Question:

Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?

Answer:

Answer:

1. Ortho nitrophenol is an aromatic compound containing a nitro group (-NO₂) which is a strongly electron-withdrawing group.

2. The nitro group causes the electron density of the aromatic ring to be reduced, resulting in a more negative charge on the oxygen atom of the hydroxyl group.

3. This increased negative charge on the oxygen atom makes the hydroxyl group more acidic than the hydroxyl group of ortho methoxyphenol, which does not contain a nitro group.

Question:

Illustrate with examples the limitations of Williamson’s synthesis for the preparation of certain types of ethers.

Answer:

1. Williamson’s synthesis is a method of synthesizing ethers by reacting an alkyl halide with an alkoxide ion. The ether formed is a symmetrical ether, meaning that the two alkyl groups attached to the oxygen atom are the same.

2. Limitations of Williamson’s synthesis for the preparation of certain types of ethers include:

a. It is not suitable for the synthesis of unsymmetrical ethers, meaning ethers where the two alkyl groups attached to the oxygen atom are different. For example, the synthesis of ethyl methyl ether (CH₃OCH₂CH₃) cannot be achieved using Williamson’s synthesis.

b. It is not suitable for the synthesis of aryl ethers, meaning ethers where one of the alkyl groups is replaced by an aromatic group. For example, the synthesis of phenyl ethyl ether (C₆H₅OCH₂CH₃) cannot be achieved using Williamson’s synthesis.

c. It is not suitable for the synthesis of ethers with a tertiary alkyl halide, meaning an alkyl halide where the carbon atom attached to the halogen is bonded to three other carbon atoms. For example, the synthesis of tert-butyl ethyl ether (CH₃CH(OCH₃)CH₃) cannot be achieved using Williamson’s synthesis.

Question:

While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile? Give reason.

Answer:

Answer: The isomer which will be steam volatile is para nitrophenol. This is because para nitrophenol has a lower boiling point compared to ortho nitrophenol, which makes it more suitable for steam distillation.

Question:

Write the mechanism of acid dehydration of ethanol to yield ethene.

Answer:

1. Protonation of ethanol molecule by the acid catalyst: CH₃CH₂OH + H⁺ → CH₃CH₂OH₂+

2. Deprotonation of protonated ethanol molecule by the acid catalyst: CH₃CH₂OH₂+ + OH^- → CH₃CH₂O- + H₂O

3. Loss of water molecule from protonated ethanol molecule: CH₃CH₂O- → CH₂=CH₂ + H₂O

Question:

Explain why propanol has higher boiling point than that of the butane?

Answer:

Answer:

1. Propanol is a type of alcohol, which has an oxygen atom in its molecular structure.
2. This oxygen atom forms a strong intermolecular hydrogen bond with other molecules of propanol, which increases the boiling point of propanol.
3. Butane, on the other hand, is a hydrocarbon and does not contain an oxygen atom in its molecular structure.
4. Therefore, it does not form any hydrogen bonds and has a lower boiling point than propanol.

Question:

Write structures of the compounds whose IUPAC names are as follows: (i) 2-Methylbutan-2-ol (ii) 1-Phenylpropan-2-ol (iii) 3,5-Dimethylhexane 1, 3, 5-triol (iv) 2,3 Diethylphenol (v) 1 Ethoxypropane (vi) 2-Ethoxy-3-methylpentane (vii) Cyclohexylmethanol (viii) 3-Cyclohexylpentan-3-ol (ix) Cyclopent-3-en-1-ol (x) 4-Chloro-3-ethylbutan-1-ol.

Answer:

(i) 2-Methylbutan-2-ol Structure: CH₃CH(OH)CH₂CH₃

(ii) 1-Phenylpropan-2-ol Structure: C₆H₅CH₂CH(OH)CH₃

(iii) 3,5-Dimethylhexane 1, 3, 5-triol Structure: CH₃(CH₂)4CH(OH)CH(OH)CH(OH)CH₂CH₃

