Step 1: Analyzing the Reactants and Reagents

The given reaction involves $2\text{-bromo-2-methylbutane}$ reacting with $\text{sodium ethoxide } (\text{C}_2\text{H}_5\text{ONa})$ in the presence of $\text{ethanol } (\text{C}_2\text{H}_5\text{OH})$.

• Substrate: $2\text{-bromo-2-methylbutane}$ is a tertiary ($3^\circ$) alkyl halide.
• Reagent: Sodium ethoxide is a strong base and a strong nucleophile.
• Conditions: Heat and a protic solvent (ethanol) strongly favor the elimination pathway over substitution.

Step 2: Determining the Mechanism

Because the substrate is a tertiary alkyl halide and the reagent is a strong base, the reaction proceeds via the $E2$ (Elimination Bimolecular) mechanism [Per Zaitsev's Rule]. In $E2$ reactions, the base abstracts a $\beta\text{-hydrogen}$ to form an alkene. Since $3^\circ$ halides are sterically hindered, $S_N2$ substitution is suppressed, and elimination becomes the dominant pathway.

Step 3: Identification of $\beta\text{-hydrogens}$ and Regioselectivity

Structure of $2\text{-bromo-2-methylbutane}$:

The $\alpha\text{-carbon}$ is attached to $C_1$ (methyl), $C_3$ (methylene), and another methyl group. We have two distinct types of $\beta\text{-hydrogens}$:

• Pathway A: Removal of H from the $\text{CH}_3$ group.
• Pathway B: Removal of H from the $\text{CH}_2$ group.

Step 4: Application of Zaitsev's Rule

According to Zaitsev's Rule, the major product in dehydrohalogenation is the most highly substituted alkene [Thermodynamic stability of alkenes].

Comparing the possible products:

Step 5: Final Reaction Equation

The base abstracts the proton from the $C_3$ position to form the more stable trisubstituted alkene: $2\text{-methylbut-2-ene}$.

$\text{CH}_3\text{CH}_2\text{C(CH}_3\text{)(Br)CH}_3 + \text{C}_2\text{H}_5\text{O}^- \rightarrow \text{CH}_3\text{CH=C(CH}_3)_2 + \text{C}_2\text{H}_5\text{OH} + \text{Br}^-$

Final Solution: The major organic product is 2-methylbut-2-ene.

Given Variables & Initial Setup

The objective is to convert $CH_3CH_2CH=CH_2$ (But-1-ene) to $CH_3CH=CHCH_3$ (But-2-ene). This transformation requires a two-step process: first, the addition of a hydrogen halide to form a secondary haloalkane, followed by dehydrohalogenation to achieve the more stable, substituted alkene.

Step 1: Hydrohalogenation of But-1-ene

React But-1-ene with Hydrogen Bromide ($HBr$). According to Markovnikov's Rule, the electrophilic hydrogen adds to the carbon atom with the greater number of hydrogen atoms, while the nucleophilic bromide attaches to the more substituted carbon atom.

$CH_3CH_2CH=CH_2 + HBr \xrightarrow{} CH_3CH_2CH(Br)CH_3$

[Per Markovnikov's Rule: The electrophile $H^+$ adds to $C_1$, creating the more stable secondary carbocation intermediate at $C_2$.]

Step 2: Dehydrohalogenation

React 2-bromobutane with alcoholic potassium hydroxide ($KOH_{(alc)}$). This is a $\beta$-elimination reaction (or dehydrohalogenation).

$CH_3CH_2CH(Br)CH_3 + KOH_{(alc)} \xrightarrow{\Delta} CH_3CH=CHCH_3 + KBr + H_2O$

[Per Saytzeff's Rule: In elimination reactions, the alkene that is more highly substituted (more alkyl groups attached to the double-bonded carbons) is the major product due to increased hyperconjugative stabilization.]

Visual Representation of the Conversion

Summary of Reaction Conditions

Final Solution: The conversion of But-1-ene to But-2-ene is achieved via hydrobromination to produce 2-bromobutane, followed by base-catalyzed dehydrohalogenation to yield the thermodynamically more stable But-2-ene.

Step 1: Analyzing the Nomenclature and Molecular Framework

The compound p-bromochlorobenzene is a disubstituted benzene derivative. The name comprises three structural components:

• Benzene Ring: The parent aromatic hydrocarbon core ($C_6H_6$).
• Chloro group ($-Cl$): A chlorine atom attached to the ring.
• Bromo group ($-Br$): A bromine atom attached to the ring.
• Prefix "p-" (Para): This indicates the relative positions of the substituents. In a benzene ring, the 1,4-positions are referred to as para positions [Per IUPAC rules for aromatic substitution].

Step 2: Determining Substituent Positioning

To construct the structure, we assign the first substituent to position $1$ and the second substituent to the opposite position $4$ on the hexagonal carbon ring:

Step 3: Visualization of Molecular Geometry

The following SVG represents the para-substituted configuration of the benzene ring, maintaining 120-degree bond angles for $sp^2$ hybridized carbon atoms.

Step 4: Finalizing the Chemical Structure

The molecular formula for p-bromochlorobenzene is $C_6H_4BrCl$. The structural representation shows the bromine atom substituted at the carbon atom directly opposite to the carbon atom substituted by the chlorine atom.

