Question 1

Q1. Give the IUPAC names of the following compounds:
(i) CH₃—CH—CH—CH₂OH
          |   |
         CH₃ OH
(ii) CH₃—CH₂—CH—CH₂—CH—CH₃
                  |       |
                 O

Accepted Answer

The IUPAC names of the given compounds are as follows:

(i) **2,3-Dimethylbutan-1-ol**
    *   Longest carbon chain containing -OH group is 4 carbons (butane).
    *   Numbering starts from the end closer to -OH group (carbon 1).
    *   Methyl groups are at positions 2 and 3.

(ii) **3-Ethylhexane-3-ol**
    *   Longest carbon chain containing -OH group is 6 carbons (hexane).
    *   Numbering starts from the end closer to -OH group (carbon 3).
    *   Ethyl group is also at position 3.

(iii) **4-Chloro-2-methylpentan-1-ol**
    *   Longest carbon chain containing -OH group is 5 carbons (pentane).
    *   Numbering starts from the end closer to -OH group (carbon 1).
    *   Methyl group at position 2, chloro group at position 4.
    *   Alphabetical order for substituents: chloro before methyl.

(iv) **2-Methylpentane-1,3,5-triol**
    *   Longest carbon chain containing all -OH groups is 5 carbons (pentane).
    *   Numbering starts from the end closer to an -OH group, giving the lowest sum of locants for all -OH groups (carbon 1).
    *   Methyl group at position 2.

(v) **2,3,5-Trimethylhexan-3-ol**
    *   Longest carbon chain containing -OH group is 6 carbons (hexane).
    *   Numbering starts from the end closer to -OH group (carbon 3).
    *   Methyl groups at positions 2, 3, and 5.

(vi) **Cyclohexylmethanol**
    *   This is a common name. The IUPAC name is derived from the methanol parent chain with a cyclohexyl substituent.

(vii) **3-Cyclohexylpentan-3-ol**
    *   This is already an IUPAC name. The parent chain is pentane with an -OH group at carbon 3 and a cyclohexyl group also at carbon 3.

(viii) **Cyclopent-3-en-1-ol**
    *   This is already an IUPAC name. The parent ring is cyclopentene with a double bond at position 3 and an -OH group at position 1.

(ix) **4-Methylphenol**
    *   This is a common name (p-cresol). The IUPAC name uses phenol as the parent, with a methyl group at position 4 (para position).
    *   Alternatively, it can be named 4-methylbenzenol.

Question 2

Q2. Write structures of the compounds whose IUPAC names are as follows:
(i) 2-Methylbutan-2-ol
(ii) 1-Phenylpropan-1-ol
(iii) 3,5-Dimethylhexane-1,3,5-triol
(iv) Cyclohex-3-en-1-ol

Accepted Answer

The structures of the compounds with the given IUPAC names are as follows:

(i) **2-Methylbutan-2-ol**
    *   Parent chain: Butane (4 carbons).
    *   Hydroxyl group (-OH) at carbon 2.
    *   Methyl group (-CH₃) at carbon 2.
    *   Structure: CH₃—C(CH₃)(OH)—CH₂—CH₃

(ii) **1-Phenylpropan-1-ol**
    *   Parent chain: Propan-1-ol (3 carbons, -OH at carbon 1).
    *   Phenyl group (-C₆H₅) at carbon 1.
    *   Structure: C₆H₅—CH(OH)—CH₂—CH₃

(iii) **3,5-Dimethylhexane-1,3,5-triol**
    *   Parent chain: Hexane (6 carbons).
    *   Three hydroxyl groups (-OH) at carbons 1, 3, and 5.
    *   Methyl groups (-CH₃) at carbons 3 and 5.
    *   Structure: HO—CH₂—CH₂—C(CH₃)(OH)—CH₂—C(CH₃)(OH)—CH₂OH

(iv) **Cyclohex-3-en-1-ol**
    *   Parent ring: Cyclohexene (6-membered ring with one double bond).
    *   Double bond between carbons 3 and 4 (if -OH is at C1).
    *   Hydroxyl group (-OH) at carbon 1.
    *   Structure: A cyclohexene ring with a double bond between C3 and C4, and an -OH group at C1.

(v) **4-Ethylpent-3-en-1-ol**
    *   Parent chain: Pent-3-en-1-ol (5 carbons, double bond between C3 and C4, -OH at C1).
    *   Ethyl group (-CH₂CH₃) at carbon 4.
    *   Structure: HO—CH₂—CH₂—C(CH₂CH₃)=CH—CH₃

(vi) **Bicyclo[2.2.1]heptan-2-ol**
    *   Parent structure: Bicyclo[2.2.1]heptane (a bicyclic system with 7 carbons, two bridges of 2 carbons and one bridge of 1 carbon).
    *   Hydroxyl group (-OH) at carbon 2 (one of the bridgehead carbons is C1, and the carbons are numbered sequentially along the longest bridge first).
    *   Structure: A bicyclo[2.2.1]heptane skeleton with an -OH group attached to one of the carbons adjacent to a bridgehead carbon.

Question 3

Q3. (i) Draw the structures of the eight isomeric alcohols of molecular formula C₅H₁₂O.
(ii) Give their IUPAC names.
(iii) Classify them as primary, secondary and tertiary alcohols.

Accepted Answer

### (i) Structures of the eight isomeric alcohols of molecular formula C₅H₁₂O

To find all isomers, we systematically vary the carbon skeleton and the position of the -OH group. The molecular formula C₅H₁₂O indicates a saturated acyclic alcohol (degree of unsaturation = (2C+2-H)/2 = (2*5+2-12)/2 = 0). There are 8 such isomers.

1.  **Pentane chain isomers:**
    *   **Pent-1-ol:** CH₃—CH₂—CH₂—CH₂—CH₂—OH
    *   **Pent-2-ol:** CH₃—CH₂—CH₂—CH(OH)—CH₃
    *   **Pent-3-ol:** CH₃—CH₂—CH(OH)—CH₂—CH₃

2.  **Butane chain isomers (with one methyl branch):**
    *   **2-Methylbutan-1-ol:** CH₃—CH₂—CH(CH₃)—CH₂—OH
    *   **3-Methylbutan-1-ol:** (CH₃)₂CH—CH₂—CH₂—OH
    *   **2-Methylbutan-2-ol:** CH₃—CH₂—C(CH₃)(OH)—CH₃
    *   **3-Methylbutan-2-ol:** (CH₃)₂CH—CH(OH)—CH₃

3.  **Propane chain isomers (with two methyl branches):**
    *   **2,2-Dimethylpropan-1-ol:** (CH₃)₃C—CH₂—OH

### (ii) IUPAC Names

1.  **Pent-1-ol**
2.  **Pent-2-ol**
3.  **Pent-3-ol**
4.  **2-Methylbutan-1-ol**
5.  **3-Methylbutan-1-ol**
6.  **2-Methylbutan-2-ol**
7.  **3-Methylbutan-2-ol**
8.  **2,2-Dimethylpropan-1-ol**

### (iii) Classification as Primary (1°), Secondary (2°), and Tertiary (3°) alcohols

Alcohols are classified based on the number of alkyl groups attached to the carbon atom bearing the hydroxyl group.

*   **Primary (1°) alcohol:** The carbon atom bearing the -OH group is attached to only one alkyl group (or no alkyl groups, as in methanol).
*   **Secondary (2°) alcohol:** The carbon atom bearing the -OH group is attached to two alkyl groups.
*   **Tertiary (3°) alcohol:** The carbon atom bearing the -OH group is attached to three alkyl groups.

Based on this classification:

1.  **Pent-1-ol:** Primary (1°) -OH attached to -CH₂-
2.  **Pent-2-ol:** Secondary (2°) -OH attached to -CH-
3.  **Pent-3-ol:** Secondary (2°) -OH attached to -CH-
4.  **2-Methylbutan-1-ol:** Primary (1°) -OH attached to -CH₂-
5.  **3-Methylbutan-1-ol:** Primary (1°) -OH attached to -CH₂-
6.  **2-Methylbutan-2-ol:** Tertiary (3°) -OH attached to -C-
7.  **3-Methylbutan-2-ol:** Secondary (2°) -OH attached to -CH-
8.  **2,2-Dimethylpropan-1-ol:** Primary (1°) -OH attached to -CH₂-

Question 4

Q4. Explain why propanol has higher boiling point than that of hydrocarbon, butane?

Accepted Answer

Propanol (CH₃CH₂CH₂OH) has a significantly higher boiling point (97 °C) than butane (CH₃CH₂CH₂CH₃) (0 °C), despite having comparable molecular masses (propanol: 60 g/mol; butane: 58 g/mol). This difference can be explained by the types of intermolecular forces present in each compound.

1.  **Intermolecular Forces in Butane:** Butane is a nonpolar hydrocarbon. The only intermolecular forces present are weak London dispersion forces (also known as van der Waals forces). These forces arise from temporary fluctuations in electron distribution, creating instantaneous dipoles. They are relatively weak and require little energy to overcome, leading to a low boiling point.

2.  **Intermolecular Forces in Propanol:** Propanol is an alcohol, containing a hydroxyl (-OH) group. The oxygen atom in the -OH group is highly electronegative, creating a polar O-H bond. This polarity allows propanol molecules to form strong **hydrogen bonds** with each other. Hydrogen bonding is a special type of dipole-dipole interaction where a hydrogen atom covalently bonded to a highly electronegative atom (like O, N, or F) is attracted to another electronegative atom in a different molecule.

