Step 1: Structural Expansion and Functional Group Identification

To determine the IUPAC name of the compound with the condensed formula $Cl_2CHCH_2OH$, we first expand the structure to identify the carbon backbone and the principal functional group:

The structure is: $Cl - \overset{\phantom{Cl}}{CH} (Cl) - CH_2 - OH$

• Principal Functional Group: The hydroxyl group ($-OH$) is present, which assigns the suffix -ol to the parent alkane [Per IUPAC nomenclature rules for alcohols].
• Substituents: There are two chlorine atoms ($-Cl$) attached to the first carbon atom of the chain. These are identified as chloro substituents.
• Carbon Chain: The longest carbon chain containing the hydroxyl group consists of two carbon atoms, which corresponds to the parent alkane ethane.

Step 2: Numbering the Carbon Chain

According to the IUPAC priority rules, the principal functional group ($-OH$) must be assigned the lowest possible locant [Per IUPAC Blue Book P-14.4].

Thus, the carbon attached to the hydroxyl group is designated as C-1. Consequently, the carbon attached to the two chlorine atoms is designated as C-2.

Step 3: Constructing the IUPAC Name

We assemble the components following the format: [Locants of substituents]-[Prefixes]-[Parent Alkane]-[Locant of functional group]-[Suffix].

• Substituent locants and names: Both chlorine atoms are on carbon 2, denoted as 2,2-dichloro.
• Parent chain: Ethane (retaining the 'e' only if the suffix starts with a consonant, but here it starts with 'o', so we use 'ethan').
• Suffix: Since the $-OH$ is on carbon 1, we add -1-ol.

Step 4: Final Nomenclature Synthesis

Combining the elements: 2,2-dichloro + ethan + 1 + ol.

Result: 2,2-dichloroethanol

Final Solution: 2,2-dichloroethanol

1. Theoretical Principle of Distillation

Distillation is a fundamental physical separation technique employed to purify liquids or to separate components of a liquid mixture based on differences in their boiling points. The process involves heating a liquid mixture until it vaporizes, followed by the immediate cooling and condensation of the vapors back into a liquid state in a separate vessel. [Per Dalton’s Law of Partial Pressures and Raoult’s Law, the composition of the vapor phase differs from that of the liquid phase if the components have distinct vapor pressures at a given temperature].

2. Mechanism and Apparatus

The separation relies on the fact that the liquid with the lower boiling point (more volatile) transitions into the vapor state at a lower temperature than the liquid with the higher boiling point. By carefully controlling the temperature, the volatile component is collected as a distillate, while the non-volatile impurities or the less volatile component remain in the original flask.

3. Exemplary Case: Purification of Aniline and Water

Consider a mixture of a liquid organic compound and a non-volatile impurity (e.g., Aniline mixed with salt/non-volatile contaminants).

• Step A: Vaporization: The crude mixture is placed in a round-bottomed flask and heated. As the temperature reaches the boiling point of the organic component (Aniline, $BP \approx 184^\circ C$), it transforms into a gaseous state.
• Step B: Condensation: The hot vapors enter the water-cooled condenser. The cold jacket of the condenser absorbs latent heat, facilitating the phase change from vapor to liquid.
• Step C: Collection: The purified liquid (distillate) is collected in a separate receiver flask, effectively separating it from the non-volatile residue remaining in the distillation flask.

4. Summary of Conditions

Final Solution: Distillation is a technique used to separate a volatile liquid from non-volatile impurities or to separate two or more liquids with significantly different boiling points by the process of differential vaporization and subsequent condensation.

Step 1: Theoretical Foundation of Resonance in Phenol

Phenol ($C_6H_5OH$) exhibits resonance due to the interaction between the lone pair of electrons on the oxygen atom of the hydroxyl ($-OH$) group and the $\pi$-electron system of the benzene ring. This phenomenon is governed by the +R (Resonance) effect, where the $-OH$ group acts as an electron-donating group. [Per the principles of Conjugated Systems and Delocalization].

