• (a) 3-bromocyclohexene: This is an Allylic Halide. The halogen atom ($\text{Br}$) is bonded to an $sp^3$-hybridized carbon atom that is located directly adjacent to a carbon-carbon double bond ($C=C$).
• (b) 1-chlorocyclohexene: This is a Vinylic Halide. The halogen atom ($\text{Cl}$) is bonded directly to one of the $sp^2$-hybridized carbon atoms of the carbon-carbon double bond ($C=C$).
• (c) (1-chloroethyl)benzene: This is a Benzylic Halide. The halogen atom ($\text{Cl}$) is attached to an $sp^3$-hybridized carbon atom that is bonded directly to an aromatic benzene ring.

(i) $(\text{CH}_3)_3\text{CCH}_2\text{Br}$:

The longest continuous carbon chain containing the halogen has 3 carbons (propane). There is a bromine atom at carbon-1 and two methyl groups at carbon-2.

IUPAC Name: 1-Bromo-2,2-dimethylpropane

(ii) $\text{CH}_3\text{CH}=\text{C(Cl)CH}_2\text{CH(CH}_3)_2$:

We number the carbon chain from left to right to give the double bond the lowest possible locant (positions 2 and 3):

• $\text{C}_1$: Methyl group
• $\text{C}_2, \text{C}_3$: Alkene double bond ($\text{-CH}=\text{C-}$)
• $\text{C}_3$: Chlorine atom (chlorido substituent)
• $\text{C}_5$: Methyl group

IUPAC Name: 3-Chloro-5-methylhex-2-ene

Reason 1: Hybridization state of carbon.

In chlorobenzene, the chlorine atom is attached to an $sp^2$-hybridized carbon atom of the aromatic ring. In cyclohexyl chloride, it is attached to an $sp^3$-hybridized carbon atom. Since $sp^2$ carbon has more $s$-character ($33\%$) than $sp^3$ carbon ($25\%$), it is more electronegative, which decreases the shared electron displacement towards the chlorine atom, reducing the $C-Cl$ bond polarity.

Reason 2: Resonance stabilization.

Due to the conjugation of chlorine's lone pairs with the $\pi$-cloud of the benzene ring, the $C-Cl$ bond acquires partial double bond character. This resonance shortening of the bond reduces the overall bond length ($d$). Since dipole moment is a product of charge and distance ($\mu = q \times d$), this drop in bond distance decreases chlorobenzene's dipole moment.

Thermodynamic Explanation:

For a solute to dissolve in a solvent, the energy released when new solute-solvent interactions are formed must be sufficient to overcome both the solute-solute and solvent-solvent intermolecular forces.

When haloalkanes are added to water, the new attractions formed between haloalkanes and water molecules are weak dipole-dipole interactions. These are not strong enough to break the exceptionally strong hydrogen bonds that bind pure water molecules together. Since the hydration energy released is far less than the energy required to break water's hydrogen bonds, haloalkanes remain insoluble in water.

Preferred Method:

The reaction of primary alcohols with thionyl chloride is as follows:

Reason: During this reaction, the other by-products formed (sulfur dioxide, $\text{SO}_2$, and hydrogen chloride, $\text{HCl}$) are both **gases**. They spontaneously bubble out and escape from the reaction mixture, leaving behind the pure liquid alkyl chloride. This eliminates the need for expensive and complex distillation or purification processes.

Reaction Products:

Chlorination of butane yields two structural isomers:

1. 2-Chlorobutane ($\text{CH}_3\text{CH(Cl)CH}_2\text{CH}_3$) — Major Product ($\approx 64\%$)
2. 1-Chlorobutane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl}$) — Minor Product ($\approx 36\%$)

Mechanism Explanation:

The reaction proceeds via a free-radical intermediate. The abstraction of a hydrogen atom from the secondary ($\text{C}_2$ or $\text{C}_3$) carbon generates a secondary ($2^\circ$) butyl radical, whereas abstraction from a methyl ($\text{C}_1$ or $\text{C}_4$) carbon generates a primary ($1^\circ$) butyl radical.

Because secondary free radicals are significantly more stable than primary ones due to hyperconjugation and inductive effects, the formation of the secondary radical is faster, making 2-chlorobutane the major product.

