Question 1

1 Write IUPAC names of the following compounds:
(i) (ii)
(iii) (iv)
(v) (vi) (vii) (viii)
(ix) (x) C6H5–O–C2H5
(xi) C6H5–O–C7H15 (n–) (xii)

Accepted Answer

Since the structures are not fully given in the text, the general approach to naming ethers is to identify the alkyl groups attached to the oxygen and name them alphabetically followed by 'ether'. For example, C6H5–O–C2H5 is ethoxybenzene (phenyl ethyl ether). Similarly, C6H5–O–C7H15 (n–) is n-heptyloxybenzene. For other compounds, the IUPAC names depend on the alkyl groups attached to oxygen. The student should write the IUPAC names accordingly. — IUPAC naming of ethers involves naming the two alkyl or aryl groups attached to oxygen as substituents and adding 'ether' or using the alkoxy prefix. For example, ethoxybenzene is phenyl ethyl ether. The student should identify the groups and write the correct names.

Question 2

2 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.

Accepted Answer

The student should draw the structures corresponding to each IUPAC name:
(i) 2-Methylbutan-2-ol: A four-carbon chain with a methyl group and an OH at carbon 2.
(ii) 1-Phenylpropan-2-ol: A three-carbon chain with a phenyl group at carbon 1 and OH at carbon 2.
(iii) 3,5-Dimethylhexane-1,3,5-triol: Six-carbon chain with methyl groups at carbons 3 and 5 and hydroxyl groups at carbons 1, 3, and 5.
(iv) 2,3-Diethylphenol: Phenol ring with ethyl groups at positions 2 and 3.
(v) 1-Ethoxypropane: Propane with an ethoxy group at carbon 1.
(vi) 2-Ethoxy-3-methylpentane: Five-carbon chain with ethoxy at carbon 2 and methyl at carbon 3.
(vii) Cyclohexylmethanol: Cyclohexane ring attached to a methanol group.
(viii) 3-Cyclohexylpentan-3-ol: Pentane chain with cyclohexyl and hydroxyl groups at carbon 3.
(ix) Cyclopent-3-en-1-ol: Cyclopentene with a double bond at carbon 3 and hydroxyl at carbon 1.
(x) 4-Chloro-3-ethylbutan-1-ol: Four-carbon chain with chloro at carbon 4, ethyl at carbon 3, and hydroxyl at carbon 1. — The structures are drawn by following IUPAC rules: numbering the longest chain, placing substituents and functional groups at correct positions, and representing the groups as per their names.

Question 3

3 (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
(ii) Classify the isomers of alcohols in question 7.3 (i) as primary, secondary and tertiary alcohols.

Accepted Answer

(i) The isomeric alcohols with molecular formula C5H12O are:
1. Pentan-1-ol (primary)
2. Pentan-2-ol (secondary)
3. Pentan-3-ol (secondary)
4. 2-Methylbutan-1-ol (primary)
5. 3-Methylbutan-1-ol (primary)
6. 2-Methylbutan-2-ol (tertiary)
7. 3-Methylbutan-2-ol (secondary)
8. 2,2-Dimethylpropan-1-ol (primary)

(ii) Classification:
Primary alcohols: Pentan-1-ol, 2-Methylbutan-1-ol, 3-Methylbutan-1-ol, 2,2-Dimethylpropan-1-ol
Secondary alcohols: Pentan-2-ol, Pentan-3-ol, 3-Methylbutan-2-ol
Tertiary alcohols: 2-Methylbutan-2-ol — Isomers are drawn by arranging the carbon skeleton differently and placing the hydroxyl group at different carbons. Classification depends on the carbon to which –OH is attached: primary (attached to carbon with one alkyl group), secondary (two alkyl groups), tertiary (three alkyl groups).

Question 4

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

Accepted Answer

Propanol has higher boiling point than butane because propanol molecules can form hydrogen bonds due to the presence of –OH group, which are strong intermolecular forces. Butane, being a hydrocarbon, has only weak Van der Waals forces (London dispersion forces). Hydrogen bonding requires more energy to break, thus propanol has a higher boiling point. — Hydrogen bonding occurs in alcohols due to –OH group, increasing intermolecular attraction and boiling point. Hydrocarbons lack this and have weaker forces, resulting in lower boiling points.

Question 5

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

Accepted Answer

Alcohols are more soluble in water because they can form hydrogen bonds with water molecules due to the presence of the polar –OH group. Hydrocarbons are non-polar and cannot form hydrogen bonds with water, so they are less soluble. The polar –OH group interacts strongly with water, increasing solubility. — Solubility depends on the ability to form hydrogen bonds. Alcohols have –OH groups that interact with water molecules, while hydrocarbons do not, leading to higher solubility of alcohols.

Question 6

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

Accepted Answer

Hydroboration-oxidation is a two-step reaction that converts alkenes into alcohols. In the first step, borane (BH3) adds across the double bond (hydroboration) in an anti-Markovnikov manner. In the second step, oxidation with hydrogen peroxide (H2O2) in alkaline medium replaces the boron atom with a hydroxyl group, yielding an alcohol.
Example:
CH3–CH=CH2 + BH3 → CH3–CH2–CH2–BH2 (hydroboration)
Then,
CH3–CH2–CH2–BH2 + H2O2/NaOH → CH3–CH2–CH2–OH (1-propanol) — Hydroboration adds boron and hydrogen across the alkene double bond syn and anti-Markovnikov. Oxidation replaces boron with OH, producing alcohol with OH at less substituted carbon.