Final Solution: The structure of p-Bromochlorobenzene is a benzene ring with a Chloro group ($-Cl$) at position 1 and a Bromo group ($-Br$) at position 4.

Representation: $Cl-C_6H_4-Br$ (para-isomer)

Step 1: Conversion of Benzyl Alcohol to Benzyl Chloride

The first objective is to replace the hydroxyl ($-OH$) group with a chlorine atom to facilitate nucleophilic substitution. We utilize thionyl chloride ($SOCl_2$) in the presence of pyridine [Per the Darzens reaction mechanism].

$C_6H_5CH_2OH + SOCl_2 \xrightarrow{\text{Pyridine}} C_6H_5CH_2Cl + SO_2(g) + HCl(g)$

Justification: This method is preferred as the by-products are gaseous, leaving behind pure benzyl chloride.

Step 2: Nucleophilic Substitution to Benzyl Cyanide

To extend the carbon chain, we introduce a cyanide ion ($CN^-$) via a nucleophilic substitution reaction ($S_N2$). We treat benzyl chloride with ethanolic potassium cyanide ($KCN$).

$C_6H_5CH_2Cl + KCN(\text{alc.}) \xrightarrow{\Delta} C_6H_5CH_2CN + KCl$

Justification: Cyanide is an ambident nucleophile that prefers carbon-bonding in $S_N2$ conditions, resulting in the formation of phenylacetonitrile (benzyl cyanide) [Per the kinetics of $S_N2$ reactions].

Step 3: Acidic Hydrolysis of Benzyl Cyanide

The final step involves the conversion of the nitrile functional group ($-CN$) into a carboxylic acid group ($-COOH$) through complete acid-catalyzed hydrolysis.

$C_6H_5CH_2CN + 2H_2O + H^+ \xrightarrow{\Delta} C_6H_5CH_2COOH + NH_4^+$

Justification: Nitriles undergo stepwise hydrolysis; the first step produces an amide, which subsequently hydrolyzes to the carboxylic acid upon heating with an aqueous mineral acid [Per the chemical properties of nitriles].

Summary of Reaction Sequence

Final Solution: The conversion is achieved via a three-step process: chlorination using thionyl chloride, cyanation using ethanolic KCN, and final acid-catalyzed hydrolysis to yield 2-phenylethanoic acid.

Given Variables & Initial Setup

The objective is to synthesize 1-iodobutane ($CH_3CH_2CH_2CH_2I$) from but-1-ene ($CH_3CH_2CH=CH_2$). This transformation requires a two-step sequence: first, anti-Markovnikov addition of hydrogen bromide to place the bromine at the terminal carbon, followed by a nucleophilic substitution (Finkelstein reaction) to replace the bromide with an iodide.

Step 1: Anti-Markovnikov Addition of HBr

To convert but-1-ene into 1-bromobutane, we utilize the Kharasch effect (peroxide effect). When hydrogen bromide is added to an asymmetrical alkene in the presence of an organic peroxide ($R_2O_2$), the addition proceeds via a free radical mechanism rather than the standard electrophilic addition.

$CH_3CH_2CH=CH_2 + HBr \xrightarrow{Peroxide} CH_3CH_2CH_2CH_2Br$

[Justification: The free radical mechanism ensures the bromine atom attaches to the less substituted carbon atom, following the regioselectivity of the peroxide effect.]

Step 2: Finkelstein Reaction (Halogen Exchange)

To obtain 1-iodobutane, 1-bromobutane is reacted with sodium iodide ($NaI$) in dry acetone. This is a classic $S_N2$ reaction.

$CH_3CH_2CH_2CH_2Br + NaI \xrightarrow{Acetone, \Delta} CH_3CH_2CH_2CH_2I + NaBr \downarrow$

[Justification: Per the Finkelstein reaction mechanism, $NaI$ is used to exchange the halogen. Sodium bromide ($NaBr$) is insoluble in dry acetone and precipitates, which, according to Le Chatelier’s Principle, drives the equilibrium forward to completion.]

Reaction Sequence Visualization

Summary of Chemical Transformation

Final Solution: The preparation involves the anti-Markovnikov addition of $HBr$ to but-1-ene followed by a Finkelstein reaction: $CH_3CH_2CH=CH_2 \xrightarrow{HBr, Peroxide} CH_3CH_2CH_2CH_2Br \xrightarrow{NaI, Acetone} CH_3CH_2CH_2CH_2I$.

Step 1: Theoretical Foundation of the $S_{N}2$ Mechanism

The $S_{N}2$ (Substitution Nucleophilic Bimolecular) reaction is a concerted mechanism where the rate-determining step involves the simultaneous attack of the nucleophile and the departure of the leaving group. The transition state is characterized by the carbon atom undergoing a pentacoordinate geometry where the bond to the nucleophile is forming while the bond to the leaving group is breaking. [Per the kinetic theory of $S_{N}2$ reactions, the rate of reaction is proportional to the ability of the leaving group to stabilize the negative charge as it departs].

Step 2: Analysis of Leaving Group Ability

The rate of $S_{N}2$ reaction is heavily dependent on the nature of the leaving group. A better leaving group is one that is a weaker base, which corresponds to the conjugate base of a strong acid. The bond dissociation enthalpy (BDE) also plays a critical role in the carbon-halogen bond breaking energy.