Step 2: Electron Displacement Mechanics

The oxygen atom possesses two lone pairs. One lone pair participates in forming a $\pi$-bond with the adjacent carbon of the aromatic ring, triggering the migration of existing $\pi$-bonds across the ring structure. This creates a series of canonical structures (resonance contributors) where the ortho and para positions of the ring develop partial negative charge density.

Step 3: Analytical Derivation of Resonance Contributors

Step 4: Summary of Electron Migration

• First Transition: A lone pair from Oxygen forms a double bond with the aromatic carbon, pushing the ortho-double bond onto the ortho-carbon atom as a lone pair [Resonance hybridization rule].
• Subsequent Transitions: The negative charge at the ortho position shifts to the para position, and eventually to the other ortho position via conjugation, maintaining the double bond between the Oxygen and the ring carbon.
• Final Configuration: The hybrid structure shows partial double bond character between $C-O$ and partial negative charges distributed across the $ortho$ and $para$ positions of the benzene ring.

Final Solution: The resonance structures for Phenol involve the delocalization of the oxygen lone pair into the ring, resulting in five contributing structures (I-V) where the ortho and para positions possess negative charge density and the $C-O$ bond acquires partial double-bond character.

Theoretical Foundation: Dalton’s Law of Partial Pressures

Steam distillation is a method employed for the isolation and purification of organic compounds that are immiscible with water and possess high molecular weights or high boiling points. The fundamental principle governing this process is the application of Dalton’s Law of Partial Pressures.

According to Dalton's Law, in a mixture of two immiscible liquids, the total vapor pressure exerted by the system is the sum of the individual vapor pressures of the pure components at that specific temperature:

$P_{total} = p_1 + p_2$

Where:

• $P_{total}$ is the total vapor pressure of the system.
• $p_1$ is the vapor pressure of water.
• $p_2$ is the vapor pressure of the organic compound.

Step 1: Thermodynamic Analysis of Boiling

Boiling occurs when the total vapor pressure of the liquid system becomes equal to the external atmospheric pressure ($P_{atm}$):

$P_{total} = P_{atm}$

[Per the definition of boiling point]

Step 2: Derivation of the Lowered Boiling Point

For an organic liquid that is immiscible with water, the condition for boiling is achieved when:

$p_1 + p_2 = P_{atm}$

Since both the water and the organic liquid contribute to the total pressure, the vapor pressure of the organic liquid ($p_2$) alone does not need to reach $P_{atm}$ for vaporization to occur. Instead, it only needs to reach $P_{atm} - p_1$.

Because $p_1 > 0$ at any temperature where water exists as a liquid, it follows that:

$p_2 < P_{atm}$

Consequently, the temperature at which $p_2 = (P_{atm} - p_1)$ is necessarily lower than the temperature at which $p_2 = P_{atm}$ (which is the normal boiling point of the organic substance).

Visual Representation of Vapor Pressure Summation

Summary of Variables and Logic

Final Solution: In steam distillation, the organic liquid vaporizes at a temperature below its boiling point because the total pressure of the system (sum of the partial pressures of water and the organic liquid) reaches the atmospheric pressure at a lower temperature than the individual vapor pressure of the organic liquid would alone.

1. Isopropyl alcohol (Propan-2-ol)

Isopropyl alcohol, systematically named propan-2-ol, consists of a three-carbon chain where the hydroxyl functional group ($-OH$) is attached to the central carbon atom. [Per IUPAC nomenclature for alkanols].

Structure: $CH_3-CH(OH)-CH_3$

2. 2,3-Dimethylbutanal

The parent chain is butane (4 carbons). The aldehyde group ($-CHO$) defines carbon 1. Methyl substituents are located at positions 2 and 3. [Per IUPAC priority rules for carbonyl compounds].

Structure: $CH_3-CH(CH_3)-CH(CH_3)-CHO$

3. Heptan-4-one

Heptan-4-one implies a seven-carbon saturated chain with a ketone functional group ($C=O$) positioned at the fourth carbon atom. [Per the suffix -one for ketones].

