Grignard reagents are decomposed by water as

$$\text{RMgI} + \text{H}_2\text{O} \longrightarrow \text{RH (g)} + \text{Mg(OH)I}$$

Hence the only gaseous product is the corresponding alkane $$\text{RH}$$.

Step 1: Determine the molar mass of the liberated gas

Given: the gas occupies $$1.4\,\text{dm}^3$$ per gram at STP.
Density of the gas, $$d = \dfrac{1\,\text{g}}{1.4\,\text{dm}^3} = 0.714\,\text{g dm}^{-3}$$.

Molar mass, $$M = d \times V_\text{m} = 0.714 \times 22.4 \approx 16.0\,\text{g mol}^{-1}$$.

No alkane other than methane has a molar mass close to $$16\,\text{g mol}^{-1}$$, so

$$\text{RH} \equiv \text{CH}_4,\qquad R = \text{CH}_3\!-\!$$

Therefore the original Grignard reagent is $$\text{CH}_3\text{MgI}$$.

Step 2: Reaction with $$\text{I}_2/\text{HIO}_3$$

$$\text{CH}_4 + \text{I}_2 \xrightarrow[\text{HIO}_3]{} \text{CH}_3\text{I} + \text{HI}$$

The oxidant $$\text{HIO}_3$$ regenerates $$\text{I}_2$$ from the HI formed, driving the iodination forward. Hence

Compound $$(X) = \text{CH}_3\text{I}$$ (iodomethane).

Step 3: Wurtz coupling of (X)

In dry ether, sodium couples two molecules of an alkyl halide:

$$2\,\text{CH}_3\text{I} + 2\,\text{Na} \longrightarrow \text{CH}_3{-}\text{CH}_3 + 2\,\text{NaI}$$

Thus compound $$(Y) = \text{C}_2\text{H}_6$$ (ethane).

Step 4: Molar mass of (Y)

$$M_Y = 2(12.0) + 6(1.0) = 30.0\,\text{g mol}^{-1}$$.

The required molar mass of compound (Y) (nearest integer) is 30.

The given reaction sequence converts an alcohol first into three different alkides P, Q and R in which the carbon is attached to three different hetero-atoms X.
For alcoholic substrates the most common choices of X that are introduced by standard laboratory reagents are the halogens F, Cl and Br.
Hence we set

$$P : \;\; \text{R-Cl}, \qquad Q : \;\; \text{R-Br}, \qquad R : \;\; \text{R-F}$$

All four statements can now be examined one by one.

Case 1 - C-X bond length
Within one row the bond length increases with the size (atomic radius) of the halogen:
$$\text{C-F} (1.35\;\text{Å}) \lt \text{C-Cl} (1.78\;\text{Å}) \lt \text{C-Br} (1.94\;\text{Å}).$$ Therefore $$\text{C-X length : } Q(\text{Br}) \gt P(\text{Cl}) \gt R(\text{F}).$$ Option A claims a different order (Q > R > P), hence A is incorrect.

Case 2 - C-X bond enthalpy
Bond enthalpy falls as the halogen becomes larger and the orbital overlap weakens: $$\text{C-F} \;(485\;\text{kJ mol}^{-1}) \gt \text{C-Cl} \;(327\;\text{kJ mol}^{-1}) \gt \text{C-Br} \;(285\;\text{kJ mol}^{-1}).$$ Thus $$\text{C-X enthalpy : } R(\text{F}) \gt P(\text{Cl}) \gt Q(\text{Br}).$$ This matches Option B exactly, so B is correct.

Case 3 - Relative S$$\mathbf{N}$$2 reactivity
The rate of an S$$\mathbf{N}$$2 reaction depends mainly on how good the leaving group X$$^-$$ is. A better leaving group corresponds to a weaker C-X bond and a stronger conjugate acid HX (lower p$$K_a$$). Hence $$\text{Br}^- \;(\text{best}) \gt \text{Cl}^- \gg \text{F}^-.$$ Consequently $$\text{Reactivity : } Q(\text{Br}) \gt P(\text{Cl}) \gg R(\text{F}).$$ Option C proposes P > R > Q, which is the reverse trend, so C is incorrect.

Case 4 - p$$K_a$$ of the conjugate acids
$$pK_a(\text{HF}) \approx 3.2, \qquad pK_a(\text{HCl}) \approx -7, \qquad pK_a(\text{HBr}) \approx -9.$$ Larger (less negative) p$$K_a$$ means the acid is weaker: $$pK_a : R(\text{HF}) \gt P(\text{HCl}) \gt Q(\text{HBr}).$$ Option D expects R > Q > P, so it is also incorrect.

Only Option B satisfies the correct order in its entirety.

Answer - Option B which is: C-X bond enthalpy in P, Q and R follows the order R > P > Q.

• (A) Iodoform ($$\text{CHI}_3$$): Widely used as an antiseptic for dressing wounds due to the liberation of free iodine.
  $$\implies \text{(A)} \rightarrow \text{(III)}$$
• (B) Carbon tetrachloride ($$\text{CCl}_4$$): Historically used as an industrial fire extinguisher (under the name Pyrene).
  $$\implies \text{(B)} \rightarrow \text{(I)}$$
• (C) CFC (Chlorofluorocarbons): Extensively utilized as synthetic refrigerants and propellants.
  $$\implies \text{(C)} \rightarrow \text{(IV)}$$
• (D) DDT (Dichlorodiphenyltrichloroethane): A well-known organochlorine compound applied as an insecticide.
  $$\implies \text{(D)} \rightarrow \text{(II)}$$

Conclusion:

Aligning the compounds with their standard applications gives the matching sequence: A-III, B-I, C-IV, D-II.

Answer: Option B — A-III, B-I, C-IV, D-II

1. Option A: (Bromomethyl)cyclobutane
   • The carbon attached to $$-\text{Br}$$ is directly bonded to only one other carbon (the cyclobutane ring carbon).
   • This is a primary ($$1^\circ$$) alkyl halide. It has the least steric hindrance, leaving the backside wide open for a nucleophilic attack.
2. Option B:
   • The carbon attached to $$-\text{Br}$$ is bonded to two other carbons (the ring and a methyl group).
   • This is a secondary ($$2^\circ$$) alkyl halide. Increased steric hindrance slows the reaction down.
3. Option C: 1-Bromo-1-methylcyclobutane
   • The carbon attached to $$-\text{Br}$$ is a tertiary ring carbon.
   • This is a tertiary ($$3^\circ$$) alkyl halide. Tertiary halides are too crowded for backside attack and generally do not undergo $$\text{S}_\text{N}2$$ reactions at all.
4. Option D:
   • The carbon attached to $$-\text{Br}$$ is a tertiary ($$3^\circ$$) alkyl halide attached to a bulky neopentyl-like branch system. Extremely sterically hindered.

The general rate of reactivity for $$\text{S}_\text{N}2$$ reactions follows the order:

$$\text{Methyl} > 1^\circ > 2^\circ > 3^\circ$$

We need to evaluate the Assertion (A) and Reason (R) about $$S_N2$$ reactions of benzyl bromide versus ethyl bromide.

Considering Assertion (A): "$$S_N2$$ reaction of $$C_6H_5CH_2Br$$ (benzyl bromide) occurs more readily than the $$S_N2$$ reaction of $$CH_3CH_2Br$$ (ethyl bromide)."

This assertion is correct. Benzyl bromide undergoes $$S_N2$$ reactions faster than ethyl bromide. The relative reactivity order for $$S_N2$$ reactions is: benzyl > allyl > primary > secondary. The enhanced reactivity of benzyl halides in $$S_N2$$ is due to stabilization of the transition state (explained below).

Considering Reason (R): "The partially bonded unhybridized p-orbital that develops in the trigonal bipyramidal transition state is stabilized by conjugation with the phenyl ring."

This reason is correct and provides the correct explanation for Assertion (A). Here is why:

In an $$S_N2$$ mechanism, the nucleophile attacks the substrate carbon from the backside while the leaving group departs simultaneously. The transition state has a trigonal bipyramidal geometry around the central carbon.

In this transition state, the central carbon changes from $$sp^3$$ hybridization (in the reactant) toward $$sp^2$$-like character. The carbon develops a partially bonded, unhybridized p-orbital that is perpendicular to the plane formed by the three remaining substituents.

In benzyl bromide, this developing p-orbital at the transition state can overlap (conjugate) with the $$\pi$$-electron system of the adjacent phenyl ring. This conjugation delocalizes electron density and stabilizes the transition state, thereby lowering the activation energy.

In ethyl bromide, there is no such conjugation possible with the methyl group, so the transition state is not stabilized to the same extent.

Since both (A) and (R) are correct, and (R) is the correct explanation for (A), the correct answer is Option (4): Both (A) and (R) are correct and (R) is the correct explanation of (A).

Mechanism: $$\text{E}2$$ Elimination. Tertiary alkyl halides preferentially undergo elimination over substitution when treated with a strong base.

Formation of Product A (Elimination)

• Reaction: Bromoethane ($$\text{CH}_3\text{CH}_2\text{Br}$$) reacting with $$\text{NaOH}$$ in alcoholic medium ($$\text{C}_2\text{H}_5\text{OH}$$).
• Mechanism: Alcoholic $$\text{NaOH}$$ acts as a strong base, favoring $$\text{E}2$$ dehydrohalogenation.
• Product A: Ethene ($$\text{CH}_2=\text{CH}_2$$)
• Hydrogen Count in A: 4

2. Formation of Product B (Substitution)

• Reaction: Bromoethane ($$\text{CH}_3\text{CH}_2\text{Br}$$) reacting with $$\text{NaOH}$$ in aqueous medium ($$\text{H}_2\text{O}$$).
• Mechanism: Aqueous $$\text{NaOH}$$ provides strong nucleophilic hydroxyl ions ($$\text{OH}^-$$), favoring $$\text{S}_\text{N}2$$ substitution.
• Product B: Ethanol ($$\text{CH}_3\text{CH}_2\text{OH}$$)
• Hydrogen Count in B: 6

We need to identify the correct statement about freons.

Define freons.

Freons are the trade name for a group of chlorofluorocarbon (CFC) compounds. They are organic compounds that contain carbon, chlorine, and fluorine atoms. Common examples include:

- Freon-12 ($$CCl_2F_2$$, dichlorodifluoromethane)

- Freon-11 ($$CCl_3F$$, trichlorofluoromethane)

- Freon-22 ($$CHClF_2$$, chlorodifluoromethane)

Properties and uses.

Freons are chemically inert, non-toxic, non-flammable, and easily liquefiable gases. They were widely used as refrigerants, propellants in aerosol cans, and as solvents. However, they are now largely banned because they cause ozone layer depletion when they reach the stratosphere.

Evaluate the options.

Option (1): "These are chlorofluorocarbon compounds" - This is the correct and precise definition of freons.

Option (2): "These are chemicals causing skin cancer" - This is misleading. While ozone depletion caused by freons can lead to increased UV radiation (which may cause skin cancer), freons themselves are not directly chemicals causing skin cancer.

Option (3): "These are radicals of chlorine and chlorine monoxide" - This is incorrect. Radicals of Cl and ClO are formed when freons decompose in the stratosphere, but freons themselves are not radicals.

Option (4): "All radicals are called freons" - This is completely incorrect.

The correct answer is Option (1): These are chlorofluorocarbon compounds.

We need to find the isomeric deuterated bromide with molecular formula $$C_4H_8DBr$$ that has two chiral carbon atoms.

We start by noting that

Option 1: 2-Bromo-1-deuterobutane ($$CH_2D-CHBr-CH_2-CH_3$$)

C-1 is bonded to: H, H, D, and $$CHBr-CH_2CH_3$$. Since C-1 has two identical H atoms, it is not a chiral centre.

C-2 is bonded to: $$CH_2D$$, Br, H, and $$CH_2CH_3$$. These are four different groups, so C-2 is chiral.

This compound has only 1 chiral centre. Not the answer.

Option 2: 2-Bromo-2-deuterobutane ($$CH_3-CBrD-CH_2-CH_3$$)

C-2 is bonded to: $$CH_3$$, Br, D, and $$CH_2CH_3$$. These are four different groups, so C-2 is chiral.

No other carbon has four different groups attached.

This compound has only 1 chiral centre. Not the answer.

Option 3: 2-Bromo-3-deuterobutane ($$CH_3-CHBr-CHD-CH_3$$)

C-2 is bonded to: $$CH_3$$, H, Br, and $$CHD-CH_3$$. These are four different groups, so C-2 is chiral.

C-3 is bonded to: $$CH_3$$, H, D, and $$CHBr-CH_3$$. Since D and H are different atoms (deuterium is an isotope of hydrogen), these are four different groups, so C-3 is chiral.

This compound has 2 chiral centres. This is our answer.

Option 4: 2-Bromo-1-deutero-2-methylpropane ($$CH_2D-CBr(CH_3)-CH_3$$)

C-2 is bonded to: $$CH_2D$$, Br, $$CH_3$$, and $$CH_3$$. Since two of the groups ($$CH_3$$) are identical, C-2 is not chiral.

This compound has 0 chiral centres. Not the answer.

Deuterium (D) and hydrogen (H) are different isotopes and are treated as different substituents for chirality analysis. This makes C-3 in 2-bromo-3-deuterobutane a chiral centre.

The correct answer is Option (3): 2-Bromo-3-deuterobutane.

The substrate is triphenylmethyl chloride, $$\left(C_6H_5\right)_3CCl$$.

Due to the presence of three bulky phenyl groups around the carbon atom, backside attack by the nucleophile is highly hindered. Therefore, an $$S_N2$$ mechanism is not feasible.

When the carbon-chlorine bond breaks, the triphenylmethyl carbocation,

$$\left(C_6H_5\right)_3C^+$$

is formed.

This carbocation is highly stable because the positive charge is extensively delocalized over the three phenyl rings through resonance.

Hence, the reaction proceeds through an $$S_N1$$ mechanism.

In an $$S_N1$$ reaction, the slow and rate-determining step is the formation of the carbocation.

Therefore, the rate depends only on the concentration of the substrate and is independent of the concentration of the nucleophile.

The rate law is

$$\text{Rate}=k\left[\left(C_6H_5\right)_3CCl\right]$$

Thus, the reaction is first order with respect to the substrate and zero order with respect to $$OH^-$$.

A plot of rate versus substrate concentration must therefore be a straight line passing through the origin.

A plot of rate versus $$OH^-$$ concentration must be a horizontal line because the rate is independent of the nucleophile concentration.

If Option A shows a horizontal line against

$$\left[\left(C_6H_5\right)_3CCl\right]$$

then it cannot be correct, since that would imply zero-order dependence on the substrate.

The graph that correctly represents the rate law

$$\text{Rate}=k\left[\left(C_6H_5\right)_3CCl\right]$$

is the straight-line graph of rate versus substrate concentration.

Hence, the correct graph is Option C.

The given reaction is the Finkelstein reaction, which proceeds through an $$S_N2$$ (bimolecular nucleophilic substitution) mechanism.

In this reaction, an alkyl bromide ($RCH_2Br$) reacts with an iodide ion ($I^-$) in the presence of acetone to form an alkyl iodide ($RCH_2I$).

Consider each statement individually.

For option (C):

Initially, the reactants consist of a neutral alkyl halide and a localized iodide ion ($I^-$), where the entire negative charge resides on a single iodine atom.

During the $$S_N2$$ reaction, the nucleophile attacks the carbon atom while the leaving group departs simultaneously, producing the transition state

$$[I^{\delta-}\cdots CH_2(R)\cdots Br^{\delta-}]^{\ddagger}.$$

In this transition state, the negative charge is distributed over the incoming iodine atom, the central carbon atom, and the leaving bromine atom rather than being localized on a single atom.

As a result, the transition state has a lower charge density and is therefore less polar than the isolated iodide ion.

Hence, statement (C) is correct.

For option (A):

In the Finkelstein reaction, acetone dissolves sodium iodide but not sodium bromide.

As the reaction proceeds,

$$NaBr$$

precipitates out of the solution, removing $$Br^-$$ ions from the reaction mixture.

Therefore, $$Br^-$$ cannot effectively act as a competing nucleophile.

Hence, statement (A) is incorrect.

For option (B):

Acetic acid is a polar protic solvent.

Polar protic solvents strongly solvate nucleophilic anions such as $$I^-$$ through hydrogen bonding, significantly reducing their nucleophilicity and slowing the $$S_N2$$ reaction.

The Finkelstein reaction requires a polar aprotic solvent such as acetone.

Hence, statement (B) is incorrect.

For option (D):

Acetone is a polar aprotic solvent and does not strongly solvate anions because it cannot form hydrogen bonds with them.

Moreover, the rate-determining step involves a transition state rather than freely solvated ions.

Therefore, the statement that the solvent solvates the ions formed in the rate-determining step is incorrect.

Hence, statement (D) is incorrect.

Therefore, the only correct statement is

(C) The transition state is less polar than the iodide ion.

Counting chlorine atoms in each compound:

Chloral: CCl$$_3$$CHO — contains 3 chlorine atoms

Gammaxene (Lindane/BHC): C$$_6$$H$$_6$$Cl$$_6$$ — contains 6 chlorine atoms

Chloropicrin: CCl$$_3$$NO$$_2$$ — contains 3 chlorine atoms

Freon-12: CCl$$_2$$F$$_2$$ — contains 2 chlorine atoms

The compound with the maximum number of chlorine atoms is Gammaxene (C$$_6$$H$$_6$$Cl$$_6$$) with 6 chlorine atoms.