• The strength of the $C-X$ bond decreases as the size of the halogen atom ($X$) increases.
• Atomic radii: $I > Br > Cl > F$.
• Bond dissociation enthalpy order: $C-I < C-Br < C-Cl < C-F$.

[By the Periodic Trend of Bond Strength, $C-I$ has the longest bond length and the lowest bond dissociation energy, making the iodide ion ($I^-$) a significantly better leaving group than the bromide ion ($Br^-$)].

Step 3: Comparative Kinetic Analysis

In the reaction with $OH^-$, we compare the transition states:

Since the $I^-$ ion is a weaker base and a better leaving group, the activation energy ($E_a$) for the displacement of $I^-$ is lower than that of $Br^-$. [Per the Arrhenius Equation ($k = Ae^{-E_a/RT}$), a lower activation energy results in a higher rate constant $k$].

Step 4: Visualizing the Transition State

Final Solution: $CH_{3}I$ will react faster than $CH_{3}Br$ because the iodide ion ($I^-$) is a better leaving group than the bromide ion ($Br^-$) due to the weaker $C-I$ bond and the high polarizability of the iodine atom.

Step 1: Chemical Transformation Identification

The conversion of aniline ($C_6H_5NH_2$) to phenylisocyanide ($C_6H_5NC$) is classified as a Carbylamine Reaction (or Isocyanide Test). This reaction is specific to primary amines ($R-NH_2$) and is utilized as a diagnostic test for their identification. [Per the mechanism of Carbylamine reaction].

Step 2: Defining the Reaction Components

The transformation requires the interaction of the primary amine with chloroform in the presence of a strong base (potassium hydroxide).

• Reactant: Aniline ($C_6H_5NH_2$)
• Reagents: Chloroform ($CHCl_3$) and alcoholic Potassium Hydroxide ($KOH$)
• Product: Phenylisocyanide ($C_6H_5NC$)
• By-products: Potassium chloride ($KCl$) and Water ($H_2O$)

Step 3: Balanced Chemical Equation

The reaction stoichiometry is represented as follows:

$C_6H_5NH_2 + CHCl_3 + 3KOH_{(alc)} \xrightarrow{\Delta} C_6H_5NC + 3KCl + 3H_2O$

Step 4: Mechanistic Overview

1. Generation of Dichlorocarbene: The base ($KOH$) abstracts a proton from chloroform ($CHCl_3$), leading to the formation of the highly reactive electrophilic intermediate, dichlorocarbene ($:CCl_2$). [By $\alpha$-elimination mechanism].
2. Nucleophilic Attack: The lone pair on the nitrogen of the aniline attacks the electron-deficient carbon of the dichlorocarbene.
3. Elimination: Subsequent loss of two molecules of $HCl$ (neutralized by $KOH$) results in the formation of the isocyanide linkage ($-NC$).

Visual Representation of Molecule Structure

Summary of Conditions

Final Solution: The conversion of aniline to phenylisocyanide is achieved by heating aniline with chloroform in the presence of alcoholic potassium hydroxide, yielding phenylisocyanide, potassium chloride, and water as the final products.

Step 1: Structural Expansion and Parent Chain Identification

The given molecular formula is $(CH_{3})_{2}CHCH(Cl)CH_{3}$. To identify the IUPAC name, we expand the condensed structure:

$CH_{3} - CH(CH_{3}) - CH(Cl) - CH_{3}$

Following IUPAC nomenclature rules [Per IUPAC Blue Book for Organic Chemistry]:

• Identify the longest carbon chain: The longest chain containing the halogen substituent consists of 4 carbon atoms. Thus, the parent alkane is butane.
• Numbering the chain: Numbering must start from the end that gives the lowest possible locants to substituents. Numbering from right to left assigns position 2 to the chlorine atom and position 3 to the methyl group.

Step 2: Assigning Substituents and Nomenclature

Based on the numbering (1 to 4 from right to left):

• Chloro group is at position $2$.
• Methyl group is at position $3$.

Alphabetical order for naming substituents dictates that 'chloro' precedes 'methyl'. Therefore, the IUPAC name is 2-chloro-3-methylbutane.

Step 3: Classification of the Halide

To classify the compound, we examine the carbon atom directly bonded to the chlorine atom ($C-Cl$ bond):

The structure is $R-CH(Cl)-R'$. The carbon atom attached to the chlorine is bonded to two other carbon atoms. By definition, a carbon atom bonded to two other carbons is a secondary ($2^\circ$) carbon [Classification based on Substitution Grade]. Since the halogen is attached to an alkyl chain and is not adjacent to a double bond, it is classified as an alkyl halide.

Visual Representation

Data Summary Table

Final Solution: The IUPAC name is 2-chloro-3-methylbutane and it is classified as a secondary ($2^\circ$) alkyl halide.

Step 1: Analyzing Reactants and Products

The starting material is 2-methyl-1-propene ($CH_3-C(CH_3)=CH_2$), which is an alkene with a terminal double bond. The target product is 2-chloro-2-methylpropane ($CH_3-C(Cl)(CH_3)-CH_3$), which is a tertiary alkyl halide. To convert the alkene to the alkyl halide, we must add a molecule of $HCl$ across the double bond.