(iv) 2,3 Diethylphenol Structure: C₆H4(C2H₅)2OH

(v) 1 Ethoxypropane Structure: CH₃CH₂OCH₂CH₃

(vi) 2-Ethoxy-3-methylpentane Structure: CH₃CH₂CH₂OCH₂CH(CH₃)CH₃

(vii) Cyclohexylmethanol Structure: C₆H11OH

(viii) 3-Cyclohexylpentan-3-ol Structure: C₆H11CH(OH)CH₂CH₂CH₂CH₃

(ix) Cyclopent-3-en-1-ol Structure: C₅H9OH

(x) 4-Chloro-3-ethylbutan-1-ol Structure: ClCH₂CH₂CH₂CH(OH)CH₃

Question:

Give equations of the following reactions: (i) Oxidation of propan-1-ol with alkaline KMnO4​ solution. (ii) Bromine in CS₂ with phenol. (iii) Dilute HNO₃​ with phenol. (iv) Treating phenol wih chloroform in presence of aqueous NaOH.

Answer:

(i) C₃H₇OH + 2KMnO4 + 3H₂O → 2MnO₂ + KOH + 3H₂CO3

(ii) C₆H₅OH + Br₂ → C₆H₅Br + HBr

(iii) C₆H₅OH + 3HNO₃ → C₆H₅NO₃ + 3H₂O

(iv) C₆H₅OH + CCl₃ + 3NaOH → C₆H₅ONa + 3NaCl + 3H₂O

Question:

Give reason for the higher boiling point of ethanol in comparison to methoxymethane.

Answer:

Step 1: Boiling point is the temperature at which a liquid changes to its gaseous state.

Step 2: Ethanol has a higher boiling point than methoxymethane because it has a higher molecular weight and more hydrogen bonding.

Step 3: Hydrogen bonding is a type of intermolecular force of attraction between molecules that have hydrogen atoms covalently bonded to a highly electronegative atom, such as oxygen or nitrogen. The hydrogen atom in ethanol is covalently bonded to an oxygen atom, which is more electronegative than the carbon atom in methoxymethane. This causes the hydrogen atom to be more strongly attracted to the oxygen atom, resulting in a stronger intermolecular force of attraction.

Step 4: The stronger intermolecular force of attraction in ethanol results in a higher boiling point than methoxymethane.

Question:

Write the mechanism of the reaction of HI with methoxymethane.

Answer:

Step 1: The reaction of HI with methoxymethane is an S N₂ reaction.

Step 2: In this reaction, the hydrogen atom of HI is replaced by a methoxy group.

Step 3: The reaction mechanism involves a nucleophilic attack by the lone pair of electrons on the carbon atom of the HI molecule.

Step 4: The lone pair of electrons on the oxygen atom of the methoxy group is used to form a bond with the carbon atom of HI.

Step 5: The hydrogen atom of HI is displaced and the hydrogen ion is released.

Step 6: The reaction is completed by a proton transfer from the oxygen atom of the methoxy group to the hydrogen ion.

Question:

Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.

Answer:

Structures:

1.

IUPAC name: 2-Methylphenol

2.

IUPAC name: 3-Methylphenol

3.

IUPAC name: 4-Methylphenol

Question:

Explain the following with an example. (i) Kolbes reaction. (ii) Reimer-Tiemann reaction. (iii) Williamson ether synthesis. (iv) Unsymmetrical ether.

Answer:

(i) Kolbe’s Reaction: Kolbe’s reaction is an organic reaction in which a carboxylic acid reacts with sodium hydroxide to form an alkali metal salt of the carboxylic acid and water. For example, the reaction of acetic acid and sodium hydroxide produces sodium acetate and water: CH₃COOH + NaOH → CH₃COONa + H₂O

(ii) Reimer-Tiemann Reaction: The Reimer-Tiemann reaction is an organic reaction in which aromatic aldehydes are converted to phenols by reaction with chloroform and an alkali. For example, the reaction of benzaldehyde and sodium hydroxide produces sodium benzylate and phenol: C₆H₅CHO + NaOH + CHCl₃ → C₆H₅CH₂ONa + C₆H₅OH

(iii) Williamson Ether Synthesis: Williamson ether synthesis is an organic reaction in which an alkoxide ion reacts with an alkyl halide to form an ether. For example, the reaction of sodium ethoxide and methyl bromide produces diethyl ether: CH₃Br + NaOCH₂CH₃ → CH₃OCH₂CH₂CH₃ + NaBr

(iv) Unsymmetrical Ether: An unsymmetrical ether is an ether molecule that contains two different alkyl groups on either side of the oxygen atom. For example, the molecule diethyl ether is an unsymmetrical ether because it contains two different ethyl groups: CH₃CH₂OCH₂CH₃

Question:

(i) Draw the structures of all isomeric alcohols of molecular formula C₅​H₁₂​O and give their IUPAC names. (ii) Classify the isomers of alcohols in above question as primary, secondaryand tertiary alcohols.

