Theoretical Foundation: Williamson's Synthesis

Williamson's synthesis is a nucleophilic substitution reaction ($S_N2$) involving the reaction of an alkoxide ion ($R-O^-$) with a primary alkyl halide ($R'-X$). To prepare an asymmetric ether like 1-methoxyethane ($CH_3-O-CH_2CH_3$), one must select the reactants such that the alkyl halide is primary to minimize steric hindrance and maximize the reaction rate [Per $S_N2$ mechanism kinetics].

Step 1: Strategic Selection of Reactants

To synthesize 1-methoxyethane ($CH_3-O-CH_2CH_3$), we consider two possible pathways:

1. Sodium methoxide ($CH_3ONa$) + Chloroethane ($CH_3CH_2Cl$)
2. Sodium ethoxide ($CH_3CH_2ONa$) + Chloromethane ($CH_3Cl$)

Both pathways are viable; however, the use of a primary halide ($CH_3CH_2Cl$ or $CH_3Cl$) is mandatory to avoid elimination ($E2$) side reactions that occur with secondary or tertiary halides.

Step 2: Chemical Equations

Using the first pathway, the reaction proceeds as follows:

$CH_3ONa + CH_3CH_2Cl \rightarrow CH_3-O-CH_2CH_3 + NaCl$

Using the second pathway:

$CH_3CH_2ONa + CH_3Cl \rightarrow CH_3-O-CH_2CH_3 + NaCl$

Step 3: Visual Representation of the Mechanism

Analytical Justification

The choice of sodium methoxide and chloroethane is preferred in industrial applications due to the high reactivity of primary alkyl halides towards the $S_N2$ attack by the methoxide nucleophile [By $S_N2$ Transition State Geometry].

Final Solution: The reagents required for the Williamson synthesis of 1-Methoxyethane are Sodium methoxide ($CH_3ONa$) and Chloroethane ($CH_3CH_2Cl$). The reaction equation is: $CH_3ONa + CH_3CH_2Cl \rightarrow CH_3OCH_2CH_3 + NaCl$.

Step 1: Chemical Reagents and Substrate Analysis

The substrate is Anisole (methoxybenzene), represented by the formula $C_6H_5OCH_3$. The methoxy group ($-OCH_3$) is a strongly activating group and an ortho/para-directing substituent due to the resonance effect [Per the +R effect of the $-OCH_3$ group]. The reagents for Friedel-Crafts acetylation are Acetyl chloride ($CH_3COCl$) and an anhydrous Lewis acid catalyst, typically Aluminum chloride ($AlCl_3$).

Step 2: Generation of the Electrophile

The reaction initiates with the interaction between the acetyl chloride and the Lewis acid catalyst to generate the electrophilic acylium ion:

$CH_3COCl + AlCl_3 \rightarrow CH_3C^+O + [AlCl_4]^-$

[Justification: The Lewis acid accepts a lone pair from the chlorine atom, weakening the $C-Cl$ bond and facilitating the formation of the resonance-stabilized acylium ion, $CH_3-C \equiv O^+ \leftrightarrow CH_3-C^+=O$.]

Step 3: Electrophilic Aromatic Substitution Mechanism

The acylium ion attacks the electron-rich aromatic ring at the ortho and para positions. Because of steric hindrance at the ortho position, the para-isomer is the major product.

Step 4: Balanced Chemical Equation

The overall reaction is represented as follows:

$C_6H_5OCH_3 + CH_3COCl \xrightarrow{Anhydrous AlCl_3} CH_3OC_6H_4COCH_3 + HCl$

Final Solution: The Friedel-Crafts acetylation of anisole yields 4-methoxyacetophenone as the major product and 2-methoxyacetophenone as the minor product, accompanied by the evolution of Hydrogen Chloride gas.

Step 1: Chemical Structure and Substituent Analysis

To determine the relative acidity of ortho-nitrophenol and ortho-methoxyphenol, we must analyze the electronic effects of the substituent groups ($-\text{NO}_2$ and $-\text{OCH}_3$) positioned at the ortho position relative to the phenolic hydroxyl ($-OH$) group.