The melting point of a compound depends heavily on its crystal lattice packing efficiency and molecular symmetry, rather than just its dipole moment.
para-Dichlorobenzene (Highest Melting Point): The para isomer is highly symmetrical. This symmetry allows the molecules to pack very closely and tightly into the crystal lattice.
ortho-Dichlorobenzene (Intermediate Melting Point): The ortho isomer has a high dipole moment due to the close proximity of the two chlorine atoms, leading to relatively strong dipole-dipole interactions, which gives it a higher melting point.
meta-Dichlorobenzene (Lowest Melting Point): The meta isomer lacks the perfect symmetry of the para form and doesn't possess a dipole moment as strong as the ortho form. This results in weaker packing efficiency and weaker intermolecular forces.

Therefore, Option D is correct.

• (A) Boiling Point: Correct. For homologous series (like alkyl chlorides), boiling point increases with increasing molecular mass (chain length) due to stronger van der Waals forces.
  • Ethyl chloride < Propyl chloride < Butyl chloride
• (B) Density: Incorrect. Density generally increases with the atomic mass of the halogen (Cl < Br < I). The order shown suggests a bromide is less dense than a chloride, which is incorrect for similar alkyl groups.
• (C) Boiling Point: Correct. Boiling point increases with molecular mass. A polyhalogenated compound like ($$CBr_4$$ or $$CHBr_3$$) would have a much higher boiling point than ethyl bromide.
• (D) Density: Incorrect. Density increases with the number of halogen atoms and the mass of the halogen. A compound with both Iodine and Bromine would typically be denser than one with just Chlorine and Bromine.
• (E) Boiling Point: Correct. For isomeric compounds, branching decreases the surface area, which weakens van der Waals forces and lowers the boiling point.
  • n-butyl chloride > isobutyl chloride > tert-butyl chloride

Rule: The rate of an SN2 reaction decreases as the steric hindrance around the leaving group increases:
Primary 1 > Secondary 2>Tertiary 3 degree
hence a is correct.

For SN1 reaction, The rate of an  reaction depends directly on the stability of the carbocation intermediate formed during the rate-determining step, Benzylic=Allylic > Tertiary > Secondary > Primary
hence B is correct

Electrophilic Aromatic Substitution (EAS)

• Rule: Electron-donating groups (EDGs) activate the aromatic ring and increase the reaction rate, while electron-withdrawing groups (EWGs) deactivate the ring and decrease the reaction rate
  Chlorobenzene >1-chloro-4-nitrobenzene
  hence C is correct
  Nucleophilic Aromatic Substitution
  The presence of strong electron-withdrawing groups (-NO2) at the ortho or para positions relative to the leaving group stabilizes the intermediate Meisenheimer complex via resonance, significantly boosting reactivity.
  hence d correct

Firstly the Carbocation is formed near the ring. The ring expands to relieve steric strain and forms a ring of 5 carbons. Now, the carbocation gets attached to the Br- molecule and product C is formed.

1-Bromopropane (CH₃CH₂CH₂Br) reacts with different reagents:

A. KOH (alc): Alcoholic KOH causes elimination (dehydrohalogenation) → produces alkene (propene) → Matches III

B. KCN (alc): CN⁻ from KCN acts as a nucleophile attacking through carbon → produces nitrile (butanenitrile) → Matches I

C. AgNO₂: Silver nitrite reacts with alkyl halides through nitrogen (Ag favors O-attack but with alkyl halides, AgNO₂ gives nitroalkane via N) → Actually, AgNO₂ gives nitroalkane (R-NO₂) because Ag⁺ being a soft acid favors the N-bound product → Matches IV (Nitroalkane)

D. CH₃COOAg: Silver acetate reacts with alkyl halide → produces ester (propyl acetate) → Matches II

The matching is: A-III, B-I, C-IV, D-II

This matches option 1.

4-Methyloct-1-ene is $$C_9H_{18}$$ with molar mass
$$M_{\text{alkene}} = 9(12) + 18(1) = 126\text{ g mol}^{-1}$$.

Moles of the alkene taken:
$$n_{\text{alkene}} = \frac{2.52\text{ g}}{126\text{ g mol}^{-1}} = 0.020\text{ mol}$$.

In the presence of peroxide, $$HBr$$ adds anti-Markovnikov. Hence the major product is the primary bromide $$\text{1-bromo-4-methyloctane}$$; the minor is the corresponding secondary bromide. Given overall yield $$= 50\%$$, the moles of both bromides actually obtained are

$$n_{\text{bromides}} = 0.020 \times 0.50 = 0.010\text{ mol}$$.

The isomer ratio is $$9:1$$ (primary : secondary). Therefore moles of the primary bromide are

$$n_{\text{primary}} = 0.010 \times \frac{9}{10} = 0.009\text{ mol}$$.

This entire amount is allowed to react with diethylamine $$(C_2H_5)_2NH$$. An $$S_N2$$ substitution gives the tertiary amine

$$S = C_9H_{19}N(C_2H_5)_2 \;=\; C_{13}H_{29}N$$.

Its molar mass is
$$M_S = 13(12) + 29(1) + 14 = 156 + 29 + 14 = 199\text{ g mol}^{-1}$$.

Because the reaction yield of this step is 100 %:

$$m_S = n_{\text{primary}}\;M_S = 0.009\text{ mol}\;\times\;199\text{ g mol}^{-1} = 1.791\text{ g}$$.

Converting to milligrams:
$$1.791\text{ g} = 1791\text{ mg}$$.

Hence, the mass of the non-ionic product $$S$$ obtained is 1791 mg.

Compound (D): Has three strong electron-withdrawing nitro groups (NO2) at the two ortho positions and the para position. This deactivates the ring to the maximum extent towards electrons but makes it highly reactive toward nucleophilic attack.
Compound (B): Has one strong electron-withdrawing nitro group (NO2) at the ortho position, which activates the ring for nucleophilic substitution.

Compound (A): Chlorobenzene has no extra activating or deactivating groups on the ring. It undergoes substitution only under very drastic conditions

Compound (C): Has a strong electron-donating methoxy group (-OMe) at the ortho position. Via its +M resonance effect, it increases electron density on the ring, making the nucleophilic attack extremely difficult. This is the least reactive compound

hence , D>B>A>C

The given reactions involve substitution of bromine by the $$OH^{-}$$ ion. Whether the product keeps, inverts or partially loses the original configuration depends on (i) the mechanism ($$S_N2$$ or $$S_N1$$) and (ii) the position of the stereogenic centre(s) in the molecule.

Case P : $$(-)\text{-1-Bromo-2-ethylpentane}\xrightarrow[S_N2]{aq.\,NaOH}$$

• In $$S_N2$$ the attack is backside and therefore the carbon that bears the leaving group undergoes inversion. 
• Here the leaving group is on C-1, which is not a stereogenic centre (it is $$CH_2Br$$). 
• The only stereogenic centre is C-2. Because that carbon is not touched during the reaction, its configuration is retained.
Hence the optical purity of the starting material is preserved: the product shows retention of configuration.

⇒ P → (2)

Case Q : $$(-)\text{-2-Bromopentane}\xrightarrow[S_N2]{aq.\,NaOH}$$

• The bromine is on the stereogenic C-2. 
• An $$S_N2$$ displacement on this carbon forces backside attack and gives one step inversion. 
Therefore the configuration is converted into the opposite enantiomer: the product shows inversion of configuration.

⇒ Q → (1)

Case R : $$(-)\text{-3-Bromo-3-methylhexane}\xrightarrow[S_N1]{aq.\,NaOH}$$

• C-3 is tertiary; ionisation of $$C-Br$$ gives a planar carbocation. 
• Nucleophilic attack by $$OH^{-}$$ is equally probable from either face. 
• No other stereogenic centre is present. Thus the product is obtained as an equal (50 : 50) pair of enantiomers, i.e. a racemic mixture.

⇒ R → (3)

Case S : (The substrate possesses the leaving group on one stereogenic centre and contains at least one more untouched stereogenic centre.) In an $$S_N1$$ process the reacting centre racemises while the remote centre retains its configuration. Consequently the product set contains stereoisomers that are not mirror images but differ in the configuration of only one of the centres — they are diastereomers.

⇒ S → (5)

Collecting all matches:
P → 2; Q → 1; R → 3; S → 5

Therefore the correct option is
Option B which is: P → 2; Q → 1; R → 3; S → 5

The structural formula of 2,2,5,5-tetramethylhexane can be written as:

$$\text{(CH}_3)_3\text{C} - \text{CH}_2 - \text{CH}_2 - \text{C(CH}_3)_3$$

This molecule is highly symmetrical, with a plane of symmetry right down the middle of the C3-C4 bond.

There are only two positions containing hydrogen atoms where chlorination can occur:

• Position A (Methyl groups): The 12 hydrogens on the four terminal $$-\text{CH}_3$$ groups attached to the quaternary carbons (C2 and C5) are completely equivalent.
• Position B (Methylene groups): The 4 hydrogens on the two $$-\text{CH}_2-$$ groups (C3 and C4) are also completely equivalent.

Chlorination at Position A (Methyl Groups)

Substituting one hydrogen from any of the methyl groups gives:

$$\text{Cl-CH}_2 - \text{C(CH}_3)_2 - \text{CH}_2 - \text{CH}_2 - \text{C(CH}_3)_3$$

• Chirality: The carbon attached to the $$-\text{CH}_2\text{Cl}$$ group is bonded to two identical methyl groups ($$-\text{CH}_3$$). Therefore, this molecule has no chiral center and is achiral.
• Isomer Count: 1 (structural isomer)
  

Chlorination at Position B (Methylene Groups)

Substituting one hydrogen from either C3 or C4 gives:

$$\text{(CH}_3)_3\text{C} - \text{CH(Cl)} - \text{CH}_2 - \text{C(CH}_3)_3$$

• Chirality: Let's look at the groups attached to the chlorinated carbon (C3):
  1. $$-\text{H}$$
  2. $$-\text{Cl}$$
  3. $$-\text{C(CH}_3)_3$$ (tert-butyl group)
  4. $$-\text{CH}_2-\text{C(CH}_3)_3$$ (neopentyl group)

  Since all 4 groups attached to C3 are completely different, C3 is a chiral (asymmetric) carbon center.

• Isomer Count: Because it contains one chiral center, it exists as a pair of enantiomers ($d$- and $l$-forms). This gives 2 stereoisomers.

$$\text{Total Isomeric Monochloro Derivatives} = 1 \text{ (achiral)} + 2 \text{ (enantiomers)} = \mathbf{3}$$

This reaction involves the precipitation of silver chloride from an organic molecule when treated with silver nitrate. The number of moles of formed depends on the reactivity of each chlorine atom, specifically their ability to leave and form a stable carbocation.

In the given molecule, we have four different chlorine environments:

• Vinylic Chlorine: Attached to a double bond; unreactive because the C-Cl bond has partial double-bond character due to resonance.
• Aromatic (Chlorobenzene) Chlorine: Directly on the benzene ring; unreactive because of the strong sp2 C-Cl bond and resonance.
• Benzylic Chlorine: On the carbon adjacent to the ring. This is highly reactive because the resulting carbocation is resonance-stabilized by the benzene ring.
• Tertiary Chlorine: On the far right. This is also reactive because it forms a stable carbocation.

Ethylidene chloride is a common name for a compound where two chlorine atoms are attached to the same carbon of ethane.

Identify the structure of ethylidene chloride: The prefix "ethylidene" refers to the $$CH_3CH=$$ group (a divalent group derived from ethane by removing two hydrogens from the same carbon). When two chlorine atoms attach to this group, we get:

$$CH_3CHCl_2$$

Apply IUPAC nomenclature: The parent chain is ethane (2 carbons). Both chlorine atoms are on carbon-1 (the carbon bearing the most substituents when numbering to give the lowest locants).

$$\Rightarrow$$ 1, 1-dichloroethane

Evaluate the options: Option A: 1-chloroethene — This would be vinyl chloride ($$CH_2=CHCl$$), which is incorrect.

Option B: 1, 2-dichloroethane — This is ethylene dichloride ($$ClCH_2CH_2Cl$$), where chlorines are on different carbons. Incorrect.

Option C: 1, 1-dichloroethane — This matches $$CH_3CHCl_2$$. Correct.

Option D: 1-chloroethyne — This would be $$HC \equiv CCl$$, which is incorrect.

The correct answer is Option C: 1, 1-dichloroethane.

Explanation: The final product is triphenylmethanol $$\mathrm{Ph_3C-OH}$$.

In the reaction of an ester with a Grignard reagent, the ester provides the carbonyl carbon and one substituent attached to it.

Methyl benzoate $$\mathrm{Ph-CO-OCH_3}$$ already contains one phenyl group.

Therefore, the remaining two phenyl groups must come from the Grignard reagent.

Hence, the Grignard reagent formed in step (i) must be phenylmagnesium bromide $$\mathrm{Ph-MgBr}$$.

The reactant required to form this Grignard reagent is bromobenzene $$\mathrm{Ph-Br}$$.

The first equivalent of $$\mathrm{Ph-MgBr}$$ converts the ester into benzophenone $$\mathrm{Ph-CO-Ph}$$.

The second equivalent attacks benzophenone to form the alkoxide $$\mathrm{Ph_3C-O^-}$$.

Acidic hydrolysis then gives triphenylmethanol $$\mathrm{Ph_3C-OH}$$.

Therefore, reactant $$\mathrm{A}$$ is bromobenzene $$\mathrm{Ph-Br}$$.

The final product is 4-bromophenylacetic acid.

$$Br-C_6H_4-CH_2COOH$$

Step 3 involves acidic hydrolysis using $$H_2O/H^+/\Delta$$. Nitriles undergo hydrolysis under these conditions to form carboxylic acids.

$$R-C\equiv N \xrightarrow{H_2O/H^+,\Delta} R-COOH$$

Therefore, the compound before hydrolysis must be 4-bromophenylacetonitrile.

$$Br-C_6H_4-CH_2CN$$

Step 2 involves treatment with $$CN^-$$, which undergoes nucleophilic substitution and replaces a halogen atom.

$$R-CH_2Cl \xrightarrow{CN^-} R-CH_2CN$$

Hence, the precursor for this step must be 4-bromobenzyl chloride.

$$Br-C_6H_4-CH_2Cl$$

Step 1 involves $$Cl_2/h\nu$$, which carries out free-radical substitution at the benzylic position.

$$Ar-CH_3 \xrightarrow{Cl_2/h\nu} Ar-CH_2Cl$$

Therefore, the starting compound $$A$$ must be 4-bromotoluene.

The reaction sequence is

$$Br-C_6H_4-CH_3 \xrightarrow{Cl_2/h\nu} Br-C_6H_4-CH_2Cl$$

$$Br-C_6H_4-CH_2Cl \xrightarrow{CN^-} Br-C_6H_4-CH_2CN$$

$$Br-C_6H_4-CH_2CN \xrightarrow{H_2O/H^+,\Delta} Br-C_6H_4-CH_2COOH$$

Hence, compound $$A$$ is 4-bromotoluene.

Therefore, the correct answer is Option (C).

In the first step, the alkene reacts with bromine in the presence of methanol.

$$\mathrm{Br_2/CH_3OH}$$

The $$\mathrm{\pi}$$ electrons of the double bond attack $$\mathrm{Br_2}$$ to form a cyclic bromonium ion intermediate.

Since methanol is present in excess, it acts as the nucleophile instead of $$\mathrm{Br^-}$$.

Methanol attacks the more substituted carbon because it carries greater partial positive character.

After deprotonation, intermediate $$\mathrm{A}$$ is formed.

$$\mathrm{Alkene \xrightarrow{Br_2/CH_3OH} Bromomethoxy\ Compound\ (A)}$$

Intermediate $$\mathrm{A}$$ contains:

1. A methoxy group $$\mathrm{(-OCH_3)}$$ on the tertiary carbon

2. A bromine atom on the adjacent secondary carbon

In the second step, compound $$\mathrm{A}$$ reacts with hydroiodic acid.

$$\mathrm{HI}$$ protonates the ether oxygen, converting it into a good leaving group.

$$\mathrm{R-OCH_3 + H^+ \longrightarrow R-O^+HCH_3}$$

Since the oxygen is attached to a tertiary carbon, cleavage occurs through an $$\mathrm{S_N1}$$ mechanism.

Methanol leaves, generating a stable tertiary carbocation.

$$\mathrm{R-O^+HCH_3 \longrightarrow R^+ + CH_3OH}$$

The iodide ion then attacks the tertiary carbocation.

$$\mathrm{R^+ + I^- \longrightarrow R-I}$$

The bromine atom on the neighbouring carbon remains unchanged throughout the reaction.

Thus, in the final product $$\mathrm{B}$$:

1. The bromine atom remains on the secondary carbon

2. The methoxy group is replaced by iodine

Therefore, the correct structure is:

$$\boxed{\mathrm{Option\ containing\ Br\ and\ I\ on\ adjacent\ carbons}}$$

We need to find the total number of isomers (including stereoisomers) obtained on monochlorination of methylcyclohexane.