Step 2: Theoretical Mechanism (Electrophilic Addition)

The reaction follows the mechanism of Electrophilic Addition to an alkene. According to Markownikoff’s Rule [Per the Markovnikov rule, in the addition of a protic acid $HX$ to an asymmetric alkene, the acid hydrogen $H$ adds to the carbon with more hydrogen atoms, and the halide $X$ adds to the carbon with fewer hydrogen atoms].

The reaction proceeds via the formation of a carbocation intermediate:

1. Protonation: The double bond attacks the $H^+$ ion from $HCl$, resulting in the formation of the more stable carbocation.
2. Stability Assessment: The tertiary carbocation $[(CH_3)_3C^+]$ is formed, which is significantly more stable than the primary carbocation $[(CH_3)_2CH-CH_2^+]$ due to the +I effect (inductive effect) and hyperconjugation of nine $\alpha$-hydrogens.
3. Nucleophilic Attack: The chloride ion $(Cl^-)$ attacks the tertiary carbocation to form the final product.

Step 3: Chemical Equation

The reaction is represented as follows:

$CH_3-C(CH_3)=CH_2 + HCl \rightarrow CH_3-C(Cl)(CH_3)-CH_3$

Visual Representation of the Molecule

Step 4: Summary of Reaction Conditions

Final Solution: The conversion of 2-methyl-1-propene to 2-chloro-2-methylpropane is achieved by the addition of Hydrogen Chloride ($HCl$) to the alkene, which proceeds via the formation of a stable tertiary carbocation following Markownikoff's Rule.

Step 1: Analyzing the Molecular Formula and Structural Constraints

The given primary alkyl halide has the molecular formula $C_4H_9Br$. Since it is a primary alkyl halide, the bromine atom is attached to a carbon atom that is bonded to only one other carbon atom. The possible primary isomers of $C_4H_9Br$ are:

• n-butyl bromide: $CH_3CH_2CH_2CH_2Br$
• Isobutyl bromide: $(CH_3)_2CHCH_2Br$

We are told that (a) reacts with sodium (Wurtz reaction) to give (d), $C_8H_{18}$, which is different from the product formed by n-butyl bromide (n-octane). Therefore, (a) must be isobutyl bromide ($(CH_3)_2CHCH_2Br$).

Step 2: Identifying Compounds (a), (b), (c), and (d)

Based on the analysis:

• (a): Isobutyl bromide, $(CH_3)_2CHCH_2Br$
• (b): Elimination of $HBr$ using alcoholic $KOH$ yields Isobutylene, $(CH_3)_2C=CH_2$ [Per Dehydrohalogenation mechanism].
• (c): Addition of $HBr$ to isobutylene follows Markovnikov's rule to give t-butyl bromide, $(CH_3)_3CBr$, which is an isomer of (a).
• (d): Wurtz reaction of isobutyl bromide: $2(CH_3)_2CHCH_2Br + 2Na \rightarrow (CH_3)_2CHCH_2CH_2CH(CH_3)_2 + 2NaBr$, resulting in 2,5-dimethylhexane.

Step 3: Chemical Equations

1. Formation of (b) from (a):

$(CH_3)_2CHCH_2Br + alc. KOH \rightarrow (CH_3)_2C=CH_2 + KBr + H_2O$

2. Formation of (c) from (b):

$(CH_3)_2C=CH_2 + HBr \rightarrow (CH_3)_3CBr$

3. Formation of (d) from (a) (Wurtz Reaction):

$2(CH_3)_2CHCH_2Br + 2Na \xrightarrow{dry ether} (CH_3)_2CHCH_2CH_2CH(CH_3)_2 + 2NaBr$

Structural Representation

Final Solution: Compound (a) is isobutyl bromide (1-bromo-2-methylpropane). The equations are: (a) $\rightarrow$ (b) : $(CH_3)_2CHCH_2Br \xrightarrow{alc. KOH} (CH_3)_2C=CH_2 + KBr + H_2O$; (b) $\rightarrow$ (c) : $(CH_3)_2C=CH_2 \xrightarrow{HBr} (CH_3)_3CBr$; (a) $\rightarrow$ (d) : $2(CH_3)_2CHCH_2Br + 2Na \xrightarrow{dry ether} (CH_3)_2CHCH_2CH_2CH(CH_3)_2 + 2NaBr$.

Step 1: Chemical Transformation Overview

The conversion of Benzene ($\text{C}_6\text{H}_6$) to Biphenyl ($\text{C}_6\text{H}_5-\text{C}_6\text{H}_5$) is achieved through a multi-step sequence involving Electrophilic Aromatic Substitution followed by a Coupling reaction. The most efficient pathway involves the halogenation of benzene followed by a Wurtz-Fittig or Fittig reaction.

Step 2: Step-by-Step Reaction Mechanism

Step 2.1: Electrophilic Aromatic Bromination

Benzene is reacted with bromine ($\text{Br}_2$) in the presence of a Lewis acid catalyst, such as ferric bromide ($\text{FeBr}_3$), to form bromobenzene. This is a standard electrophilic substitution [Per the mechanism of electrophilic aromatic substitution].