Answer:

(i) Structures of all isomeric alcohols of molecular formula C₅H₁₂O:

1. Pentan-1-ol (IUPAC name: 1-Pentanol)

2. Pentan-2-ol (IUPAC name: 2-Pentanol)

3. Pentan-3-ol (IUPAC name: 3-Pentanol)

4. 2-Methylbutan-1-ol (IUPAC name: 2-Methyl-1-butanol)

5. 2-Methylbutan-2-ol (IUPAC name: 2-Methyl-2-butanol)

(ii) Classification of isomers of alcohols:

1. Pentan-1-ol: Primary Alcohol

2. Pentan-2-ol: Secondary Alcohol

3. Pentan-3-ol: Tertiary Alcohol

4. 2-Methylbutan-1-ol: Primary Alcohol

5. 2-Methylbutan-2-ol: Secondary Alcohol

Question:

Show how will you synthesise: (i) 1-phenylethanol from a suitable alkene. (ii) cyclohexylmethanol using an alkyl halide by an SN​2 reaction. (iii) pentan-1-ol using a suitable alkyl halide.

Answer:

(i) 1-Phenylethanol Synthesis: Step 1: Treat the alkene with a strong base such as sodium hydroxide to form an alkoxide. Step 2: Treat the alkoxide with aqueous acid to form the alcohol.

(ii) Cyclohexylmethanol Synthesis: Step 1: Treat the alkyl halide with a strong nucleophile such as sodium hydroxide to form an intermediate alkyl halide. Step 2: Treat the alkyl halide with a strong base such as sodium hydroxide to form an alkoxide. Step 3: Treat the alkoxide with aqueous acid to form the alcohol.

(iii) Pentan-1-ol Synthesis: Step 1: Treat the alkyl halide with a strong nucleophile such as sodium hydroxide to form an intermediate alkyl halide. Step 2: Treat the alkyl halide with a strong base such as sodium hydroxide to form an alkoxide. Step 3: Treat the alkoxide with aqueous acid to form the alcohol.

Question:

Write equations of the following reactions: (i) Friedel-Crafts reaction alkylation of anisole. (ii) Nitration of anisole. (iii) Bromination of anisole in ethanoic acid medium. (iv) Friedel-Crafts acetylation of anisole.

Answer:

(i) C8H₇O + R-Cl → C8H₇OR + Cl^-

(ii) C8H₇O + HNO₃ → C8H₇NO₂ + H₂O

(iii) C8H₇O + Br₂ + CH₃COOH → C8H₇OBr + CH₃COO^-

(iv) C8H₇O + CH₃COCl → C8H₇OCOCH₃ + HCl

Question:

How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction.

Answer:

Answer:

1-Propoxypropane can be synthesized from propan-1-ol by a process known as alcoholysis. The reaction mechanism is as follows:

Step 1: Propan-1-ol undergoes nucleophilic attack by a hydroxide ion, forming a tetrahedral intermediate.

Step 2: The hydroxide ion departs, and the intermediate collapses to form a carbocation.

Step 3: The carbocation undergoes a proton transfer with the hydroxide ion, forming 1-propoxypropane.

Question:

Write the equation of the reaction of hydrogen iodide with: (i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether.

Answer:

(i) 1-Propoxypropane: C₃H₇O + HI → C₃H₇I + H₂O

(ii) Methoxybenzene: C₆H₅OCH₃ + HI → C₆H₅I + H₂O

(iii) Benzyl ethyl ether: C₆H₅CH₂OCH₂CH₃ + HI → C₆H₅CH₂I + H₂O

Question:

What is meant by hydroboration-oxidation reaction? Illustrate it with an example.

Answer:

1. Hydroboration-oxidation reaction is an organic reaction that involves the addition of a boron hydride (BH₃) and a proton (H⁺) to an alkene, followed by the addition of an oxidizing agent (such as H₂O₂).

