Nitrobenzyl alcohol

Answer:

(i) Propanone to Propene: Oxidation and Hydrogenation (ii) Benzoic acid to Benzaldehyde: Oxidation (iii) Ethanol to 3-Hydroxybutanal: Oxidation (iv) Benzene to m-Nitroacetophenone: Nitration (v) Benzaldehyde to Benzophenone: Oxidation (vi) Bromobenzene to 1-Phenylethanol: Hydrolysis (vii) Benzaldehyde to 3-Phenylpropan-1-ol: Hydrogenation (viii) Benazaldehyde to α-Hydroxyphenylacetic acid: Oxidation and Hydrolysis (ix) Benzoic acid to m- Nitrobenzyl alcohol: Nitration and Reduction

Question:

Arrange 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) CH₃​CH₂​CH(Br)COOH, CH₃​CH(Br)CH₂​COOH, (CH₃​)2​CHCOOH, CH₃​CH₂​CH₂​COOH (acid strength) (iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid,4-Methoxybenzoic acid (acid strength)

Answer:

(i) Di-tert-butyl ketone, Methyl tert-butyl ketone, Acetone, Acetaldehyde

(ii) (CH₃​)2​CHCOOH, CH₃​CH₂​CH₂​COOH, CH₃​CH(Br)CH₂​COOH, CH₃​CH₂​CH(Br)COOH

(iii) Benzoic acid, 4-Methoxybenzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid

Question:

How 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.

Answer:

(i) Methyl benzoate: Step 1: React benzene with methanol in the presence of an acid catalyst to form methyl benzoate.

(ii) m-Nitrobenzoic acid: Step 1: React benzene with nitric acid to form nitrobenzene. Step 2: React nitrobenzene with concentrated sulfuric acid to form m-nitrobenzoic acid.

(iii) p-Nitrobenzoic acid: Step 1: React benzene with nitric acid to form nitrobenzene. Step 2: React nitrobenzene with concentrated sulfuric acid and heating to form p-nitrobenzoic acid.

(iv) Phenylacetic acid: Step 1: React benzene with acetic acid in the presence of an acid catalyst to form phenylacetaldehyde. Step 2: React phenylacetaldehyde with concentrated sulfuric acid to form phenylacetic acid.

(v) p-Nitrobenzaldehyde: Step 1: React benzene with nitric acid to form nitrobenzene. Step 2: React nitrobenzene with sodium hydroxide and heating to form p-nitrobenzaldehyde.

Question:

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

Answer:

1. Phenoxide ion is an anion formed when a phenol molecule loses a hydrogen atom, while carboxylate ion is an anion formed when a carboxylic acid molecule loses a hydrogen atom.

2. Phenoxide ion has more number of resonating structures, which means that the electrons in the molecule are more delocalised. This makes the phenoxide ion less stable and less able to donate protons, making phenol a weaker acid than carboxylic acid.

3. On the other hand, carboxylate ion has fewer resonating structures, which means that the electrons in the molecule are more localized. This makes the carboxylate ion more stable and more able to donate protons, making carboxylic acid a stronger acid than phenol.

Question:

Draw 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 acid

Answer:

(i) 3-Methylbutanal

Structure:

H3C-CH₂-CH₂-CH₂-CHO

(ii) p-Nitropropiophenone

Structure:

O-NO2-C₆H₄-CH(CH₃)-CO-CH₃

(iii) p-Methylbenzaldehyde

Structure:

O-CH₃-C₆H₄-CHO

(iv) 4-Methylpent-3-en-2-one

Structure:

H3C-CH₂-CH=CH-CH₂-CH₂-CO-CH₃

(v) 4-Chloropentan-2-one

Structure:

Cl-CH₂-CH₂-CH₂-CH₂-CO-CH₃

(vi) 3-Bromo-4-phenylpentanoic acid

Structure:

Br-C₆H5-CH₂-CH₂-CH₂-CH₂-COOH

(vii) p,p-Dihydroxybenzophenone

Structure:

O-C₆H₄-C(OH)-C(OH)-CO-C₆H₄

(viii) Hex-2-en-4-ynoic acid

Structure:

H₂C=CH-CH₂-CH=CH-COOH

Question:

Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names. A CH₃​CO(CH₂​)4​CH₃​ B CH₃​CH₂​CHBrCH₂​CH(CH₃​)CHO C CH₃​(CH₂​)5​CHO D Ph−CH=CH−CHO

Answer:

A: Pentan-2-one (Valeraldehyde) B: 2-Bromo-3-methylpentanal (2-Bromo-3-methylvaleraldehyde) C: Hexanal (Caproaldehyde) D: Phenylacetaldehyde

Question:

Write 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.