• ortho-Nitrophenol: The $-\text{NO}_2$ group exerts both a strong negative inductive effect ($-I$) and a strong negative mesomeric/resonance effect ($-M$).
• ortho-Methoxyphenol: The $-\text{OCH}_3$ group exerts a negative inductive effect ($-I$) but a strong positive mesomeric/resonance effect ($+M$) due to the presence of lone pairs on the oxygen atom.

Step 2: Stability of the Conjugate Base (Phenoxide Ion)

Acidity is directly proportional to the stability of the conjugate base (phenoxide ion) formed after the release of a proton ($H^+$). Stability is achieved through the delocalization and dispersal of the negative charge on the phenoxide oxygen [Per Brønsted-Lowry Theory].

Step 3: Visual Representation of Substituent Effects

Step 4: Evaluating the Acidity Differential

The acidity of a phenol is defined by the equilibrium constant $K_a$ of the dissociation reaction: $C_6H_4(X)OH \rightleftharpoons C_6H_4(X)O^- + H^+$.

• In ortho-nitrophenol, the $-M$ effect of the nitro group is highly effective at withdrawing electron density from the benzene ring and the oxygen atom of the phenoxide ion, significantly increasing the acidity [By Resonance Theory].
• In ortho-methoxyphenol, the lone pair on the oxygen of the methoxy group is donated into the ring ($+M$ effect), which increases the electron density on the oxygen of the phenoxide ion, thereby destabilizing it and decreasing the acidity [By Electronic Displacement Effects].

Final Solution: Ortho-nitrophenol is more acidic than ortho-methoxyphenol because the $-\text{NO}_2$ group provides intense electronic stabilization to the conjugate base through its $-I$ and $-M$ effects, whereas the $-\text{OCH}_3$ group destabilizes the conjugate base through its $+M$ electron-donating effect.

1. Theoretical Foundation: Intermolecular Forces

The solubility of a substance in water is primarily governed by the principle of "like dissolves like." Solubility is determined by the ability of a solute to form intermolecular forces with the solvent that are strong enough to overcome the solute-solute and solvent-solvent attractions. [Per the Solvation Principle].

2. Comparative Analysis of Molecular Interactions

To analyze the solubility difference between alcohols ($R-OH$) and hydrocarbons ($R-H$), we must evaluate the nature of their interactions with water molecules ($H_2O$):

• Alcohols: The hydroxyl ($-OH$) group is highly polar due to the large electronegativity difference between oxygen ($\approx 3.44$) and hydrogen ($\approx 2.20$). Consequently, alcohols can form intermolecular hydrogen bonds with water molecules.
• Hydrocarbons: These are non-polar compounds. They lack polar functional groups and are incapable of forming hydrogen bonds. They interact with water only through weak London Dispersion Forces, which are insufficient to disrupt the strong Hydrogen-Bonding network of water. [Per the Gibbs Free Energy Equation $\Delta G = \Delta H - T\Delta S$].

3. Visualizing the Hydrogen Bonding Interaction

4. Step-by-Step Mechanistic Explanation

Step 1: Disruption of Solvent Lattice: Water molecules are held together by an extensive network of hydrogen bonds. To dissolve a solute, this network must be partially broken, which requires energy.

Step 2: Formation of Solute-Solvent Bonds: In the case of alcohols, the energy required to break the water-water hydrogen bonds is compensated by the energy released upon the formation of new alcohol-water hydrogen bonds.

Step 3: Entropy Considerations: Hydrocarbons create a "caged" structure (clathrate) around themselves in water to minimize surface area contact, which leads to a significant decrease in entropy ($\Delta S < 0$), making the process thermodynamically non-spontaneous. Alcohols do not impose this restriction as they integrate into the hydrogen-bonded network.

Final Solution: Alcohols are more soluble in water because they possess a polar hydroxyl group capable of forming intermolecular hydrogen bonds with water molecules, whereas hydrocarbons are non-polar and lack the capacity to form such bonds, making them hydrophobic.