$\text{C}_6\text{H}_6 + \text{Br}_2 \xrightarrow{\text{FeBr}_3} \text{C}_6\text{H}_5\text{Br} + \text{HBr}$

Step 2.2: Fittig Reaction (Coupling)

Bromobenzene is treated with metallic sodium in the presence of dry ether. Two molecules of bromobenzene couple to form biphenyl. This is an application of the Fittig reaction, specifically designed for the synthesis of diaryls from aryl halides [By the Fittig Reaction Mechanism].

$2\text{C}_6\text{H}_5\text{Br} + 2\text{Na} \xrightarrow{\text{Dry ether}} \text{C}_6\text{H}_5-\text{C}_6\text{H}_5 + 2\text{NaBr}$

Step 3: Visual Representation of the Fittig Coupling

Summary of Reaction Conditions

Final Solution: The conversion is performed by first brominating benzene to produce bromobenzene, followed by a Fittig coupling reaction with sodium in dry ether to yield biphenyl ($\text{C}_6\text{H}_5-\text{C}_6\text{H}_5$).

Step 1: Identifying the Reactants and Reagent

The given reaction involves $CH_3CH_2CH(Cl)CH_2CH_3$ (3-chloropentane) reacting with $KOH$ in the presence of $ethanol$ (alcoholic $KOH$).

Substrate: $CH_3-CH_2-CHCl-CH_2-CH_3$ (A secondary alkyl halide).

Reagent: $KOH$ (alc.). Alcoholic $KOH$ acts as a source of ethoxide ions ($C_2H_5O^-$) or hydroxide ions ($OH^-$), which function as strong bases. [Per the mechanism of $\beta$-elimination (dehydrohalogenation)].

Step 2: Determining the Reaction Mechanism

The reaction proceeds via the $E2$ mechanism (Bimolecular Elimination). The base abstracts a proton ($\beta$-hydrogen) from the carbon adjacent to the carbon bonded to the chlorine atom, leading to the formation of a carbon-carbon double bond and the elimination of $HCl$.

Step 3: Analyzing Regioselectivity (Saytzeff's Rule)

In 3-chloropentane, the chlorine atom is on the $C_3$ position. The $\beta$-carbons are the $C_2$ and $C_4$ positions.
Structure: $CH_3^{(1)} - CH_2^{(2)} - CH^{(3)}(Cl) - CH_2^{(4)} - CH_3^{(5)}$
Since the molecule is symmetrical, removing a hydrogen from $C_2$ or $C_4$ yields the same product.

Upon elimination of $HCl$, the double bond forms between $C_2$ and $C_3$ (or $C_3$ and $C_4$):
$CH_3-CH_2-CHCl-CH_2-CH_3 \xrightarrow{KOH(alc.), \Delta} CH_3-CH=CH-CH_2-CH_3 + KCl + H_2O$

Step 4: Final Product Analysis

The major product is pent-2-ene.

• Structure: $CH_3-CH=CH-CH_2-CH_3$
• Classification: Disubstituted alkene.
• Justification: [Per Saytzeff's Rule, the more highly substituted alkene is the major product in dehydrohalogenation reactions].

Final Solution: The major organic product is Pent-2-ene ($CH_3CH=CHCH_2CH_3$).

Step 1: Structural Expansion and Molecular Representation

The given molecular formula is $(CCl_{3})_{3}CCl$. To determine the IUPAC name, we must expand the condensed formula to identify the longest continuous carbon chain and the substituents attached to it.

The central carbon atom is bonded to three $-CCl_{3}$ groups and one chlorine atom ($-Cl$). Structurally, this is expressed as:

Central Carbon $C$ is connected to:

• One Chlorine atom ($-Cl$)
• Three trichloromethyl groups ($-CCl_{3}$)

Expanded visual representation:

Step 2: Identifying the Principal Carbon Chain

Per the IUPAC rules for nomenclature, we must select the longest carbon chain containing the principal functional group. In this symmetric molecule, any choice of two carbon atoms connected through the central carbon will form the longest chain of 2 carbon atoms (an ethane derivative).

If we select a chain of 2 carbons ($C-C$), the remaining components are treated as substituents.

Step 3: Numbering and Substitution Analysis

Let the chain be $C_1-C_2$. One carbon is bonded to three chlorine atoms (the $-CCl_{3}$ group). The other carbon is bonded to three chlorine atoms (from the second $-CCl_{3}$ group), one chlorine atom, and the remaining $-CCl_{3}$ group.

Wait, evaluating the structure $C(CCl_3)_4$ (if simplified) vs the specific question $(CCl_3)_3CCl$: The central carbon is attached to 3 $-CCl_3$ groups and 1 $-Cl$. This molecule is $2,2,2$-trichloro-$1,1,1$-tris(trichloromethyl)ethane? No, that is incorrect. Let us look at the central carbon bonded to three $-CCl_3$ units and one $Cl$. The longest chain is only 2 carbons long (one $C$ of the $CCl_3$ group and the central $C$).

Actually, the structure is $1,1,1,3,3,3$-hexachloro-$2,2$-bis(trichloromethyl)propane? No. Let's re-evaluate:

Central $C$ is attached to:

• $-Cl$
• $-CCl_{3}$ (Carbon 1)
• $-CCl_{3}$ (Carbon 2)
• $-CCl_{3}$ (Carbon 3)

This molecule is 2-chloro-1,1,1,3,3,3-hexachloro-2-(trichloromethyl)propane is not optimal. The IUPAC name is 1,1,1,3,3,3-hexachloro-2-(trichloromethyl)propan-2-yl chloride or more simply: 2-chloro-1,1,1,2,3,3,3-heptachloropropane.