Methylcyclohexane has a cyclohexane ring with a methyl group at C1. The distinct hydrogen positions are:

- $$CH_3$$ group (methyl hydrogens): 3 equivalent H atoms

- C1 (tertiary hydrogen): 1 H atom

- C2 and C6 (equivalent by symmetry): each has 2 H atoms

- C3 and C5 (equivalent by symmetry): each has 2 H atoms

- C4: 2 H atoms

(i) Chlorination at the methyl group:

Replacing one H on $$CH_3$$ gives a $$CH_2Cl$$ group attached to the ring.

No chiral center is created. This gives 1 product.

(ii) Chlorination at C1 (tertiary position):

Cl replaces the tertiary H. C1 now bears $$CH_3$$, Cl, and two ring carbons. The molecule has a plane of symmetry through C1-C4, so no chirality.

This gives 1 product.

(iii) Chlorination at C2 (or C6):

This produces 1-methyl-2-chlorocyclohexane. Both C1 and C2 become stereocenters.

- cis-1-methyl-2-chlorocyclohexane: The two substituents are on the same side of the ring. C1 and C2 are both chiral centers with no internal plane of symmetry, giving a pair of enantiomers = 2 stereoisomers.

- trans-1-methyl-2-chlorocyclohexane: The two substituents are on opposite sides. Again, C1 and C2 are chiral centers giving a pair of enantiomers = 2 stereoisomers.

Subtotal from C2/C6 position: 4 stereoisomers.

(iv) Chlorination at C3 (or C5):

This produces 1-methyl-3-chlorocyclohexane. Both C1 and C3 are stereocenters.

- cis-1-methyl-3-chlorocyclohexane: Enantiomeric pair = 2 stereoisomers.

- trans-1-methyl-3-chlorocyclohexane: Enantiomeric pair = 2 stereoisomers.

Subtotal from C3/C5 position: 4 stereoisomers.

(v) Chlorination at C4:

This produces 1-methyl-4-chlorocyclohexane. The molecule has a plane of symmetry through C1 and C4.

- cis-1-methyl-4-chlorocyclohexane: Has a plane of symmetry, so it is achiral = 1 stereoisomer.

- trans-1-methyl-4-chlorocyclohexane: Also has a plane of symmetry, so it is achiral = 1 stereoisomer.

Subtotal from C4 position: 2 stereoisomers.

Therefore, the total number of isomers (including stereoisomers) is $$\boxed{12}$$.

An ambident nucleophile is a nucleophile that has two different nucleophilic sites through which it can attack. The key pairs to consider are those where the silver salt and potassium salt of the same anion react through different atoms.

(A) In the case of $$\text{CN}^-$$, the cyanide ion can attack through either carbon or nitrogen. $$\text{KCN}$$ typically gives cyanides (attack through carbon), while $$\text{AgCN}$$ gives isocyanides (attack through nitrogen). This demonstrates the ambident nature of $$\text{CN}^-$$.

(B) The carboxylate ion $$\text{RCOO}^-$$ has two equivalent oxygen atoms, but both give the same product (an ester). The silver and potassium salts do not produce fundamentally different products based on different nucleophilic sites, so this is not a typical ambident nucleophile pair.

(C) The nitrite ion $$\text{NO}_2^-$$ can attack through either nitrogen or oxygen. $$\text{KNO}_2$$ gives nitro compounds (attack through nitrogen), while $$\text{AgNO}_2$$ gives nitrite esters (attack through oxygen). This demonstrates ambident behavior.

(D) The iodide ion $$\text{I}^-$$ has only one nucleophilic site, so it is not an ambident nucleophile.

Therefore, the correct pairs of ambident nucleophiles are (A) and (C) only.

The gas described in the question is unreactive in the lower atmosphere (troposphere), non-toxic and non-flammable, and is able to reach the stratosphere where it undergoes photochemical dissociation by ultraviolet radiation.

These are the characteristic properties of chlorofluorocarbons (CFCs), commonly known as Freons. The most common example discussed in standard textbooks is Freon-12, $$\mathrm{CF_2Cl_2}$$. Because CFCs are chemically inert in the lower atmosphere, they survive long enough to drift into the stratosphere.

Once CFCs reach the stratosphere, they are exposed to high-energy ultraviolet (UV) radiation. This radiation provides sufficient energy to cause homolytic cleavage of their chemical bonds.

Since the carbon-chlorine bond is weaker than the carbon-fluorine bond, the $$\mathrm{C-Cl}$$ bond breaks preferentially.

The photochemical reaction is

$$\mathrm{CF_2Cl_2(g)\xrightarrow{UV}Cl^\bullet(g)+CF_2Cl^\bullet(g)}$$

Here,

• $$\mathrm{Cl^\bullet}$$ is the chlorine free radical.
• $$\mathrm{CF_2Cl^\bullet}$$ is the chlorodifluoromethyl free radical.

The chlorine free radical subsequently reacts with ozone ($$\mathrm{O_3}$$), initiating a chain reaction that leads to ozone depletion.

Therefore, the initial intermediates formed by the UV dissociation of gas A ($$\mathrm{CF_2Cl_2}$$) are $$\mathrm{Cl^\bullet}$$ and $$\mathrm{CF_2Cl^\bullet}$$.

Hence, the correct answer is $$\boxed{\text{Option (B)}}$$.

To determine which compound forms a yellow precipitate under the given conditions, we need to understand the role of each reagent and the reactivity of the different carbon-halogen bonds present in the compounds.

The first reagent is $$\mathrm{NaOH}$$.

Aqueous $$\mathrm{NaOH}$$ acts as a nucleophile and displaces reactive halogen atoms through nucleophilic substitution reactions, releasing halide ions into the solution.

The second reagent is dilute $$\mathrm{HNO_3}$$.

Its purpose is to neutralize the excess $$\mathrm{NaOH}$$ so that silver oxide does not form when silver nitrate is added.

The third reagent is $$\mathrm{AgNO_3}$$.

It reacts with the liberated halide ions to form insoluble silver halides.

The colours of the precipitates are:

• $$\mathrm{AgCl}$$ : White precipitate
• $$\mathrm{AgBr}$$ : Pale yellow precipitate
• $$\mathrm{AgI}$$ : Yellow precipitate

Therefore, a yellow precipitate will be obtained only if iodide ions $$\left(\mathrm{I^-}\right)$$ are released into the solution.

Now let us examine the nature of the carbon-halogen bonds in the given compounds.

Halogens directly attached to a benzene ring are aryl halides. These are highly resistant to nucleophilic substitution because the carbon-halogen bond acquires partial double bond character through resonance.

On the other hand, alkyl and benzylic halides undergo nucleophilic substitution readily. Benzylic halides are particularly reactive because the intermediate formed during substitution is resonance stabilized.

Compound I contains chlorine attached directly to the benzene ring. It is an aryl halide and does not react with $$\mathrm{NaOH}$$.

Compound II contains two aryl chlorides. Neither of them undergoes substitution under these conditions.

Compound III contains bromine directly attached to the benzene ring. It is also an aryl halide and remains unreactive.

Compound IV contains both an aryl chlorine and a benzylic iodine atom.

The benzylic carbon undergoes nucleophilic substitution easily.

$$\mathrm{Ar{-}CH_2I + NaOH \rightarrow Ar{-}CH_2OH + NaI}$$

The iodide ions released then react with silver nitrate.

$$\mathrm{Ag^+ + I^- \rightarrow AgI,(yellow\ precipitate)}$$

Hence, Compound IV is the only compound that produces the characteristic yellow precipitate.

Therefore, the correct answer is Option (D) IV.

Density of aromatic halogen compounds increases with:

1. Increase in number of halogen atoms

2. Increase in atomic mass of halogen atoms

This is due to increase in molar mass.

(A)

$$\mathrm{C_6H_6}$$

(Benzene)

Contains no halogen atom.

Hence, it has the lowest density.

(B)

$$\mathrm{C_6H_5Cl}$$

(Chlorobenzene)

One hydrogen atom is replaced by chlorine.

Since chlorine is heavier than hydrogen, density increases.

(C)

$$\mathrm{C_6H_4Cl_2}$$

(1,3-Dichlorobenzene)

Contains two chlorine atoms.

Hence, density is greater than chlorobenzene.

(D)

$$\mathrm{C_6H_4BrCl}$$

(1-Bromo-3-chlorobenzene)

Contains one bromine and one chlorine atom.

Since bromine has much larger atomic mass than chlorine:

$$\mathrm{Br > Cl}$$

this compound has the highest density.

Therefore, decreasing order of density is:

$$\mathrm{(D) > (C) > (B) > (A)}$$

Correct Order: (i) < (ii) < (iii) < (iv)

The Core Concept: Nucleophilic Aromatic Substitution (SNAr) Normally, aryl halides (like chlorobenzene) are highly unreactive towards nucleophilic substitution. This is because the carbon-chlorine bond has partial double-bond character due to resonance, making it hard to break. Furthermore, the electron-rich benzene ring naturally repels incoming nucleophiles.

The Role of Electron-Withdrawing Groups To make an aryl halide reactive, you need strong electron-withdrawing groups (EWGs), such as the nitro group (-NO₂), attached to the ring.

• Location matters: These groups are most effective when located at the ortho and para positions relative to the halogen.
• Mechanism: During the reaction, an intermediate with a negative charge (a carbanion) is formed. The -NO₂ groups at the ortho and para positions help stabilize this negative charge through strong resonance (-R) and inductive (-I) effects. The more stabilized this intermediate is, the faster the reaction happens.

Evaluating the Compounds: The tendency to undergo nucleophilic substitution increases directly with the number of electron-withdrawing -NO₂ groups at the ortho and para positions:

• (i) Chlorobenzene: Has zero -NO₂ groups. The ring is electron-rich, making it the least reactive.
• (ii) 1-Chloro-4-nitrobenzene: Has one -NO₂ group at the para position. It is more reactive than chlorobenzene.
• (iii) 1-Chloro-2,4-dinitrobenzene: Has two -NO₂ groups (one ortho, one para). The combined withdrawing effect makes it significantly more reactive than (ii).
• (iv) 1-Chloro-2,4,6-trinitrobenzene: Has three -NO₂ groups (two ortho, one para). This ring is extremely electron-deficient, making the intermediate highly stable. This is the most reactive compound of the set.

Therefore, the reactivity steadily increases as you add more nitro groups: (i) < (ii) < (iii) < (iv).

First reaction proceeds via $$E_1$$ reaction allowing for carbocation rearrangement. Second reaction proceeds via $$E_2$$ due to a bulky base and thus no carbocation is formed.

Alkyl iodides are often prepared by the reaction of alkyl chlorides/bromides with NaI in dry acetone. This reaction is known as the Finkelstein reaction.

R-X + NaI → R-I + NaX

(where X = Cl, Br)

First reaction is addition of HBr in presence of peroxide which takes place according to anti-markonikov's rule.

The synthesis of alkyl fluorides is best accomplished by heating an alkyl chloride/bromide in the presence of a metallic fluoride such as AgF, Hg₂F₂, CoF₂ or SbF₃. The reaction is termed as Swarts reaction.

$$H_3C-Br + AgF \longrightarrow H_3C-F + AgBr$$, however this does not effect the bromine directly attached to the benzene ring.

The substrate given in the question is $$(S)\;CH_3-CH(Br)-CH_2-CH_3$$ ( $$(S)$$-2-bromobutane ). It is treated with aqueous hydroxide ion, and the statement in the question specifically says that the substitution proceeds by an $$S_N2$$ mechanism.

Step 1 : Nature of the mechanism
An $$S_N2$$ (bimolecular nucleophilic substitution) reaction is characterised by
  • a single, concerted transition state and
  • backside attack of the nucleophile on the carbon that bears the leaving group.
Because of the backside attack, the configuration at the stereogenic carbon is inverted. This stereochemical change is called Walden inversion.

Step 2 : Applying inversion to the present molecule
The nucleophile here is $$HO^-$$ and the leaving group is $$Br^-$$. The stereogenic carbon is $$C_2$$ of the butane chain. Since the starting configuration about $$C_2$$ is $$(S)$$, after complete inversion the configuration must become $$(R)$$.

Step 3 : Writing the product
Replacing $$Br$$ by $$OH$$ with inversion therefore gives $$(R)\;CH_3-CH(OH)-CH_2-CH_3,$$ i.e. $$(R)$$-butan-2-ol.

Thus the major product is $$(R)$$-2-butanol.

Option A which is: $$(\mathbf{R})$$-butan-2-ol

Grignard Reagent is formed in presence of anhydrous solvent.

The starting material is 3-bromo-2,2-dimethylbutane. Numbering the butane backbone from left to right (C1-C2-C3-C4), carbon C2 bears two methyl substituents and carbon C3 bears the bromine. The explicit structure is $$\text{CH}_3\text{-C(CH}_3)_2\text{-CHBr-CH}_3$$, which has a total of six carbons including the two methyl groups on C2.

When treated with ethanol (C₂H₅OH), a weak nucleophile and weak base, the reaction proceeds via an SN1 mechanism. The C-Br bond ionizes to give a secondary carbocation at C3: $$\text{CH}_3\text{-C(CH}_3)_2\text{-}\overset{+}{\text{CH}}\text{-CH}_3$$.

This secondary carbocation undergoes a 1,2-methyl shift from the adjacent C2 to the cationic C3, generating the more stable tertiary carbocation at C2: $$\text{CH}_3\text{-}\overset{+}{\text{C}}\text{(CH}_3)\text{-CH(CH}_3)\text{-CH}_3$$. In this rearranged cation, C2 now carries only one methyl group (the original one at C2 that did not migrate), the migrated methyl is at C3, and the positive charge sits at C2.

Ethanol attacks this tertiary carbocation at C2 as a nucleophile, and loss of a proton gives the ether product: $$\text{CH}_3\text{-C(OC}_2\text{H}_5\text{)(CH}_3)\text{-CH(CH}_3)\text{-CH}_3$$. This is 2-ethoxy-2,3-dimethylbutane, which corresponds to Option 3.

The correct order of reactivity for these chlorides with acetate in acetic acid is governed by the

$$S_N1$$

solvolysis mechanism, where the rate-determining step is the departure of the chloride leaving group to form a carbocation intermediate. 3-chloro-4-methylcyclohexene forms a resonance-stabilized secondary allylic carbocation; its ionization is uniquely accelerated by ground-state steric relief (B-strain) between the adjacent bulky

$$-CH_3$$ and $$-Cl$$ groups, making its reaction the fastest. 3-chloro-1-methylcyclohexene lacks this severe ground-state steric strain but forms a highly stable allylic carbocation that benefits from tertiary resonance character, placing it closely second in reactivity. 3-(chloromethyl)cyclohexene ionizes to form a primary carbocation, but it reacts much faster than a typical primary or secondary halide due to significant anchimeric assistance (neighboring group participation) from the adjacent

$$\pi$$-bond, which stabilizes the transition state via a non-classical homoallylic carbocation. Finally, 5-chloro-3-methylcyclohexene forms an isolated, non-allylic secondary carbocation that completely lacks any resonance stabilization or

$$\pi$$-bond assistance, rendering it the least stable intermediate and consequently the least reactive chloride in the series.

The reaction occurs in presence of $$\mathrm{NaOH/EtOH}$$, which favours:

$$\mathrm{E2\ Elimination}$$

The base abstracts a $$\mathrm{\beta}$$-hydrogen adjacent to the carbon bearing chlorine.

Removal of the vinylic hydrogen is unfavourable, so the proton is removed from the neighbouring $$\mathrm{sp^3}$$ hybridised carbon containing the methyl group.

Simultaneously, the chloride ion leaves and a new $$\mathrm{\pi}$$ bond is formed.

This elimination produces a:

$$\mathrm{Conjugated\ Diene}$$

Conjugated dienes are highly stable due to delocalisation of $$\mathrm{\pi}$$ electrons.

Hence, the major product formed is the conjugated methylcyclopentadiene structure.

Correct Option: $$\mathrm{B}$$

Since the medium is acidic and the $$-OH$$ group is on the $$2^0$$ carbon, the reaction process can be $$SN_1$$. So there will be a carbocation intermediate formed. Rearrangement also takes place as follows:

The $$-Br$$ gets added to the carbon with the positive charge. Thus, the right option is B.

To determine which compound has the shortest $$C-Cl$$ bond, we examine the hybridization of the carbon atom bonded to chlorine and the effect of resonance on the bond.

The bond length of a $$C-Cl$$ bond depends primarily on two factors. First, as the $$s$$-character of the carbon hybrid orbital increases, the bond becomes shorter. Thus, the trend is

$$sp<sp^2<sp^3$$

in terms of bond length. Second, if resonance imparts partial double bond character to the $$C-Cl$$ bond, the bond becomes shorter and stronger than a normal single bond.

In tert-butyl chloride, the carbon attached to chlorine is $$sp^3$$ hybridized, so the $$C-Cl$$ bond is a normal single bond and relatively long.

In allyl chloride, although the molecule contains a double bond, the carbon directly bonded to chlorine is still $$sp^3$$ hybridized. Therefore, the $$C-Cl$$ bond remains essentially a single bond.

In methyl chloride, the carbon bonded to chlorine is also $$sp^3$$ hybridized, giving a normal single $$C-Cl$$ bond.