**Conclusion:** The presence of strong hydrogen bonds in propanol requires a much greater amount of energy to overcome during boiling compared to the weak London dispersion forces in butane. Therefore, propanol has a much higher boiling point than butane.

Question 5

Q5. Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.

Accepted Answer

Alcohols are significantly more soluble in water than hydrocarbons of comparable molecular masses due to the ability of alcohols to form hydrogen bonds with water molecules.

1.  **Nature of Water:** Water is a highly polar solvent, and its molecules are extensively hydrogen-bonded to each other. For a substance to dissolve in water, it must be able to disrupt these existing hydrogen bonds between water molecules and form new, favorable interactions with water.

2.  **Interactions of Hydrocarbons with Water:** Hydrocarbons are nonpolar molecules. They primarily interact through weak London dispersion forces. When a hydrocarbon is added to water, it cannot form hydrogen bonds with water molecules. Instead, it disrupts the existing hydrogen bonds between water molecules, which is an energetically unfavorable process. Water molecules then tend to reorient themselves to maximize their hydrogen bonding with each other, effectively 'squeezing out' the nonpolar hydrocarbon molecules. This leads to the immiscibility of hydrocarbons in water.

3.  **Interactions of Alcohols with Water:** Alcohols contain a polar hydroxyl (-OH) group. The oxygen atom in the -OH group can act as a hydrogen bond acceptor, and the hydrogen atom of the -OH group can act as a hydrogen bond donor. This allows alcohol molecules to form strong **hydrogen bonds** with water molecules. The energy released by the formation of these new alcohol-water hydrogen bonds compensates for the energy required to break the existing hydrogen bonds in water and the weak intermolecular forces in the alcohol itself. This favorable interaction makes lower molecular mass alcohols (like methanol, ethanol, propanol) completely miscible with water.

**Conclusion:** The ability of the hydroxyl group in alcohols to form strong hydrogen bonds with water molecules is the primary reason for their higher solubility in water compared to nonpolar hydrocarbons of similar molecular masses.

Question 6

Q6. Give the structures of the products of the following reactions:
(i) CH3–CH2–CH2–O–CH3 + HBr →
(ii)
(iii) (CH3)3C–OC2H5 + HI →
(iv) CH3–CH2–CH2–OH + SOCl2 →
(v)
(vi) CH3–CH2–C

Accepted Answer

The structures of the products for the given reactions are as follows:

**(i) CH3–CH2–CH2–O–CH3 + HBr → CH3–CH2–CH2–Br + CH3–OH**
Explanation: This is an ether cleavage reaction with HBr. Since the alkyl groups are primary and methyl, the reaction proceeds via SN2 mechanism. The smaller alkyl group (methyl) forms alcohol, and the larger alkyl group (propyl) forms alkyl halide.

**(ii)**
**Reactant:** Phenol reacting with concentrated HNO3 in the presence of concentrated H2SO4.
**Product:** 2,4,6-Trinitrophenol (Picric acid)
Explanation: Phenol undergoes electrophilic substitution. The -OH group is an ortho-para directing and strongly activating group. With concentrated nitric acid and sulfuric acid, multiple nitration occurs at ortho and para positions, leading to the formation of picric acid.

**(iii) (CH3)3C–OC2H5 + HI → (CH3)3C–I + CH3–CH2–OH**
Explanation: This is an ether cleavage reaction with HI. One alkyl group is tertiary (t-butyl) and the other is primary (ethyl). The reaction proceeds via SN1 mechanism because a stable tertiary carbocation can be formed. The tertiary alkyl group forms the alkyl halide, and the primary alkyl group forms the alcohol.

**(iv) CH3–CH2–CH2–OH + SOCl2 → CH3–CH2–CH2–Cl + SO2 + HCl**
Explanation: This is the conversion of a primary alcohol to an alkyl chloride using thionyl chloride (SOCl2). This reaction is excellent for preparing alkyl chlorides because the by-products (SO2 and HCl) are gases and escape, leaving a pure alkyl chloride.

**(v)**
**Reactant:** Anisole (methoxybenzene) reacting with HI.
**Product:** Phenol + CH3–I
Explanation: This is an ether cleavage reaction of an aryl alkyl ether. The aryl-oxygen bond has partial double bond character due to resonance and is stronger, hence it does not cleave. The alkyl-oxygen bond cleaves, leading to the formation of phenol and methyl iodide.

**(vi) CH3–CH2–CH=CH2 + H2O H+ → CH3–CH2–CH(OH)–CH3**
Explanation: This is the acid-catalyzed hydration of an alkene (But-1-ene). The reaction follows Markovnikov's rule, where the -OH group adds to the carbon atom that has fewer hydrogen atoms (the more substituted carbon) of the double bond. This leads to the formation of Butan-2-ol.

Question 7

Q7. Write the mechanism of hydration of ethene to yield ethanol.

Accepted Answer

The hydration of ethene to yield ethanol is an acid-catalyzed electrophilic addition reaction. The mechanism involves three main steps:

### Mechanism of Hydration of Ethene to Ethanol

**Step 1: Protonation of Ethene to form Carbocation**
Ethene, being an electron-rich alkene, acts as a nucleophile and attacks the proton (H+) supplied by the acid catalyst (e.g., H2SO4). This results in the formation of a primary carbocation. The proton adds to one of the sp2 carbons, and the other sp2 carbon becomes positively charged.

**Step 2: Nucleophilic Attack by Water**
The carbocation formed in Step 1 is highly reactive. A water molecule, acting as a nucleophile, attacks the electron-deficient carbocation. This forms an oxonium ion, where the oxygen atom of water is positively charged.

**Step 3: Deprotonation to form Ethanol**
The oxonium ion is unstable due to the positive charge on oxygen. Another water molecule (or the conjugate base of the acid catalyst) acts as a base and removes a proton from the oxonium ion. This deprotonation step regenerates the acid catalyst and yields the final product, ethanol.

This mechanism is consistent with Markovnikov's rule, although for ethene (a symmetrical alkene), the regioselectivity is not apparent. For unsymmetrical alkenes, the more stable carbocation would be formed, leading to the addition of -OH to the more substituted carbon.

Question 8

Q8. Show how will you synthesise the following alcohols from appropriate alkenes.
(i) 1-methylcyclohexanol
(ii) 1-methylcyclopentanol
(iii) 2-phenylpropan-2-ol

Accepted Answer

The synthesis of alcohols from alkenes typically involves hydration, which can be achieved through acid-catalyzed hydration or oxymercuration-demercuration. Both methods follow Markovnikov's rule, meaning the -OH group adds to the more substituted carbon of the double bond. We will use acid-catalyzed hydration (H2O/H+) for these syntheses.

**(i) Synthesis of 1-methylcyclohexanol**
To synthesize 1-methylcyclohexanol, we need an alkene that, upon hydration, will place the -OH group at the tertiary carbon (C1) bearing the methyl group. The appropriate alkene is 1-methylcyclohexene.

**Reaction:**
1-Methylcyclohexene + H2O (in presence of H+) → 1-Methylcyclohexanol

**(ii) Synthesis of 1-methylcyclopentanol**
Similarly, to synthesize 1-methylcyclopentanol, the alkene should be 1-methylcyclopentene. Hydration of 1-methylcyclopentene will place the -OH group at the tertiary carbon (C1) bearing the methyl group, following Markovnikov's rule.

**Reaction:**
1-Methylcyclopentene + H2O (in presence of H+) → 1-Methylcyclopentanol

**(iii) Synthesis of 2-phenylpropan-2-ol**
For 2-phenylpropan-2-ol, the -OH group is on the second carbon of the propane chain, which is also attached to a phenyl group and two methyl groups. This is a tertiary alcohol. The corresponding alkene would be 2-phenylpropene (α-methylstyrene). Hydration of 2-phenylpropene will place the -OH group at the tertiary carbon, following Markovnikov's rule.

**Reaction:**
2-Phenylpropene + H2O (in presence of H+) → 2-Phenylpropan-2-ol

Question 9

Q9. Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.

Accepted Answer

### Reactions Showing Acidic Nature of Phenol
Phenol is acidic in nature, although weaker than carboxylic acids, it is significantly more acidic than alcohols. Its acidic character is due to the direct attachment of the -OH group to the sp2 hybridized carbon of the benzene ring, which allows for resonance stabilization of the phenoxide ion.

**1. Reaction with Active Metals (e.g., Sodium):**
Phenol reacts with active metals like sodium to liberate hydrogen gas, similar to alcohols, but the reaction is more vigorous for phenol, indicating its stronger acidic nature compared to simple alcohols.

**Reaction:**
2C6H5OH (Phenol) + 2Na (Sodium) → 2C6H5O-Na+ (Sodium phenoxide) + H2 (Hydrogen gas)

**2. Reaction with Sodium Hydroxide (NaOH):**
Phenol reacts with strong bases like sodium hydroxide to form sodium phenoxide and water. This reaction demonstrates its acidic nature, as it can neutralize a strong base. Ethanol, on the other hand, does not react with NaOH.

**Reaction:**
C6H5OH (Phenol) + NaOH (Sodium hydroxide) → C6H5O-Na+ (Sodium phenoxide) + H2O (Water)

### Comparison of Acidity: Phenol vs. Ethanol
Phenol is significantly more acidic than ethanol. This difference in acidity can be explained by the stability of their respective conjugate bases.

**1. Acidity of Ethanol:**
Ethanol (CH3CH2OH) is a simple alcohol. When it loses a proton, it forms an ethoxide ion (CH3CH2O-). The negative charge on the oxygen atom in the ethoxide ion is localized. The alkyl group (ethyl group) is an electron-donating group (+I effect), which destabilizes the ethoxide ion by intensifying the negative charge on the oxygen. This makes the ethoxide ion less stable and ethanol a very weak acid (pKa ≈ 16-18), comparable to water.