2. An example of this reaction is the hydroboration-oxidation of 2-methyl-2-butene to form 2,3-dimethyl-2-butanol. The reaction proceeds as follows:

Step 1: Addition of BH₃ and H⁺: 2-methyl-2-butene + BH₃ + H⁺ → 2-methyl-2-butylborane

Step 2: Addition of H₂O₂: 2-methyl-2-butylborane + H₂O₂ → 2,3-dimethyl-2-butanol + H₂O

The hydroboration-oxidation reaction is used to convert alkenes into alcohols.

Question:

Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.

Answer:

1. Alcohols are organic compounds that contain an hydroxyl group (-OH) attached to a carbon atom.

2. This hydroxyl group is polar in nature, which means that it has a positive charge on one side and a negative charge on the other side.

3. Water is also a polar molecule, with a slightly positive charge on one side and a slightly negative charge on the other side.

4. Due to the polarity of both alcohols and water, the two molecules are attracted to each other. This attraction is known as hydrogen bonding.

5. The hydrogen bonds between water and alcohol molecules are stronger than the weak Van der Waals forces between the hydrocarbon molecules.

6. As a result, alcohols are more soluble in water than hydrocarbons of comparable molecular masses.

Question:

Explain how does the OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?

Answer:

Step 1: An OH group attached to a carbon of a benzene ring is known as a phenol group.

Step 2: Phenol groups can activate the benzene ring towards electrophilic substitution due to the presence of the electron-withdrawing oxygen atom.

Step 3: The oxygen atom in the phenol group can withdraw electrons from the benzene ring, making it more susceptible to electrophilic attack.

Step 4: This creates an electrophilic centre on the benzene ring, which can be attacked by an electrophile, resulting in a substitution reaction.

Question:

Write the mechanism of hydration of ethene to yield ethanol.

Answer:

Step 1: Ethene reacts with water molecules to form a carbocation intermediate.

Step 2: The carbocation intermediate then reacts with a hydroxide ion (OH^-) to form an oxonium ion intermediate.

Step 3: The oxonium ion then reacts with a hydrogen ion (H⁺) to form ethanol.

Question:

Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.

Answer:

1. Reaction of phenol with sodium hydroxide: Phenol reacts with sodium hydroxide to form sodium phenoxide and water. This reaction is an example of the acidic nature of phenol.

2. Reaction of phenol with sodium bicarbonate: Phenol reacts with sodium bicarbonate to form sodium phenoxide and carbon dioxide. This reaction is also an example of the acidic nature of phenol.

The acidity of phenol is greater than that of ethanol. This is because phenol has a higher pKa value than ethanol. Phenol has a pKa value of about 10, while ethanol has a pKa value of about 15.

Question:

Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis: (i) 1-Propoxypropane (ii) Ethoxybenzene (iii) 2-Methoxy-2-methylpropane (iv) 1-Methoxyethane

Answer:

(i) 1-Propoxypropane Reagent: Sodium metal + Propyl bromide Equation: CH₃CH₂CH₂Br + Na → CH₃CH₂CH₂OCH₂CH₃ + NaBr

(ii) Ethoxybenzene Reagent: Sodium metal + Ethyl bromide Equation: C2H₅Br + Na → C2H₅OC₆H₅ + NaBr

(iii) 2-Methoxy-2-methylpropane Reagent: Sodium metal + 2-Methylpropyl bromide Equation: CH₃CH(CH₃)Br + Na → CH₃CH(CH₃)OCH₃ + NaBr

(iv) 1-Methoxyethane Reagent: Sodium metal + Ethyl methyl ether Equation: CH₃OCH₂CH₃ + Na → CH₃OCH₂CH₂CH₃ + NaBr

Question:

Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.

Answer:

1. Ethers are formed by the reaction of an alcohol with an alkyl halide in the presence of an acid catalyst, such as sulfuric acid.

2. Secondary and tertiary alcohols have less acidic hydrogen atoms than primary alcohols, and therefore, they are not easily dehydrated by acid catalysis.

3. Furthermore, secondary and tertiary alcohols are more sterically hindered than primary alcohols, and thus, they are less reactive towards the alkyl halide.

4. Therefore, preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method because the reaction is not efficient enough.

Question:

Write chemical reaction for the preparation of phenol from chlorobenzene.

