(i) $\text{CH}_3\text{CH}=\text{C(CH}_3)\text{CH(OH)CH}_2\text{OH}$:

We select the longest carbon chain containing the maximum principal functional groups (hydroxyl groups):

• Number the chain from right to left to give the hydroxyl groups the lowest locants (positions 1 and 2): $$ \text{C}_1\text{H}_2\text{(OH)} - \text{C}_2\text{H(OH)} - \text{C}_3\text{(CH}_3) = \text{C}_4\text{H} - \text{C}_5\text{H}_3 $$
• Substituent: Methyl group at Carbon-3 ($\text{-CH}_3$).
• Alkene double bond: Position 3 ($\text{pent-3-ene}$).
• Hydroxyl groups: Positions 1 and 2 ($\text{-1,2-diol}$).

IUPAC Name: 3-Methylpent-3-ene-1,2-diol

(ii) $\text{CH}_3\text{-O-CH(CH}_3)_2$:

This is an unsymmetrical ether. The larger alkyl group is propane (chain length of 3 carbons), and the smaller group is methyl ($\text{-CH}_3$) combined with oxygen as the alkoxy group ($\text{methoxy-}$ substituent) at position 2 of the propane chain.

IUPAC Name: 2-Methoxypropane

Reason: Propan-1-ol molecules contain a highly polar hydroxyl ($\text{-OH}$) group. This allows the molecules to form strong **intermolecular hydrogen bonds** in the liquid state.

Conversely, butane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$) is a non-polar hydrocarbon whose molecules are held together only by weak, transient **Van der Waals dispersion forces**. Because hydrogen bonds are substantially stronger than Van der Waals forces, propan-1-ol requires much more thermal energy to vaporize, leading to a much higher boiling point.

Explanation: The solubility of lower alcohols in water is due to their ability to form hydrogen bonds with water molecules via their polar hydroxyl ($\text{-OH}$) groups.

An alcohol molecule consists of two parts: a polar, hydrophilic (water-loving) $\text{-OH}$ group and a non-polar, hydrophobic (water-repelling) alkyl group. As the molecular mass of the alcohol increases, the **size of the non-polar hydrophobic alkyl group** grows larger. This bulky hydrocarbon part prevents water molecules from effectively hydrogen-bonding with the polar $\text{-OH}$ group, decreasing its overall solubility in water.

Lucas' Reagent: A solution containing a mixture of **concentrated hydrochloric acid ($\text{HCl}$) and anhydrous zinc chloride ($\text{ZnCl}_2$)**.

Chemical Principle: Alcohols react with Lucas' reagent to produce alkyl chlorides ($\text{R-Cl}$), which are insoluble in water and form a cloudy, turbid suspension (turbidity).

Distinguishing Test at Room Temperature:

1. Tertiary ($3^\circ$) Alcohols: Form turbidity **immediately** because they react rapidly via a highly stable tertiary carbocation intermediate.
2. Secondary ($2^\circ$) Alcohols: Form turbidity within **5 minutes**.
3. Primary ($1^\circ$) Alcohols: Do **not** produce any turbidity at room temperature. Turbidity only appears upon heating.

Reason: As the carbon chain becomes more branched, the molecular shape shifts from a linear structure to a more compact, spherical geometry.

This decrease in length reduces the overall **molecular surface area**. Because intermolecular Van der Waals forces are directly proportional to the surface area of contact, branching weakens these attractive forces. Consequently, less heat is needed to separate the molecules, causing the boiling point to decrease (e.g., $\text{n-butyl alcohol} \gt \text{isobutyl alcohol} \gt \text{tert-butyl alcohol}$).

The higher acidity of phenols compared to aliphatic alcohols is explained by two primary factors:

1. Resonance Stabilization of the Phenoxide Ion: When phenol loses a proton ($\text{H}^{+}$), it forms the phenoxide ion ($\text{C}_6\text{H}_5\text{O}^{-}$). The negative charge on oxygen is conjugated with the $\pi$-electron system of the aromatic ring. This charge is delocalized over the ortho and para positions through resonance:
   $$ \text{C}_6\text{H}_5\text{O}^{-} \leftrightarrow \text{Resonance stabilization over aromatic ring} $$
   This makes the phenoxide ion exceptionally stable. Conversely, when an alcohol loses a proton, it forms an alkoxide ion ($\text{R-O}^{-}$). The negative charge remains localized on the oxygen atom because there is no resonance system to delocalize it, making the alkoxide ion unstable.
2. Inductive Effect: In phenol, the hydroxyl group ($\text{-OH}$) is attached to an $sp^2$-hybridized carbon atom of the benzene ring, which is highly electronegative and withdraws electron density from oxygen. In alcohols, the $\text{-OH}$ group is attached to an $sp^3$-hybridized carbon, which is less electronegative and pushes electron density toward oxygen via the $+I$ inductive effect, destabilizing the negative charge on the alkoxide oxygen.