Structure: $CH_3-CH_2-CH_2-C(=O)-CH_2-CH_2-CH_3$

Final Solution: The bond line structures have been rendered sequentially based on IUPAC nomenclature requirements for identifying the parent carbon chain and functional group placement.

Step 1: Analysis of Bond Cleavage

In the chemical process provided, we are examining the cleavage of a $C-Br$ bond where the bonding electrons are transferred to the more electronegative atom (Bromine). The reaction is represented as follows:

$CH_3-CH_2-Br \longrightarrow CH_3-CH_2^+ + :Br^-$

Step 2: Identification of Electron Flow

To represent the electron flow using curved-arrow notation, a full-headed arrow ($\rightarrow$) is drawn from the center of the covalent bond toward the atom receiving the electron pair. This indicates the heterolytic movement of a shared electron pair.

Step 3: Classification of Cleavage and Intermediate

• Type of Cleavage: This process is classified as heterolysis. [Per IUPAC definition, heterolytic fission occurs when a covalent bond breaks such that the shared pair of electrons remains with one of the fragments, resulting in ions].
• Reactive Intermediate: The resulting species $CH_3-CH_2^+$ possesses a carbon atom with a positive charge and a sextet of valence electrons. This is identified as a carbocation (specifically, an ethyl carbocation).

Step 4: Comparative Summary Table

Final Solution: The bond cleavage is a heterolysis resulting in the formation of an ethyl carbocation ($CH_3CH_2^+$) and a bromide anion ($Br^-$).

Step 1: Establishing Given Parameters

• Mass of organic compound ($m$) = $0.50\text{ g}$
• Volume of $H_2SO_4$ taken ($V_{total}$) = $50\text{ mL} = 0.050\text{ L}$
• Molarity of $H_2SO_4$ ($M_{acid}$) = $0.5\text{ M}$
• Volume of $NaOH$ used for back titration ($V_{NaOH}$) = $60\text{ mL} = 0.060\text{ L}$
• Molarity of $NaOH$ ($M_{base}$) = $0.5\text{ M}$

Step 2: Calculating Total Moles of Acid Provided

The total millimoles of $H_2SO_4$ added is calculated by: $n = M \times V(\text{in mL})$.

$\text{Total mmol of } H_2SO_4 = 0.5 \times 50 = 25\text{ mmol}$ [Definition of Molarity].

Step 3: Determining Residual Acid Neutralized by NaOH

The reaction between $H_2SO_4$ and $NaOH$ is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.

Millimoles of $NaOH$ used = $M \times V = 0.5 \times 60 = 30\text{ mmol}$.

Per the stoichiometry of the reaction, 2 moles of $NaOH$ neutralize 1 mole of $H_2SO_4$.

$\text{Residual mmol of } H_2SO_4 = \frac{\text{mmol of } NaOH}{2} = \frac{30}{2} = 15\text{ mmol}$.

Step 4: Calculating Acid Consumed by Ammonia

The amount of acid consumed by the ammonia ($NH_3$) evolved from the sample is the difference between total acid and residual acid:

$\text{mmol of } H_2SO_4 \text{ reacted with } NH_3 = 25 - 15 = 10\text{ mmol}$.

Step 5: Stoichiometric Relation to Nitrogen

The neutralization reaction is: $2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$.

Since 1 mole of $H_2SO_4$ reacts with 2 moles of $NH_3$, the moles of $NH_3$ (and thus nitrogen atoms) are:

$\text{mmol of } NH_3 = 2 \times (\text{mmol of } H_2SO_4 \text{ reacted}) = 2 \times 10 = 20\text{ mmol}$.

Mass of Nitrogen = $\text{moles} \times \text{atomic mass} = 20 \times 10^{-3} \text{ mol} \times 14\text{ g/mol} = 0.28\text{ g}$.

Step 6: Calculating Percentage Composition

$\text{Percentage of Nitrogen} = \frac{\text{Mass of Nitrogen}}{\text{Mass of Organic Compound}} \times 100$

$\text{Percentage} = \frac{0.28}{0.50} \times 100 = 56\%$

Final Solution: The percentage composition of nitrogen in the compound is 56%.