In vinyl chloride, the chlorine atom is directly attached to an $$sp^2$$ hybridized carbon of a double bond. Moreover, the lone pair on chlorine participates in conjugation with the adjacent $$\pi$$ bond, leading to resonance as shown below:

$$CH_2=CH-\ddot{Cl};\leftrightarrow;{}^{-}CH_2-CH=\overset{+}{Cl}.$$

This resonance imparts partial double bond character to the $$C-Cl$$ bond, making it significantly shorter than a normal single bond.

Therefore, vinyl chloride (Option D) has the shortest $$C-Cl$$ bond because the carbon is $$sp^2$$ hybridized and the bond possesses partial double bond character due to resonance.

We are asked to arrange the four given halides in the order in which they react with an ethanolic solution of silver nitrate, $$AgNO_3$$. Silver-nitrate in ethanol brings about ionisation of the C-X bond and thus follows an $$S_N1$$ mechanism. In the $$S_N1$$ pathway the rate-determining step is the formation of a carbocation, so the more stable the carbocation, the faster the reaction. Hence we simply have to compare the stabilities of the carbocations that would be produced from the four substrates.

First, let us write the general ionisation step which every substrate must undergo:

$$R-X \;\xrightarrow[\text{EtOH}]{AgNO_3}\; R^{+} \;+\; AgX\!\downarrow$$

The precipitate, $$AgX$$, appears only after the carbocation $$R^{+}$$ has formed, so the relative rate of precipitation mirrors the relative ease of carbocation formation. Now we look at each compound.

Compound (B) possesses a benzylic chloride. When the C-Cl bond breaks, the resulting benzylic carbocation is resonance-stabilised over the aromatic ring. We can draw:

$$\overset{+}{CH_2}-C_6H_5 \;\longleftrightarrow\; CH_2= C_6H_5^{+}$$

Because several equivalent resonance structures distribute the positive charge, the benzylic cation is extremely stable, so (B) will ionise fastest.

Compound (A) is a tertiary alkyl chloride. Tertiary carbocations are stabilised by the +I (electron-releasing) effect and hyperconjugation from three neighbouring alkyl groups. They are very stable, but still slightly less so than the resonance-delocalised benzylic cation. Therefore (A) comes after (B).

Compound (C) is an allylic chloride. Cleavage of the C-Cl bond furnishes an allylic carbocation that is resonance-stabilised over the adjacent C=C bond:

$$\overset{+}{CH_2}-CH=CH_2 \;\longleftrightarrow\; CH_2=CH-CH_2^{+}$$

This resonance gives appreciable stability, but not as much as in a benzylic or tertiary situation, so (C) reacts more slowly than (A).

Compound (D) is a simple primary alkyl chloride without any resonance or strong inductive stabilisation. The primary carbocation generated would be highly unstable, making ionisation very slow. Consequently (D) is the least reactive of the four.

Collecting these observations, the order of decreasing reactivity is

$$\text{(B) > (A) > (C) > (D)}$$

This exactly matches Option 4 in the given list.

Hence, the correct answer is Option 4.

First, let us recall a general fact from chemical bonding: $$\text{Bond length} \propto \dfrac1{\text{Bond order}}.$$

So, if any factor increases the bond order of the $$\text{C-Cl}$$ bond (that is, gives it some double-bond character), the bond will become shorter. Conversely, anything that decreases the bond order will make the bond longer.

In vinyl chlorides the lone pair on chlorine can take part in resonance with the adjacent $$\mathrm C=C$$ double bond. We write the two principal resonance structures as

$$$Cl{-}CH=CH\!-\!R \;\longleftrightarrow\; Cl^{+}=CH{-}CH^{-}\!-\!R$$$

where $$R$$ stands for the group attached to the second carbon ($$CH_2,\ NO_2,\ CH_3,\ OCH_3$$ in the four alternatives). In the right-hand structure the $$\text{C-Cl}$$ bond is a double bond ($$Cl^{+}=C$$), so a greater contribution of this structure means a higher bond order and a shorter bond.

Now we must analyse how the substituent $$R$$ affects the contribution of this resonance form.

• If $$R$$ is an electron-withdrawing group (strong −I / −M effect), it pulls electron density away from the $$\pi$$-system. The negative charge on the second carbon in the resonance form $$Cl^{+}=CH{-}CH^{-}\!-\!R$$ is then stabilised because the group can accommodate or disperse that negative charge. This stabilisation increases the importance of the double-bond structure, raises the bond order of $$\text{C-Cl}$$, and consequently shortens the bond.

• If $$R$$ is an electron-donating group (+I or +M), it pushes electrons toward the double bond. That makes the negatively charged resonance form less favourable, decreases the double-bond character between C and Cl, lowers the bond order, and lengthens the bond.

Let us classify each substituent provided:

$$$\begin{aligned} R &= CH_2 \quad &(\text{no strong effect}) \\ R &= NO_2 \quad &(\text{strong } -I \text{ and } -M,\ \text{very electron-withdrawing}) \\ R &= CH_3 \quad &(\text{weak } +I,\ \text{slightly electron-releasing}) \\ R &= OCH_3 \quad &(\text{strong } +M,\ \text{electron-donating by resonance}) \end{aligned}$$$

Among these, $$NO_2$$ is by far the strongest electron-withdrawing group. Therefore the resonance form containing $$Cl^{+}=C$$ is most stabilised when $$R = NO_2$$, giving the $$\text{C-Cl}$$ bond its greatest double-bond character and the least length.

Writing this conclusion explicitly:

$$$\text{Bond order of C-Cl: } NO_2 > CH_2 \approx CH_3 > OCH_3$$$

Hence, the $$\text{C-Cl}$$ bond is shortest in $$Cl{-}CH=CH{-}NO_2$$, which corresponds to Option B.

Hence, the correct answer is Option 2.

We need to compare the relative rates of substitution ($$k_s$$) and elimination ($$k_e$$) for a primary alkyl halide reacting with two different alkoxide bases: Ethoxide ion (A) and tert-Butoxide ion (B).

Core Chemical Principles:

• Ethoxide ion ($$\text{CH}_3\text{CH}_2\text{O}^\ominus$$ - A): A small, sterically unhindered nucleophile/base. It easily attacks the carbon center, making substitution ($$\text{S}_\text{N}2$$) the major pathway. Thus, $$k_s$$ is high and $$\mu_A = \frac{k_s}{k_e}$$ is large.
• tert-Butoxide ion ($$(\text{CH}_3)_3\text{C--O}^\ominus$$ - B): A bulky, highly sterically hindered base. The steric crowding prevents it from attacking the carbon center for substitution, forcing it to abstract a peripheral proton to undergo elimination ($$\text{E}2$$). Thus, elimination is strongly favored, meaning $$k_e(B) > k_e(A)$$ and $$\mu_B = \frac{k_s}{k_e}$$ is very small.

Comparison Results:

1. Since substitution is favored with the unhindered base A and elimination dominates with the bulky base B, we have: $$\mu_A > \mu_B$$
2. Because tert-butoxide is a stronger, bulkier base designed specifically to drive elimination pathways, its elimination rate constant is significantly higher: $$k_e(B) > k_e(A)$$

Conclusion:

Steric hindrance in the tert-butoxide base suppresses substitution while accelerating elimination compared to the unhindered ethoxide base.

Answer: Option B — $$\mu_A > \mu_B$$ and $$k_e(B) > k_e(A)$$

An $$\text{E}1$$ elimination reaction proceeds via a two-step mechanism where the first and rate-determining step (RDS) is the loss of the leaving group ($$\text{Cl}^-$$) to form a carbocation intermediate:

$$\text{Reactivity toward E1} \propto \text{Stability of the Carbocation intermediate}$$

The general order of carbocation stability is: Allylic / Benzylic (resonance stabilized) > Tertiary ($$3^\circ$$) > Secondary ($$2^\circ$$) > Primary ($$1^\circ$$).

Step-by-Step Analysis of the Generated Carbocations:

• Compound (B): Primary Allylic Chloride

  Loss of $$\text{Cl}^-$$ yields a primary allylic carbocation ($$\text{CH}_2=\text{CH--CH}_2^+\text{--CH}_2\text{--CH}_3$$ or similar chain system). This carbocation is highly stable because the positive charge is directly delocalized via resonance with the adjacent carbon-carbon double bond. Hence, it is exceptionally reactive.




• Compound (D): Secondary Allylic Chloride

  Loss of $$\text{Cl}^-$$ forms a secondary allylic-like carbocation. While it has resonance stability, it undergoes minor structural or steric constraint compared to the primary unhindered linear conjugation of B, making it slightly less favored than B but far more stable than non-conjugated carbocations.




• Compound (C): Secondary Alkyl Chloride

  Loss of $$\text{Cl}^-$$ creates a simple secondary ($$2^\circ$$) carbocation ($$\text{CH}_3\text{--CH}^+\text{--CH}_2\text{--CH}_3$$). It is stabilized purely by hyperconjugation and inductive effects, which are weaker than resonance. It is less reactive than the allylic systems (B and D).




• Compound (A): Primary Alkyl Chloride

  Loss of $$\text{Cl}^-$$ yields a primary ($$1^\circ$$) carbocation ($$\text{CH}_3\text{--CH}_2\text{--CH}_2\text{--CH}_2\text{--CH}_2^+$$). Primary carbocations are extremely unstable and difficult to form under standard conditions. Thus, it is the least reactive compound.

Conclusion:

Arranging the compounds based on the stability profile of their respective carbocations yields the following decreasing order of reactivity:

$$\mathbf{B > D > C > A}$$

Answer: Option — B > D > C > A

To determine the order of boiling points, we first predict the major products A, B, and C formed in the given addition reactions. All three products have the molecular formula $$C_4H_9Br$$ and are structural isomers.

Reaction A (Anti-Markovnikov Addition)

But-1-ene reacts with $$HBr$$ in the presence of benzoyl peroxide. The peroxide initiates a free-radical mechanism, leading to the anti-Markovnikov product in which bromine adds to the less substituted carbon.

Product A: 1-bromobutane (a straight-chain primary alkyl halide).

Reaction B (Markovnikov Addition)

2-Methylprop-1-ene reacts with $$HBr$$ through the normal electrophilic addition mechanism. Protonation forms the most stable tertiary carbocation, followed by bromide attack.

Product B: 2-bromo-2-methylpropane (tert-butyl bromide), a highly branched tertiary alkyl halide.

Reaction C (Markovnikov Addition)

But-2-ene reacts with $$HBr.$$ Since the alkene is symmetrical, addition from either side produces the same secondary carbocation.

Product C: 2-bromobutane (a branched secondary alkyl halide).

For these isomeric compounds, the boiling point depends primarily on the strength of intermolecular van der Waals (dispersion) forces, which in turn depend on the molecular surface area.

Straight-chain molecules such as Product A possess a larger surface area, allowing stronger intermolecular attractions and, therefore, a higher boiling point.

Branching makes molecules more compact and spherical, reducing the effective surface area and weakening dispersion forces. Product B is more highly branched than Product C and consequently has the lowest boiling point.

Hence, the boiling point order is:

• Lowest: B (most branched)
• Intermediate: C (less branched)
• Highest: A (straight-chain)

Therefore, the increasing order of boiling points is

$$
B<C<A.
$$

This matches option A.

The two elementary steps hidden in the statement are of different molecularities.
Let us first recall the general rate‐laws.

Case 1: $$S_N1/E1$$ mechanism (unimolecular)
Rate-determining step is the slow, first-order ionisation of the alkyl halide to give a carbocation.
Hence $$\text{Rate} = k\,[\text{substrate}]$$ and is independent of the concentration or basic strength of the nucleophile / base present in solution.

Case 2: $$S_N2/E2$$ mechanism (bimolecular)
Here the nucleophile / base attacks in the rate-determining transition state.
Hence $$\text{Rate} = k\,[\text{substrate}]\,[\text{base}]$$ and it does depend on both the substrate and the base. A stronger base such as $$^{\ominus}OR$$ will generally accelerate the step compared with $$OH^{\ominus}$$.

Looking at the options we infer:

• Statement A — says changing $$OH^{\ominus}$$ to $$^{\ominus}OR$$ “will have no effect on reaction (2)”. If reaction (2) is bimolecular, its rate will certainly change with the strength of the base, so A is false.

• Statement B — says changing the concentration of base will have no effect on reaction (1). If reaction (1) proceeds by an $$S_N1/E1$$ path, the rate is independent of base concentration, so B is true.

• Statement C — says doubling base concentration doubles the rate of both reactions. This is wrong because reaction (1) is zero-order in base.

• Statement D — says changing base concentration has no effect on reaction (2). Because reaction (2) is bimolecular, this is again false.

Therefore the only correct statement is:

Option B which is: Changing the concentration of base will have no effect on reaction (1).

We need to determine the structures of the intermediates or products in the given intramolecular coupling reaction mediated by sodium in ether.

Reaction Mechanism: Intramolecular Wurtz-like Coupling

1. Radical Formation:

   When the starting material containing two alkyl bromide groups ($$\text{--CH}_2\text{Br}$$ and the side-chain terminal $$\text{--Br}$$) is treated with metallic sodium ($$\text{Na}$$) in an anhydrous solvent like ether, sodium donates electrons to the carbon-halogen bonds. This generates reactive radical or organosodium intermediates at both alkyl carbon centers.




2. Intramolecular Ring Closure (Cyclization):

   The two highly reactive radical centers (or the carbanion and the remaining alkyl halide) undergo a rapid, favorable intramolecular homolytic/heterolytic carbon-carbon ($$\text{C--C}$$) bond coupling. This brings the two side chains together to close a new six-membered saturated ring fused directly to the existing benzene ring:

   $$\text{Bis-brominated Precursor} \xrightarrow{\text{Na / Ether}} \text{Fused Bicyclic Tetralin Core} + 2\text{NaBr}$$

Conclusion:

The reaction is a classic example of an intramolecular Wurtz coupling that efficiently constructs a stable six-membered ring fused to the phenolic framework, yielding a tetrahydronaphthol derivative.

Answer: The option showing the corresponding structures matching the intramolecular cyclization pathway (Option C).

For an $$\mathrm{S_N1}$$ (unimolecular nucleophilic substitution) process, the mechanism proceeds in two distinct kinetic steps. First, the substrate $$R\!-\!X$$ ionises to form a carbocation $$R^{\oplus}$$ and a leaving group $$X^{\ominus}$$. Second, the nucleophile $$Y^{\ominus}$$ attacks the planar carbocation. Symbolically, the elementary events are

$$R - X \;\longrightarrow\; R^{\oplus} + X^{\ominus}$$

followed by

$$R^{\oplus} + Y^{\ominus} \;\longrightarrow\; R - Y.$$

The experiment-based rate law for an $$\mathrm{S_N1}$$ reaction is stated first:

$$\text{rate} \;=\; k\,[R\!-\!X]$$

because the slow step is the formation of the carbocation. This rate law contains no concentration term for the nucleophile, so increasing or decreasing nucleophile strength hardly alters the rate.

Now we judge each student observation one by one using the above principles.

(a) “The reaction is favoured by weak nucleophiles.” Since the nucleophile does not appear in the rate law, even very weak nucleophiles react readily after the carbocation is formed. Therefore the statement is true.

(b) “$$R^{\oplus}$$ would be easily formed if the substituents are bulky.” Bulky alkyl substituents mean more alkyl groups attached to the positively charged carbon. More alkyl groups stabilise a carbocation by the inductive effect and by hyperconjugation. The order of carbocation stability is $$\text{tertiary} > \text{secondary} > \text{primary} > \text{methyl}.$$ Hence a bulky (tertiary) substrate ionises faster, so this statement is also true.

(c) “The reaction is accompanied by racemization.” After the leaving group departs, the carbocation centre is trigonal planar. The nucleophile $$Y^{\ominus}$$ can attack from either side with almost equal probability, giving both inversion and retention products. The net result is racemisation (usually partial but often substantial). Thus statement (c) is true.

(d) “The reaction is favoured by non-polar solvents.” Formation of a carbocation and a free anion creates highly polar species. Polar protic solvents (e.g. water, alcohols) stabilise these ions through solvation and therefore accelerate the reaction. Non-polar solvents cannot stabilise ions efficiently, so the reaction rate drops. Thus statement (d) is false.

Collecting the results, (a), (b) and (c) are correct, while (d) is incorrect.

Hence, the correct answer is Option C.

We begin by recalling that the word retention means that the absolute configuration (R or S) of the stereogenic centre remains the same after the substitution has taken place.

There are three usual mechanistic possibilities for a nucleophilic substitution on a chiral carbon bearing a leaving group X:

1. In a one-step $$S_N2$$ process the nucleophile approaches from the side opposite to the leaving group, causing a single inversion of configuration.
2. In a two-step $$S_N1$$ process the leaving group departs first, giving a planar carbocation. The nucleophile can then attack from either side, giving a racemic mixture (50 % retention + 50 % inversion in the ideal case).
3. In a mechanism involving neighbouring-group participation the atom of a neighbouring group with a lone pair (often another halogen) first displaces the leaving group intramolecularly. This produces a three-membered cyclic ion. The nucleophile then opens this ring from the back side. Each of the two intramolecular steps proceeds with inversion, so the overall result is double inversion => net retention.

With these facts we examine every option.

Option A is $$CH_3-\underset{\displaystyle *}{C(Br)(H)}-C_6H_{13}$$. Here the stereogenic carbon is directly bonded to Br, H, CH3 and C6H13. There is no neighbouring atom with a lone pair capable of participating. Hence the reaction will proceed either by $$S_N2$$ (one inversion) or, less likely, by $$S_N1$$ (racemisation). Retention is not expected.