1. Finkelstein Reaction (Preparation of Alkyl Iodides):

This reaction involves reacting an alkyl chloride or bromide with sodium iodide ($\text{NaI}$) dissolved in dry acetone:

The reaction is favored in the forward direction because sodium chloride ($\text{NaCl}$) is insoluble in dry acetone and precipitates, shifting the equilibrium according to Le Chatelier's principle.

2. Swarts Reaction (Preparation of Alkyl Fluorides):

This reaction is the best method for synthesizing alkyl fluorides. It involves heating an alkyl bromide or chloride in the presence of metallic fluorides like silver fluoride ($\text{AgF}$), cobalt fluoride ($\text{CoF}_2$), or antimony trifluoride ($\text{SbF}_3$):

The Sandmeyer reaction is carried out in two distinct synthetic steps:

Step 1: Diazotization.

Aniline ($\text{C}_6\text{H}_5\text{NH}_2$) is dissolved or suspended in cold dilute mineral acid and treated with an aqueous solution of sodium nitrite ($\text{NaNO}_2$) at $273\text{-}278\text{ K}$ ($0\text{-}5^\circ\text{C}$) to form a stable diazonium salt:

Step 2: Replacement of Diazonium Group.

The freshly prepared benzene diazonium chloride solution is mixed with cuprous chloride ($\text{Cu}_2\text{Cl}_2$) dissolved in $\text{HCl}$ to replace the diazonium group with a chlorine atom, releasing nitrogen gas:

Correct Option: (a)

Explanation: The rate of an $S_N2$ reaction is governed by **steric hindrance** around the carbon atom holding the leaving group. Since the nucleophile must attack from the backside, bulky groups block its path, slowing down the reaction. The order of steric hindrance in these alkyl halides is:

• (i) 2-Bromo-2-methylpropane: Tertiary ($3^\circ$) halide — most sterically hindered (slowest).
• (ii) 2-Bromobutane: Secondary ($2^\circ$) halide — moderately hindered.
• (iii) 1-Bromo-2-methylpropane: Primary ($1^\circ$) halide but has a bulky methyl group on the adjacent ($\beta$) carbon.
• (iv) 1-Bromobutane: Primary ($1^\circ$) straight-chain halide — least sterically hindered (fastest).

Thus, the increasing order of reactivity is:

Faster Reactant: **$\text{C}_6\text{H}_5\text{CH}_2\text{Cl}$ (Benzyl chloride)** reacts significantly faster.

Detailed Scientific Reason:

An $S_N1$ reaction proceeds through a two-step mechanism, where the first, rate-determining step involves the slow ionization of the substrate to form a carbocation intermediate.

• In benzyl chloride, the loss of chloride yields the **benzyl carbocation ($\text{C}_6\text{H}_5\text{CH}_2^{+}$)**. This carbocation is exceptionally stable because the positive charge is delocalized over the ortho and para positions of the aromatic ring through **resonance**:
  $$ \text{C}_6\text{H}_5\text{CH}_2^{+} \leftrightarrow \text{Resonance structures delocalizing positive charge} $$
• In ethyl chloride, ionization yields the primary ethyl carbocation ($\text{CH}_3\text{CH}_2^{+}$), which is unstable and only weakly stabilized by the $+I$ inductive effect and minor hyperconjugation.

Because the benzyl carbocation is highly resonance-stabilized, its formation has a much lower activation energy, allowing benzyl chloride to undergo $S_N1$ hydrolysis much faster.

(i) Chiral Carbon Atom: A carbon atom that is bonded to four completely different atoms or groups of atoms is called a chiral (or asymmetric) carbon. It lacks an internal plane of symmetry, making the molecule non-superimposable on its mirror image.

(ii) Enantiomers: Stereoisomers that are non-superimposable mirror images of each other are called enantiomers. They share identical physical properties (such as boiling point, density, and solubility) but rotate plane-polarized light in equal and opposite directions.

(iii) Racemization: The process of converting an optically active compound (a pure dextrorotatory, $d$, or levorotatory, $l$, enantiomer) into an equimolar mixture of both forms, resulting in an optically inactive racemic mixture ($\pm$ or $dl$) due to external compensation.