Step 1: Decomposition of the IUPAC Name

To determine the chemical structure of 1-Phenylpropan-2-ol, we decompose the nomenclature into its structural components based on IUPAC conventions [Per IUPAC Gold Book nomenclature rules]:

• Parent Chain (Root): "Propan-" denotes a three-carbon saturated alkane backbone ($C-C-C$).
• Principal Functional Group: "-2-ol" indicates a hydroxyl group ($-OH$) attached to the second carbon atom of the propane chain.
• Substituent: "1-Phenyl-" indicates a phenyl group ($C_6H_5-$) attached to the first carbon atom of the propane chain.

Step 2: Skeletal Construction

Following the rules for structural representation, we arrange the atoms as follows:

1. Draw a three-carbon chain: $C_1-C_2-C_3$.
2. Attach the phenyl group to $C_1$: $C_6H_5-CH_2-CH-CH_3$.
3. Attach the hydroxyl group to $C_2$: $C_6H_5-CH_2-CH(OH)-CH_3$.

Step 3: Visual Representation

Step 4: Final Condensed Structural Formula

By balancing the valency of carbon (tetravalency) with hydrogen atoms, the complete structure is derived as:

$C_6H_5-CH_2-CH(OH)-CH_3$

Breakdown of valencies:

Final Solution: The structure of 1-Phenylpropan-2-ol is $C_6H_5CH_2CH(OH)CH_3$.

Theoretical Basis of Alcohol Oxidation

The transformation of a primary alcohol ($R-CH_2OH$) to a carboxylic acid ($R-COOH$) is a two-step oxidation process. The primary alcohol is first oxidized to an aldehyde ($R-CHO$), which is then further oxidized to the corresponding carboxylic acid. This reaction requires strong oxidizing agents capable of providing atomic oxygen under acidic conditions.

Required Reagents

To achieve the complete oxidation of a primary alcohol to a carboxylic acid, one of the following strong oxidizing agents is utilized:

• Acidified Potassium Permanganate ($KMnO_4 / H^+$): A powerful oxidizing agent.
• Acidified Potassium Dichromate ($K_2Cr_2O_7 / H^+$): Commonly used in laboratory settings for this specific conversion. [Per the Oxidation of Alcohols mechanism].

Reaction Stoichiometry

The reaction proceeds according to the following general chemical equation:

$R-CH_2OH + 2[O] \xrightarrow{KMnO_4/H^+ \text{ or } K_2Cr_2O_7/H^+} R-COOH + H_2O$

Visual Representation of the Oxidation Pathway

Analytical Justification

Final Solution: The reagents used for the oxidation of a primary alcohol to a carboxylic acid are acidified potassium permanganate ($KMnO_4 / H^+$) or acidified potassium dichromate ($K_2Cr_2O_7 / H^+$).

Given Molecular Structure

The provided chemical structure is: $CH_3-CH(CH_3)-CH_2-OH$

Step 1: Identification of the Principal Functional Group

The presence of the hydroxyl group ($-OH$) dictates that the compound is an alcohol. The suffix for an alcohol in IUPAC nomenclature is -ol. [Per IUPAC Nomenclature Rules for Alcohols].

Step 2: Determining the Longest Carbon Chain

Identify the longest continuous carbon chain containing the $-OH$ group.

• Chain structure: $C^3-C^2-C^1$
• $C^1$ is bonded to the $-OH$ group.
• The chain consists of 3 carbon atoms. The parent alkane is propane.

Step 3: Numbering the Chain

Number the chain starting from the end that gives the lowest locant to the functional group:

• $C^1$: $CH_2-OH$
• $C^2$: $CH-CH_3$ (This carbon carries a methyl substituent)
• $C^3$: $CH_3$

Step 4: Identifying and Locating Substituents

A methyl group ($-CH_3$) is attached to the second carbon atom ($C^2$). This is designated as a 2-methyl substituent.