Increasing Order of Acidity:

Scientific Explanation:

• Electron-Withdrawing Groups ($-\text{NO}_2$): The nitro group is a powerful electron-withdrawing group via both resonance ($-M$) and inductive ($-I$) effects. At the ortho position, it pulls electron density away from the phenoxide oxygen, stabilizing the negative charge and making **$o$-nitrophenol** the strongest acid.
• Unsubstituted Phenol: Serves as the baseline reference.
• Electron-Donating Methyl Group ($-\text{CH}_3$): In **$o$-cresol**, the methyl group is electron-donating via inductive effect ($+I$) and hyperconjugation. This increases the negative charge density on the oxygen atom, destabilizing the phenoxide ion and making it weaker than phenol.
• Methoxy Group ($-\text{OCH}_3$): In **$o$-methoxyphenol**, the methoxy group acts as a strong electron-donating group due to resonance ($+M$) when in the ortho position, destabilizing the phenoxide ion even further and making it the weakest acid of the group.

1. $o$-Nitrophenol (Intramolecular Hydrogen Bonding):

In $o$-nitrophenol, the nitro ($-\text{NO}_2$) and hydroxyl ($-\text{OH}$) groups are adjacent to each other. This proximity allows them to form **intramolecular hydrogen bonds** (within the same molecule):

Because the hydrogen bonding is internal, the molecules do not associate strongly with neighboring molecules. This keeps it volatile and easily vaporized by steam.

2. $p$-Nitrophenol (Intermolecular Hydrogen Bonding):

In $p$-nitrophenol, these groups are far apart at opposite ends of the ring, preventing internal bonding. Instead, the molecules form strong **intermolecular hydrogen bonds** with adjacent molecules, linking them in a large polymeric network. Because separating these associated molecules requires significant thermal energy, $p$-nitrophenol is non-volatile and has a much higher boiling point.

The acid-catalyzed hydration of ethene to yield ethanol proceeds via the following three main mechanistic steps:

Step 1: Protonation of ethene to form a carbocation.

Acid dissolves in water to form hydronium ions ($\text{H}_3\text{O}^{+}$). Ethene acts as a nucleophile, using its $\pi$-electrons to attack a proton from the hydronium ion, generating a highly reactive ethyl carbocation intermediate:

Step 2: Nucleophilic attack of water on the carbocation.

Water acts as a nucleophile, using one of oxygen's lone pairs to attack the electrophilic carbon of the carbocation, forming a protonated alcohol intermediate (ethyl oxonium ion):

Step 3: Deprotonation to form ethanol.

A water molecule abstracts a proton from the oxonium ion, regenerating the hydronium acid catalyst and yielding pure ethanol:

The elimination of water from ethanol in the presence of concentrated sulfuric acid ($\text{H}_2\text{SO}_4$) at $443\text{ K}$ proceeds via the following three-step mechanism:

Step 1: Protonation of ethanol to form ethyl oxonium ion.

The oxygen atom of the alcohol uses its lone pair of electrons to accept a proton from the acid catalyst, converting the poor leaving group ($\text{-OH}$) into an excellent leaving group ($\text{-H}_2\text{O}^{+}$):

Step 2: Formation of the carbocation.

The $C-O$ bond cleaves heterolytically, releasing a neutral water molecule and generating a reactive ethyl carbocation intermediate. This is the **slowest, rate-determining step** of the reaction:

Step 3: Elimination of a proton to form ethene.

A conjugate base (like water or hydrogen sulfate) abstracts a proton from the $\beta$-carbon, allowing the $\text{C-H}$ bonding electrons to form a $\pi$-bond and yield ethene, while regenerating the acid catalyst:

Kolbe's Reaction: Phenol is treated with sodium hydroxide ($\text{NaOH}$) to generate the highly reactive **sodium phenoxide ion**. This phenoxide ion is then treated with a weak electrophile, carbon dioxide ($\text{CO}_2$), at $400\text{ K}$ and $4\text{-}7\text{ atm}$ pressure, followed by acidification to produce $o$-hydroxybenzoic acid (**salicylic acid**) as the major product.

Chemical Equations:

Where the final product is **salicylic acid**.

Reimer-Tiemann Reaction: When phenol is treated with chloroform ($\text{CHCl}_3$) in the presence of aqueous sodium hydroxide ($\text{NaOH}$) at $340\text{ K}$, a formyl group ($\text{-CHO}$) is introduced at the ortho position of the ring. Acidification of this intermediate yields $o$-hydroxybenzaldehyde (**salicylaldehyde**).

Chemical Equations:

(i) Distillation with Zinc Dust (Reduction):

Phenol is reduced to benzene when heated with zinc dust. Zinc acts as a reducing agent, removing the oxygen atom to form zinc oxide:

(ii) Oxidation with Acidified Sodium Dichromate:

Phenol undergoes oxidation with sodium dichromate in the presence of sulfuric acid to form a conjugated diketone called **benzoquinone** (specifically $p$-benzoquinone):

Williamson's Ether Synthesis: This is a versatile laboratory method for preparing symmetrical and unsymmetrical ethers. It involves reacting an **alkyl halide** with a **sodium alkoxide**:

Reaction Mechanism:

The reaction proceeds via an **$S_N2$ mechanism**. The nucleophilic alkoxide ion ($\text{R'-O}^{-}$) attacks the electrophilic carbon of the primary alkyl halide from the backside in a single concerted step, displacing the halide leaving group.