1. Analysis of the Homologous Series

The given compound is $H-CH=CH_2$, which is ethene ($C_2H_4$). This compound belongs to the alkene homologous series, characterized by the general molecular formula $C_nH_{2n}$ and the presence of a carbon-carbon double bond ($C=C$). Each successive member of the series differs by a $CH_2$ unit.

2. Determining the First Five Members

To identify the first five members, we start with $n=2$ (as the double bond requires a minimum of two carbon atoms) and increment the carbon chain length.

3. Structural Representation (Ethene to Propene)

The following diagram illustrates the structural geometry of the first two members, demonstrating the $sp^2$ hybridization at the double-bonded carbons [Per VSEPR Theory].

4. Derivation Logic

• Homology Definition: A series of organic compounds with the same functional group and similar chemical properties, where members differ by a constant increment of $-CH_2-$. [Based on IUPAC nomenclature principles].
• Structural Integrity: By adding successive $CH_2$ groups to the terminal methyl end of the parent alkene chain, we maintain the terminal double bond configuration ($R-CH=CH_2$), ensuring all members belong to the 1-alkene subclass.

Final Solution: The first five members of the series are Ethene ($CH_2=CH_2$), Propene ($CH_3CH=CH_2$), But-1-ene ($CH_3CH_2CH=CH_2$), Pent-1-ene ($CH_3(CH_2)_2CH=CH_2$), and Hex-1-ene ($CH_3(CH_2)_3CH=CH_2$).

Theoretical Foundation: Principles of Partition Chromatography

Paper chromatography is a specific application of partition chromatography. The fundamental principle governing this technique is the differential distribution (partitioning) of components of a mixture between two immiscible phases: a stationary phase and a mobile phase [Per the Nernst Distribution Law].

Step 1: Definition of Phases

• Stationary Phase: A strip of chromatographic paper made of cellulose. Cellulose fibers contain adsorbed water molecules, which serve as the immobilized liquid phase.
• Mobile Phase: A suitable organic solvent or mixture of solvents that travels up the paper strip via capillary action.

Step 2: Mechanism of Separation

The mixture to be separated is applied as a spot near the bottom of the cellulose paper. When the solvent (mobile phase) rises through the paper by capillary action, the individual components of the mixture partition themselves between the water held in the paper (stationary phase) and the moving solvent (mobile phase) [Based on the principle of differential solubility].

Step 3: Determination via Retardation Factor ($R_f$)

The extent to which each component moves is constant for a given set of experimental conditions. This is quantified by the Retardation Factor ($R_f$), defined by the ratio of the distance traveled by the substance to the distance traveled by the solvent front [Per IUPAC chromatographic conventions].

The mathematical expression is:

$R_f = \frac{\text{Distance travelled by the substance from the baseline}}{\text{Distance travelled by the solvent front from the baseline}}$

Analytical Summary

Final Solution: Paper chromatography relies on the partition of mixture components between a stationary aqueous phase held in cellulose fibers and a mobile organic solvent phase, causing separation based on varying solubilities and resulting in characteristic $R_f$ values for each component.

Step 1: Structural Identification

The provided image displays a branched alkane. To determine the IUPAC name, we must first identify the longest continuous carbon chain (the parent chain) and the substituents attached to it.

The structure is as follows: A central carbon bonded to three methyl groups ($CH_3$) and one carbon atom that is part of a longer chain. Let us expand the condensed structure:

$CH_3 - CH_2 - CH(CH_3) - CH_2 - CH_3$

Analyzing the backbone:

• The longest chain consists of 5 carbon atoms.
• This classifies the parent alkane as pentane.

Step 2: Numbering the Chain

To assign the lowest possible locants to substituents [Per IUPAC Rule of Lowest Locants], we number the chain from the end that gives the substituent the lowest number:

By numbering from left to right, the methyl group is located at position 3.

Step 3: Visual Representation

Step 4: Assembling the IUPAC Name

Following the standard nomenclature format: [Substituent Locant]-[Substituent Name][Parent Alkane Name] [Per IUPAC Blue Book P-14.4]:

• The substituent is a methyl group at position 3.
• The parent chain is pentane.
• Result: 3-methylpentane.