Step 5: Constructing the IUPAC Name

The name is constructed by combining the substituent, the parent chain, and the functional group suffix:
Structure: [Substituent Locant]-[Substituent Name][Parent Alkane]-[Functional Group Locant]-[Suffix]
Result: 2-methyl + propane + 1 + ol = 2-Methylpropan-1-ol

Visual Representation of Structure

Final Solution: 2-Methylpropan-1-ol

1. Theoretical Foundation of Williamson’s Synthesis

Williamson’s ether synthesis involves the nucleophilic substitution ($S_N2$) of an alkyl halide by an alkoxide or phenoxide ion. The general reaction is represented as: $R-O^-Na^+ + R'-X \rightarrow R-O-R' + NaX$. For the synthesis of an aromatic ether like Ethoxybenzene (phenetole), we must select reagents such that the nucleophile is the phenoxide ion ($C_6H_5O^-$) and the electrophile is an alkyl halide (ethyl halide), as the aryl halide ($C_6H_5-X$) does not undergo $S_N2$ reactions due to partial double bond character of the $C-X$ bond and the instability of the phenyl cation [Ref: Nucleophilic Aromatic Substitution limitations].

2. Selected Reagents

• Phenol ($C_6H_5OH$): Reacted with $NaOH$ to generate the nucleophile.
• Sodium Hydroxide ($NaOH$): Used to deprotonate phenol to form Sodium Phenoxide ($C_6H_5O^-Na^+$).
• Ethyl Bromide ($CH_3CH_2Br$): The alkyl halide electrophile ($S_N2$ substrate).

3. Step-by-Step Reaction Mechanism

Step 1: Formation of Sodium Phenoxide
$C_6H_5OH + NaOH \rightarrow C_6H_5O^-Na^+ + H_2O$

Step 2: Nucleophilic Attack ($S_N2$)
The phenoxide ion attacks the primary carbon of the ethyl bromide, displacing the bromide ion.

4. Visual Representation of the Synthesis

5. Summary of Reaction Conditions

Final Solution: The preparation of Ethoxybenzene via Williamson’s synthesis is achieved by reacting Sodium Phenoxide with Ethyl Bromide (or Ethyl Iodide). The balanced chemical equation is: $C_6H_5ONa + C_2H_5Br \rightarrow C_6H_5OC_2H_5 + NaBr$.

Step 1: Chemical Reaction Context & Theoretical Foundation

The reaction of phenol with chloroform ($CHCl_3$) in the presence of an aqueous base (typically $NaOH$) is known as the Reimer-Tiemann Reaction. This reaction results in the introduction of a formyl group ($-CHO$) at the ortho position of the benzene ring, leading to the formation of o-hydroxybenzaldehyde, commonly known as salicylaldehyde. [Per the mechanism of electrophilic aromatic substitution involving a carbene intermediate].

Step 2: Reaction Mechanism Overview

• Generation of Dichlorocarbene: The aqueous $NaOH$ abstracts a proton from $CHCl_3$ to form a trichloromethyl carbanion, which subsequently eliminates a chloride ion to generate the electrophilic species: dichlorocarbene ($:CCl_2$).
• Electrophilic Attack: The phenoxide ion (formed by the reaction of phenol with $NaOH$) attacks the $:CCl_2$ species.
• Hydrolysis: Subsequent hydrolysis and acidification yield the final salicylaldehyde product.

Step 3: Chemical Equation

The overall stoichiometric transformation is represented as follows:

$C_6H_5OH + CHCl_3 + 3NaOH_{(aq)} \rightarrow C_6H_4(OH)(CHO) + 3NaCl + 2H_2O$

Step 4: Visual Representation of the Chemical Transformation

Step 5: Step-by-Step Reaction Summary

Final Solution: The treatment of phenol with chloroform in the presence of aqueous $NaOH$ follows the Reimer-Tiemann reaction path to produce salicylaldehyde (o-hydroxybenzaldehyde) as the major product.

Reaction 1: Reaction with Active Metals

Phenol reacts with reactive metals such as sodium to form sodium phenoxide and hydrogen gas. This reaction is a characteristic property of acidic hydrogen atoms [Per the Brønsted-Lowry acid-base theory].

$2C_6H_5OH + 2Na \rightarrow 2C_6H_5ONa + H_2(g)$

Reaction 2: Reaction with Strong Bases

Phenol reacts with aqueous sodium hydroxide to form sodium phenoxide and water. The formation of a salt from an acid and a base is a fundamental neutralization reaction [Per the Arrhenius acid-base definition].