Applying the rule of IUPAC systematically:

Step 4: Final Nomenclature Synthesis

Applying the locants and alphabetical ordering (all are chloro):

The name is derived by identifying the 3-carbon chain. Carbon 2 holds one chlorine and the remaining trichloromethyl groups. Thus, there are three chlorines on $C_1$, one chlorine on $C_2$, and three chlorines on $C_3$.

This equates to: 1,1,1,2,3,3,3-Heptachloropropane.

Final Solution: 1,1,1,2,3,3,3-Heptachloropropane

Step 1: Structural Expansion and Identification of the Principal Carbon Chain

The chemical formula is $(CH_3)_3CCH=CClC_6H_4I-p$. We first expand the structure to identify the longest carbon chain containing the principal functional group (the double bond):

• The $(CH_3)_3C-$ group is a tert-butyl group attached to a carbon atom.
• The structure is: $(CH_3)_3C - CH = C(Cl) - (C_6H_4I)$.
• The $C_6H_4I-p$ indicates a para-iodophenyl group attached to the double-bonded carbon.

Expanding the main chain: $CH_3-C(CH_3)_2-CH=C(Cl)-(4-IC_6H_4)$.

Step 2: Determining IUPAC Nomenclature Priority

Per the IUPAC rules for alkenes with substituents:

1. The parent chain must include the double bond. The longest chain containing the double bond is a 4-carbon chain (but-2-ene derivative).
2. Numbering starts from the end that gives the double bond the lowest possible locant. If we number from the left (near the tert-butyl group), the double bond starts at position 2.

Step 3: Assigning Locants and Substituents

Chain structure: $C^1H_3-C(CH_3)_2-C^2H=C^3(Cl)-C^4(4-iodophenyl)$.

However, strictly speaking, the principal chain involves the alkene. Let us name the substituents:

• Two methyl groups at position 3: $3,3\text{-dimethyl}$.
• One chloro group at position 2: $2\text{-chloro}$.
• One $4\text{-iodophenyl}$ group at position 1: $1\text{-(4-iodophenyl)}$.

Refining the chain for the longest path containing the double bond: $1\text{-(4-iodophenyl)-2-chloro-3,3-dimethylbut-1-ene}$.

Step 4: Verification of Substituent Alphabetical Order

The substituents are: Chloro, Dimethyl, Iodo (as part of the iodophenyl group). According to IUPAC rules, alphabetical order is followed:

• Chloro
• Iodophenyl
• Methyl

Assembling the name: 2-chloro-3,3-dimethyl-1-(4-iodophenyl)but-1-ene.

Structural Representation

Final Solution: 2-chloro-3,3-dimethyl-1-(4-iodophenyl)but-1-ene

Step 1: Reaction Identification and Analysis

The reaction provided is the nucleophilic substitution of $n$-butyl bromide ($CH_3CH_2CH_2CH_2Br$) with potassium cyanide ($KCN$) to form pentanenitrile ($CH_3CH_2CH_2CH_2CN$). Given the primary nature of the alkyl halide and the presence of an ambident nucleophile ($CN^-$), this reaction proceeds via the $S_N2$ (Substitution Nucleophilic Bimolecular) mechanism.

Step 2: Mechanism Workflow

The $S_N2$ mechanism is a single-step concerted process characterized by the simultaneous bond formation and bond breaking [Per Kinetic Theory of $S_N2$ reactions].

• Nucleophilic Attack: The cyanide ion ($:CN^-$), being a strong nucleophile, attacks the electrophilic carbon atom bearing the bromine atom from the backside (180° to the leaving group) to minimize steric repulsion.
• Transition State: A pentacoordinate, high-energy transition state is formed where the carbon is partially bonded to both the incoming nucleophile ($CN^-$) and the leaving group ($Br^-$).
• Departure of Leaving Group: As the $C-CN$ bond strengthens, the $C-Br$ bond weakens and eventually breaks, releasing the bromide ion ($Br^-$).

Step 3: Chemical Equation Representation

The concerted mechanism can be represented as follows:

$[:N \equiv C]^⁻ + CH_3CH_2CH_2CH_2-Br \rightarrow [N \equiv C \cdots CH_2CH_2CH_2CH_3 \cdots Br]^\ddagger \rightarrow CH_3CH_2CH_2CH_2-C \equiv N + Br^-$

Step 4: Visual Mechanism Representation

Step 5: Justification of Stereochemical and Kinetic Outcome

Because the nucleophile attacks from the side opposite to the bromine, this reaction results in Walden Inversion [Per the stereochemical requirements of $S_N2$]. The reaction follows second-order kinetics, where the rate law is defined as:
$Rate = k[CH_3CH_2CH_2CH_2Br]^1[CN^-]^1$
The rate depends on the concentration of both the substrate and the nucleophile, confirming the bimolecular nature of the reaction.

Final Solution: The reaction follows an $S_N2$ mechanism involving a single-step backside attack by the cyanide nucleophile, resulting in the formation of pentanenitrile and the displacement of the bromide ion through a pentacoordinate transition state.