Final Solution: 3-methylpentane

Step 1: Analyzing the Molecular Components

The given reaction involves the transformation of an alkyl halide (iodoethane) into an alcohol (ethanol) in the presence of an aqueous potassium hydroxide solution.

• Substrate: $CH_3CH_2I$ (Iodoethane)
• Reagent: $KOH(aq)$
• Product: $CH_3CH_2OH$ (Ethanol)
• By-product: $KI$ (Potassium iodide)

Step 2: Identifying the Chemical Mechanism

In this transformation, the hydroxide ion ($OH^-$), acting as a nucleophile provided by the aqueous $KOH$, attacks the electrophilic carbon atom bonded to the iodine atom. The iodide ion ($I^-$), being a superior leaving group, is displaced from the carbon chain. [Per the Nucleophilic Substitution mechanism].

Step 3: Justification of Classification

The reaction is categorized as follows:

1. Nucleophilic Substitution: Because the nucleophile ($OH^-$) substitutes the leaving group ($I^-$).
2. Bimolecular Substitution ($S_N2$): Since the substrate is a primary alkyl halide, the mechanism proceeds via a concerted transition state where the rate depends on both the concentration of the substrate and the nucleophile. [Per Ingold's Mechanism Theory].
3. Hydrolysis: Since the reagent is $KOH(aq)$, the process effectively introduces a hydroxyl group at the expense of the halide, often referred to as alkaline hydrolysis.

Step 4: Visualizing the Substitution Process

Analytical Comparison

Final Solution: The reaction is classified as a Nucleophilic Substitution reaction (specifically, an $S_N2$ substitution or alkaline hydrolysis).

Step 1: Analyzing the Reaction Mechanism

The given reaction is: $CH_3COOH + \mathbf{HO^-} \rightarrow CH_3COO^- + H_2O$

In this chemical transformation, the species $\mathbf{HO^-}$ (hydroxide ion) is interacting with the carboxylic acid ($CH_3COOH$). Specifically, the oxygen atom in the hydroxide ion possesses a formal negative charge and three lone pairs of electrons, rendering it electron-rich [Per the Lewis Definition of Bases].

Step 2: Identifying Electron Density and Reactivity

To classify $\mathbf{HO^-}$, we evaluate its role in the electronic interaction:

• Nucleophile (Nucleus-loving): A species that donates an electron pair to an electrophile to form a chemical bond. These are typically negatively charged or possess lone pairs of electrons [Per the Lewis Theory of Acids and Bases].
• Electrophile (Electron-loving): A species that accepts an electron pair from a nucleophile to form a chemical bond. These are typically electron-deficient or positively charged.

Step 3: Justification of the Role of $\mathbf{HO^-}$

The hydroxide ion ($\mathbf{HO^-}$) acts as a base/nucleophile. It attacks the proton ($H^+$) attached to the oxygen atom of the acetic acid molecule. By donating its lone pair of electrons to the hydrogen atom, it forms a covalent bond, resulting in the formation of a water molecule ($H_2O$).

Step 4: Classification Analysis

Since the hydroxide ion provides an electron pair to form the $O-H$ bond in $H_2O$, it satisfies the criteria for a nucleophile according to the IUPAC definition of nucleophilic species.

Final Solution: The reagent $\mathbf{HO^-}$ acts as a nucleophile because it possesses a lone pair of electrons and a negative charge, which it donates to the electrophilic hydrogen center of the acetic acid molecule.

Step 1: Structural Representation of the Proposed Names

To determine the correct IUPAC name, we must first construct the carbon skeletons based on the provided names: 2,4,7-trimethyloctane and 2,5,7-trimethyloctane. Both represent an octane parent chain ($C_8$ backbone).

Step 2: Applying the Lowest Locant Rule

According to the IUPAC nomenclature rules for alkanes, when multiple substituents are present, the chain must be numbered such that the set of locants is the lowest possible [Per IUPAC Blue Book Rule P-14.4].