Option B is $$CH_3-\underset{\displaystyle *}{CH(Br)}-C_6H_5$$ (benzyl bromide). The benzyl cation derived from it is highly stabilised, so an $$S_N1$$ pathway is favoured, producing racemisation. Again, no net retention.

Option C is $$CH_3-\underset{\displaystyle *}{CH(Br)}-CH_3$$ (sec-butyl bromide). Being a simple secondary bromide with no assisting group, it reacts mainly by $$S_N2$$ with a single inversion. Net retention is impossible.

Option D is $$CH_3-\underset{\displaystyle *}{CH(C_2H_5)}-CH_2Br$$. Here the stereogenic carbon is α to a $$CH_2Br$$ group. The bromine atom on the neighbouring carbon possesses lone pairs that can assist. Let us write every algebraic step.

Step 1: Intramolecular displacement (first inversion).
The lone pair on the neighbouring bromine attacks the stereogenic centre while the leaving bromide departs:

$$\text{(R or S)-CH}_3-\overset{*}{CH(C_2H_5)}-CH_2Br \;\xrightarrow{\text{l.p. on Br}}\; \bigl[\text{bicyclic bromonium ion}\bigr]^+ + Br^-$$

This internal attack must come from the back side of the C-Br bond that is breaking, so the stereogenic centre undergoes one inversion.

Step 2: Nucleophilic attack by $$OH^-$$ (second inversion).
The $$OH^-$$ ion now opens the three-membered bromonium ring from the side opposite to the bridging bromine:

$$\bigl[\text{bromonium ion}\bigr]^+ + OH^- \;\longrightarrow\; CH_3-\overset{*}{CH(C_2H_5)}-CH_2OH + Br^-$$

This ring opening is also a backside attack, so a second inversion occurs. Performing two consecutive inversions brings the configuration back to its original sense, i.e. overall retention.

Therefore only Option D provides the necessary neighbouring-group participation that gives double inversion and hence net retention.

Hence, the correct answer is Option 4.

The conversion begins with 1-chloro-4-(2-chloropropyl)benzene undergoing an E2 dehydrohalogenation reaction when treated with alcoholic potassium hydroxide. The strong hydroxide base selectively abstracts a highly acidic benzylic proton from the aliphatic side chain, forcing the elimination of the adjacent chloride ion. This step yields the thermodynamically favoured intermediate, 1-chloro-4-(prop-1-en-1-yl)benzene (also known as p-chloropropenylbenzene), where the newly formed alkene double bond is highly stabilized through extended conjugation with the aromatic ring; the aryl chloride attached directly to the benzene ring remains completely unreactive due to its partial double-bond character. In the second step, this conjugated alkene intermediate undergoes free radical polymerization initiated by a thermal radical source like benzoyl peroxide or AIBN. The radical initiator adds to the alkene monomer, breaking its $\pi$-bond to generate a reactive benzylic radical that successively attacks subsequent monomer units. This continuous propagation chain link builds a high-molecular-weight polymer backbone of alternating carbon atoms, resulting in the final product: poly[1-chloro-4-(prop-1-en-1-yl)benzene]

Precipitate is formed if the formation of Carbocation intermediate is favourable in any given reactant. In option B the ring becomes aromatic making it highly favourable to form a Carbocation hence, it will give a precipitate.

We first recall the standard qualitative test: when an alkyl (or benzyl/allyl) halide is treated with ethanolic silver nitrate, the halide ion $$X^-$$ that departs from the molecule combines with the silver ion $$Ag^+$$ to form a curdy white precipitate of silver halide $$AgX$$.

This observation is explained by the nucleophilic substitution mechanism. The rate of substitution for chlorides follows the order

$$\text{tert-alkyl chloride} > \text{sec-alkyl chloride} > \text{prim-alkyl chloride} \gg \text{vinyl or aryl chloride}.$$

Now we analyse each option one by one.

Option A : $$CH_2 = CH{-}Cl$$ (vinyl chloride)
The chlorine atom here is directly attached to an $$sp^2$$ hybridised carbon of a C=C double bond. Such $$C{-}Cl$$ bonds are very strong and do not undergo the usual $$S_N1$$ or $$S_N2$$ processes in ethanolic $$AgNO_3$$. Hence no chloride ion is liberated, so no $$AgCl$$ precipitate forms.

Option B : $$(CH_3)_3CCl$$ (tert-butyl chloride)
This chloride is tertiary. In ethanolic medium the $$C{-}Cl$$ bond ionises readily to give the stable tert-butyl carbocation $$\,(CH_3)_3C^+\,$$ and the chloride ion $$Cl^-$$:

$$ (CH_3)_3CCl \;\longrightarrow\; (CH_3)_3C^+ + Cl^- $$

The released $$Cl^-$$ immediately reacts with silver ion to give an insoluble precipitate:

$$ Ag^+ + Cl^- \;\longrightarrow\; AgCl\downarrow $$

So a white precipitate appears quickly.

Option C : $$CCl_4$$ (carbon tetrachloride)
Although $$CCl_4$$ contains four chlorine atoms, the central carbon has no empty orbital for coordination and the molecule is quite resistant to ionisation in protic solvents. Therefore $$Cl^-$$ is not produced and no precipitate is observed.

Option D : $$CHCl_3$$ (chloroform)
Chloroform is also rather inert toward substitution under these conditions; the $$C{-}Cl$$ bonds are not cleaved by ethanolic $$AgNO_3$$, so again no $$Cl^-$$ is released.

Among the four compounds, only the tertiary chloride $$(CH_3)_3CCl$$ readily furnishes chloride ions that can precipitate silver as $$AgCl$$.

Hence, the correct answer is Option 2.

We have the substrate $$CH_3CHClCH_2CO_2CH_2CH_3$$ (ethyl 3-chlorobutanoate). The chlorine atom is attached to the carbon that is α to the ester carbonyl, while two different sets of β-hydrogen atoms are available for an E2 elimination.

Because the reagent is $$NaOEt$$ in ethanol and the reaction is carried out under heat, we immediately recognise that a strong base is present and that the reaction will proceed predominantly through the E2 mechanism rather than substitution. In an E2 process the base removes a β-hydrogen that is anti-periplanar to the leaving group, the C-H and C-Cl bonds break simultaneously and a C=C double bond is formed. The general E2 equation can be written as

$$\text{Base} + RCH_2CH_2Cl \;\longrightarrow\; RCH=CH_2 + Cl^- + \text{Base-H}$$

In our molecule we label the carbon bearing chlorine as $$C_\alpha$$ and the two adjacent carbons as $$C_{\beta 1}$$ (the CH2 group next to the ester) and $$C_{\beta 2}$$ (the terminal CH3 group). Hence the two distinct elimination possibilities are:

1. Removal of a hydrogen from $$C_{\beta 1}$$ (the CH2 between the Cl-bearing carbon and the carbonyl).
2. Removal of a hydrogen from $$C_{\beta 2}$$ (the terminal CH3).

We now write both possible eliminations explicitly.

Path 1 : β-H from $$C_{\beta 1}$$

The base abstracts one of the two hydrogens on $$C_{\beta 1}$$:

$$\begin{aligned} CH_3CHClCH_2CO_2CH_2CH_3 &\xrightarrow[\text{E2}]{NaOEt/\,\Delta} CH_3CH = CHCO_2CH_2CH_3 + Cl^- + HOEt \end{aligned}$$

The double bond produced is between $$C_\alpha$$ and $$C_{\beta 1}$$, giving us ethyl but-2-enoate (also called ethyl crotonate). This alkene is (i) trisubstituted and therefore more highly substituted, and (ii) conjugated with the carbonyl group of the ester, giving it extra resonance stabilisation. The resonance can be shown as

$$CH_3CH = CHCO_2Et \;\rightleftharpoons\; CH_3CH^- - CH = C^+O_2Et$$

Such conjugation lowers the overall energy of the molecule, favouring its formation.

Path 2 : β-H from $$C_{\beta 2}$$

If instead the base removes a hydrogen from the terminal CH3, we obtain

$$\begin{aligned} CH_3CHClCH_2CO_2CH_2CH_3 &\xrightarrow[\text{E2}]{NaOEt/\,\Delta} CH_3CH_2CH = CHCO_2CH_2CH_3 + Cl^- + HOEt \end{aligned}$$

This product is ethyl but-3-enoate (a terminal, monosubstituted alkene). It is less substituted and it is not conjugated with the carbonyl, so it is thermodynamically less stable.

According to Zaitsev’s rule, “in an elimination the more substituted (and therefore more stable) alkene is formed preferentially.” Furthermore, conjugation with the ester carbonyl provides an additional driving force. Therefore Path 1 overwhelmingly dominates.

Thus the major product is

$$\boxed{CH_3CH = CHCO_2CH_2CH_3}$$

This structure matches Option 3 in the given list.

Hence, the correct answer is Option 3.

We need to determine the major product formed when the given dibromo compound undergoes dehydrohalogenation with alcoholic $$\text{KOH}$$ under heating ($$\Delta$$).

Reaction Mechanism:

1. Double $$\text{E2}$$ Elimination ($$-2\text{HBr}$$):

   Alcoholic $$\text{KOH}$$ is a strong base used to induce elimination reactions. Since the substrate contains two bromine leaving groups on adjacent carbons along with accessible $$\beta$$-hydrogens, it undergoes a double dehydrohalogenation process, losing two molecules of $$\text{HBr}$$.




2. Thermodynamic Driving Force (Conjugation):

   The elimination is highly regioselective and follows Saytzeff's rule to produce the most thermodynamically stable alkene system possible. By eliminating both bromine atoms systematically, a highly extended conjugated diene system is formed.

   $$\text{Dibromo Substrate} \xrightarrow{\text{Alc. KOH, }\Delta} \text{Highly Conjugated Diene Product} + 2\text{HBr}$$



3. Resonance Stabilization with the Benzene Ring:

   The newly formed pair of carbon-carbon double bonds ($$\text{C=C}$$) aligns perfectly with the adjacent phenyl ($$\text{Ph}$$) ring. This creates an uninterrupted, delocalized $$\pi$$-electron cloud spanning across the entire molecular framework, providing immense resonance stabilization energy.

Conclusion:

The reaction yields a stable, extended conjugated diene system directly attached to the benzene ring due to the thermodynamic advantage of resonance delocalization.

Answer: Option C

The ease of SN-1 reaction depends on the most stable Carbocation. +M and +I effect increase the stability of the carbocation formed. Moreover, more methyl substituted carbocation is more stable compared to lesser methyl substitutions due to +I effect of Alkyl groups.

1. The reaction begins with the generation of an initial carbocation intermediate (typically via the protonation and loss of a leaving group like a hydroxyl group or via electrophilic addition to an alkene). The carbocation forms.




2. Nucleophilic Attack by Chloride Ion ($$\text{Cl}^\ominus$$):

   The chloride ion ($$\text{Cl}^\ominus$$) acts as a nucleophile. It attacks the electron-deficient sp2 hybridized carbocation carbon:

   $$\text{R}^\oplus + \text{Cl}^\ominus \rightarrow \text{R--Cl}$$

Analyzing the Final Product Structure:

Based on the provided mechanism diagram:

• The parent framework consists of a fused bicyclic ring system (a 6-membered ring fused to a 5-membered or 6-membered ring).
• An exocyclic group contains an isopropyl-like unit where the rearrangement takes place.
• The chloride ion attacks the tertiary carbon center bearing the two methyl ($$\text{--CH}_3$$) groups, resulting in a tertiary alkyl chloride derivative as the final stable major product: $$\text{--C(Cl)(CH}_3)_2$$.

Conclusion:

The major product is formed via a thermodynamic pathway favored by the generation of a stable tertiary carbocation intermediate, followed by direct nucleophilic capture by the chloride ion.

Answer: The option corresponding to the structure containing the chlorine atom attached to the tertiary carbon carrying the two methyl groups on the bicyclic framework (Option D).

An $$S_N1$$ substitution proceeds in TWO distinct steps:
  1. Slow ionisation of the alkyl halide to give a carbocation and a halide ion.
  2. Fast attack of the nucleophile on this carbocation.

Potential-energy consequences of these steps:

• Two elementary steps ⇒ two transition states ⇒ the energy profile must show TWO maxima.
• The first step (ionisation) is rate-determining, so its activation energy is larger; therefore the first maximum is higher than the second.
• Between the two maxima lies a local minimum that represents the carbocation intermediate.

Among the given diagrams, only Option A shows
  • two humps (two transition states),
  • the first hump higher than the second, and
  • an intermediate valley between them.
Hence it correctly depicts the $$S_N1$$ mechanism.

Option A which is: the diagram with two maxima (first higher) and a carbocation intermediate.

+I and +M effect must be considered for finding the most stable Carbocation formed in SN-1 reaction.

Since CH3OH is a weak nucleophile and there is no heating given that means we will have and SN-1 reaction taking place and your CH3OH is a polar protic solvent. Hence  a carbocation would be formed first then rearrangement would take place leading to more substituted carbon. Then -OCH3 will come and attach to the carbocation hence forming product in option A.

Intramolecular Alkylation 

• Reagent: Potassium tert-butoxide is a strong, bulky base.
• Action: It deprotonates the alpha-carbon of the ketone (the carbon between the benzene ring and the carbonyl). This creates a nucleophilic enolate.
• Cyclization: The enolate then undergoes an intramolecular SN2 reaction, attacking the carbon attached to the chlorine atom at the end of the chain.
• Result: This forms a five-membered or six-membered ring attached to the original chain. In this specific structure, the chain length favours the formation of a six-membered ring, creating a tetralone skeleton.

2. Acid-Catalyzed Cyclization/Dehydration 

• Reagent: Concentrated sulfuric acid and heat.
• Action: This promotes an electrophilic aromatic substitution (Friedel-Crafts-like cyclization).
• Mechanism: The acid protonates the carbonyl oxygen, making the carbonyl carbon highly electrophilic. This carbon then attacks the electron-rich benzene ring (specifically at the ortho position relative to the alkyl chain).
• Dehydration: Under heating and acidic conditions, the resulting intermediate loses a water molecule to restore aromaticity and complete the bicyclic system.

We have to check two separate things - first, whether the assertion about vinyl (alkenyl) halides is true, and second, whether the given reason is itself true and whether it really explains the assertion.

Vinyl halide means a molecule in which the halogen atom is directly attached to an $$\mathrm{sp^2}$$-hybridised carbon atom of a carbon-carbon double bond, for example $$CH_{2}=CH-Cl$$. Let us recall the two main nucleophilic substitution pathways:

1. $$\mathrm{S_N1}$$ mechanism 2. $$\mathrm{S_N2}$$ mechanism

For an $$\mathrm{S_N1}$$ process, the first step is:

$$R-X\; \longrightarrow \; R^{+ + X^{-}}$$

That is, heterolytic cleavage of the C-X bond to give a carbocation. The stability of the intermediate carbocation determines whether this step is easy or difficult. A vinyl carbocation would look like

$$CH_{2}=CH^{+}$$

The positive charge resides on an $$\mathrm{sp^2}$$ carbon which possesses $$50\%$$ $$s$$-character. Greater $$s$$-character holds the electrons more tightly to the nucleus, so the carbon is already electron-deficient; adding a full positive charge makes the ion highly unstable. Therefore:

$$\text{Stability: } \mathrm{vinyl\; carbocation} \ll \mathrm{allyl\; or\; tert\; carbocation}$$

Because the required carbocation is extremely unstable, the $$\mathrm{S_N1}$$ path is practically blocked.

For an $$\mathrm{S_N2}$$ process, the nucleophile must perform a backside attack on the carbon that carries the leaving group, while the C-X bond breaks simultaneously. In vinyl halides, that carbon is part of a $$\pi$$ bond, i.e.

$$C(sp^2)$$

Geometrically, the $$\pi$$ bond occupies the region exactly where the nucleophile would need to approach, so backside overlap is impossible. Hence the $$\mathrm{S_N2}$$ path is also blocked.

Thus, vinyl halides do not undergo nucleophilic substitution easily. The assertion (A) is therefore true.

Now let us analyse the reason (R). It says: “Even though the intermediate carbocation is stabilized by loosely held $$\pi$$-electrons, the cleavage is difficult because of the strong bonding.”

We have already seen that the vinyl carbocation is not stabilised; the $$\pi$$-electrons are not “loosely held”. Instead, the high $$s$$-character makes the positive charge on an $$\mathrm{sp^2}$$ carbon especially unstable. Hence the first part of (R) is wrong. The second part, about strong C-X bonding, is also inaccurate; the principal difficulty is not an exceptionally strong $$\sigma$$ bond but rather the inability to form a stable carbocation or permit backside attack.

Therefore, the reason (R) is false, and it certainly does not explain the (true) assertion.

So, we have:

$$\text{Assertion (A): true}, \qquad \text{Reason (R): false}$$

Hence, the correct answer is Option D.

In presence of DMSO an SN2 reaction will take place which will first replace the -I group by -CN from KCN. Now, in catalytic Hydrogenation this would reduce the ketone to alcohol and Cyanide to amine. Hence, option C would be the answer 

1. Addition-Fragmentation Pathway:

   The reaction proceeds via a radical addition-fragmentation chain transfer mechanism. A free radical initiator attacks the activated, electron-deficient chloroalkene double bond ($$-\text{C}(\text{Cl})=\text{CH}_2$$), creating a stable intermediate radical species.