Limitation (Why Tertiary Halides Fail):

Because the mechanism is strictly $S_N2$, the reaction is highly sensitive to steric hindrance. If a **tertiary ($3^\circ$) alkyl halide** is used (such as $\text{tert-butyl chloride}$), the bulky methyl groups block the backside attack of the alkoxide. Instead, the strongly basic alkoxide ion abstracts a $\beta$-proton, causing an **elimination ($E2$) reaction** to yield an **alkene** as the major product, rather than an ether:

The reaction of anisole with $\text{HI}$ proceeds through the following step-by-step mechanism:

Step 1: Protonation of Anisole.

Anisole acts as a base, using one of oxygen's lone pairs to accept a proton from the strong acid $\text{HI}$, forming a protonated ether (methylphenyl oxonium ion) intermediate:

Step 2: Nucleophilic Attack of Iodide Ion ($\text{I}^{-}$).

The iodide ion ($\text{I}^{-}$) acts as a nucleophile and must attack one of the carbon atoms attached to the oxonium oxygen to cleave the ether bond.

• Why the bond to the phenyl ring does not cleave: In protonated anisole, the bond between oxygen and the $sp^2$-hybridized phenyl carbon has partial double-bond character due to resonance. This makes the $\text{C(phenyl)-O}$ bond exceptionally strong and resistant to nucleophilic cleavage.
• Why the bond to the methyl group cleaves: The bond between oxygen and the $sp^3$-hybridized methyl carbon is a normal, weaker single covalent bond. Additionally, the methyl group is small and sterically unhindered, allowing the iodide ion to easily undergo an **$S_N2$ backside attack** on the methyl carbon, forming methyl iodide and releasing phenol:
  $$ \text{I}^{-} + \text{CH}_3\text{-}\text{O}^{+}(\text{H})\text{-C}_6\text{H}_5 \rightarrow \text{CH}_3\text{I} + \text{C}_6\text{H}_5\text{OH} \text{ (Phenol)} $$

Because the $\text{C(phenyl)-O}$ bond cannot be cleaved, iodobenzene and methanol are never formed.

(a) With Cold, Equimolar $\text{HI}$ (Partial Cleavage):

The reaction yields one molecule of **ethanol** and one molecule of **ethyl iodide**:

(b) With Hot, Excess $\text{HI}$ (Complete Cleavage):

The excess $\text{HI}$ reacts with the newly formed ethanol from step (a) to convert it into a second molecule of ethyl iodide, yielding **two molecules of ethyl iodide** and water:

(i) Propene to Propan-2-ol:

This conversion can be achieved in a single step using acid-catalyzed hydration. The addition of water across the double bond follows Markovnikov's rule:

(ii) Benzyl chloride to Benzyl alcohol:

This conversion involves nucleophilic substitution of the chloride leaving group. Because benzyl chloride is highly reactive, treatment with aqueous sodium hydroxide ($\text{NaOH}$) easily yields benzyl alcohol:

(i) Propan-1-ol (Primary Alcohol Synthesis):

To synthesize a primary alcohol, react a Grignard reagent with **formaldehyde ($\text{HCHO}$)**. To obtain the 3-carbon chain of propan-1-ol, react ethyl magnesium chloride ($\text{CH}_3\text{CH}_2\text{MgCl}$) with formaldehyde, followed by acid hydrolysis:

(ii) 2-Methylpropan-2-ol (Tertiary Alcohol Synthesis):

To synthesize a tertiary alcohol, react a Grignard reagent with a **ketone**. Reacting methyl magnesium bromide ($\text{CH}_3\text{MgBr}$) with **acetone ($\text{CH}_3\text{COCH}_3$)**, followed by acid hydrolysis, yields the desired product:

Correct Option: (c)

Explanation: Pyridinium chlorochromate ($\text{PCC}$) is a mild, selective oxidizing agent that oxidizes primary alcohols to aldehydes and secondary alcohols to ketones. Unlike strong oxidizing agents like $\text{KMnO}_4$ or $\text{K}_2\text{Cr}_2\text{O}_7$, which oxidize aldehydes further into carboxylic acids, $\text{PCC}$ stops the reaction at the aldehyde stage.

Correct Option: (a)

Explanation: For protonation to occur, the oxygen atom must donate a lone pair of electrons to an incoming proton ($\text{H}^{+}$). In phenol, the lone pairs on oxygen are in conjugation with the benzene ring, which delocalizes them into the ring through resonance. This decreases the electron density on oxygen, making the lone pair less available for donation. In aliphatic alcohols, the lone pairs are fully localized on oxygen, allowing them to protonate much more readily. Thus, both statements are true and the reason provides the correct explanation.