**2. Acidity of Phenol:**
Phenol (C6H5OH) loses a proton to form a phenoxide ion (C6H5O-). The phenoxide ion is stabilized by resonance. The negative charge on the oxygen atom can be delocalized over the benzene ring through resonance. This delocalization spreads the negative charge, making the phenoxide ion more stable than the ethoxide ion. The greater stability of the conjugate base (phenoxide ion) makes phenol a stronger acid than ethanol (pKa ≈ 10).

**Resonance Structures of Phenoxide Ion:**
The negative charge on the oxygen atom of the phenoxide ion can be delocalized to the ortho and para positions of the benzene ring, as shown by the following resonance structures:
(i) Oxygen with negative charge
(ii) Negative charge at ortho position
(iii) Negative charge at para position
(iv) Negative charge at other ortho position
(v) Oxygen with negative charge (return to original)

**Conclusion:**
Due to the resonance stabilization of the phenoxide ion, phenol is a much stronger acid than ethanol, where the ethoxide ion is destabilized by the electron-donating inductive effect of the ethyl group. Phenol can react with strong bases like NaOH, while ethanol cannot.

Question 10

Q10. Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?

Accepted Answer

To compare the acidity of ortho nitrophenol and ortho methoxyphenol, we need to consider the stability of their respective conjugate bases (phenoxide ions) formed after losing a proton. The more stable the conjugate base, the stronger the acid.

### Acidity of Ortho Nitrophenol
Ortho nitrophenol contains a nitro group (-NO2) at the ortho position. The nitro group is a strong electron-withdrawing group due to both its strong negative inductive effect (-I effect) and strong negative resonance effect (-R effect).

When ortho nitrophenol loses a proton, it forms the ortho nitrophenoxide ion. The electron-withdrawing nitro group helps to delocalize and disperse the negative charge on the phenoxide oxygen, thereby stabilizing the conjugate base. This stabilization makes ortho nitrophenol a stronger acid.

Furthermore, intramolecular hydrogen bonding can occur between the phenolic -OH and the nitro group in ortho nitrophenol. While this might slightly reduce the availability of the proton for donation, the dominant effect contributing to increased acidity is the strong electron-withdrawing nature of the nitro group stabilizing the conjugate base.

### Acidity of Ortho Methoxyphenol
Ortho methoxyphenol contains a methoxy group (-OCH3) at the ortho position. The methoxy group is an electron-donating group due to its positive resonance effect (+R effect) (lone pair on oxygen can delocalize into the ring). It also has a negative inductive effect (-I effect) due to the electronegativity of oxygen, but the +R effect usually dominates in aromatic systems, especially at ortho/para positions.

When ortho methoxyphenol loses a proton, it forms the ortho methoxyphenoxide ion. The electron-donating methoxy group intensifies the negative charge on the phenoxide oxygen, thereby destabilizing the conjugate base. This destabilization makes ortho methoxyphenol a weaker acid.

### Conclusion
Comparing the two, the strong electron-withdrawing nature of the nitro group in ortho nitrophenol effectively stabilizes its conjugate base (ortho nitrophenoxide ion) by delocalizing the negative charge. In contrast, the electron-donating nature of the methoxy group in ortho methoxyphenol destabilizes its conjugate base (ortho methoxyphenoxide ion) by intensifying the negative charge. Therefore, **ortho nitrophenol is more acidic than ortho methoxyphenol**.

Question 11

Q11. Give the structures of the products of the following reactions:
(i) CH₃–CH₂–CH₂–O–CH₃ + HBr →
(ii) (C₂H₅)₃C–O–CH₃ + HBr →
(iii) CH₃–CH₂–O–C(CH₃)₃ + HBr →
(iv) CH₃–CH₂–CH₂–O–CH(

Accepted Answer

The cleavage of ethers by HBr (or other hydrogen halides, HX) proceeds via different mechanisms depending on the nature of the alkyl groups attached to the oxygen atom.

**General Principles:**
1.  **SN2 Mechanism:** If both alkyl groups are primary or secondary, the smaller alkyl group forms the alkyl halide, and the larger alkyl group forms the alcohol. The nucleophile (Br⁻) attacks the less sterically hindered carbon atom.
2.  **SN1 Mechanism:** If one of the alkyl groups is tertiary, benzylic, or allylic, the reaction proceeds via an SN1 mechanism. The carbocation formed is stable, and the tertiary/benzylic/allylic group forms the alkyl halide, while the other group forms the alcohol.
3.  **Phenols:** If one of the groups is an aryl group (e.g., in anisole), the aryl-oxygen bond is not cleaved because of its partial double bond character due to resonance. The alkyl group forms the alkyl halide, and phenol is formed.

Let's apply these principles to the given reactions:

**(i) CH₃–CH₂–CH₂–O–CH₃ + HBr →**
*   Reactant: Propyl methyl ether. Both alkyl groups (propyl and methyl) are primary.
*   Mechanism: SN2. Bromide ion attacks the less sterically hindered methyl carbon.
*   Products: CH₃Br (Bromomethane) + CH₃CH₂CH₂OH (Propan-1-ol)

**(ii) (C₂H₅)₃C–O–CH₃ + HBr →**
*   Reactant: Triethylmethyl methyl ether (assuming (C₂H₅)₃C represents a tertiary carbon center). One alkyl group is tertiary, and the other is primary.
*   Mechanism: SN1. A stable tertiary carbocation is formed.
*   Products: (C₂H₅)₃CBr (1-Bromo-1-ethyl-1-methylpropane, or more simply, a tertiary alkyl bromide) + CH₃OH (Methanol)

**(iii) CH₃–CH₂–O–C(CH₃)₃ + HBr →**
*   Reactant: Ethyl tert-butyl ether. One alkyl group is tertiary (tert-butyl), and the other is primary (ethyl).
*   Mechanism: SN1. A stable tertiary carbocation is formed.
*   Products: (CH₃)₃CBr (2-Bromo-2-methylpropane or tert-butyl bromide) + CH₃CH₂OH (Ethanol)

**(iv) CH₃–CH₂–CH₂–O–CH(CH₃)₂ + HBr →**
*   Reactant: Propyl isopropyl ether. One alkyl group is primary (propyl), and the other is secondary (isopropyl).
*   Mechanism: SN2. Bromide ion attacks the less sterically hindered primary carbon.
*   Products: CH₃CH₂CH₂Br (1-Bromopropane) + CH₃CH(OH)CH₃ (Propan-2-ol)

**(v) CH₃–CH₂–C(CH₃)₂–O–CH₂–CH₃ + HBr →**
*   Reactant: 2-Methyl-2-butyl ethyl ether. One alkyl group is tertiary (2-methyl-2-butyl), and the other is primary (ethyl).
*   Mechanism: SN1. A stable tertiary carbocation is formed.
*   Products: CH₃CH₂C(CH₃)₂Br (2-Bromo-2-methylbutane) + CH₃CH₂OH (Ethanol)

Question 12

Q12. Write the mechanism of hydration of ethene to yield ethanol.

Accepted Answer

The hydration of ethene to yield ethanol is an acid-catalyzed electrophilic addition reaction. The mechanism involves three main steps:

**Step 1: Protonation of ethene to form carbocation by an electrophilic attack of H₃O⁺.**
Ethene, being an electron-rich alkene, acts as a nucleophile and attacks the electrophilic proton from the hydronium ion (H₃O⁺). This leads to the formation of a primary carbocation and water.

**Step 2: Nucleophilic attack of water on the carbocation.**
The water molecule, acting as a nucleophile, attacks the positively charged carbon atom of the carbocation. This forms a protonated alcohol (an oxonium ion).

**Step 3: Deprotonation to yield ethanol.**
Another water molecule acts as a base and removes a proton from the protonated alcohol. This regenerates the acid catalyst (H₃O⁺) and yields the final product, ethanol.

This mechanism follows Markovnikov's rule for unsymmetrical alkenes, but for ethene, which is symmetrical, only one product is possible.

Question 13

Q13. How are the following conversions carried out?
(i) Propene to Propan-1-ol
(ii) Ethanol to But-1-yne
(iii) 1-Bromopropane to 2-Bromopropane
(iv) Toluene to Benzyl alcohol
(v) P

Accepted Answer

Here are the step-by-step conversions for each given reaction:

**(i) Propene to Propan-1-ol**
This conversion requires an anti-Markovnikov addition of water to propene, which is achieved through hydroboration-oxidation.

**(ii) Ethanol to But-1-yne**
This conversion involves multiple steps: converting ethanol to an alkyl halide, then to an alkyne, and finally extending the carbon chain using a terminal alkyne's acidic proton.

**(iii) 1-Bromopropane to 2-Bromopropane**
This conversion involves an elimination reaction followed by an addition reaction, leading to a change in the position of the bromine atom.

**(iv) Toluene to Benzyl alcohol**
This conversion involves free radical halogenation of the methyl group, followed by nucleophilic substitution.

**(v) Phenol to 2,4,6-Tribromophenol**
Phenol is highly activated towards electrophilic substitution due to the strong electron-donating effect of the -OH group. Reaction with bromine water leads to tribromination.

**(vi) Anisole to Phenol**
Anisole (methoxybenzene) is an ether. Cleavage of aryl alkyl ethers with strong acids like HI or HBr yields phenol and an alkyl halide.