2. Elimination of Chlorine:

   Instead of typical linear addition, the intermediate eliminates a chlorine radical ($$\text{Cl}^\bullet$$) to restore the double bond configuration or assist in building the polymeric chain. The expelled chlorine radical combines with available hydrogen sources to form the structure-stabilized $$\text{HCl}$$ byproduct.

The reaction is a free-radical chlorination carried out with $$Cl_2$$ under ultraviolet light. Such halogenations proceed through the chain mechanism
  Initiation : $$Cl_2 \xrightarrow{h\nu}\; 2\,Cl^{\bullet}$$
  Propagation : $$Cl^{\bullet}+RH \rightarrow R^{\bullet}+HCl$$ followed by $$R^{\bullet}+Cl_2 \rightarrow RCl+Cl^{\bullet}$$.

In free-radical halogenation, the site from which the hydrogen is abstracted is decided mainly by the stability of the carbon radical that would be formed. The order of radical stability is
$$3^{\circ}\;(\text{tertiary}) \;\gt\; 2^{\circ}\;(\text{secondary}) \;\gt\; 1^{\circ}\;(\text{primary})$$.

The given alkane contains one set of secondary hydrogens and another set of primary hydrogens. Abstraction of a secondary hydrogen produces a secondary radical, which is appreciably more stable than a primary radical. Hence the chain propagation step occurs predominantly at the secondary carbon.

Consequently, chlorine substitutes that secondary hydrogen, giving the monochloride in which the halogen is bonded to the secondary carbon. That compound is shown in Option A, whereas Options B, C and D correspond to products that would arise from less favourable (primary) radical pathways and therefore are obtained only in minor amounts.

Thus the major product is the secondary alkyl chloride represented in Option A.

Answer : Option A which is: the secondary (2-chloro) derivative.

The given reaction is a free-radical (chain) halogenation of an alkane with molecular halogen in the presence of ultraviolet light. Such reactions proceed through three elementary stages: initiation, propagation and termination.

Step 1 - Initiation
UV light $$\left(h\nu\right)$$ homolytically cleaves the $$\mathrm{Cl-Cl}$$ bond:
$$\mathrm{Cl_2}\;\xrightarrow{h\nu}\;2\,\mathrm{Cl^{\bullet}}$$

Step 2 - Propagation
1. A chlorine radical abstracts one hydrogen atom to give hydrochloric acid and an alkyl radical.
    For n-butane $$\left(\mathrm{CH_3-CH_2-CH_2-CH_3}\right)$$ there are two nonequivalent kinds of H-atoms:
    • primary H (on C-1 or C-4)    • secondary H (on C-2 or C-3)

The rate of abstraction depends on radical stability: $$\text{secondary} \gt \text{primary}$$.
Hence formation of the secondary butyl radical $$\left(\mathrm{CH_3-\dot{C}H-CH_2-CH_3}\right)$$ is favoured.

2. The alkyl radical then combines with $$\mathrm{Cl_2}$$ to give the chlorinated product and regenerate $$\mathrm{Cl^{\bullet}}$$.

Relative product distribution
Number of abstractable H-atoms:
    Primary H : 6  (3 on C-1 + 3 on C-4)
    Secondary H : 4  (2 on C-2 + 2 on C-3)
Although there are more primary hydrogens, the secondary radical is significantly more stable, so the overall rate of secondary substitution dominates. Therefore the chief product is the one arising from replacement of a secondary hydrogen, namely 2-chlorobutane.

Major product
$$\mathrm{CH_3-CHCl-CH_2-CH_3}\;(\text{2-chlorobutane})$$

Thus the major product corresponds to Option C.

$$t-BuOK$$ with heat ($$\Delta$$): The bulky tert-butyl group makes $$t-BuOK$$ a very poor nucleophile for substitution ($$S_N2$$) but a powerful base for elimination.

Hence, the major product is the alkene $$C_6H_5CH=CHC_6H_5$$. The substitution products (ethers) shown in options B, C, and D are minor and not formed in major quantities due to the high steric hindrance of the base.

In an $$S_N1$$ reaction, reactivity depends on the stability of the carbocation formed: $$3^\circ > 2^\circ > 1^\circ$$

(II) is a $$1^\circ$$ alkyl halide, forming an unstable primary carbocation → least reactive.

(I) is a $$2^\circ$$ alkyl halide, forming a more stable secondary carbocation.

(III) is a benzylic halide. The carbocation formed is highly stabilized by resonance and the $$+M$$ effect of the $$-OCH_3$$ group.

Therefore, the increasing order of $$S_N1$$ reactivity is: $$(II)<(I)<(III)$$

The reaction involves two consecutive $$\beta$$-eliminations (dehydro-halogenations) of a vicinal dihalide with alcoholic $$KOH$$. Each elimination removes one molecule of $$HBr$$ and generates a new $$C=C$$ bond. After the second elimination three different arrangements of the two $$\pi$$ bonds are possible:

• conjugated diene : $$\; CH_3CH = CH - CH = CHCH_3$$ (double bonds separated by one single bond)
• isolated diene    : $$\; CH_2 = CHCH_2CH = CHCH_3$$ or $$CH_2 = CHCH = CHCH_2CH_3$$ (double bonds separated by two or more single bonds)
• cumulated diene : $$\; CH_3CH = C = CHCH_2CH_3$$ (two consecutive double bonds, an allene)

Step 1 - Relative stability of the three possibilities
The heat of hydrogenation (or equivalently, the enthalpy of formation) follows the order

$$\text{conjugated} \; \lt \; \text{isolated} \; \lt \; \text{cumulated}$$

This means conjugated dienes are the most stable because the two $$\pi$$ bonds can overlap with the intermediate $$\sigma$$ bond, giving an extended delocalised $$\pi$$-system with extra resonance stabilisation.

Step 2 - Product distribution under elimination conditions
E2 eliminations are under thermodynamic control when they are carried out with a strong base at high temperature in alcoholic medium. Under such conditions the most stable alkene (or diene) predominates. Consequently the conjugated diene is formed in the greatest amount, while the less-stable isolated and cumulated dienes are minor or negligible.

Step 3 - Matching with the given options
Only Option A, $$CH_3CH = CH - CH = CHCH_3$$, is the conjugated diene. Options B and D are isolated dienes, and Option C is the cumulated diene (allene). Therefore Option A corresponds to the thermodynamically favoured major product.

Hence the major product is obtained as

Option A which is: $$CH_3CH = CH - CH = CHCH_3$$

The given substrate is a tertiary alkyl bromide. In the presence of $$C_2H_5ONa$$/$$C_2H_5OH$$, which is a strong base, β-elimination (E2) reaction takes place. It follows the Saytzeff's rule, and formation of a more substituted alkene is favoured.

Thus, the major product is option A

We start with compound I, whose molecular formula is given as $$C_3H_6Cl_2$$. We need to decide which dichloropropane isomer it is. The whole decision will be based on the behaviour of the products formed in the subsequent steps.

First, I is treated with aqueous $$KOH$$. A well-known result of treating a gem-dichloride (both chlorines on the same carbon) with aqueous alkali is hydrolysis to a gem-diol, which then immediately loses water to give a carbonyl compound. The overall transformation is

$$R_2CCl_2 + 2\,KOH_{(aq)} \;\rightarrow\; R_2C(OH)_2 + 2\,KCl$$

followed by

$$R_2C(OH)_2 \;\rightarrow\; R_2C=O + H_2O$$

If I were the gem-dichloride $$CH_3-C(Cl)_2-CH_3$$ (2,2-dichloropropane), the reaction would be

$$CH_3-C(Cl)_2-CH_3 + 2\,KOH_{(aq)} \;\rightarrow\; CH_3-C(OH)_2-CH_3 + 2\,KCl$$

and immediately

$$CH_3-C(OH)_2-CH_3 \;\rightarrow\; CH_3-CO-CH_3 + H_2O$$

Thus compound II would be $$CH_3-CO-CH_3$$, that is, acetone (propan-2-one).

Now II is treated with methylmagnesium bromide and then hydrolysed. Before using the reagent, we recall the Grignard addition formula to a ketone:

$$R_2C=O + R'MgX \;\rightarrow\; R_2C(OMgX)R' \;\overset{H_2O/H^+}{\longrightarrow}\; R_2C(OH)R' + Mg(OH)X$$

Applying it to acetone:

$$CH_3-CO-CH_3 + CH_3MgBr \;\rightarrow\; (CH_3)_3C-O^-MgBr$$

and acid work-up gives

$$ (CH_3)_3C-O^-MgBr + H_2O/H^+ \;\rightarrow\; (CH_3)_3C-OH + MgBr(OH)$$

Therefore compound III is $$ (CH_3)_3C-OH$$, namely tert-butyl alcohol (2-methyl-2-propanol).

Finally, III is tested with Lucas reagent (anhydrous $$ZnCl_2$$ + concentrated $$HCl$$). The Lucas test relies on the following:

- Tertiary alcohols give turbidity immediately.
- Secondary alcohols turn turbid in about 5 min.
- Primary alcohols remain clear for a long time.

Because turbidity appears at once, III must be a tertiary alcohol. Our deduction already shows III is tert-butyl alcohol, perfectly matching the observation.

Hence compound I must be the gem-dichloride $$CH_3-C(Cl)(Cl)-CH_3$$, that is, 2,2-dichloropropane.

Hence, the correct answer is Option A.

To show decolourisation, the compound must be unsaturated.

Options A, B, and C will undergo elimination reaction to form unsaturated compound.

Option D will undergo nucleophilic substitution and hence, will have no unsaturation. Hence, it cannot decolourise $$Br_2$$ water.

Elimination requires a strong base to remove a $$\beta$$-hydrogen and form an alkene.

$$NaOEt/EtOH$$ and $$NaOH$$ (with heat or alcohol) act as strong bases, so they favor elimination.

$$NaI$$ provides $$I^-$$, which is a good nucleophile but a weak base. Hence, it undergoes substitution instead of elimination.

Therefore, $$NaI$$ is not suitable for the reaction.

Both elimination and substitution reactions are possible.

Hence, all the three products given are possible.

We begin by recalling the well-known named reactions that are commonly taught in organic chemistry for the replacement of halogen atoms.

First, the Swarts reaction is a halogen-exchange reaction which specifically converts an alkyl chloride, bromide or iodide into the corresponding alkyl fluoride. The general form of the reaction is written as

$$R{-}X \;+\; AgF \;\longrightarrow\; R{-}F \;+\; AgX,$$

where $$R$$ represents any alkyl group and $$X$$ can be $$Cl$$, $$Br$$ or $$I$$. Silver fluoride, antimony trifluoride $$\big(SbF_3\big)$$ or other metal fluorides are typically used because fluorine itself is too reactive to be handled safely in most laboratory situations. Thus, Swarts reaction is deliberately designed to give alkyl fluorides smoothly and selectively.

Next, we examine free-radical fluorination. Although it can, in principle, introduce a fluorine atom into an alkane, it proceeds via a chain-radical mechanism that is extremely vigorous. Side reactions, over-fluorination and even explosions are frequent. Therefore, it is not the method “best accomplished” for controlled synthesis of specific alkyl fluorides.

Sandmeyer’s reaction is employed for converting diazonium salts into chlorides, bromides or cyanides, not fluorides. Hence Sandmeyer’s reaction does not suit the present need.

Finkelstein reaction is another halogen-exchange reaction, but it is used for interconverting alkyl chlorides or bromides and alkyl iodides, typically with sodium iodide in acetone:

$$R{-}Cl \;+\; NaI \;\longrightarrow\; R{-}I \;+\; NaCl.$$

This reaction does not generate fluorides either.

Comparing all four possibilities, only the Swarts reaction is both selective and practical for preparing alkyl fluorides from other alkyl halides. Therefore, we conclude that the best method for synthesising alkyl fluorides is indeed the Swarts reaction.

Hence, the correct answer is Option A.

We have to find what happens when $$\text{1,1,1-trichloroethane}$$, whose condensed formula is $$CH_3CCl_3$$, is treated with finely divided silver metal. Silver in the zero-valent state is a strong reducing agent that removes halogen atoms as silver halide, $$AgCl$$. Whenever all the halogen atoms that are attached to the same carbon (a geminal arrangement) are removed in this way, the carbon momentarily attains a divalent “carbene” state, which then couples with another such species. The overall result is the formation of a carbon-carbon triple bond.

First, write the reduction of one molecule of the trihalide by three atoms of silver.

$$CH_3CCl_3 + 3\,Ag \;\rightarrow\; CH_3C: \;+\; 3\,AgCl$$

In this equation $$CH_3C:$$ represents the carbene (a divalent carbon bearing only six electrons). Such a highly reactive species never stays free for long; two of them immediately combine by forming a new carbon-carbon bond. That coupling step is

$$2\,CH_3C: \;\rightarrow\; CH_3C \equiv CCH_3$$

Combining the two half-steps for two starting molecules gives the overall stoichiometric equation

$$2\,CH_3CCl_3 + 6\,Ag \;\rightarrow\; CH_3C \equiv CCH_3 + 6\,AgCl$$

The product $$CH_3C \equiv CCH_3$$ is 2-butyne, an alkyne containing four carbon atoms with the triple bond between the second and third carbons (counting from either end).

No alkene can appear because removal of all three halogens from a geminal trihalide forces the carbon to use the remaining valence electrons to make a triple bond after coupling. Therefore the major organic compound is the alkyne 2-butyne.

Hence, the correct answer is Option 3.

The compound given is 1,1,1-trichloropropane, which has the structure $$CH_{3}-CH_{2}-CCl_{3}$$. Here, the carbon atom at position 1 (the terminal carbon) is bonded to three chlorine atoms.

When treated with aqueous potassium hydroxide (KOH), this compound undergoes hydrolysis. Aqueous KOH provides hydroxide ions ($$OH-$$) that act as a nucleophile. The reaction involves the substitution of chlorine atoms by hydroxyl groups, but due to the presence of three chlorine atoms on the same carbon, the process leads to the formation of a carboxylic acid.

The mechanism proceeds as follows:

First, one hydroxide ion attacks the carbon atom bearing the chlorine atoms, replacing one chlorine atom to form an intermediate:

$$CH_{3}-CH_{2}-CCl_{3} + OH- -\gt CH_{3}-CH_{2}-CCl_{2}OH + Cl-$$

This intermediate, $$CH_{3}-CH_{2}-CCl_{2}OH$$, is unstable because it has two chlorine atoms and a hydroxyl group on the same carbon. It readily loses a chloride ion and rearranges to form a dichlorocarbene-like species, but in aqueous conditions, it undergoes further hydrolysis.

The intermediate reacts with another hydroxide ion:

$$CH_{3}-CH_{2}-CCl_{2}OH + OH- -\gt CH_{3}-CH_{2}-CCl(OH)2 + Cl-$$

Now, we have $$CH_{3}-CH_{2}-CCl(OH)2$$, which is a geminal diol with one chlorine. This compound is highly unstable and undergoes rapid hydrolysis. The chlorine atom is replaced by another hydroxyl group:

$$CH_{3}-CH_{2}-CCl(OH)2 + OH- -\gt CH_{3}-CH_{2}-C(OH)3 + Cl-$$

The resulting compound, $$CH_{3}-CH_{2}-C(OH)3$$, is a geminal triol. Geminal triols are unstable and spontaneously lose a water molecule to form the corresponding carboxylic acid:

$$CH_{3}-CH_{2}-C(OH)3 -\gt CH_{3}-CH_{2}-COOH + H_{2}O$$

Thus, the overall reaction is:

$$CH_{3}-CH_{2}-CCl_{3} + 2KOH -\gt CH_{3}-CH_{2}-COOH + 2KCl + H_{2}O$$

The product is propionic acid, which has the formula $$CH_{3}CH_{2}COOH$$.

Now, comparing with the options:

• Option A: 2-Propanol is $$CH_{3}CH(OH)CH_{3}$$
• Option B: Propyne is $$CH_{3}C \equiv CH$$
• Option C: Propionic acid is $$CH_{3}CH_{2}COOH$$
• Option D: 1-Propanol is $$CH_{3}CH_{2}CH_{2}OH$$

The major product is propionic acid, corresponding to option C.

Hence, the correct answer is Option C.

To determine the correct order of increasing carbon-halogen bond length in the compounds CH$$_3$$F, CH$$_3$$Cl, CH$$_3$$Br, and CH$$_3$$I, we need to understand how bond length relates to the halogen atoms. Bond length is the distance between the nuclei of two bonded atoms. Since the carbon atom is the same in all these compounds, the bond length depends primarily on the size of the halogen atom bonded to it.

Halogens belong to Group 17 of the periodic table. As we move down the group, the atomic size increases due to the addition of electron shells. The atomic radii increase in the order: fluorine (F) < chlorine (Cl) < bromine (Br) < iodine (I). Therefore, the covalent radius of fluorine is the smallest, and iodine is the largest.

In a covalent bond, the bond length is approximately the sum of the covalent radii of the bonded atoms. Since carbon has a fixed covalent radius, the carbon-halogen bond length will increase as the covalent radius of the halogen increases. Thus, the C-F bond will be the shortest, and the C-I bond will be the longest.