$C_6H_5OH + NaOH(aq) \rightarrow C_6H_5ONa + H_2O$

Comparative Analysis: Phenol vs. Ethanol

The acidity of these compounds is determined by the stability of their respective conjugate bases (the alkoxide or phenoxide ion) formed after the release of a proton ($H^+$).

Conclusion on Acidity

The phenoxide ion is significantly more stable than the ethoxide ion. Consequently, the equilibrium for the dissociation of phenol ($C_6H_5OH \rightleftharpoons C_6H_5O^- + H^+$) lies further to the right compared to the dissociation of ethanol. Therefore, phenol is a stronger acid than ethanol.

Final Solution: Phenol exhibits acidity through reaction with sodium metal and sodium hydroxide. It is more acidic than ethanol due to the resonance stabilization of the phenoxide ion compared to the destabilizing electron-donating inductive effect in the ethoxide ion.

Given Reactants & Theoretical Foundation

The reaction involves the cleavage of benzyl ethyl ether ($C_6H_5CH_2OCH_2CH_3$) with concentrated Hydrogen Iodide ($HI$).

The cleavage of unsymmetrical ethers follows an $S_N1$ or $S_N2$ mechanism based on the stability of the resulting carbocation. In this specific case, the benzyl group provides a highly stable resonance-stabilized carbocation intermediate: the benzyl cation ($C_6H_5CH_2^+$).

Step 1: Protonation of the Ether Oxygen

The ether oxygen possesses lone pairs, which act as a Lewis base. The hydrogen ion ($H^+$) from the dissociation of $HI$ attacks the oxygen atom to form an alkyloxonium ion.

Reaction: $C_6H_5CH_2-O-CH_2CH_3 + HI \rightarrow [C_6H_5CH_2-O^+(H)-CH_2CH_3] + I^-$

[Per the Brønsted-Lowry acid-base theory, the oxygen acts as the proton acceptor.]

Step 2: Nucleophilic Attack and Cleavage

Because the benzyl carbon ($C_6H_5CH_2-$) can form a resonance-stabilized carbocation, the $C-O$ bond on the benzyl side undergoes cleavage via an $S_N1$ mechanism. The iodide ion ($I^-$) acts as a nucleophile.

• The $C-O$ bond breaks, releasing ethanol ($CH_3CH_2OH$) and a benzyl carbocation ($C_6H_5CH_2^+$).
• The $I^-$ ion rapidly attacks the benzyl carbocation to form benzyl iodide ($C_6H_5CH_2I$).

[Justification: The benzyl carbocation is stabilized by delocalization of the positive charge into the benzene ring, facilitating the heterolytic cleavage of the bond adjacent to the phenyl group.]

Visual Representation of Mechanism

Chemical Equation

The balanced chemical equation for the reaction is:

$C_6H_5CH_2-O-CH_2CH_3 + HI \xrightarrow{\Delta} C_6H_5CH_2I + CH_3CH_2OH$

Final Solution: The reaction of benzyl ethyl ether with hydrogen iodide yields benzyl iodide ($C_6H_5CH_2I$) and ethanol ($CH_3CH_2OH$).

1. Analysis of Molecular Structure: $CH_3-CH(CH_3)-CH_2-OH$

The structure represents a branched primary alcohol. To determine the IUPAC name, we follow the IUPAC nomenclature rules for Alkanols:

• Identification of the Principal Functional Group: The hydroxyl group ($-OH$) dictates that the compound is an alcohol. The suffix is -ol.
• Selection of the Principal Chain: The longest carbon chain containing the carbon atom attached to the $-OH$ group consists of 3 carbon atoms. This is the propane derivative.
• Numbering the Chain: Numbering starts from the carbon attached to the $-OH$ group to assign the lowest possible locant: $C1$ ($CH_2-OH$), $C2$ ($CH$), and $C3$ ($CH_3$).
• Substituent Identification: There is a methyl group ($-CH_3$) attached to $C2$.

2. Systematic IUPAC Formulation

Based on the structure, we identify:

[Per the IUPAC rules for nomenclature of branched alcohols, the locant for the hydroxyl group takes precedence over alkyl substituents.]