Stereochemistry of $S_N2$:

An $S_N2$ reaction is a concerted, one-step process that does not form any intermediate. The incoming nucleophile approaches the electrophilic carbon atom from the **backside** (exactly $180^\circ$ opposite to the leaving halide group). This pathway minimizes electrostatic repulsion between the negative charges of the nucleophile and the leaving halide group.

As the nucleophile begins to share its electrons with carbon, the carbon-halogen bond simultaneously weakens, forming a transition state where the central carbon is temporarily bonded to both groups. As the halide ion finally departs from the frontside, the three remaining substituents are pushed to the opposite side (similar to an umbrella turning inside out in a strong wind). This results in a product with a completely **inverted configuration** relative to the starting material.

1. Resonance Effect: The lone pairs of electrons on the halogen atom are in conjugation with the $\pi$-system of the benzene ring. This delocalization imparts a **partial double bond character** to the $C-X$ bond, making it shorter, stronger, and significantly harder to cleave than a single $C-X$ bond in haloalkanes.
2. Hybridization of Carbon: In haloarenes, the halogen is attached to an $sp^2$-hybridized carbon atom, which is more electronegative than the $sp^3$-hybridized carbon in haloalkanes. This increases the electronegativity of the carbon, causing it to hold the electron pair of the $C-X$ bond more tightly and resist nucleophilic cleavage.
3. Instability of Phenyl Cation: In an $S_N1$ mechanism, self-ionization would require the formation of a phenyl cation. Because a positive charge on an $sp^2$-hybridized carbon cannot be stabilized by resonance, the phenyl cation is highly unstable, preventing the reaction from proceeding via this pathway.

The cyanide ion ($\text{CN}^{-}$) is an **ambidentate nucleophile** that can coordinate to a carbon electrophile through either its carbon ($\text{C}$) or nitrogen ($\text{N}$) atom:

• Reaction with $\text{KCN}$: Potassium cyanide is an ionic compound that completely dissociates in solution to provide free, nucleophilic cyanide ($\text{C}\equiv\text{N}^{-}$) ions. Although both carbon and nitrogen are nucleophilic, the attack occurs preferentially through the **carbon atom** because the resulting $C-C$ covalent bond is thermodynamically much stronger and more stable than the alternative $C-N$ bond, forming **alkyl cyanides** (nitriles).
• Reaction with $\text{AgCN}$: Silver cyanide is a predominantly covalent compound. The silver-carbon bond ($\text{Ag-C}$) remains intact, meaning the carbon atom is not free. Thus, only the lone pair on the **nitrogen atom** is available to act as a nucleophile. The attack occurs through nitrogen, producing **alkyl isocyanides** (isonitriles) as the major product.

Reaction Mechanism:

Nucleophilic aromatic substitution ($S_N\text{Ar}$) proceeds via a bimolecular pathway where the nucleophile attacks, forming a resonance-stabilized carbanion intermediate (Meisenheimer complex) before the leaving halide group departs.

• At Ortho and Para Positions: When the nucleophile attacks a haloarene with a nitro group at the ortho or para positions, the negative charge generated in the intermediate is delocalized onto the carbon atom holding the nitro group. Since the nitro group ($\text{-NO}_2$) is a powerful electron-withdrawing group (via both resonance and inductive effects), it directly stabilizes this negative charge by pulling it towards its highly electronegative oxygen atoms.
• At Meta Position: If the nitro group is at the meta position, the negative charge generated during delocalization **never** resides on the carbon atom holding the nitro group (as shown by drawing the resonance structures). Therefore, the nitro group cannot stabilize the carbanion intermediate via its resonance effect, resulting in no significant increase in reactivity.

(a) With Aqueous $\text{KOH}$ (Substitution):

The major product is **Ethanol ($\text{CH}_3\text{CH}_2\text{OH}$)**:

Explanation: In an aqueous solution, the potassium hydroxide dissociates to provide highly hydrated hydroxide ions ($\text{OH}^{-}$). These behave as strong, relatively small nucleophiles and undergo a substitution reaction ($S_N2$), displacing the bromide ion.

(b) With Alcoholic $\text{KOH}$ (Elimination):

The major product is **Ethene ($\text{CH}_2=\text{CH}_2$)**:

Explanation: In an alcoholic solution (ethanol), the hydroxide ions react with solvent molecules to form **ethoxide ions ($\text{C}_2\text{H}_5\text{O}^{-}$)**, which are extremely strong, sterically bulky bases. Instead of attacking the carbon as a nucleophile, the bulky ethoxide ion abstracts a proton from the $\beta$-carbon, triggering a $\beta$-elimination ($E2$) reaction to form an alkene.