**(vii) Ethanol to Ethanal**
This is an oxidation reaction. Mild oxidizing agents are used to prevent further oxidation to carboxylic acid.

**(viii) Phenol to Phenyl ethanoate**
This is an esterification reaction where phenol reacts with an acylating agent.

**(ix) Chlorobenzene to Phenol**
This conversion is typically carried out via the Dow's process, which involves harsh conditions for nucleophilic aromatic substitution.

**(x) 2-Propanol to Propanone**
This is an oxidation reaction of a secondary alcohol to a ketone. Mild oxidizing agents are suitable.

Question 14

Q14. Show how will you synthesise the following alcohols from appropriate alkenes.
(i) 1-Propanol
(ii) 2-Butanol
(iii) 2-Methylpropan-2-ol

Accepted Answer

Alcohols can be synthesized from alkenes primarily through two methods: acid-catalyzed hydration (following Markovnikov's rule) and hydroboration-oxidation (following anti-Markovnikov's rule). The choice of method depends on the desired regiochemistry of the alcohol.

**(i) 1-Propanol**
1-Propanol is a primary alcohol. To obtain a primary alcohol from propene, an anti-Markovnikov addition of water is required. This is achieved through hydroboration-oxidation.

**(ii) 2-Butanol**
2-Butanol is a secondary alcohol. It can be obtained from but-1-ene or but-2-ene. From but-1-ene, it would be a Markovnikov addition of water. From but-2-ene, it would also be a Markovnikov addition, but since but-2-ene is symmetrical with respect to the double bond, only 2-butanol is formed.

**(iii) 2-Methylpropan-2-ol**
2-Methylpropan-2-ol is a tertiary alcohol. To obtain a tertiary alcohol from an alkene, a Markovnikov addition of water is required. This is achieved through acid-catalyzed hydration.

Question 15

Q15. Give the structures and IUPAC names of the products expected from the following reactions:
(a) Catalytic reduction of butanal
(b) Hydration of propene in the presence of dilute s

Accepted Answer

Here are the structures and IUPAC names of the products for each reaction:

**(a) Catalytic reduction of butanal**
Butanal is an aldehyde. Catalytic reduction (e.g., with H2/Pd, Pt, or Ni) converts aldehydes to primary alcohols.
Reaction:
CH3CH2CH2CHO  +  H2  --[Pd/Pt/Ni]-->  CH3CH2CH2CH2OH
(Butanal)                                     (Butan-1-ol)

Product Structure: CH3—CH2—CH2—CH2—OH
IUPAC Name: Butan-1-ol

**(b) Hydration of propene in the presence of dilute sulphuric acid**
Hydration of alkenes in the presence of dilute sulphuric acid follows Markovnikov's rule, where the -OH group adds to the carbon atom with fewer hydrogen atoms.
Reaction:
CH3—CH=CH2  +  H2O  --[H2SO4 (dilute)]-->  CH3—CH—CH3
(Propene)                                        |
                                                 OH
                                             (Propan-2-ol)

Product Structure: CH3—CH(OH)—CH3
IUPAC Name: Propan-2-ol

**(c) Reaction of propanone with methylmagnesium bromide followed by hydrolysis**
Propanone is a ketone. Grignard reagents react with ketones to form tertiary alcohols after hydrolysis.
Reaction:
CH3—CO—CH3  +  CH3MgBr  --[1. Dry Ether]-->  (CH3)3C—OMgBr  --[2. H2O/H+]-->  (CH3)3C—OH
(Propanone)                                                                         (2-Methylpropan-2-ol)

Product Structure: (CH3)3C—OH (or CH3—C(OH)(CH3)—CH3)
IUPAC Name: 2-Methylpropan-2-ol

**(d) Hydroboration-oxidation of propene**
Hydroboration-oxidation of alkenes is an anti-Markovnikov hydration, resulting in the addition of the -OH group to the carbon atom with more hydrogen atoms. It is a syn addition.
Reaction:
CH3—CH=CH2  --[1. BH3, THF / 2. H2O2, OH-]-->  CH3—CH2—CH2—OH
(Propene)                                            (Propan-1-ol)

Product Structure: CH3—CH2—CH2—OH
IUPAC Name: Propan-1-ol

Question 16

Q16. Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal.
(i) CH3—CH—CH2OH
          |
          CH3
(ii) CH3—CH2—CH2OH

Accepted Answer

Grignard reagents (RMgX) react with methanal (formaldehyde, HCHO) to produce primary alcohols after hydrolysis. The alkyl group (R) from the Grignard reagent becomes part of the alcohol product. The general reaction is:

HCHO + RMgX --(1. Dry Ether)--> RCH2OMgX --(2. H2O/H+)--> RCH2OH

We need to identify the 'R' group for each target alcohol.

**(i) Preparation of 2-Methylpropan-1-ol (CH3—CH(CH3)—CH2OH)**

The target alcohol is 2-Methylpropan-1-ol. It is a primary alcohol with the structure (CH3)2CH-CH2OH. Comparing this with RCH2OH, the 'R' group must be (CH3)2CH- (isopropyl group).

Therefore, the suitable Grignard reagent is isopropylmagnesium halide (e.g., isopropylmagnesium bromide).

Reaction:

CH3
|
H—C=O  +  CH—MgBr  --[1. Dry Ether]-->  CH3—CH—CH2—OMgBr  --[2. H2O/H+]-->  CH3—CH—CH2OH
|
CH3
(Methanal)  (Isopropylmagnesium bromide)                                       (2-Methylpropan-1-ol)

**(ii) Preparation of Propan-1-ol (CH3—CH2—CH2OH)**

The target alcohol is Propan-1-ol, with the structure CH3CH2CH2OH. Comparing this with RCH2OH, the 'R' group must be CH3CH2- (ethyl group).

Therefore, the suitable Grignard reagent is ethylmagnesium halide (e.g., ethylmagnesium bromide).

Reaction:

H—C=O  +  CH3CH2—MgBr  --[1. Dry Ether]-->  CH3CH2—CH2—OMgBr  --[2. H2O/H+]-->  CH3CH2—CH2OH
|
(Methanal)  (Ethylmagnesium bromide)                                          (Propan-1-ol)

Question 17

Q17. Write the mechanism of hydration of ethene to yield ethanol.

Accepted Answer

The hydration of ethene to yield ethanol is an acid-catalyzed reaction. It proceeds via a carbocation intermediate and follows Markovnikov's rule (though for ethene, which is symmetrical, this rule doesn't differentiate products). The mechanism involves three main steps:

### Mechanism of Hydration of Ethene to Ethanol

**Step 1: Protonation of Ethene to form Carbocation**

*   The π-electrons of the ethene double bond act as a nucleophile and attack the proton (H+) supplied by the acid (e.g., H3O+ from H2SO4 in water). This is the slowest and rate-determining step.
*   A primary carbocation is formed. Although primary carbocations are less stable, in the case of ethene, it's the only possible carbocation.

**Step 2: Nucleophilic Attack by Water**

*   The carbocation formed in Step 1 is highly electrophilic.
*   A water molecule, acting as a nucleophile, attacks the positively charged carbon atom of the carbocation.
*   This forms a protonated alcohol (an oxonium ion).

**Step 3: Deprotonation to form Ethanol**

*   The protonated alcohol (oxonium ion) loses a proton (H+) to a water molecule (which acts as a base).
*   This regenerates the acid catalyst (H3O+) and forms the final product, ethanol.

### Summary of Steps with Reactions:

**Step 1: Formation of Carbocation**

CH2=CH2  +  H—O+—H  ⇌  CH3—CH2+  +  H2O
               |
               H
(Ethene)   (Hydronium ion)  (Ethyl carbocation)

**Step 2: Nucleophilic Attack by Water**

CH3—CH2+  +  H2O  ⇌  CH3—CH2—O+—H
                               |
                               H
(Ethyl carbocation)           (Water)       (Protonated ethanol)

**Step 3: Deprotonation**

CH3—CH2—O+—H  +  H2O  ⇌  CH3—CH2—OH  +  H3O+
               |
               H
(Protonated ethanol)           (Water)       (Ethanol)      (Hydronium ion)

The overall reaction is:
CH2=CH2  +  H2O  --[H+/H2SO4]-->  CH3CH2OH

Question 18

Q18. Show how will you synthesise:
(i) 1-phenylethanol from an appropriate alkene.
(ii) Cyclohexylmethanol using an alkyl halide by an SN2 reaction.
(iii) Pentan-1-ol using an alkyl

Accepted Answer

Here are the synthesis pathways for the given alcohols:

**(i) Synthesis of 1-phenylethanol from an appropriate alkene**

1-Phenylethanol is a secondary alcohol. It can be prepared from styrene (phenylethene) via hydration. Since 1-phenylethanol is a Markovnikov product (the -OH group is on the more substituted carbon), acid-catalyzed hydration is suitable.

**Starting Alkene:** Styrene (Phenylethene)

**Reaction:** Acid-catalyzed hydration

C6H5—CH=CH2  +  H2O  --[H+/H2SO4]-->  C6H5—CH(OH)—CH3
(Styrene)                                         (1-Phenylethanol)

Alternatively, it can also be prepared via oxymercuration-demercuration, which also follows Markovnikov's rule.

**(ii) Synthesis of Cyclohexylmethanol using an alkyl halide by an SN2 reaction**

Cyclohexylmethanol is a primary alcohol (R-CH2OH). To synthesize it via an SN2 reaction from an alkyl halide, we need a primary alkyl halide and a strong nucleophile (OH-). The carbon skeleton of cyclohexylmethanol is (Cyclohexyl)-CH2-OH. This means the alkyl halide should be (Cyclohexyl)-CH2-X, where X is a good leaving group (e.g., Br, Cl, I).