Step 3: Comparative Analysis

The "Lowest Locant Rule" dictates that we compare the locant sets term-by-term starting from the first point of difference:

• Compare the first locant: $2$ vs $2$ (Equal)
• Compare the second locant: $4$ vs $5$

Since $4 < 5$, the set $\{2, 4, 7\}$ is the lower set of locants. Therefore, the numbering that results in 2,4,7-trimethyloctane is preferred over 2,5,7-trimethyloctane.

Step 4: Final IUPAC Verification

By selecting the 2,4,7-trimethyl configuration, we adhere to the IUPAC requirement of assigning the lowest possible numbers to substituents to ensure systematic, unambiguous identification of the organic structure.

Final Solution: The correct IUPAC name is 2,4,7-Trimethyloctane.

Step 1: Analyzing the Reaction Components

The given chemical equation is:

$(CH_3)_2C = CH_2 + HCl \rightarrow (CH_3)_2ClC - CH_3$

In this transformation, we observe two distinct chemical phenomena:

• Substrate: 2-Methylpropene, an unsaturated hydrocarbon containing a carbon-carbon double bond ($C=C$). [Per IUPAC nomenclature for alkenes].
• Reagent: Hydrogen chloride ($HCl$), a polar electrophilic reagent.
• Product: 2-Chloro-2-methylpropane, a saturated haloalkane.

Step 2: Identifying the Reaction Mechanism

The reaction involves the breaking of the $\pi$-bond of the alkene and the addition of the $H^+$ and $Cl^-$ species across the double bond. This is a characteristic Electrophilic Addition Reaction [Per the mechanism of electrophilic addition to alkenes].

The process follows these sub-stages:

1. Formation of Electrophile: The double bond attacks the electrophilic proton ($H^+$) from $HCl$.
2. Carbocation Formation: The proton adds to the terminal carbon to form a tertiary carbocation $(CH_3)_3C^+$, which is highly stable due to the inductive effect and hyperconjugation.
3. Nucleophilic Attack: The chloride ion ($Cl^-$) acts as a nucleophile and attacks the tertiary carbocation to form the final product.

Step 3: Visualizing the Molecular Geometry

Step 4: Theoretical Justification

According to Markovnikov's Rule, in the addition of a protic acid to an asymmetric alkene, the acid hydrogen ($H$) becomes attached to the carbon with fewer alkyl substituents (or the carbon with more hydrogens), and the halide group ($Cl$) attaches to the carbon with more alkyl substituents [Per Markovnikov’s Regioselectivity Principle].

Final Solution: The reaction is classified as an Electrophilic Addition Reaction.

Step 1: Analyzing the Molecular Structure and Bond Nature

The given structure represents an ethyl group bonded to a bromine atom ($CH_3-CH_2-Br$). The bond cleavage occurs at the $C-Br$ covalent bond. Bromine is more electronegative ($\chi \approx 2.96$) than carbon ($\chi \approx 2.55$), which polarizes the bond, placing partial negative charge on bromine and partial positive charge on carbon [Per Electronegativity Difference Principles].

Step 2: Mechanism of Bond Cleavage (Curved-Arrow Notation)

To show the heterolytic cleavage of the $C-Br$ bond, a double-headed curved arrow is drawn starting from the center of the shared bond pointing towards the more electronegative bromine atom. This indicates the migration of the entire electron pair (2 electrons) to the bromine atom.

Step 3: Classification of Cleavage and Reactive Intermediate

• Classification: The process is defined as Heterolysis (or Heterolytic Cleavage). This is because the shared electron pair is distributed unequally between the two fragments [Per IUPAC definition of heterolytic fission].
• Reactive Intermediate: The carbon atom loses its shared electron to bromine, resulting in an electron-deficient species with a positive charge on the carbon atom ($CH_3CH_2^+$). This species is classified as a carbocation (specifically, an ethyl carbocation).

Step 4: Summary Table of Transformation

Final Solution: The bond cleavage is classified as heterolysis. The reactive intermediate produced is a carbocation ($CH_3CH_2^+$).