Question 7

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

Accepted Answer

Monohydric phenols with molecular formula C7H8O are methylphenols (cresols). There are three isomers:
1. o-Cresol (2-methylphenol)
2. m-Cresol (3-methylphenol)
3. p-Cresol (4-methylphenol)
Structures: A benzene ring with one hydroxyl group and one methyl group at ortho, meta, or para positions respectively. — The molecular formula corresponds to phenol with one methyl substituent. The position of methyl group defines the isomer.

Question 8

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

Accepted Answer

Para-nitrophenol is steam volatile because it exists predominantly in the molecular (non-associated) form due to intramolecular hydrogen bonding, which lowers its boiling point and makes it volatile with steam. Ortho-nitrophenol forms intermolecular hydrogen bonds leading to higher boiling point and less volatility. — Intramolecular hydrogen bonding in para-nitrophenol reduces intermolecular forces, increasing volatility. Ortho isomer forms stronger intermolecular hydrogen bonds, reducing volatility.

Question 9

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

Accepted Answer

Preparation of phenol from cumene involves the cumene hydroperoxide process:
Step 1: Cumene (isopropylbenzene) is oxidized by oxygen to form cumene hydroperoxide:
C6H5–CH(CH3)2 + O2 → C6H5–C(CH3)2–OOH
Step 2: Acid-catalyzed cleavage of cumene hydroperoxide yields phenol and acetone:
C6H5–C(CH3)2–OOH + H+ → C6H5–OH + (CH3)2CO — Cumene is oxidized to hydroperoxide which upon acid cleavage gives phenol and acetone, an industrial method for phenol production.

Question 10

10 Write chemical reaction for the preparation of phenol from chlorobenzene.

Accepted Answer

Phenol can be prepared from chlorobenzene by nucleophilic aromatic substitution:
C6H5Cl + 2NaOH (at 623 K and high pressure) → C6H5OH + NaCl
This reaction involves heating chlorobenzene with aqueous sodium hydroxide under pressure to substitute Cl by OH. — Chlorobenzene undergoes nucleophilic substitution with hydroxide ion at high temperature and pressure to form phenol.

Question 11

11 Write the mechanism of hydration of ethene to yield ethanol.

Accepted Answer

Mechanism of acid-catalyzed hydration of ethene:
Step 1: Protonation of ethene double bond by H+ forms a carbocation intermediate.
CH2=CH2 + H+ → CH3–CH2+ (ethyl carbocation)
Step 2: Nucleophilic attack by water on carbocation forms protonated ethanol.
CH3–CH2+ + H2O → CH3–CH2–OH2+
Step 3: Deprotonation of protonated ethanol yields ethanol.
CH3–CH2–OH2+ → CH3–CH2–OH + H+
Overall, ethene reacts with water in presence of acid to give ethanol. — The reaction proceeds via electrophilic addition of H+ to alkene, formation of carbocation, nucleophilic attack by water, and deprotonation to yield alcohol.

Question 12

12 You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.

Accepted Answer

Step 1: Nitration of benzene using conc. H2SO4 and HNO3 to form nitrobenzene (not given but implied).
Step 2: Reduction of nitrobenzene to aniline.
Step 3: Diazotization of aniline with NaNO2 and HCl at 0-5°C to form diazonium salt.
Step 4: Heating diazonium salt with aqueous NaOH to yield phenol.
Overall:
C6H6 → C6H5NO2 → C6H5NH2 → C6H5N2+Cl– → C6H5OH — Phenol is prepared from benzene via nitration, reduction, diazotization, and hydrolysis of diazonium salt with NaOH.

Question 13

13 Show how will you synthesise:
(i) 1-phenylethanol from a suitable alkene.
(ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.
(iii) pentan-1-ol using a suitable alkyl halide?

Accepted Answer

(i) 1-Phenylethanol can be synthesized by acid-catalyzed hydration of styrene (phenylethene):
C6H5–CH=CH2 + H2O (H+) → C6H5–CH(OH)–CH3

(ii) Cyclohexylmethanol can be prepared by nucleophilic substitution (SN2) of bromomethylcyclohexane with aqueous NaOH:
C6H11–CH2–Br + NaOH → C6H11–CH2–OH + NaBr

(iii) Pentan-1-ol can be prepared by nucleophilic substitution of 1-bromopentane with aqueous NaOH:
CH3–(CH2)4–Br + NaOH → CH3–(CH2)4–OH + NaBr — Hydration of alkene yields alcohol (Markovnikov addition). SN2 reaction involves substitution of halide by OH– to give alcohol.

Question 14

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

Accepted Answer

Two reactions showing acidic nature of phenol:
1. Reaction with sodium metal:
C6H5OH + Na → C6H5ONa + 1/2 H2
2. Reaction with sodium bicarbonate (CO2 source):
Phenol does not react with NaHCO3 (unlike carboxylic acids), but reacts with NaOH to form phenoxide ion.
Acidity comparison:
Phenol is more acidic than ethanol because the phenoxide ion formed after deprotonation is resonance stabilized by the aromatic ring, whereas ethoxide ion is not stabilized. Hence, phenol loses proton more easily. — Phenol's acidity is due to resonance stabilization of its conjugate base; ethanol lacks this stabilization, making it less acidic.

Question 15

15 Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?

Accepted Answer

Ortho nitrophenol is more acidic because the nitro group is an electron withdrawing group via -I and -M effects, stabilizing the phenoxide ion formed after deprotonation. Ortho methoxyphenol has a methoxy group which is electron donating (+M effect), destabilizing the phenoxide ion and decreasing acidity. — Electron withdrawing groups stabilize the conjugate base increasing acidity; electron donating groups destabilize it, decreasing acidity.

and Ethers

and Ethers — Study Notes

Introduction

Nomenclature of Ethers

Preparation of Ethers

Practice Questions — and Ethers

All 5 Chapters in Chemistry-II