**Starting Alkyl Halide:** Cyclohexylmethyl halide (e.g., Cyclohexylmethyl bromide)

**Reaction:** SN2 reaction with aqueous KOH/NaOH

C6H11—CH2—Br  +  KOH(aq)  --[SN2]-->  C6H11—CH2—OH  +  KBr
(Cyclohexylmethyl bromide)                  (Cyclohexylmethanol)

**(iii) Synthesis of Pentan-1-ol using an alkyl halide**

Pentan-1-ol is a primary alcohol (CH3CH2CH2CH2CH2OH). To synthesize it from an alkyl halide, we need a primary alkyl halide with the same carbon skeleton and a strong nucleophile (OH-). The alkyl halide should be 1-halopentane.

**Starting Alkyl Halide:** 1-Halopentane (e.g., 1-Bromopentane)

**Reaction:** SN2 reaction with aqueous KOH/NaOH

CH3CH2CH2CH2CH2—Br  +  KOH(aq)  --[SN2]-->  CH3CH2CH2CH2CH2—OH  +  KBr
(1-Bromopentane)                                (Pentan-1-ol)

Question 19

Q19. Give the structures of the products of the following reactions:
(i) CH3–CH2–CH2–O–CH3 + HBr →
(ii) (C2H5)3C–O–CH3 + HI →
(iii) CH3–CH(CH3)–O–C2H5 + HBr →

Accepted Answer

The cleavage of ethers by hydrogen halides (HBr, HI) proceeds via either SN1 or SN2 mechanism, depending on the nature of the alkyl groups attached to the oxygen atom.

**(i) CH3–CH2–CH2–O–CH3 + HBr →**
In this ether, one alkyl group is n-propyl (primary) and the other is methyl (primary). Both are primary alkyl groups. The reaction proceeds via an SN2 mechanism, where the halide ion (Br-) attacks the less sterically hindered carbon atom. In this case, the methyl group is less hindered than the n-propyl group.
Products: CH3Br (Bromomethane) and CH3CH2CH2OH (Propan-1-ol)

**(ii) (C2H5)3C–O–CH3 + HI →**
Here, one alkyl group is triethylmethyl (a tertiary alkyl group) and the other is methyl (primary). When one of the alkyl groups is tertiary, the reaction proceeds via an SN1 mechanism, forming a stable tertiary carbocation. The halide ion (I-) then attacks this carbocation.
Products: (C2H5)3C–I (1-Iodo-1,1-diethylpropane) and CH3OH (Methanol)

**(iii) CH3–CH(CH3)–O–C2H5 + HBr →**
In this ether, one alkyl group is isopropyl (secondary) and the other is ethyl (primary). When one of the alkyl groups is secondary, the reaction generally proceeds via an SN1 mechanism if a relatively stable carbocation can be formed. The isopropyl carbocation is more stable than the ethyl carbocation. Therefore, the halide ion (Br-) attacks the isopropyl carbocation.
Products: CH3–CH(CH3)–Br (2-Bromopropane) and C2H5OH (Ethanol)

Question 20

Q20. Write the mechanism of the reaction of HI with methoxymethane.

Accepted Answer

The reaction of HI with methoxymethane (CH3–O–CH3) involves the cleavage of the ether linkage. Since both alkyl groups are methyl (primary), the reaction proceeds via an SN2 mechanism. If HI is in excess, both methyl groups are converted to methyl iodide.

**Mechanism:**

**Step 1: Protonation of the ether oxygen.**
The oxygen atom of methoxymethane, being a Lewis base, gets protonated by the strong acid HI to form an oxonium ion. This step makes the oxygen a better leaving group.

CH3–O–CH3 + H–I  ⇌  CH3–O+H–CH3 + I-

**Step 2: Nucleophilic attack by iodide ion.**
The iodide ion (I-), a strong nucleophile, attacks one of the methyl carbon atoms from the backside, displacing the methanol molecule. This is an SN2 reaction, as the methyl carbon is sterically unhindered.

I- + CH3–O+H–CH3  →  CH3–I + CH3OH

**Step 3: Further reaction of methanol (if HI is in excess).**
If HI is present in excess, the methanol formed in Step 2 can further react with HI. Methanol is first protonated, and then the iodide ion attacks the methyl carbon, displacing water.

CH3OH + H–I  ⇌  CH3–O+H2 + I-

I- + CH3–O+H2  →  CH3–I + H2O

Thus, the overall products when HI is in excess are methyl iodide (CH3I) and water (H2O). If HI is not in excess, the products would be methyl iodide and methanol.

Question 21

Q21. Write the mechanism of hydration of ethene to yield ethanol.

Accepted Answer

The hydration of ethene to yield ethanol is an acid-catalyzed reaction. It proceeds via an electrophilic addition mechanism involving the formation of a carbocation intermediate. The overall reaction is:
CH₂=CH₂ + H₂O $\xrightarrow{H^+}$ CH₃CH₂OH

The mechanism involves three main steps:

**Step 1: Protonation of alkene to form carbocation**
Ethene, being a nucleophile, attacks the proton (H⁺) from the acid catalyst (e.g., H₃O⁺). This leads to the formation of a primary carbocation. This is the slowest step and hence the rate-determining step.

**Step 2: Nucleophilic attack of water on carbocation**
The carbocation formed in Step 1 is highly electrophilic. A water molecule, acting as a nucleophile, attacks the carbocation, forming a protonated alcohol (oxonium ion).

**Step 3: Deprotonation to form alcohol**
The protonated alcohol formed in Step 2 loses a proton to a water molecule (or the conjugate base of the acid catalyst), regenerating the acid catalyst and yielding the final product, ethanol. This step also helps in regenerating the catalyst for further reaction cycles.

Question 22

Q22. Show how will you synthesise the following alcohols from appropriate alkenes.
(i) 1-methylcyclohexanol
(ii) 1-methylcyclopentanol
(iii) 2-phenylpropan-2-ol

Accepted Answer

Alcohols can be synthesized from alkenes primarily through hydration reactions. Acid-catalyzed hydration (addition of H₂O in the presence of H⁺) follows Markovnikov's rule, where the -OH group adds to the more substituted carbon of the double bond and hydrogen adds to the less substituted carbon. Oxymercuration-demercuration also follows Markovnikov's rule without carbocation rearrangements. For the given alcohols, we will identify the alkene that would yield the desired alcohol upon Markovnikov addition of water.

**(i) 1-methylcyclohexanol**
To get 1-methylcyclohexanol, the -OH group must be on the carbon bearing the methyl group in the cyclohexane ring. This carbon must have been part of a double bond. Therefore, the appropriate alkene is 1-methylcyclohexene. Upon hydration, the -OH adds to the more substituted carbon (C1) and H adds to C2.

**Reaction:** 1-methylcyclohexene $\xrightarrow{H_2O, H^+}$ 1-methylcyclohexanol

**(ii) 1-methylcyclopentanol**
Similarly, for 1-methylcyclopentanol, the -OH group is on the carbon bearing the methyl group in the cyclopentane ring. The corresponding alkene is 1-methylcyclopentene. Hydration will place the -OH on the more substituted carbon (C1) and H on C2.

**Reaction:** 1-methylcyclopentene $\xrightarrow{H_2O, H^+}$ 1-methylcyclopentanol

**(iii) 2-phenylpropan-2-ol**
In 2-phenylpropan-2-ol, the -OH group is on the second carbon of the propane chain, which is also attached to a phenyl group and two methyl groups. This carbon is tertiary. To obtain this alcohol via hydration, the double bond must be between C1 and C2, or C2 and C3. If the double bond is between C1 and C2 (i.e., 2-phenylpropene), the -OH will add to C2 (the more substituted carbon) and H to C1. If the double bond is between C2 and C3, it would be 1-phenylpropene, which would give 1-phenylpropan-1-ol or 2-phenylpropan-1-ol depending on the regioselectivity. Thus, 2-phenylpropene is the appropriate alkene.

**Reaction:** 2-phenylpropene $\xrightarrow{H_2O, H^+}$ 2-phenylpropan-2-ol

Question 23

Q23. When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:

Show the mechanism of this reaction.

Accepted Answer

The reaction of 3-methylbutan-2-ol with HBr to form 2-bromo-2-methylbutane involves an SN1 mechanism with a carbocation rearrangement. The starting alcohol is secondary, but the product is a tertiary alkyl bromide, indicating a rearrangement from a less stable secondary carbocation to a more stable tertiary carbocation.

**Overall Reaction:**
CH₃-CH(CH₃)-CH(OH)-CH₃ + HBr $\longrightarrow$ CH₃-C(CH₃)(Br)-CH₂-CH₃ + H₂O
(3-methylbutan-2-ol)                                 (2-bromo-2-methylbutane)

**Mechanism:**

**Step 1: Protonation of the alcohol**
The oxygen atom of the alcohol is protonated by HBr (or H₃O⁺ formed from HBr and water), forming a good leaving group, protonated alcohol (an oxonium ion).

**Step 2: Formation of carbocation**
The protonated alcohol loses a water molecule, forming a secondary carbocation (3-methylbutan-2-yl carbocation). This is the rate-determining step.

**Step 3: Carbocation rearrangement (Hydride shift)**
The secondary carbocation formed in Step 2 is less stable than a possible tertiary carbocation. A 1,2-hydride shift occurs, where a hydrogen atom with its bonding electrons migrates from the adjacent carbon (C3) to the positively charged carbon (C2). This forms a more stable tertiary carbocation (2-methylbutan-2-yl carbocation).