Let us list the bonds:

• CH$$_3$$F has a C-F bond
• CH$$_3$$Cl has a C-Cl bond
• CH$$_3$$Br has a C-Br bond
• CH$$_3$$I has a C-I bond

Based on the halogen sizes, the bond lengths should follow the order: C-F < C-Cl < C-Br < C-I. Therefore, the compounds should be ordered as: CH$$_3$$F < CH$$_3$$Cl < CH$$_3$$Br < CH$$_3$$I.

Now, comparing with the given options:

• Option A: CH$$_3$$F < CH$$_3$$Br < CH$$_3$$Cl < CH$$_3$$I — This is incorrect because C-Br bond length is greater than C-Cl, so CH$$_3$$Br should come after CH$$_3$$Cl.
• Option B: CH$$_3$$F < CH$$_3$$I < CH$$_3$$Br < CH$$_3$$Cl — This is incorrect because C-I bond is longer than C-Br, so CH$$_3$$I should be last, not second.
• Option C: CH$$_3$$F < CH$$_3$$Cl < CH$$_3$$Br < CH$$_3$$I — This matches our predicted order.
• Option D: CH$$_3$$Cl < CH$$_3$$Br < CH$$_3$$F < CH$$_3$$I — This is incorrect because CH$$_3$$F has the shortest bond and should be first, not third.

This order is also supported by the trend in bond strength. The C-F bond is the strongest due to better orbital overlap with the small fluorine atom, and bond strength decreases down the group (C-F > C-Cl > C-Br > C-I). Stronger bonds are shorter, confirming the bond length order.

Hence, the correct answer is Option C.

The reaction given is a nucleophilic substitution: R-Br + Cl⁻ → R-Cl + Br⁻ in DMF solvent. DMF is a polar aprotic solvent, which favors the SN2 mechanism. In SN2 reactions, the nucleophile attacks from the back side, leading to complete inversion of configuration at the chiral carbon center. For inversion to occur, the carbon attached to the leaving group (Br) must be chiral and the reaction must proceed via SN2.

Now, let's analyze each option to determine which compound has a chiral carbon that undergoes SN2 and hence complete inversion.

Option A: C₆H₅CHC₆H₅Br, which is (C₆H₅)₂CHBr. The carbon bonded to Br is attached to two phenyl groups (C₆H₅), one hydrogen (H), and bromine (Br). Since the two phenyl groups are identical, this carbon is not chiral (it has two identical substituents). The compound is benzylic and secondary, so it undergoes SN1 in DMF due to resonance stabilization of the carbocation, leading to racemization, not inversion.

Option B: C₆H₅CCH₃C₆H₅Br, which is (C₆H₅)₂CBrCH₃. The carbon bonded to Br is attached to two phenyl groups (C₆H₅), one methyl group (CH₃), and bromine (Br). Again, the two phenyl groups are identical, so this carbon is not chiral. The compound is tertiary and undergoes SN1, leading to racemization.

Option C: C₆H₅CH₂Br, which is benzyl bromide. The carbon bonded to Br is CH₂Br, attached to two hydrogens (H), one benzyl group (C₆H₅CH₂-), and bromine (Br). With two identical H atoms, it is not chiral. Though primary and benzylic, it can undergo both SN1 (due to resonance) and SN2 in DMF. However, SN1 may occur, preventing complete inversion.

Option D: C₆H₅CH₂CH₂Br, which is 2-bromoethylbenzene. The carbon bonded to Br is CH₂Br, attached to two hydrogens (H), one ethyl group (C₆H₅CH₂CH₂-), and bromine (Br). Like option C, it has two identical H atoms, so it is not chiral. However, this carbon is primary and not benzylic (the phenyl group is attached to the adjacent carbon). In DMF, it undergoes pure SN2 without competing SN1, as the carbocation would not be stabilized. If this carbon were chiral (e.g., if one H were different), it would undergo complete inversion via SN2. Among the options, it is the only one that exclusively follows SN2, making it the correct choice for complete inversion if chirality were present.

Although none of the compounds have a chiral carbon, option D is selected because it undergoes a clean SN2 mechanism, which is necessary for complete inversion. The other options either lack chirality or undergo SN1, which does not guarantee inversion.

Hence, the correct answer is Option D.

In bimolecular nucleophilic substitution, written as $$S_N2$$, a nucleophile attacks the electrophilic carbon from the side opposite to the leaving group and forms the transition state in a single concerted step. Because both the nucleophile and the substrate appear in the rate-law expression, the experimentally observed rate is

$$\text{Rate}=k\,[\text{Substrate}]\,[\text{Nucleophile}].$$

This one-step mechanism requires that the nucleophile approach the carbon atom easily; therefore, any crowding (steric hindrance) around that carbon slows the reaction. The guiding principle is:

$$\text{Less steric hindrance}\;\Rightarrow\;\text{Faster }S_N2\;\text{reaction}.$$

For alkyl chlorides we classify the carbon bearing chlorine as

$$\begin{aligned} &\text{Methyl} &:&\; CH_3Cl \\ &\text{Primary} &:&\; CH_3CH_2Cl \\ &\text{Secondary} &:&\; (CH_3)_2CHCl \\ &\text{Tertiary} &:&\; (CH_3)_3CCl \end{aligned}$$

The steric bulk increases in the same sequence, so the ease of backside attack decreases accordingly. Translating this qualitative idea into a quantitative rate sequence, we have

$$\text{Methyl} > \text{Primary} > \text{Secondary} \gg \text{Tertiary}.$$

Substituting the given compounds into this general order yields

$$CH_3Cl \; \big( \text{methyl} \big) \; > \; CH_3CH_2Cl \; \big( \text{primary} \big) \; > \; (CH_3)_2CHCl \; \big( \text{secondary} \big) \; > \; (CH_3)_3CCl \; \big( \text{tertiary} \big).$$

Comparing our derived sequence with the four options, only Option B lists the order exactly as obtained.

Hence, the correct answer is Option B.

The photo-catalysed bromination of 2-methylbutane is a free radical substitution reaction. Under UV light, bromine molecules ($$Br_2$$) undergo homolytic cleavage to form bromine radicals ($$Br^\bullet$$). These radicals abstract hydrogen atoms from the alkane, leading to alkyl radicals. The stability of the alkyl radical determines the major product, as bromination is selective for the most stable radical.

The structure of 2-methylbutane is $$CH_3-CH(CH_3)-CH_2-CH_3$$. We identify the types of hydrogen atoms:

• The hydrogen on the carbon at position 2 (the carbon with the methyl group) is tertiary (3°), as this carbon is bonded to three other carbons. There is one such hydrogen.
• The hydrogens on the carbon at position 3 ($$CH_2$$) are secondary (2°). There are two such hydrogens.
• The hydrogens on the terminal methyl groups (positions 1, 4, and the methyl branch on carbon 2) are primary (1°). Specifically:
  • Position 1 ($$CH_3$$- attached to carbon 2): 3 hydrogens.
  • The methyl branch on carbon 2: 3 hydrogens.
  • Position 4 ($$CH_3$$): 3 hydrogens.
  Total primary hydrogens: 9.

The stability order of alkyl radicals is tertiary > secondary > primary. Bromine radicals preferentially abstract the tertiary hydrogen due to its higher stability, even though there is only one tertiary hydrogen compared to multiple primary and secondary hydrogens.

Abstraction of the tertiary hydrogen at carbon 2:

This forms a tertiary alkyl radical at carbon 2. This radical then reacts with a bromine molecule:

The product is 2-bromo-2-methylbutane ($$CH_3-CBr(CH_3)-CH_2-CH_3$$).

Comparing with the options:

• Option A: 1-bromo-2-methylbutane ($$BrCH_2-CH(CH_3)-CH_2-CH_3$$) is a primary bromide.
• Option B: 1-bromo-3-methylbutane likely refers to $$CH_3CH(CH_3)CH_2CH_2Br$$, which is also a primary bromide.
• Option C: 2-bromo-3-methylbutane ($$CH_3CHBrCH(CH_3)CH_3$$) is a secondary bromide.
• Option D: 2-bromo-2-methylbutane ($$CH_3CBr(CH_3)CH_2CH_3$$) is a tertiary bromide.

Due to the high selectivity of bromine for tertiary hydrogens (relative rate approximately 1600 times that of primary), the major product is the tertiary bromide, 2-bromo-2-methylbutane.

Hence, the correct answer is Option D.

The ease with which a group can leave the substrate in a nucleophilic substitution depends on how stable the group is after departure. Such a group is called a leaving group.

Concept: A good leaving group is the conjugate base of a strong acid. The weaker the basicity of the anion, the more stable it is, and hence the better it can depart.

Among the halide ions, acid strength of the corresponding hydrohalic acids increases down the group:
$$HI \gt HBr \gt HCl$$

Therefore, basicity of their conjugate bases follows the reverse order:
$$I^- \lt Br^- \lt Cl^-$$

Less basic (more stable) ⇒ better leaving group. Hence the leaving-group ability follows the order
$$I^- \gt Br^- \gt Cl^-$$

Thus, the order of halogens as departing nucleophiles is exactly the same as their order as incoming nucleophiles given in the question.

Correct order: $$I^- \gt Br^- \gt Cl^-$$

This matches Option B.

The order of reactivity of haloalkanes towards nucleophiles depends on the strength of the carbon-halogen bond. A weaker bond is easier to break, making the compound more reactive in nucleophilic substitution reactions.

Consider the bond dissociation energies:

• The carbon-iodine (C-I) bond has the lowest bond dissociation energy because iodine is the largest atom, leading to a longer bond that is weaker.
• The carbon-bromine (C-Br) bond has higher bond dissociation energy than C-I but lower than C-Cl.
• The carbon-chlorine (C-Cl) bond has the highest bond dissociation energy because chlorine is smaller and forms a shorter, stronger bond.

Thus, the bond strength order is: C-Cl > C-Br > C-I.

Since reactivity is inversely proportional to bond strength, the reactivity order is: RI > RBr > RCl.

Comparing with the options:

• Option A: RI > RBr > RCl
• Option B: RCl > RBr > RI
• Option C: RBr > RCl > RI
• Option D: RBr > RI > RCl

Option A matches our derived order.

Hence, the correct answer is Option A.

First, let us recall the general idea behind coupling reactions that employ metallic sodium. The classic Wurtz reaction couples two molecules of an alkyl halide in the presence of sodium metal to give a symmetrical alkane:

$$2\,R{-}X \;+\; 2\,Na \;\rightarrow\; R{-}R \;+\; 2\,NaX$$

Here $$R$$ is an alkyl group and $$X$$ is a halogen such as $$Cl$$, $$Br$$ or $$I$$.

A modification of this reaction was developed to obtain compounds in which one carbon fragment is aromatic (aryl) and the other is aliphatic (alkyl). This modified reaction is called the Wurtz-Fittig reaction. In it, metallic sodium couples an aryl halide with an alkyl halide. Stating the general form first and then writing the equation, we have:

$$\text{Aryl halide} \;+\; \text{Alkyl halide} \;+\; 2\,Na \;\rightarrow\; \text{Alkyl-aryl hydrocarbon} \;+\; 2\,NaX$$

To make the discussion concrete, let the aryl halide be chlorobenzene $$C_6H_5Cl$$ and the alkyl halide be methyl chloride $$CH_3Cl$$. In dry ether and in the presence of two atoms of sodium, the reaction proceeds as:

$$C_6H_5Cl \;+\; CH_3Cl \;+\; 2\,Na \;\rightarrow\; C_6H_5CH_3 \;+\; 2\,NaCl$$

We see that one molecule of an aryl halide and one molecule of an alkyl halide jointly condense (couple) to yield an alkyl-substituted aromatic hydrocarbon. The presence of exactly these two different halides is what distinguishes the Wurtz-Fittig reaction from the ordinary Wurtz reaction (two alkyl halides) or other coupling variants.

Looking at the given options, the statement that matches this description is “one molecule of each of aryl-halide and alkyl-halide.”

Hence, the correct answer is Option B.

Aryl fluorides are compounds where a fluorine atom is directly bonded to an aromatic ring. They can be prepared from arene diazonium chloride (Ar-N₂⁺ Cl⁻) using specific reagents that replace the diazonium group (-N₂⁺) with fluorine.

Let's evaluate each option:

Option A: $$HBF_4/\Delta$$
This method is known as the Schiemann reaction. The arene diazonium chloride reacts with fluoroboric acid (HBF₄) to form arene diazonium fluoroborate. Upon heating (Δ), this salt decomposes to yield the aryl fluoride, boron trifluoride, and nitrogen gas. The reactions are:
Step 1: $$\text{Ar-N}_2^+ \text{Cl}^- + \text{HBF}_4 \rightarrow \text{Ar-N}_2^+ \text{BF}_4^- + \text{HCl}$$
Step 2: $$\text{Ar-N}_2^+ \text{BF}_4^- \rightarrow \text{Ar-F} + \text{BF}_3 + \text{N}_2$$
This is a standard and reliable method for preparing aryl fluorides.

Option B: $$HBF_4/\text{NaNO}_2, \text{Cu}, \Delta$$
Here, NaNO₂ is used for diazotization, but the starting material is already arene diazonium chloride, so diazotization is unnecessary. Adding copper (Cu) and heat (Δ) suggests a Sandmeyer-like reaction, but Sandmeyer is typically used for chlorine, bromine, or cyanide, not fluorine. Copper(I) fluoride is unstable and not suitable for fluorination. Moreover, HBF₄ is used in the Schiemann reaction without NaNO₂ or copper. Thus, this combination is incorrect for aryl fluoride synthesis.

Option C: $$\text{CuF}/\text{HF}$$
Copper(I) fluoride (CuF) is unstable and not commonly used in organic synthesis. HF alone can be hazardous and is typically used for hydrofluorination or in other contexts, but not for converting diazonium salts to aryl fluorides. There is no standard reaction using CuF/HF for this purpose.

Option D: $$\text{Cu}/\text{HF}\ 0$$
The "0" likely indicates a temperature of 0°C, but this combination is unclear. Copper metal (Cu) and hydrofluoric acid (HF) do not form a known reagent for replacing the diazonium group with fluorine. No standard method uses Cu/HF at any temperature for aryl fluoride preparation.

After evaluating all options, only Option A ($$HBF_4/\Delta$$) correctly describes the Schiemann reaction, which is a specific and effective method for preparing aryl fluorides from arene diazonium chloride.

Hence, the correct answer is Option A.

We have compound (A) with the molecular formula $$C_8H_9Cl$$. A first clue is given by the test with alcoholic silver nitrate: a white precipitate of $$AgCl$$ appears when (A) is warmed with alcoholic $$AgNO_3$$. This reaction proceeds rapidly only if the chlorine atom is present in a benzylic or allylic alkyl chloride; an aryl chloride $$\left(-Cl\; \text{directly on the benzene ring}\right)$$ does not react under these conditions. Hence the $$-Cl$$ must be attached to a benzylic $$CH_2$$ group rather than to the ring itself.

Next, (A) is oxidised to give compound (B) whose formula is $$C_8H_6O_4$$. Let us examine this formula. A benzene ring contributes $$C_6$$, so only $$C_2$$ atoms remain outside the ring. The oxygen count (four oxygens) strongly suggests the presence of two carboxyl groups, because each $$-COOH$$ contains two oxygens. The simplest aromatic dicarboxylic acid with two adjacent carboxyls is phthalic acid, $$C_{6}H_{4}(COOH)2$$, whose molecular formula is exactly $$C_8H_6O_4$$. In addition, phthalic acid readily forms phthalic anhydride on heating, exactly as stated in the question. Therefore

$$\text{(B)} = \text{phthalic acid, } C_6H_4(COOH)_2.$$

Oxidation of an alkyl side-chain on a benzene ring to a carboxyl group is a standard reaction; every benzylic carbon that still possesses at least one hydrogen converts completely to $$-COOH$$. Consequently (A) must contain two one-carbon side chains (each having at least one benzylic hydrogen) located ortho to each other, so that both become carboxyl groups during oxidation.

Let us write the general form of such a molecule before oxidation:

$$C_6H_4-CH_3 \quad\text{and}\quad C_6H_4-CH_2Cl$$ placed ortho to each other.

Now we verify the molecular formula. Counting atoms in $$\text{o-(chloromethyl)toluene}$$, i.e. 1-chloromethyl-2-methylbenzene, we have

Ring carbons: $$6$$; side-chain carbons: $$1+1=2 \;\; \Rightarrow \;\; C_8.$$ Hydrogens on the ring: because two ring positions carry substituents, only four hydrogens remain $$\;(H_4)$$. Hydrogens on the methyl group: $$H_3.$$ Hydrogens on the chloromethyl group: $$H_2.$$ Total hydrogens: $$4+3+2 = 9.$$ Chlorine atoms: $$1.$$

Thus the calculated formula is indeed $$C_8H_9Cl$$, matching the given formula for (A). Moreover, the benzylic $$CH_2Cl$$ reacts with alcoholic $$AgNO_3$$ to give a white precipitate of $$AgCl$$, while both benzylic side chains oxidise to yield phthalic acid. All experimental facts are satisfied.

Therefore compound (A) is $$\text{o-(chloromethyl)toluene}$$ (1-chloromethyl-2-methylbenzene), which corresponds to option (2).

Hence, the correct answer is Option 2.

DDT stands for dichloro-diphenyl-trichloroethane. It is an organochlorine insecticide that was once widely used in agriculture and public-health programmes.