Step 1: Structural Expansion and Identification of the Principal Chain

The given condensed formula is $(CH_{3}C(C_{2}H_{5})_{2}CH_{2}Br)$. Expanding this structure reveals the longest carbon chain containing the functional group ($-Br$):

The structure contains a quaternary carbon bonded to two methyl groups, two ethyl groups, and a $CH_2Br$ group. Let us identify the longest chain containing the halogen atom:

       $CH_{3}$
       |
$CH_{3}—C—CH_{2}Br$
       |
       $C_{2}H_{5}$  (Wait, re-evaluating connectivity for the parent chain)

Correct structural expansion: The central carbon atom is bonded to one $-CH_3$, two $-C_2H_5$ groups, and one $-CH_2Br$ group. The longest chain follows the path through the ethyl group: $CH_3—CH_2—C(CH_3)(CH_2CH_3)—CH_2Br$.

Step 2: IUPAC Nomenclature

1. Select the parent chain: The longest carbon chain containing the $-Br$ substituent consists of 5 carbons (pentane derivative).
2. Numbering: Number the chain starting from the carbon attached to the bromine to give it the lowest possible locant.
   Position 1: $C$ attached to $Br$.
   Position 2: The quaternary carbon atom.
   Positions 3, 4, 5: The ethyl chain extension ($C—C—C$).
3. Identifying Substituents: There is a methyl group and an ethyl group at position 2.

Following IUPAC rules [Per IUPAC Blue Book P-44], the substituents are listed alphabetically: 2-ethyl-2-methylbutyl bromide is the common name, but the IUPAC name is 1-bromo-2-ethyl-2-methylbutane.

Step 3: Classification of the Halide

To classify the halide, we observe the carbon atom directly attached to the halogen ($-Br$):

• The halogen is attached to a primary carbon ($—CH_2Br$).
• A primary carbon is defined as a carbon atom attached to only one other carbon atom [Per Carbon Classification Hierarchy].
• Since the bromine is attached to a $CH_2$ group which is directly bonded to a quaternary carbon, this is classified as a primary ($1^\circ$) alkyl halide.

Summary Data Table

Structural Diagram

Final Solution: The IUPAC name is 1-bromo-2-ethyl-2-methylbutane, and it is classified as a primary ($1^\circ$) alkyl halide.

Given Variables & Initial Setup

The objective is to convert Chloroethane ($CH_3CH_2Cl$) into Butane ($CH_3CH_2CH_2CH_3$). This transformation requires the formation of a carbon-carbon single bond, effectively doubling the length of the alkyl chain. [Per the principles of organic synthesis, the Wurtz reaction is the standard method for coupling two alkyl halide molecules].

Step 1: Chemical Reaction Selection

To synthesize a symmetrical alkane from an alkyl halide, the Wurtz Reaction is employed. This reaction involves the coupling of two molecules of an alkyl halide in the presence of metallic sodium and dry ether as a solvent.

Step 2: Reaction Stoichiometry & Mechanism

The reaction proceeds via a free-radical mechanism where sodium acts as a reducing agent, donating electrons to the alkyl group, which subsequently undergo a coupling process.

The balanced chemical equation is expressed as:

$2CH_3CH_2Cl + 2Na \xrightarrow{\text{dry ether}} CH_3CH_2CH_2CH_3 + 2NaCl$

Step 3: Visual Representation of the Coupling

Step 4: Procedural Requirements & Justifications

• Solvent Choice: The solvent must be dry ether because sodium metal reacts violently with moisture (water) to produce sodium hydroxide and hydrogen gas, which would quench the reaction prematurely [Per chemical safety and reactivity protocols].
• Coupling Mechanism: The carbon atom bearing the chlorine in chloroethane carries a partial positive charge. The sodium metal facilitates a radical transfer, leading to the dimerization of the ethyl group ($CH_3CH_2^\bullet$).

Final Summary/Result

Final Solution: The conversion of Chloroethane to Butane is achieved by reacting Chloroethane with metallic Sodium in the presence of dry ether, yielding Butane and Sodium Chloride as a byproduct according to the Wurtz reaction: $2CH_3CH_2Cl + 2Na \xrightarrow{\text{dry ether}} CH_3CH_2CH_2CH_3 + 2NaCl$.

Step 1: Structural Expansion and Identification of Principal Chain

The given molecular structure is $CH_{3}CH_{2}CH(CH_{3})CH(C_{2}H_{5})Cl$. To identify the IUPAC name, we first expand the condensed formula to determine the longest carbon chain containing the halogen atom ($-Cl$).

Expansion:

      CH3 (substituent)
       |
CH3 - CH2 - CH - CH - CH2 - CH3 (Principal Chain)
                  |
                  Cl

The longest continuous chain containing the carbon bonded to the chlorine atom consists of 6 carbon atoms. [Per IUPAC nomenclature rules, the principal chain must contain the principal functional group, in this case, the chloro substituent].

Step 2: Numbering the Carbon Chain

Numbering must be done to assign the lowest possible locants to the substituents.

• From left to right: $C_1(CH_3), C_2(CH_2), C_3(CH), C_4(CH), C_5(CH_2), C_6(CH_3)$.
• Positions: Chlorine is at $C_3$, Methyl is at $C_4$.
• Alternatively: If numbered from right to left, Chlorine is at $C_4$.