**Step 4: Nucleophilic attack by bromide ion**
The bromide ion (Br⁻), acting as a nucleophile, attacks the more stable tertiary carbocation, forming the final product, 2-bromo-2-methylbutane.

Question 24

Q24. Write the mechanism of the following reactions:
(i) CH₃CH₂OH $\xrightarrow{HBr}$ CH₃CH₂Br + H₂O
(ii) (CH₃)₃C-OH $\xrightarrow{H^+}$ (CH₃)₂C=CH₂ + H₂O

Accepted Answer

Here are the mechanisms for the given reactions:

**(i) Mechanism for the reaction of ethanol with HBr (SN2 mechanism)**
Primary alcohols react with HBr via an SN2 mechanism. In this mechanism, the nucleophilic attack and the departure of the leaving group occur simultaneously in a single step.

**Step 1: Protonation of the alcohol**
The oxygen atom of ethanol is protonated by HBr, forming a protonated alcohol (an oxonium ion), which is a good leaving group.

**Step 2: Nucleophilic attack and leaving group departure**
The bromide ion (Br⁻), acting as a nucleophile, attacks the carbon atom bearing the protonated hydroxyl group from the backside. Simultaneously, the water molecule (H₂O) departs as the leaving group, leading to the formation of bromoethane.

**(ii) Mechanism for the dehydration of tert-butyl alcohol (E1 mechanism)**
Tertiary alcohols undergo dehydration in the presence of an acid catalyst (H⁺) to form alkenes via an E1 mechanism. This mechanism involves the formation of a stable tertiary carbocation.

**Step 1: Protonation of the alcohol**
The oxygen atom of tert-butyl alcohol is protonated by the acid catalyst (H⁺), forming a protonated alcohol (an oxonium ion), which is a good leaving group.

**Step 2: Formation of carbocation**
The protonated alcohol loses a water molecule, forming a stable tertiary carbocation (tert-butyl carbocation). This is the rate-determining step.

**Step 3: Deprotonation to form alkene**
A water molecule (or the conjugate base of the acid catalyst) abstracts a proton from a carbon atom adjacent to the positively charged carbon (a $\beta$-hydrogen), leading to the formation of a double bond and regenerating the acid catalyst. This results in the formation of 2-methylpropene.

Question 25

Q25. Give the structures and IUPAC names of the products expected from the following reactions:
(a) Catalytic reduction of butanal
(b) Hydration of propene in the presence of dilute s

Accepted Answer

Here are the structures and IUPAC names of the products for the given reactions:

**(a) Catalytic reduction of butanal**
*   **Reaction:** Butanal (an aldehyde) undergoes catalytic reduction to form a primary alcohol.
*   **Product Structure:** CH₃—CH₂—CH₂—CH₂—OH
*   **IUPAC Name:** Butan-1-ol

**(b) Hydration of propene in the presence of dilute sulphuric acid**
*   **Reaction:** Propene undergoes acid-catalyzed hydration, following Markovnikov's rule, to form a secondary alcohol.
*   **Product Structure:** CH₃—CH(OH)—CH₃
*   **IUPAC Name:** Propan-2-ol

**(c) Reaction of propanone with methylmagnesium bromide followed by hydrolysis**
*   **Reaction:** Propanone (a ketone) reacts with methylmagnesium bromide (Grignard reagent) via nucleophilic addition, followed by hydrolysis, to form a tertiary alcohol.
*   **Product Structure:** (CH₃)₃C—OH
*   **IUPAC Name:** 2-Methylpropan-2-ol

**(d) Hydroboration-oxidation of propene**
*   **Reaction:** Propene undergoes hydroboration-oxidation, which is an anti-Markovnikov addition of water, to form a primary alcohol.
*   **Product Structure:** CH₃—CH₂—CH₂—OH
*   **IUPAC Name:** Propan-1-ol

**(e) Acid catalysed hydration of but-1-ene**
*   **Reaction:** But-1-ene undergoes acid-catalyzed hydration, following Markovnikov's rule, to form a secondary alcohol.
*   **Product Structure:** CH₃—CH₂—CH(OH)—CH₃
*   **IUPAC Name:** Butan-2-ol

**(f) Treatment of ethanol with conc. H₂SO₄ at 413 K**
*   **Reaction:** At 413 K (140 °C), ethanol undergoes intermolecular dehydration in the presence of concentrated H₂SO₄ to form an ether.
*   **Product Structure:** CH₃—CH₂—O—CH₂—CH₃
*   **IUPAC Name:** Ethoxyethane (or Diethyl ether)

**(g) Treatment of phenol with dilute HNO₃**
*   **Reaction:** Phenol undergoes electrophilic aromatic substitution (nitration) with dilute HNO₃. The -OH group is ortho-para directing and activating, leading to a mixture of ortho and para products.
*   **Product Structures:**
    1.  Benzene ring with -OH at C1 and -NO₂ at C2.
    2.  Benzene ring with -OH at C1 and -NO₂ at C4.
*   **IUPAC Names:** 2-Nitrophenol and 4-Nitrophenol

**(h) Treatment of phenol with Br₂(aq)**
*   **Reaction:** Phenol undergoes electrophilic aromatic substitution (bromination) with bromine water (Br₂(aq)). The -OH group is strongly activating, leading to polybromination at all available ortho and para positions.
*   **Product Structure:** Benzene ring with -OH at C1 and -Br at C2, C4, and C6.
*   **IUPAC Name:** 2,4,6-Tribromophenol

Question 26

Q26. Show how will you synthesise:
(i) 1-phenylethanol from an appropriate alkene.
(ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.
(iii) pentan-1-ol using an alkyl

Accepted Answer

Here's how to synthesize the given alcohols:

**(i) 1-phenylethanol from an appropriate alkene.**
*   **Target Product:** 1-phenylethanol (C₆H₅-CH(OH)-CH₃), a secondary alcohol.
*   **Appropriate Alkene:** Styrene (Phenylethene, C₆H₅-CH=CH₂).
*   **Reaction:** Acid-catalyzed hydration of styrene, which follows Markovnikov's rule, will yield 1-phenylethanol.
*   **Synthesis:**
    C₆H₅-CH=CH₂  +  H₂O  --[H⁺ (dil. H₂SO₄)]-->  C₆H₅-CH(OH)-CH₃
    (Styrene)                                         (1-phenylethanol)

**(ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.**
*   **Target Product:** Cyclohexylmethanol (C₆H₁₁-CH₂OH), a primary alcohol.
*   **Appropriate Alkyl Halide:** Chloromethylcyclohexane (or Bromomethylcyclohexane).
*   **Reaction:** An SN2 reaction involves the nucleophilic substitution of a halide by a hydroxide ion. For primary alcohols, a primary alkyl halide is typically used with a strong nucleophile like aqueous KOH or NaOH.
*   **Synthesis:**
    C₆H₁₁-CH₂Cl  +  KOH(aq)  --[SN2]-->  C₆H₁₁-CH₂OH  +  KCl
    (Chloromethylcyclohexane)                      (Cyclohexylmethanol)

**(iii) pentan-1-ol using an alkyl halide.**
*   **Target Product:** Pentan-1-ol (CH₃CH₂CH₂CH₂CH₂OH), a primary alcohol.
*   **Appropriate Alkyl Halide:** 1-Bromopentane (or 1-Chloropentane).
*   **Reaction:** Similar to (ii), an SN2 reaction with aqueous base will substitute the halide with a hydroxide group.
*   **Synthesis:**
    CH₃CH₂CH₂CH₂CH₂Br  +  KOH(aq)  --[SN2]-->  CH₃CH₂CH₂CH₂CH₂OH  +  KBr
    (1-Bromopentane)                                  (Pentan-1-ol)

**(iv) butan-2-ol from an appropriate alkene.**
*   **Target Product:** Butan-2-ol (CH₃CH₂CH(OH)CH₃), a secondary alcohol.
*   **Appropriate Alkene:** But-1-ene (CH₃CH₂CH=CH₂) or But-2-ene (CH₃CH=CHCH₃).
*   **Reaction:** Acid-catalyzed hydration of but-1-ene (or but-2-ene) will yield butan-2-ol, following Markovnikov's rule for but-1-ene.
*   **Synthesis (using but-1-ene):**
    CH₃CH₂CH=CH₂  +  H₂O  --[H⁺ (dil. H₂SO₄)]-->  CH₃CH₂CH(OH)CH₃
    (But-1-ene)                                         (Butan-2-ol)

Question 27

Q27. 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
Phenol is acidic due to the resonance stabilization of its conjugate base, the phenoxide ion. This acidic nature can be demonstrated by its reactions with active metals and strong bases.