Step 1: Identifying the Functional Groups in the Provided Structure

Based on the chemical structure provided in the image, we analyze the substituents attached to the benzene ring. The structure consists of a benzene ring substituted with two specific groups at the ortho ($1,2$) position.

Step 2: Structural Analysis and Nomenclature

The compound depicted is 2-hydroxybenzaldehyde, commonly known as Salicylaldehyde. We identify the groups as follows:

• Aldehyde ($-CHO$): Attached to the carbon at position 1. The carbon atom of the carbonyl group is directly bonded to the benzene ring. [Per IUPAC nomenclature for substituted benzenes].
• Hydroxyl ($-OH$): Attached to the carbon at position 2 (ortho position). Because it is directly bonded to the aromatic ring, it is categorized as a phenolic hydroxyl group. [Per the functional group classification of organic compounds].

Visual Representation of the Functional Groups

Final Solution: The functional groups present in the compound are the Aldehyde group ($-CHO$) and the Hydroxyl group ($-OH$).

Step 1: Structural Analysis of the Compound

The provided structure represents an organic molecule containing a central benzene ring attached to a substituent group. Analyzing the functional groups present requires identifying the specific atoms and their bonding patterns:

• The primary scaffold is a benzene ring ($C_6H_5-$).
• Attached to the ring is a carbonyl carbon bonded to a hydroxyl group, characterizing a Carboxylic Acid ($–COOH$).
• Attached to the ortho-position relative to the carboxyl group is a hydroxyl group ($–OH$), characterizing a Phenolic Hydroxyl group.
• Attached to the meta-position relative to the carboxyl group is an amino group ($–NH_2$), characterizing an Amine.

Step 2: Identification of Functional Groups

Based on the structural components identified, we classify the functional groups as follows:

Step 3: Visual Representation of Functional Sites

Step 4: Theoretical Justification

The molecule demonstrates the coexistence of multiple functional groups on a single aromatic nucleus.

• The Carboxylic acid group ($–COOH$) takes priority in numbering, defining the compound as a derivative of benzoic acid.
• The Hydroxyl group ($–OH$) directly bonded to the aromatic ring constitutes a phenol [By definition of aromatic alcohols].
• The Amino group ($–NH_2$) acts as a substituent, providing basic character to the molecule [Per Brønsted-Lowry acid-base theory].

Final Solution: The compound contains a Carboxylic Acid group ($-COOH$), a Phenolic Hydroxyl group ($-OH$), and an Amino group ($-NH_2$).

Given Variables & Initial Setup

The chemical reaction provided is:

$CH_3CH_2Br + HS^- \rightarrow CH_3CH_2SH + Br^-$

Here, the substrate is bromoethane ($CH_3CH_2Br$) and the attacking species is the hydrosulfide ion ($HS^-$), which acts as a nucleophile. The bromide ion ($Br^-$) acts as the leaving group.

Step 1: Identifying the Reaction Mechanism

To classify this reaction, we observe the fundamental transformation occurring at the electrophilic carbon center ($C_1$ of the ethyl group). The nucleophile ($HS^-$) possesses a lone pair of electrons and is electron-rich. It attacks the carbon atom bonded to the electronegative bromine atom, resulting in the displacement of the bromide ion. [Per the definition of Nucleophilic Substitution reactions].

Step 2: Analytical Breakdown of Atomic Movement

• Nucleophile: $HS^-$ (electron donor)
• Electrophile: $CH_3CH_2^\delta^+ - Br^{\delta^-}$ (electron acceptor at the $\alpha$-carbon)
• Leaving Group: $Br^-$ (anionic species that departs with the electron pair from the $C-Br$ bond)

The reaction follows the general form:

$Nu^- + R-LG \rightarrow R-Nu + LG^-$

where $Nu = HS^-$, $R = CH_3CH_2-$, and $LG = Br^-$. This mechanism is classically defined as a Nucleophilic Substitution reaction.