Step 3: IUPAC Naming

The substituents are chloro and methyl. Alphabetical order dictates "chloro" precedes "methyl". The parent alkane is hexane. Name: 3-chloro-4-methylhexane

Step 4: Classification of the Halide

The carbon atom directly attached to the chlorine atom ($C_3$) is bonded to two other carbon atoms ($C_2$ and $C_4$). [Per the definition of alkyl halide classification: A secondary ($2^\circ$) alkyl halide is one where the halogen is bonded to a carbon atom attached to two other carbon atoms].

Step 5: Visual Representation

Final Solution: The IUPAC name is 3-chloro-4-methylhexane, and it is classified as a secondary ($2^\circ$) alkyl halide.

Chemical Foundation of Grignard Reagents

Grignard reagents are organometallic compounds represented by the general formula $R-Mg-X$ (where $R$ is an alkyl or aryl group and $X$ is a halogen). The bond between the Carbon atom and the Magnesium atom ($C-Mg$) is highly polarized, possessing significant ionic character, which renders the carbon atom a strong nucleophile and a potent Brønsted base.

Step 1: Analyzing the Electronic Structure

Due to the electropositivity of Magnesium ($\chi \approx 1.31$) relative to Carbon ($\chi \approx 2.55$), the electron density is localized on the carbon atom [Per Pauling Electronegativity scale]. This results in a partial negative charge on the alkyl group ($\delta^-$) and a partial positive charge on the magnesium atom ($\delta^+$):

$R^{\delta-} - Mg^{\delta+} - X$

Step 2: Interaction with Protic Sources

Water ($H_2O$) is a protic solvent that contains hydrogen atoms bonded to a highly electronegative oxygen atom. These hydrogen atoms possess a partial positive charge ($H^{\delta+}$) and are acidic. When a Grignard reagent encounters even a trace amount of moisture, an acid-base reaction occurs instantaneously [Per Brønsted-Lowry Theory of Acids and Bases]:

$R-Mg-X + H_2O \rightarrow R-H + Mg(OH)X$

Step 3: Analytical Breakdown of the Reaction

• Formation of Hydrocarbon: The nucleophilic alkyl group ($R^-$) abstracts a proton ($H^+$) from the water molecule, leading to the formation of the corresponding alkane ($R-H$).
• Decomposition: The original organometallic reagent is effectively destroyed, losing its utility for subsequent nucleophilic addition or substitution reactions.
• By-product formation: The remaining fragment forms a basic magnesium hydroxy halide ($Mg(OH)X$), which is an inactive precipitate in the context of organic synthesis.

Visual Representation of Nucleophilic Attack

Conclusion on Operational Necessity

Because Grignard reagents are powerful nucleophiles, they react with any substance containing an active hydrogen atom ($O-H$, $N-H$, or $S-H$ bonds). Anhydrous conditions (absence of water) and an inert atmosphere (often diethyl ether or tetrahydrofuran) are strictly required to preserve the structural integrity of the $C-Mg$ bond. Failure to exclude moisture results in the premature protonation of the reagent, leading to yield loss and contamination of the reaction mixture with alkanes.

Final Solution: Grignard reagents are extremely sensitive to protic environments. The highly nucleophilic carbon atom in the $R-Mg-X$ bond acts as a strong base, reacting irreversibly with water to form a stable hydrocarbon ($R-H$) and basic magnesium salts, thereby decomposing the reagent.

Step 1: Chemical Transformation Identification

The conversion of aniline ($C_6H_5NH_2$) to phenylisocyanide ($C_6H_5NC$) is classified as a Carbylamine Reaction (or Isocyanide test). This reaction is specific to primary amines, where the amine reacts with chloroform in the presence of a strong base to produce an isocyanide, characterized by an offensive odor.

Step 2: Balanced Chemical Equation

The reaction involves the substitution of the hydrogen atoms on the nitrogen atom by a carbon atom derived from chloroform ($CHCl_3$). The overall balanced equation is:

$C_6H_5NH_2 + CHCl_3 + 3KOH_{(alc)} \rightarrow C_6H_5NC + 3KCl + 3H_2O$

Step 3: Reaction Mechanism Analysis

The reaction proceeds via the generation of an electrophilic intermediate, dichlorocarbene ($:CCl_2$).

• Step A (Formation of Carbene): Potassium hydroxide ($KOH$) abstracts a proton from chloroform to form a trichloromethyl anion, which subsequently loses a chloride ion to generate dichlorocarbene.
  $CHCl_3 + OH^- \rightarrow :CCl_2 + H_2O + Cl^-$ [Per the Alpha-elimination mechanism]
• Step B (Nucleophilic Attack): The lone pair on the nitrogen of aniline attacks the electron-deficient dichlorocarbene to form an intermediate species.
• Step C (Elimination): Further treatment with base leads to the elimination of two molecules of $HCl$ (neutralized by $KOH$), resulting in the formation of the isocyanide linkage ($-NC$).

Step 4: Visualizing the Structural Change

Step 5: Experimental Conditions

Final Solution: Aniline reacts with chloroform and alcoholic potassium hydroxide to yield phenylisocyanide and potassium chloride as a byproduct, according to the Carbylamine reaction: $C_6H_5NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} C_6H_5NC + 3KCl + 3H_2O$.