3. Visual Representation

4. Final Assembly of IUPAC Name

Combining the substituent locant and the parent chain name, we obtain:

$2-methylpropane-1-ol$

Final Solution: 2-Methylpropan-1-ol

Step 1: Theoretical Foundation of Cumene Oxidation

The synthesis of phenol from cumene (isopropylbenzene) is an industrial process known as the Cumene Hydroperoxide Process. It involves a two-stage reaction sequence: the air oxidation of cumene to cumene hydroperoxide, followed by an acid-catalyzed cleavage (rearrangement) to yield phenol and acetone as by-products. [Per the mechanism of Baeyer-Villiger type rearrangement].

Step 2: Stage I — Aerobic Oxidation of Cumene

Cumene ($C_6H_5-CH(CH_3)_2$) is subjected to oxidation using atmospheric oxygen at elevated temperatures in the presence of a radical initiator to form cumene hydroperoxide.

$C_6H_5-CH(CH_3)_2 + O_2 \xrightarrow{400\text{ K}} C_6H_5-C(CH_3)_2-OOH$

[Justification: The formation of a stable tertiary benzylic radical facilitates the propagation of this free-radical oxidation mechanism.]

Step 3: Stage II — Acid-Catalyzed Rearrangement

Cumene hydroperoxide is treated with dilute acid ($H_2SO_4$) to undergo protonation, followed by a 1,2-aryl shift from the carbon atom to the oxygen atom, resulting in the cleavage of the $O-O$ bond.

$C_6H_5-C(CH_3)_2-OOH \xrightarrow{H^+} C_6H_5OH + CH_3COCH_3$

Visual Representation of the Reaction Pathway

Step 4: Analytical Summary of Reagents

Final Solution: The preparation of phenol from cumene involves the formation of cumene hydroperoxide via aerobic oxidation, followed by acid-catalyzed cleavage to yield Phenol ($C_6H_5OH$) and Acetone ($CH_3COCH_3$).

Step 1: Structural Analysis and Identification of Substituents

The given ether molecule exhibits an aromatic ring attached to an oxygen atom, which is in turn linked to an alkyl group. The structure provided is: $CH_3-CH_2-CH_2-O-C_6H_5$.

According to IUPAC nomenclature for ethers, the group with the larger number of carbon atoms (or the parent aromatic structure) is treated as the parent alkane or arene, while the smaller group with the oxygen atom is treated as an alkoxy substituent [Per IUPAC Nomenclature of Organic Chemistry, Rule P-61.2.1].

• Parent Structure: The benzene ring ($C_6H_5-$) serves as the primary base for the name.
• Substituent: The chain $-O-CH_2-CH_2-CH_3$ represents a propoxy group ($CH_3CH_2CH_2O-$).

Step 2: Assigning the IUPAC Name

Combining the components identified in Step 1, we name the ether by placing the alkoxy prefix before the parent arene name.

The name is generated by concatenating "Propoxy" and "Benzene" [By convention of alkoxyarene naming].

Step 3: Visual Representation of the Molecule

Step 4: Formal Verification

Applying the rules for systematic nomenclature:

1. The parent alkane chain or aromatic ring is selected based on priority. Here, the benzene ring is the priority parent [Per IUPAC P-61.2.1].
2. The alkyl group attached to oxygen forming the ether linkage is named as an alkoxy group. Since the chain has three carbons, it is "propoxy".
3. Placement of "propoxy" before "benzene" results in "Propoxybenzene".

Final Solution: Propoxybenzene

Step 1: Analyzing the IUPAC Nomenclature

To determine the structure of 3-Cyclohexylpentan-3-ol, we decompose the name into its constituent parts:

• Parent Chain ("pentan-3-ol"): The suffix "-ol" indicates an alcohol functional group, and "pentan-" indicates a primary alkane chain of 5 carbon atoms. The "3-" denotes the position of the hydroxyl ($-OH$) group on the third carbon.
• Substituent ("3-Cyclohexyl"): A cyclohexyl group ($C_6H_{11}$) is attached to the third carbon of the pentane chain.

Step 2: Constructing the Carbon Skeleton

Following the rules of IUPAC nomenclature [Per IUPAC Gold Book guidelines for organic structure representation], we establish the backbone:

1. Draw a chain of 5 carbons: $C_1-C_2-C_3-C_4-C_5$.
2. Place the hydroxyl group ($-OH$) on $C_3$.
3. Attach the cyclohexyl ring to $C_3$.