1.  **Reaction with Active Metals (e.g., Sodium):**
    Phenol reacts with active metals like sodium to liberate hydrogen gas, similar to how acids react with active metals. This indicates the presence of an acidic hydrogen atom.
    `2 C₆H₅OH  +  2 Na  ⟶  2 C₆H₅ONa  +  H₂↑`
    (Phenol)      (Sodium)      (Sodium Phenoxide) (Hydrogen gas)

2.  **Reaction with Strong Bases (e.g., Sodium Hydroxide):**
    Phenol reacts with strong bases like sodium hydroxide to form a salt (sodium phenoxide) and water. This is a typical acid-base neutralization reaction.
    `C₆H₅OH  +  NaOH  ⟶  C₆H₅ONa  +  H₂O`
    (Phenol)      (Sodium Hydroxide) (Sodium Phenoxide) (Water)
    *Note: Phenol does not react with weaker bases like sodium bicarbonate (NaHCO₃), indicating it is a weaker acid than carbonic acid (H₂CO₃).*

### Comparison of Acidity of Phenol with Ethanol
Phenol is significantly more acidic than ethanol.

**1. Acidity of Phenol:**
*   **Reason:** Phenol (C₆H₅OH) is acidic because the phenoxide ion (C₆H₅O⁻), formed after losing a proton (H⁺), is stabilized by resonance. The negative charge on the oxygen atom is delocalized over the benzene ring. This delocalization makes the phenoxide ion more stable than the unionized phenol molecule, thus favoring the release of a proton.
*   **Resonance Structures of Phenoxide Ion:** The negative charge on oxygen can be delocalized to the ortho and para positions of the benzene ring, as shown by the following resonance structures (description):
    *   Structure 1: Negative charge on oxygen, double bond between oxygen and ring carbon.
    *   Structure 2: Negative charge at ortho position, double bond shifts.
    *   Structure 3: Negative charge at para position, double bond shifts.
    *   Structure 4: Negative charge at other ortho position, double bond shifts.
    *   Structure 5: Negative charge returns to oxygen.
*   **pKa Value:** The pKa of phenol is approximately 10.

**2. Acidity of Ethanol:**
*   **Reason:** Ethanol (CH₃CH₂OH) is a very weak acid, much weaker than phenol. The ethoxide ion (CH₃CH₂O⁻), formed after losing a proton, is not stabilized by resonance. Instead, the electron-donating ethyl group (CH₃CH₂-) intensifies the negative charge on the oxygen atom, making the ethoxide ion less stable than the unionized ethanol molecule. This destabilization makes it difficult for ethanol to release a proton.
*   **pKa Value:** The pKa of ethanol is approximately 16.

**Conclusion:**
Since the phenoxide ion is resonance-stabilized while the ethoxide ion is not (and is, in fact, destabilized by the electron-donating alkyl group), **phenol is a much stronger acid than ethanol.**

Question 28

Q28. Explain why is ortho nitrophenol more acidic than ortho methoxyphenol.

Accepted Answer

The acidity of a compound is determined by the stability of its conjugate base. A more stable conjugate base corresponds to a stronger acid.

1.  **Ortho-nitrophenol:**
    *   When ortho-nitrophenol loses a proton, it forms the ortho-nitrophenoxide ion (its conjugate base).
    *   The nitro group (-NO₂) is a strong electron-withdrawing group (EWG) due to both its strong resonance (-R) effect and inductive (-I) effect. Both effects operate effectively from the ortho position.
    *   This strong electron-withdrawing nature of the -NO₂ group helps to delocalize and disperse the negative charge on the oxygen atom of the phenoxide ion, thereby stabilizing the ortho-nitrophenoxide ion.
    *   A more stable conjugate base means ortho-nitrophenol is a stronger acid.

2.  **Ortho-methoxyphenol:**
    *   When ortho-methoxyphenol loses a proton, it forms the ortho-methoxyphenoxide ion (its conjugate base).
    *   The methoxy group (-OCH₃) is an electron-donating group (EDG) primarily due to its resonance (+R) effect, which is stronger than its inductive (-I) effect. This overall electron-donating nature is significant at the ortho position.
    *   Electron-donating groups destabilize the phenoxide ion by intensifying the negative charge on the oxygen atom, making the ortho-methoxyphenoxide ion less stable.
    *   A less stable conjugate base means ortho-methoxyphenol is a weaker acid.

**Conclusion:**
Since the electron-withdrawing nitro group stabilizes the conjugate base of ortho-nitrophenol, while the electron-donating methoxy group destabilizes the conjugate base of ortho-methoxyphenol, **ortho-nitrophenol is more acidic than ortho-methoxyphenol.**

Question 29

Q29. 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 al

Accepted Answer

The synthesis for each of the given compounds is as follows:

**(i) 1-phenylethanol from a suitable alkene**
To synthesise 1-phenylethanol, we need an alkene that upon hydration gives the desired alcohol. The suitable alkene is styrene (phenylethene). The addition of water across the double bond of styrene follows Markovnikov's rule, where the -OH group adds to the carbon atom that has fewer hydrogen atoms (the carbon attached to the phenyl group).

**Reaction:** Acid-catalysed hydration of styrene.
Styrene (C₆H₅-CH=CH₂) reacts with water in the presence of a dilute acid catalyst (like H₂SO₄) to yield 1-phenylethanol.

C₆H₅-CH=CH₂ + H₂O  (in presence of H⁺) → C₆H₅-CH(OH)-CH₃

**(ii) Cyclohexylmethanol using an alkyl halide by an Sₙ2 reaction**
To synthesise cyclohexylmethanol, we need to start with a primary alkyl halide and react it with a strong nucleophile in an Sₙ2 reaction. The appropriate alkyl halide is cyclohexylmethyl bromide (or chloride). This is a primary halide, which is ideal for Sₙ2 reactions to avoid elimination.
The nucleophile will be the hydroxide ion (OH⁻), typically from an aqueous solution of a strong base like NaOH or KOH.

**Reaction:** Nucleophilic substitution of cyclohexylmethyl bromide with aqueous KOH.
Cyclohexylmethyl bromide reacts with aqueous potassium hydroxide (KOH) via an Sₙ2 mechanism, where the hydroxide ion displaces the bromide ion to form cyclohexylmethanol.

C₆H₁₁-CH₂-Br + KOH(aq) → C₆H₁₁-CH₂-OH + KBr

**(iii) Pentan-1-ol using a suitable alkyl halide**
Similar to the previous synthesis, to obtain a primary alcohol like pentan-1-ol via nucleophilic substitution, we should start with a primary alkyl halide. The suitable alkyl halide is 1-bromopentane (or 1-chloropentane). The reaction with aqueous NaOH or KOH will proceed via an Sₙ2 mechanism.

**Reaction:** Nucleophilic substitution of 1-bromopentane with aqueous NaOH.
1-Bromopentane reacts with aqueous sodium hydroxide (NaOH), where the OH⁻ nucleophile attacks the primary carbon and displaces the bromide ion, yielding pentan-1-ol.

CH₃(CH₂)₃CH₂-Br + NaOH(aq) → CH₃(CH₂)₃CH₂-OH + NaBr

Question 30

Q30. Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.

Accepted Answer

Phenol is acidic in nature due to its ability to donate a proton (H⁺) from the hydroxyl group. Its acidity is significantly higher than that of alcohols like ethanol.

**Two Reactions Showing Acidic Nature of Phenol:**

1.  **Reaction with Active Metals:** Phenol reacts with active metals like sodium (Na) to liberate hydrogen gas and form sodium phenoxide. This reaction is characteristic of acids.
    2C₆H₅OH + 2Na → 2C₆H₅O⁻Na⁺ + H₂(g)
    (Phenol)       (Sodium)   (Sodium phenoxide)

2.  **Reaction with Strong Bases:** Phenol is a strong enough acid to react with aqueous alkalis like sodium hydroxide (NaOH) to form salt and water. This is a neutralization reaction.
    C₆H₅OH + NaOH(aq) → C₆H₅O⁻Na⁺ + H₂O
    (Phenol)   (Sodium hydroxide)  (Sodium phenoxide)
    Alcohols like ethanol do not react with NaOH, which indicates that phenol is more acidic than alcohols.

**Comparison of Acidity of Phenol and Ethanol:**

The acidity of a substance is determined by the stability of its conjugate base formed after donating a proton.

*   **Acidity of Phenol:**
    When phenol loses a proton, it forms the **phenoxide ion (C₆H₅O⁻)**. The negative charge on the oxygen atom in the phenoxide ion is delocalized over the benzene ring through resonance. This delocalization spreads the negative charge over the entire molecule, which results in significant stabilization of the phenoxide ion. The more stable the conjugate base, the stronger the acid. The resonance structures of the phenoxide ion show that the negative charge is delocalized to the ortho and para positions of the ring.

*   **Acidity of Ethanol:**
    When ethanol (CH₃CH₂OH) loses a proton, it forms the **ethoxide ion (CH₃CH₂O⁻)**. In the ethoxide ion, the ethyl group (-CH₂CH₃) is an electron-donating group due to its positive inductive effect (+I effect). This group pushes electron density towards the oxygen atom, intensifying the negative charge on it. This concentration of negative charge destabilizes the ethoxide ion.

**Conclusion:**
Because the phenoxide ion is highly stabilized by resonance, while the ethoxide ion is destabilized by the +I effect of the ethyl group, the equilibrium for the ionization of phenol lies further to the right than that for ethanol.

C₆H₅OH ⇌ C₆H₅O⁻ + H⁺ (Phenoxide ion is resonance stabilized)
CH₃CH₂OH ⇌ CH₃CH₂O⁻ + H⁺ (Ethoxide ion is destabilized by +I effect)

Therefore, **phenol is a much stronger acid than ethanol**. The pKa value of phenol is about 10, whereas the pKa value of ethanol is about 16. A lower pKa value indicates a stronger acid.

Question 31

Q31. Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.

Accepted Answer

The preparation of ethers by acid-catalysed dehydration of alcohols is a nucleophilic substitution reaction. This method is suitable for primary alcohols but not for secondary or tertiary alcohols due to competing elimination reactions.