Step 3: Visual Representation

Step 4: Theoretical Justification

The reaction involves the replacement of a bromine atom by a hydrosulfide group. Since the incoming group is a nucleophile and the departing group is a stable anion, the reaction is categorized under Nucleophilic Substitution ($S_N$ type). Specifically, as bromoethane is a primary alkyl halide, this reaction typically follows an $S_N2$ mechanism where the attack and departure occur in a concerted step [By IUPAC nomenclature standards and reaction kinetics theory].

Final Solution: The reaction is a Nucleophilic Substitution reaction.

Step 1: Structural Representation of the Molecules

To determine the correct IUPAC name, we must first construct the carbon chain for both names provided. Both names imply a pentane backbone (a chain of 5 carbon atoms).

Compound A: 2-Chloro-4-methylpentane

Compound B: 4-Chloro-2-methylpentane

Step 2: Application of IUPAC Lowest Locant Rule

According to the IUPAC nomenclature rules for aliphatic hydrocarbons [Per IUPAC Rule P-14.4], when numbering the parent chain, the set of locants must be the lowest possible at the first point of difference.

Step 3: Application of Alphabetical Priority Rule

When two different numbering sequences yield the same set of locants, the priority is given to the substituent that appears first in alphabetical order [Per IUPAC Rule P-14.5].

• Substituents: Chloro vs. Methyl
• Alphabetical Order: 'C' (Chloro) comes before 'M' (Methyl).
• Therefore, the substituent 'Chloro' must be assigned the lower locant number.

Step 4: Comparative Analysis

• In 2-chloro-4-methylpentane, Chloro is at position 2.
• In 4-chloro-2-methylpentane, Chloro is at position 4.
• Comparing position 2 and position 4, position 2 is lower.

Final Solution: The correct IUPAC name is 2-Chloro-4-methylpentane, as it assigns the lower locant to the substituent that appears first alphabetically (Chloro < Methyl).

1. Definitions of Electron Displacement Effects

Inductive Effect ($I$-Effect): The permanent displacement of shared electron pairs along a chain of carbon atoms due to the difference in electronegativity of substituent atoms or groups. It operates through $\sigma$-bonds and decreases rapidly with distance from the source [Per the principle of polarization in sigma-bonds].

• Negative Inductive Effect ($-I$): Electron-withdrawing groups (e.g., $-NO_2, -F, -Cl$).
• Positive Inductive Effect ($+I$): Electron-donating groups (e.g., alkyl groups like $-CH_3, -CH_2CH_3$).

Electromeric Effect ($E$-Effect): A temporary effect involving the complete transfer of a shared pair of $\pi$-electrons to one of the atoms joined by a multiple bond at the requirement of an attacking reagent. It is represented by a curved arrow showing electron flow [Per the mechanism of nucleophilic/electrophilic addition].

2. Analysis of the Acidity Order

The given acidity order is: $CH_3CH_2COOH > (CH_3)_2CHCOOH > (CH_3)_3C.COOH$.

To determine the acidity of carboxylic acids ($R-COOH$), we examine the stability of the conjugate base, the carboxylate ion ($R-COO^-$). An electron-donating group (EDG) destabilizes the carboxylate ion by increasing the negative charge density on the carboxylate group, thereby decreasing the acidity of the parent acid.

3. Step-by-Step Inductive Effect Analysis

Step-by-step logic:

1. The acidity of $RCOOH$ depends on the ease of proton release to form $RCOO^-$.
2. Alkyl groups exhibit a $+I$ effect. As the number of alkyl groups attached to the $\alpha$-carbon increases, the total electron-donating inductive effect increases.
3. The tert-butyl group in $(CH_3)_3C.COOH$ provides the maximum electron density to the carboxylate carbon via the $+I$ effect. This destabilizes the $RCOO^-$ anion significantly compared to the ethyl group.
4. Consequently, the conjugate base of propanoic acid ($CH_3CH_2COO^-$) is the most stable, and the conjugate base of pivalic acid ($(CH_3)_3CCOO^-$) is the least stable.

4. Visual Representation of Electron Displacement

Final Solution: The inductive effect ($+I$ effect) explains this order. As the number of electron-donating alkyl groups increases, the electron density on the carboxylate group increases, destabilizing the conjugate base and thus decreasing the acidity of the carboxylic acid.