Answer:

1. Propanal as nucleophile + Butanal as electrophile: Structural Formula: CH₃CH₂CH₂CHO + CH₃CH₂CH₂CH₂CHO → CH₃CH₂CH₂CH₂CH₂CH₂CHO Name: Pentanal

2. Propanal as electrophile + Butanal as nucleophile: Structural Formula: CH₃CH₂CH₂CHO + CH₃CH₂CH₂CH₂CHO → CH₃CH₂CH₂CH(OH)CH₂CHO Name: 3-Hydroxybutanal

3. Propanal as nucleophile + Butanal as nucleophile: Structural Formula: CH₃CH₂CH₂CHO + CH₃CH₂CH₂CH₂CHO → CH₃CH₂CH₂CH(OH)CH₂CH₂CHO Name: 3-Hydroxypentanal

4. Propanal as electrophile + Butanal as electrophile: Structural Formula: CH₃CH₂CH₂CHO + CH₃CH₂CH₂CH₂CHO → CH₃CH₂CH(OH)CH₂CH₂CH₂CHO Name: 4-Hydroxybutanal

Question:

Name the following compounds according to IUPAC system of nomenclature: (i) CH₃​CH(CH₃​)CH₂​CH₂​CHO
(ii) CH₃​CH₂​COCH(C₂​H5​)CH₂​CH₂​Cl (iii) CH₃​CH=CHCHO (iv) CH₃​COCH₂​COCH₃​ (v) CH₃​CH(CH₃​)CH₂​C(CH₃​)2​COCH₃​ (vi) (CH₃​)3​CCH₂​COOH (vii) OHCC₆H₄CHO−p

Answer:

(i) Pent-2-en-4-yn-1-ol (ii) 3-Chloro-2-ethylpent-1-ene (iii) 2-Methyl-2-propen-1-ol (iv) 2-Methylpropane-1,3-diol (v) 2,3-Dimethylbutane-1,4-diol (vi) 2,2-Dimethylpropanoic acid (vii) 4-Hydroxybenzene-1-carboxylate anion

Question:

Give 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

Answer:

(i) Propanal and Propanone: Test 1: Add a few drops of 2,4-dinitrophenylhydrazine solution to each compound. Propanal will give a yellow precipitate, while propanone will give no reaction.

(ii) Acetophenone and Benzophenone: Test 2: Add a few drops of 2,4-dinitrophenylhydrazine solution to each compound. Acetophenone will give a yellow precipitate, while benzophenone will give no reaction.

(iii) Phenol and Benzoic acid: Test 3: Add a few drops of 2,4-dinitrophenylhydrazine solution to each compound. Phenol will give a yellow precipitate, while benzoic acid will give no reaction.

(iv) Benzoic acid and Ethyl benzoate: Test 4: Add a few drops of sodium hydroxide solution to each compound. Benzoic acid will give a white precipitate, while ethyl benzoate will give no reaction.

(v) Pentan-2-one and Pentan-3-one: Test 5: Add a few drops of bromine water to each compound. Pentan-2-one will give no reaction, while pentan-3-one will decolorize the bromine water.

(vi) Benzaldehyde and Acetophenone: Test 6: Add a few drops of 2,4-dinitrophenylhydrazine solution to each compound. Benzaldehyde will give a yellow precipitate, while acetophenone will give no reaction.

(vii) Ethanal and Propanal: Test 7: Add a few drops of 2,4-dinitrophenylhydrazine solution to each compound. Ethanal will give a yellow precipitate, while propanal will give no reaction.

Question:

Predict the products formed when cyclohexane carbaldehyde reacts with the following reagents. (i) PhMgBr and then H3​O+ (ii) Tollens reagent (iii) Semicarbazide and weak acid (iv) Excess ethanol and acid (v) Zinc amalgam and dilute hydrochloric acid

Answer:

(i) PhMgBr and then H3O+: The product formed is cyclohexanol.

(ii) Tollens reagent: The product formed is silver mirror.

(iii) Semicarbazide and weak acid: The product formed is cyclohexanone.

(iv) Excess ethanol and acid: The product formed is cyclohexanone.

(v) Zinc amalgam and dilute hydrochloric acid: The product formed is cyclohexanol.