**Mechanism and Competition:**

1.  **For Primary Alcohols (SN2):** The reaction proceeds via an SN2 mechanism where one molecule of protonated alcohol is attacked by another molecule of alcohol. Elimination is not significant.
    2 R-CH₂-OH + H⁺ → R-CH₂-O-CH₂-R + H₂O

2.  **For Secondary (2°) and Tertiary (3°) Alcohols (SN1 vs E1):**
    The reaction proceeds via an SN1 mechanism. The first step is the protonation of the alcohol, followed by the loss of a water molecule to form a stable secondary or tertiary carbocation.
    (CH₃)₃C-OH + H⁺ ⇌ (CH₃)₃C-OH₂⁺ → (CH₃)₃C⁺ + H₂O

This carbocation can then undergo two competing reactions:
    *   **Substitution (SN1):** Another alcohol molecule can act as a nucleophile and attack the carbocation to form a protonated ether, which then loses a proton to form the ether. This is the desired pathway for ether synthesis.
    *   **Elimination (E1):** The carbocation can lose a proton from an adjacent carbon atom to form an alkene. This is the dehydration reaction.

**Reason for Unsuitability:**
For secondary and tertiary alcohols, the elimination (E1) reaction is highly favoured over the substitution (SN1) reaction. This is because:
*   **Carbocation Stability:** Secondary and tertiary carbocations are stable and can easily lose a proton to form a stable alkene (Saytzeff's rule).
*   **Steric Hindrance:** The bulky secondary or tertiary alkyl groups create steric hindrance, which makes the nucleophilic attack by another alcohol molecule (for ether formation) difficult. Elimination, which involves the removal of a small proton by a base (like H₂O or HSO₄⁻), is sterically less demanding.
*   **Reaction Conditions:** The acidic conditions and heat required for the reaction also favour elimination.

**Conclusion:**
Due to the predominance of the elimination reaction, acid-catalysed dehydration of secondary and tertiary alcohols results in the formation of alkenes as the major product, with only a very small yield of the desired ether. Therefore, this method is not suitable for preparing ethers from 2° or 3° alcohols. The Williamson synthesis is the preferred method in such cases.

Question 32

Q32. Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane
(ii) methoxybenzene
(iii) benzyl ethyl ether

Accepted Answer

The reaction of ethers with hydrogen iodide (HI) involves the cleavage of the C-O bond. The outcome depends on the nature of the groups attached to the oxygen atom.

**(i) 1-propoxypropane**

1-propoxypropane is a symmetrical ether with two primary propyl groups. The reaction proceeds via an SN2 mechanism. The iodide ion (I⁻) attacks one of the equivalent CH₂ groups, leading to the cleavage of a C-O bond.

**Equation:**
CH₃CH₂CH₂-O-CH₂CH₂CH₃ + HI (conc.) → CH₃CH₂CH₂-OH + CH₃CH₂CH₂-I
(1-Propoxypropane)                     (Propan-1-ol)   (1-Iodopropane)

**(ii) Methoxybenzene (Anisole)**

Methoxybenzene is an aryl alkyl ether. The oxygen atom is attached to a methyl group and a phenyl group. The O-phenyl (C₆H₅) bond is very strong due to partial double bond character from resonance and the sp² hybridization of the carbon atom in the benzene ring. Therefore, this bond does not break. The iodide ion attacks the smaller alkyl (methyl) group via an SN2 mechanism.

**Equation:**
C₆H₅-O-CH₃ + HI (conc.) → C₆H₅-OH + CH₃-I
(Methoxybenzene)             (Phenol)   (Iodomethane)

**(iii) Benzyl ethyl ether**

Benzyl ethyl ether has a benzyl group (C₆H₅CH₂) and an ethyl group (CH₃CH₂) attached to the oxygen. The cleavage of the C-O bond in this case can proceed via an SN1-like mechanism because the benzyl group can form a highly stable benzylic carbocation (C₆H₅CH₂⁺) which is stabilized by resonance. The C-O bond attached to the benzyl group breaks to form this stable carbocation, which is then attacked by the iodide ion.

**Equation:**
C₆H₅CH₂-O-CH₂CH₃ + HI (conc.) → C₆H₅CH₂-I + CH₃CH₂-OH
(Benzyl ethyl ether)                     (Benzyl iodide)   (Ethanol)

Question 33

Q33. How will you distinguish between the following pairs of compounds:
(i) Phenol and Benzoic acid
(ii) Propanal and Propanone
(iii) Phenol and Ethanol
(iv) Ethanal and Propanal

Accepted Answer

Here are the distinguishing tests for the given pairs of compounds:

**(i) Phenol and Benzoic acid**
*   **Test 1: Sodium Bicarbonate Test**
    *   **Procedure:** Add a solution of sodium bicarbonate (NaHCO₃) to each compound.
    *   **Observation:** Benzoic acid reacts with sodium bicarbonate to produce brisk effervescence due to the evolution of carbon dioxide gas. Phenol, being a weaker acid than carbonic acid, does not react with sodium bicarbonate and no effervescence is observed.
    *   **Reactions:**
        *   C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂↑ (Benzoic acid)
        *   C₆H₅OH + NaHCO₃ → No reaction (Phenol)
*   **Test 2: Ferric Chloride Test**
    *   **Procedure:** Add a few drops of neutral ferric chloride (FeCl₃) solution to an aqueous solution of each compound.
    *   **Observation:** Phenol reacts with neutral ferric chloride solution to give a characteristic violet, blue, or green coloration. Benzoic acid does not give this test.
    *   **Reaction:**
        *   6C₆H₅OH + FeCl₃ → [Fe(OC₆H₅)₆]³⁻ + 3H⁺ + 3HCl (Phenol, violet/blue/green complex)

**(ii) Propanal and Propanone**
*   **Test 1: Tollens' Test (Silver Mirror Test)**
    *   **Procedure:** Add Tollens' reagent (ammoniacal silver nitrate solution) to each compound and warm gently.
    *   **Observation:** Propanal, being an aldehyde, reduces Tollens' reagent to metallic silver, forming a silver mirror on the inner walls of the test tube. Propanone, being a ketone, does not react with Tollens' reagent.
    *   **Reactions:**
        *   CH₃CH₂CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃CH₂COO⁻ + 2Ag↓ + 4NH₃ + 2H₂O (Propanal)
        *   CH₃COCH₃ + Tollens' reagent → No reaction (Propanone)
*   **Test 2: Fehling's Test**
    *   **Procedure:** Add Fehling's solution (a mixture of Fehling's A and Fehling's B) to each compound and heat.
    *   **Observation:** Propanal reduces Fehling's solution to form a red precipitate of cuprous oxide (Cu₂O). Propanone does not react with Fehling's solution.
    *   **Reactions:**
        *   CH₃CH₂CHO + 2Cu²⁺ + 5OH⁻ → CH₃CH₂COO⁻ + Cu₂O↓ + 3H₂O (Propanal)
        *   CH₃COCH₃ + Fehling's solution → No reaction (Propanone)
*   **Test 3: Iodoform Test**
    *   **Procedure:** Add iodine solution (I₂) and dilute sodium hydroxide (NaOH) solution to each compound and warm.
    *   **Observation:** Propanone, being a methyl ketone (CH₃-CO-R), gives a yellow precipitate of iodoform (CHI₃). Propanal does not give this test.
    *   **Reaction:**
        *   CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃↓ (yellow ppt) + CH₃COONa + 3NaI + 3H₂O (Propanone)

**(iii) Phenol and Ethanol**
*   **Test 1: Ferric Chloride Test**
    *   **Procedure:** Add a few drops of neutral ferric chloride (FeCl₃) solution to an aqueous solution of each compound.
    *   **Observation:** Phenol reacts with neutral ferric chloride solution to give a characteristic violet, blue, or green coloration. Ethanol does not give this test.
    *   **Reaction:**
        *   6C₆H₅OH + FeCl₃ → [Fe(OC₆H₅)₆]³⁻ + 3H⁺ + 3HCl (Phenol, violet/blue/green complex)
*   **Test 2: Bromine Water Test**
    *   **Procedure:** Add bromine water (Br₂/H₂O) to each compound.
    *   **Observation:** Phenol reacts with bromine water to form a white precipitate of 2,4,6-tribromophenol, decolorizing the bromine water. Ethanol does not react with bromine water.
    *   **Reaction:**
        *   C₆H₅OH + 3Br₂ (aq) → C₆H₂Br₃OH↓ (white ppt) + 3HBr (Phenol)
*   **Test 3: Iodoform Test**
    *   **Procedure:** Add iodine solution (I₂) and dilute sodium hydroxide (NaOH) solution to each compound and warm.
    *   **Observation:** Ethanol, having a CH₃CH(OH)- group, gives a yellow precipitate of iodoform (CHI₃). Phenol does not give this test.
    *   **Reaction:**
        *   CH₃CH₂OH + 4I₂ + 6NaOH → CHI₃↓ (yellow ppt) + HCOONa + 5NaI + 5H₂O (Ethanol)

**(iv) Ethanal and Propanal**
*   **Test: Iodoform Test**
    *   **Procedure:** Add iodine solution (I₂) and dilute sodium hydroxide (NaOH) solution to each compound and warm.
    *   **Observation:** Ethanal (acetaldehyde), having a CH₃CHO group, gives a yellow precipitate of iodoform (CHI₃). Propanal (propionaldehyde), which does not have a CH₃CO- or CH₃CH(OH)- group, does not give this test.
    *   **Reactions:**
        *   CH₃CHO + 3I₂ + 4NaOH → CHI₃↓ (yellow ppt) + HCOONa + 3NaI + 3H₂O (Ethanal)
        *   CH₃CH₂CHO + I₂/NaOH → No reaction (Propanal)

Chapter 11: Alcohols, Phenols and Ethers