Answer:

1. Chlorobenzene + Sodium hydroxide (NaOH) → Sodium Chlorophenoxide + Water (H₂O)

2. Sodium Chlorophenoxide + Sodium hypochlorite (NaClO) → Sodium Chlorophenate + Sodium chloride (NaCl)

3. Sodium Chlorophenate + Sodium Hydroxide (NaOH) → Phenol + Sodium chloride (NaCl)

The overall reaction is:

Chlorobenzene + Sodium hydroxide (NaOH) + Sodium hypochlorite (NaClO) → Phenol + Sodium chloride (NaCl)

Question:

Give the equations of reactions for the preparation of phenol from cumene.

Answer:

1. Cumene oxidation: C₉H₁₂ + [O] → C₆H₅C(CH₃)2 + H₂O

2. Hydrolysis of the cumene hydroperoxide: C₆H₅C(CH₃)2OOH → C₆H₅OH + CH₃COOH

3. Distillation of phenol: C₆H₅OH → Phenol

Question:

How are the following conversions carried out? (i) Propene ⟶ Propan-2-ol (ii) Benzyl chloride ⟶ Benzyl alcohol (iii) Ethyl magnesium chloride ⟶ Propan-1-ol (iv) Methyl magnesium bromide ⟶ 2-Methylpropan-2-ol

Answer:

(i) Propene is first reacted with an acid catalyst to form a protonated intermediate. The protonated intermediate then reacts with water to form Propan-2-ol.

(ii) Benzyl chloride is first reacted with a base, such as sodium hydroxide, to form a protonated intermediate. The protonated intermediate then reacts with water to form Benzyl alcohol.

(iii) Ethyl magnesium chloride is first reacted with an alcohol, such as Propan-1-ol, to form a protonated intermediate. The protonated intermediate then reacts with water to form Propan-1-ol.

(iv) Methyl magnesium bromide is first reacted with an alcohol, such as 2-Methylpropan-2-ol, to form a protonated intermediate. The protonated intermediate then reacts with water to form 2-Methylpropan-2-ol.

Question:

Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.

Answer:

(i) In aryl alkyl ethers, the alkoxy group activates the benzene ring towards electrophilic substitution because the oxygen atom in the alkoxy group is more electronegative than the carbon atom in an alkyl group. This means that the oxygen atom in the alkoxy group can withdraw electrons from the benzene ring, making it more susceptible to electrophilic attack.

(ii) The alkoxy group also directs the incoming substituents to the ortho and para positions in the benzene ring because the oxygen atom in the alkoxy group is able to form a hydrogen bond with the incoming substituent, causing it to be attracted to the ortho and para positions. This is due to the fact that the ortho and para positions are the closest to the oxygen atom in the alkoxy group.

Question:

You are given benzene, conc. H₂​SO₄​ and NaOH. Write the equations for the preparation of phenol using these reagents.

Answer:

1. Step 1: Add benzene to a container.

2. Step 2: Add conc. H₂SO₄ to the container.

3. Step 3: Heat the mixture to a temperature of 80°C.

4. Step 4: Add NaOH to the mixture.

5. Step 5: The reaction of benzene with H₂SO₄ and NaOH produces phenol and sodium sulfate as products.

The equation for the preparation of phenol using these reagents is:

C₆H6 + H₂SO₄ + NaOH → C₆H₅OH + Na2SO₄

Question:

Name the reagents used in the following reactions: (i) Oxidation of a primary alcohol to carboxylic acid. (ii) Oxidation of a primary alcohol to aldehyde. (iii) Bromination of phenol to 2,4,6-tribromophenol. (iv) Benzyl alcohol to benzoic acid. (v) Dehydration of propan-2-ol to propene. (vi) Butan-2-one to butan-2-ol.

Answer:

(i) Oxidation of a primary alcohol to carboxylic acid: Sodium or Potassium Dichromate (VI) (Cr₂O₇^2-) in acid medium

(ii) Oxidation of a primary alcohol to aldehyde: Potassium Dichromate (VI) (Cr₂O₇^2-) in acid medium

(iii) Bromination of phenol to 2,4,6-tribromophenol: Bromine (Br₂)

(iv) Benzyl alcohol to benzoic acid: Sulfuric acid (H₂SO₄)

(v) Dehydration of propan-2-ol to propene: Sulfuric acid (H₂SO₄)

(vi) Butan-2-one to butan-2-ol: Sodium hydroxide (NaOH)