Step 3: Structural Visualization

The structure consists of a pentane backbone with a tertiary alcohol center at carbon-3, which also anchors the cyclohexane ring.

Step 4: Chemical Formula Derivation

The carbon-3 is quaternary in its substitution (bonded to two ethyl groups, one hydroxyl group, and one cyclohexyl group). The complete molecular formula is derived as follows:

• Pentane chain base: $C_5H_{12}$
• Cyclohexyl substituent: $C_6H_{11}$
• Minus 2H (one for $OH$ replacement, one for alkyl linkage): $C_{11}H_{21}OH$

The structural formula can be represented as: $CH_3CH_2C(OH)(C_6H_{11})CH_2CH_3$

Final Solution: The structure of 3-Cyclohexylpentan-3-ol is a pentane chain with an $-OH$ group and a $C_6H_{11}$ ring both bonded to the central carbon (C3): $CH_3-CH_2-C(OH)(C_6H_{11})-CH_2-CH_3$.

1. Theoretical Foundation: Electrophilic Aromatic Substitution

The Friedel-Crafts acetylation of anisole ($C_6H_5OCH_3$) is an electrophilic aromatic substitution reaction. Anisole features a methoxy group ($-OCH_3$), which is a strong activating group. Due to the resonance effect ($+R$ effect) of the oxygen lone pairs, electron density is increased specifically at the ortho and para positions of the benzene ring. [Per the Resonance Theory of Electrophilic Substitution].

2. Reaction Components

• Substrate: Anisole ($C_6H_5OCH_3$)
• Reagent: Acetyl chloride ($CH_3COCl$)
• Catalyst: Anhydrous Aluminium Chloride ($AlCl_3$), a Lewis acid.

3. Step-by-Step Reaction Mechanism

Step A: Generation of the Electrophile
The Lewis acid $AlCl_3$ reacts with acetyl chloride to generate the acylium ion (the electrophile):
$CH_3COCl + AlCl_3 \rightarrow [CH_3CO]^+ + [AlCl_4]^-$

Step B: Electrophilic Attack
The $[CH_3CO]^+$ ion attacks the electron-rich ortho or para positions of the anisole ring. The para isomer is major due to steric hindrance at the ortho position.

Step C: Rearomatization
Loss of a proton restores the aromaticity of the ring, forming 4-methoxyacetophenone.

4. Chemical Equation

The reaction is represented as follows:
$C_6H_5OCH_3 + CH_3COCl \xrightarrow{Anhydrous \ AlCl_3} CH_3OC_6H_4COCH_3 + HCl$

5. Visual Representation of Reaction

Final Solution: The Friedel-Crafts acetylation of anisole yields 2-methoxyacetophenone (minor) and 4-methoxyacetophenone (major) along with HCl as a byproduct.

Theoretical Foundation of Williamson Synthesis

The Williamson synthesis is a reaction of an alkyl halide ($R-X$) with a sodium alkoxide ($R'-ONa$) to yield an ether ($R-O-R'$). Mechanistically, it proceeds via an $S_N2$ substitution pathway, where the alkoxide ion acts as a strong nucleophile attacking the electrophilic carbon of the alkyl halide [Per the $S_N2$ Reaction Mechanism].

Limitation 1: Dominance of Elimination Reactions with Tertiary Alkyl Halides

Williamson synthesis requires the alkyl halide to undergo $S_N2$ substitution. Primary alkyl halides are most reactive; however, tertiary alkyl halides are sterically hindered and predominantly undergo dehydrohalogenation (elimination) rather than nucleophilic substitution. Under basic conditions, the alkoxide functions as a strong base, abstracting a proton to form an alkene.

Example: Reaction of $tert$-butyl bromide with sodium ethoxide:

$(CH_3)_3C-Br + CH_3CH_2ONa \rightarrow (CH_3)_2C=CH_2 + CH_3CH_2OH + NaBr$

Outcome: Formation of isobutylene (alkene) instead of the desired $tert$-butyl ethyl ether. [Per the competition between $S_N2$ and $E2$ reaction pathways].