Saytzeff's Rule: In a dehydrohalogenation (elimination) reaction of an alkyl halide, if more than one alkene can be formed, the preferred and major product is that alkene which has the **greater number of alkyl groups** attached to the double-bonded carbon atoms (i.e., the more highly substituted, thermodynamically stable alkene).

Dehydrohalogenation of 2-bromobutane:

2-Bromobutane contains two different types of $\beta$-carbons, each holding different hydrogen atoms:

1. Major Product ($\approx 80\%$): But-2-ene ($\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3$). This alkene is symmetrically disubstituted (having two methyl groups on the double bond) and exhibits high thermodynamic stability.
2. Minor Product ($\approx 20\%$): But-1-ene ($\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3$). This alkene is monosubstituted (having only one ethyl group on the double bond) and is less stable.

Grignard Reagents: These are organometallic compounds represented as **$\text{RMgX}$** (alkyl magnesium halides), featuring a direct, highly polar covalent bond between a carbon atom and a magnesium atom.

Synthesis reaction:

Why Anhydrous Conditions are Required:

The carbon-magnesium bond is extremely polar ($\text{R}^{\delta-}-\text{Mg}^{\delta+}\text{-X}$), making the alkyl group a very powerful base and an excellent nucleophile. Grignard reagents react rapidly with any proton source (such as water, alcohols, or amines) to produce hydrocarbons:

If moisture is present, the synthesized Grignard reagent will immediately react with water to form the corresponding alkane, ruining the synthetic step.

1. Wurtz Reaction:

Two molecules of an alkyl halide react with sodium metal in dry ether to form a symmetrical alkane containing twice the number of carbon atoms:

2. Wurtz-Fittig Reaction:

A mixture of an alkyl halide and an aryl halide reacts with sodium metal in dry ether to form an alkylated aromatic hydrocarbon (alkylbenzene):

3. Fittig Reaction:

Two molecules of an aryl halide react with sodium metal in dry ether to form diphenyl (biphenyl):

(i) Propene to 1-iodopropane:

Adding iodine to the terminal carbon of propene cannot be achieved in a single step because direct addition of $\text{HI}$ follows Markovnikov's rule to give 2-iodopropane. Thus, we use an anti-Markovnikov pathway followed by halogen exchange:

1. Anti-Markovnikov Hydrobromination: Treat propene with $\text{HBr}$ in the presence of an organic peroxide to yield 1-bromopropane:
   $$ \text{CH}_3-\text{CH}=\text{CH}_2 + \text{HBr} \xrightarrow{\text{Peroxide}} \text{CH}_3-\text{CH}_2-\text{CH}_2\text{Br} $$
2. Halogen Exchange (Finkelstein Reaction): React 1-bromopropane with sodium iodide ($\text{NaI}$) in dry acetone to obtain 1-iodopropane:
   $$ \text{CH}_3-\text{CH}_2-\text{CH}_2\text{Br} + \text{NaI} \xrightarrow{\text{dry acetone}} \text{CH}_3-\text{CH}_2-\text{CH}_2\text{I} + \text{NaBr}\downarrow $$

(ii) Ethanol to ethyl fluoride:

Direct fluorination of alcohols is highly violent. Instead, we convert the alcohol to an alkyl bromide first, then undergo metal fluoride exchange:

1. Bromination of Ethanol: React ethanol with phosphorus tribromide ($\text{PBr}_3$) to yield ethyl bromide:
   $$ 3\text{CH}_3\text{CH}_2\text{OH} + \text{PBr}_3 \rightarrow 3\text{CH}_3\text{CH}_2\text{Br} + \text{H}_3\text{PO}_3 $$
2. Halogen Exchange (Swarts Reaction): Heat ethyl bromide with silver fluoride ($\text{AgF}$) to yield ethyl fluoride:
   $$ \text{CH}_3\text{CH}_2\text{Br} + \text{AgF} \xrightarrow{\Delta} \text{CH}_3\text{CH}_2\text{F} + \text{AgBr}\downarrow $$