Limitation 2: Reactivity of Aryl Halides

Aryl halides do not undergo nucleophilic substitution under standard Williamson conditions. The $C-X$ bond in aryl halides possesses partial double-bond character due to resonance and the $sp^2$ hybridization of the carbon atom, making the carbon less electrophilic and the bond difficult to break [Per Resonance Stabilization Theory].

Example: Reaction of chlorobenzene with sodium methoxide:

$C_6H_5Cl + CH_3ONa \rightarrow \text{No Reaction}$

Visual Representation of Steric Hindrance

Summary Comparison Table

Final Solution: The Williamson synthesis is limited by the propensity of tertiary alkyl halides to undergo $E2$ elimination rather than $S_N2$ substitution, and the complete inertness of aryl halides toward nucleophilic substitution due to $C-X$ bond resonance stabilization.

Step 1: Decomposition of the IUPAC Name

To determine the structure of 1-phenylpropan-2-ol, we analyze the nomenclature based on IUPAC guidelines:

• Parent Chain: "propan" indicates a carbon skeleton of $3$ carbons ($C_1-C_2-C_3$). [Per IUPAC Nomenclature of Alkanes]
• Principal Functional Group: "-2-ol" indicates a hydroxyl group ($-OH$) attached to the second carbon atom. [Per Priority Ranking of Functional Groups]
• Substituent: "1-phenyl" indicates a phenyl group ($C_6H_5-$) attached to the first carbon atom of the propane chain. [Per Substituent Naming Conventions]

Step 2: Constructing the Carbon Skeleton

We draw a three-carbon chain and number them from left to right to satisfy the substituent positions:

$C_1 - C_2 - C_3$

Applying the components:

• At $C_2$, place the $-OH$ group.
• At $C_1$, place the $C_6H_5$ (Phenyl) ring.

Step 3: Calculating Hydrogen Saturation

Apply the tetravalency of carbon to determine hydrogen distribution:

Step 4: Structural Representation

The complete structural formula is $C_6H_5-CH_2-CH(OH)-CH_3$.

Final Solution: The structure of 1-phenylpropan-2-ol is $C_6H_5-CH_2-CH(OH)-CH_3$.

Reaction Mechanism: Acid-Catalyzed Dehydration of Ethanol

The conversion of ethanol ($CH_3CH_2OH$) to ethene ($CH_2=CH_2$) in the presence of concentrated sulfuric acid ($H_2SO_4$) at $443\,K$ is an elimination reaction ($E1$ mechanism). The reaction proceeds through three distinct stages.

Step 1: Protonation of Ethanol

The oxygen atom of the ethanol molecule possesses two lone pairs of electrons. In the acidic medium, the ethanol molecule acts as a Bronsted-Lowry base, accepting a proton ($H^+$) from the sulfuric acid to form an alkyloxonium ion. [Per the Lewis Acid-Base theory, the lone pair on Oxygen attacks the electron-deficient proton].

$CH_3-CH_2-\ddot{O}H + H^+ \rightleftharpoons CH_3-CH_2-\overset{+}{O}H_2$

Step 2: Formation of Carbocation (Rate-Determining Step)

The protonated ethanol molecule (ethyloxonium ion) loses a water molecule to form a primary carbocation. This is the slowest step of the mechanism. [By the Arrhenius activation energy principle, breaking the $C-O$ bond is endothermic].

$CH_3-CH_2-\overset{+}{O}H_2 \xrightarrow{slow} CH_3-\overset{+}{C}H_2 + H_2O$

Step 3: Deprotonation to Form Ethene

The primary carbocation loses a proton from the $\beta$-carbon atom to a water molecule (acting as a base), resulting in the formation of a double bond between the two carbon atoms. This restores the catalyst and produces the alkene.

$CH_3-\overset{+}{C}H_2 \xrightarrow{fast} CH_2=CH_2 + H^+$

Summary of Thermodynamic Conditions

Final Solution: The mechanism proceeds via protonation of the hydroxyl group to generate a good leaving group ($H_2O$), followed by unimolecular dissociation to a carbocation, and final deprotonation to yield $CH_2=CH_2$ and regenerated $H^+$.