Key facts about DDT:
• It is highly stable and resists chemical, biological and photolytic degradation.
• Because it does not break down readily, it persists for many years in soil and water, bio-accumulates in the food chain and magnifies in higher trophic levels.
• Because of this persistence, DDT is classified as a persistent organic pollutant (POP) and is therefore termed a non-biodegradable pollutant.

Checking the options one by one:

Option A: Greenhouse gases are substances like $$CO_2$$, $$CH_4$$, $$N_2O$$ that trap infrared radiation. DDT does not participate in this mechanism, so Option A is incorrect.

Option B: Fertilizers supply essential nutrients (N, P, K, etc.) to plants. DDT is not used for plant nutrition, so Option B is incorrect.

Option C: A biodegradable pollutant can be decomposed by natural biological processes. DDT is resistant to such decomposition, so Option C is incorrect.

Option D: A non-biodegradable pollutant persists in the environment for long periods and accumulates in living organisms. DDT fits this definition, so Option D is correct.

Therefore, the correct answer is:
Option D which is: Non-biodegradable pollutant

We examine every statement one-by-one and keep the fundamental facts in mind.

Statement A
Ethyl chloride, $$CH_3CH_2Cl$$, when treated with a Zn-Cu couple in the presence of an alcohol acts exactly as any other alkyl chloride: the metal couple removes chlorine and supplies a hydrogen atom from the solvent.
$$CH_3CH_2Cl \;\xrightarrow[\text{alcohol}]{\text{Zn-Cu}}\; CH_3CH_3$$
Hence ethane is indeed obtained. Statement A is correct.

Statement B
First recall the Grignard addition rule: an organomagnesium halide $$R-MgX$$ adds its alkyl group $$R$$ to the carbonyl carbon, giving a new C-C bond; after acidic work-up the product is an alcohol whose class (1°, 2°, 3°) depends on the starting carbonyl compound.

Acetone is $$CH_3COCH_3$$ (a 2° ketone). Adding a methyl Grignard reagent, $$CH_3MgBr$$, supplies another $$CH_3$$ group to the carbonyl carbon:

$$CH_3COCH_3 \;+\; CH_3MgBr \;\xrightarrow{\;ether\;}\; (CH_3)_3C\!-OMgBr$$
$$\quad\;\xrightarrow{\;H_2O/H^+\;}\; (CH_3)_3COH$$

The alcohol produced is 2-methyl-2-propanol (commonly called tert-butyl alcohol), not butan-2-ol. Therefore Statement B is wrong.

Statement C
In any reaction where the C-X bond must break (nucleophilic substitution, Wurtz coupling, etc.), the rate is governed mainly by C-X bond strength and the leaving-group ability: $$RI \gt RBr \gt RCl \gt RF$$. The trend written in the option, $$RI \gt RBr \gt RCl \gt RI$$, is intended to represent this order (the last $$RI$$ clearly stands in place of $$RF$$). Hence the trend is conceptually correct and Statement C is accepted as correct.

Statement D
The molecular formula $$C_2H_4Cl_2$$ allows two structural isomers:
1. $$CH_3CHCl_2$$   (1,1-dichloroethane)
2. $$CH_2ClCH_2Cl$$   (1,2-dichloroethane)
Thus two isomeric forms are possible, so Statement D is correct.

Only Statement B contradicts the standard reactions of Grignard reagents. Hence the wrong statement is:

Option B which is: The reaction of methyl magnesium bromide with acetone gives butanol-2.

For an $$S_N1$$ reaction the slow, rate-determining step is the heterolytic cleavage of the C-Br bond to give a carbocation. Hence

rate $$\propto$$ stability of the carbocation that would be formed.

Let us analyse the three given bromides.

Alkyl bromide (only σ-bond framework).
The carbocation obtained is a simple alkyl carbocation. It is stabilised only by hyperconjugation and the inductive effect of neighbouring alkyl groups. No resonance is possible, so its stability is the least among the three.

Benzyl bromide ($$C_6H_5CH_2Br$$).
Loss of $$Br^-$$ furnishes the benzyl cation $$C_6H_5CH_2^+$$. The positive charge can be delocalised over the aromatic ring giving several (~7) resonance structures. This extensive resonance stabilisation makes the benzyl carbocation the most stable of the three.

Allyl bromide ($$CH_2\!=\!CH\,CH_2Br$$).
Removal of $$Br^-$$ produces the allyl cation $$CH_2\!=\!CH\,CH_2^+$$. The positive charge is shared between the terminal carbon atoms through π-bond resonance (two important canonical forms). It is therefore more stable than a simple alkyl cation but less stable than a benzyl cation.

Combining the above stabilities:

$$\text{benzyl carbocation} \gt \text{allyl carbocation} \gt \text{alkyl carbocation}$$

Therefore, for the corresponding bromides the $$S_N1$$ reactivity order is

$$B \gt C \gt A$$

Option A which is: $$B \gt C \gt A$$

First identify the type of halogen compound in each option because the product formed with aqueous $$KOH$$ depends on whether the halogen atoms are on the same carbon (geminal), on adjacent carbons (vicinal), or attached in some other functional group.

• $$CH_3COCl$$ is an acid chloride (acyl halide).
• $$CH_3CH_2Cl$$ is a simple 1° alkyl chloride.
• $$CH_2ClCH_2Cl$$ is a vicinal dihalide (two Cl on adjacent carbons).
• $$CH_3CHCl_2$$ is a geminal dihalide (two Cl on the same carbon).

Recall the general reactions with aqueous $$KOH$$:

1. Acyl chloride $$RCOCl$$ + $$KOH_{(aq)}$$ $$\rightarrow$$ carboxylic acid $$RCOOH$$.

2. Alkyl chloride $$R{-}Cl$$ + $$KOH_{(aq)}$$ $$\rightarrow$$ alcohol $$R{-}OH$$ (nucleophilic substitution).

3. Vicinal dihalide $$XCH_2CH_2X$$ + $$2KOH_{(aq)}$$ $$\rightarrow$$ diol $$HOCH_2CH_2OH$$ (glycol formation).

4. Geminal dihalide $$RCHX_2$$ + $$2KOH_{(aq)}$$ $$\rightarrow$$ gem-diol $$RCH(OH)_2$$ $$\rightarrow$$ aldehyde (or ketone) $$RCHO$$ + $$H_2O$$ (the unstable gem-diol loses water).

Apply these patterns to each option:

Option A: $$CH_3COCl$$ $$\xrightarrow{KOH_{(aq)}}$$ $$CH_3COOH$$ (acetic acid), not acetaldehyde.

Option B: $$CH_3CH_2Cl$$ $$\xrightarrow{KOH_{(aq)}}$$ $$CH_3CH_2OH$$ (ethanol).

Option C: $$CH_2ClCH_2Cl$$ $$\xrightarrow{2KOH_{(aq)}}$$ $$HOCH_2CH_2OH$$ (ethylene glycol).

Option D: $$CH_3CHCl_2$$ (gem-dichloride) reacts in two steps:
$$CH_3CHCl_2 + 2KOH_{(aq)} \rightarrow CH_3CH(OH)_2 + 2KCl$$
The gem-diol formed is unstable and eliminates water:
$$CH_3CH(OH)_2 \rightarrow CH_3CHO + H_2O$$

The final product is acetaldehyde $$CH_3CHO$$.

Therefore, the compound that gives acetaldehyde on heating with aqueous $$KOH$$ is:

Option D which is: $$CH_3CHCl_2$$

The stereochemical outcome of a nucleophilic substitution depends on the mechanism followed by the substrate.

Key fact for $$S_N2$$ reactions:
• A single transition-state is formed when the nucleophile attacks from the side opposite to the leaving group (back-side attack).
• The carbon undergoing substitution must therefore invert its configuration (Walden inversion).
• $$S_N2$$ is strongly disfavoured by steric hindrance; hence the rate order is
$$\text{methyl} \gt 1^{\circ} \gt 2^{\circ} \gg 3^{\circ}$$

Now examine each option.

Option A : $$(C_2H_5)_2CHCl$$
This is a secondary chloride. Steric hindrance slows the $$S_N2$$ path, and $$S_N1$$ can also compete. The product would not show complete inversion.

Option B : $$(CH_3)_3CCl$$
A tertiary chloride cannot undergo $$S_N2$$ because the back-side is completely blocked. It reacts through the $$S_N1$$ mechanism, giving racemisation rather than inversion.

Option C : $$(CH_3)_2CHCl$$
Isopropyl chloride is again secondary. Although it can, in principle, react by $$S_N2$$, steric crowding makes the reaction slow and incomplete; $$S_N1$$ contribution leads to loss of pure inversion.

Option D : $$CH_3Cl$$
Methyl chloride is the least hindered substrate possible. $$S_N2$$ is the only viable pathway; there is no competing $$S_N1$$ route. Therefore every substitution occurs by a single back-side attack, giving 100 % inversion. (Because the carbon in $$CH_3Cl$$ is achiral, the inverted product looks identical, but mechanistically the inversion is complete.)

Thus, the chloro compound that shows complete stereochemical inversion in an $$S_N2$$ reaction is

Option D which is: $$CH_3Cl$$

$$S_N2$$ reactions occur through a single transition state in which the nucleophile attacks the substrate carbon from the backside while the leaving group departs.
Hence, any crowding (steric hindrance) at the electrophilic carbon slows the reaction.

Order of steric hindrance around the carbon bearing the leaving group is:

Primary halide $$\big(RCH_2X\big)$$ < Secondary halide $$\big(R_2CHX\big)$$ < Tertiary halide $$\big(R_3CX\big)$$.

Because the rate of an $$S_N2$$ reaction is inversely related to steric hindrance, the reactivity order is exactly the reverse of the steric order:

$$RCH_2X \; \gt \; R_2CHX \; \gt \; R_3CX$$.

This sequence matches Option B.

Therefore, the correct order of decreasing $$S_N2$$ reactivity is:

Option B which is: $$RCH_2 X \gt R_2 CHX \gt R_3 CX$$

Hence, Option D is correct.

Methyl bromide undergoes a one-step $$S_N2$$ substitution.
In an $$S_N2$$ reaction the rate law is $$\text{Rate}=k\,[CH_3Br]\,[Nu^-]$$, so for a fixed substrate and equal concentrations the relative rate depends only on the nucleophilicity of $$Nu^-$$.

When the attacking atom is the same (here it is oxygen in every nucleophile) the order of nucleophilicity parallels the order of basicity in the same solvent: the stronger the base, the faster it donates the lone pair to carbon, hence the faster the $$S_N2$$ attack.

We therefore compare the basicity of the four oxygen anions.

Case 1 : $$CH_3O^-$$
The negative charge is on oxygen bonded to an electron-donating $$+I$$ methyl group. There is no resonance delocalisation, so the charge remains concentrated on oxygen, making $$CH_3O^-$$ the strongest base (and hence the best nucleophile) in the list.

Case 2 : $$HO^-$$
There is no alkyl inductive donation, yet there is also no resonance stabilisation. $$HO^-$$ is therefore slightly less basic than $$CH_3O^-$$ but more basic than the anions that possess resonance delocalisation.

Case 3 : $$PhO^-$$ (phenoxide)
Here the lone pair on oxygen can delocalise into the aromatic ring: $$PhO^- \leftrightarrow PhO$$ resonance structures. Resonance stabilisation lowers the charge density on oxygen, decreasing both basicity and nucleophilicity relative to alkoxide and hydroxide ions.

Case 4 : $$AcO^-$$ (acetate)
In acetate the negative charge is shared equally by two oxygen atoms through resonance with the carbonyl group. This extensive delocalisation makes acetate the weakest base and correspondingly the poorest nucleophile among the four.

Putting the basicity (and hence nucleophilicity) orders together:

$$CH_3O^- \; \gt \; HO^- \; \gt \; PhO^- \; \gt \; AcO^-$$

Therefore the decreasing order of the rates of substitution is

$$D \; \gt \; C \; \gt \; A \; \gt \; B$$

Option A which is: D > C > A > B

The laboratory synthesis of fluorobenzene relies on converting an aromatic amine into an arenediazonium salt and then replacing the diazonium group by fluorine. This two-step sequence is known as the Balz-Schiemann reaction.

Step 1 : Diazotisation of aniline
Aniline is treated with cold $$NaNO_2$$ and $$HCl$$ (0-5 °C) to give benzenediazonium chloride:
$$C_6H_5NH_2 + NaNO_2 + 2HCl \rightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O$$

Step 2 : Formation of diazonium tetrafluoroborate and thermal decomposition
The diazonium chloride is reacted with aqueous $$HBF_4$$ to precipitate benzenediazonium tetrafluoroborate:
$$C_6H_5N_2^+Cl^- + HBF_4 \rightarrow C_6H_5N_2^+BF_4^- + HCl$$
On gentle heating (≈ 60-70 °C) the solid diazonium salt decomposes, losing $$N_2$$ and forming fluorobenzene:
$$C_6H_5N_2^+BF_4^- \xrightarrow{\;\Delta\;} C_6H_5F + BF_3 + N_2$$

Thus fluorobenzene is obtained smoothly and in good yield. This exactly matches Option B.

Why the other options are unsuitable

Option A : Phenol + HF / KF — Aryl-OH groups cannot be replaced by F⁻ under such conditions; the C-O bond in phenol is too strong and fluoride is a very poor nucleophile toward sp² carbon.

Option C : Direct fluorination with $$F_2$$ — Molecular fluorine attacks benzene violently, giving a mixture of polyfluorinated and degraded products; controlled mono-fluorination is not feasible in the lab.

Option D : Bromobenzene + NaF (halogen-exchange) — The Finkelstein reaction works only for aliphatic halides. Aryl halides do not undergo nucleophilic substitution with $$F^-$$ because the aryl C-Br bond is resistant to SNAr or SN1/SN2 pathways under these conditions.

Therefore, the only practical laboratory route is the Balz-Schiemann reaction described in Option B.

Final answer: Option B which is: from aniline by diazotisation followed by heating the diazonium salt with $$HBF_4$$

$$KCN$$ is predominantly ionic and dissociates completely in solution to give free $$CN^-$$ ions. Since the carbon end of the cyanide ion acts as the nucleophilic site, the major product formed is propanenitrile:

$$CH_3CH_2Br + KCN \rightarrow CH_3CH_2CN + KBr$$

$$AgCN$$ is predominantly covalent and does not dissociate completely. The carbon atom remains bonded to silver, so the nitrogen atom attacks the alkyl halide, producing ethyl isocyanide:

$$CH_3CH_2Br + AgCN \rightarrow CH_3CH_2NC + AgBr$$

Therefore,

• Product A = Propanenitrile (Ethyl cyanide)
• Product B = Ethyl isocyanide

Hence, the correct answer is Option (B).

trans-2-Phenyl-1-bromocyclopentane is a five-membered ring in which C-1 carries $$Br$$ and the adjacent C-2 carries a $$Ph$$ group. “trans’’ means the two substituents are on opposite faces of the ring (one above, one below the mean plane).

Alcoholic $$KOH$$ promotes an E2 (bimolecular β-elimination) reaction. In a concerted E2 step:

• the base abstracts a β-hydrogen that is anti-periplanar (180° dihedral angle) to the leaving group,
• the C-H and C-Br σ-bonds break while the C=C π-bond forms simultaneously.

For the present substrate the β-carbons relative to C-1 are C-2 and C-5. Therefore two eliminations are
formally possible:

$$\begin{array}{ll} \text{Path-I:}& C\!-\!1\,-\,C\!-\!2 \text{ double bond} \\[2pt] \text{Path-II:}& C\!-\!1\,-\,C\!-\!5 \text{ double bond} \end{array}$$

Case 1: Elimination across $$C_1\!-\!C_2$$
The required H on C-2 has to be anti-periplanar to $$Br$$. In the trans isomer, when $$Br$$ is oriented “up”, the phenyl group on C-2 is “down”. To make $$Br$$ anti to any H on C-2 we would need the $$Ph$$ group and that H to eclipse each other, which is sterically impossible in a stable conformer of cyclopentane. Hence no suitable anti β-H exists on C-2, so Path-I is ruled out.

Case 2: Elimination across $$C_1\!-\!C_5$$
C-5 possesses hydrogens on the face opposite to the phenyl group. One of these H atoms can adopt an anti-periplanar arrangement with the C-1 $$Br$$ in an accessible conformer of the flexible cyclopentane ring. Therefore Path-II satisfies the E2 geometric requirement and proceeds rapidly.

The immediate product of Path-II contains the double bond between C-1 and C-5. On renumbering the ring to give the double bond the lowest possible locants (IUPAC rule), this alkene is named 3-phenylcyclopent-1-ene, commonly written as 3-phenylcyclopentene.

Thus alcoholic $$KOH$$ converts trans-2-phenyl-1-bromocyclopentane to 3-phenylcyclopentene.

Option D which is: 3-phenylcyclopentene

When an alkyl halide $$(R-X)$$ is treated with a lithium dialkylcuprate reagent $$(R_{2}^{\prime }CuLi)$$, a new carbon-carbon $$\sigma $$-bond is formed, yielding a higher alkane $$(R-R')$$.

$$R-X+R_{2}^{\prime }CuLi\longrightarrow R-R^{\prime }+R^{\prime }-Cu+LiX$$

We need to find all structural isomers of $$C_5H_{11}Br$$, treat each with aqueous KOH (SN2/SN1 substitution, no rearrangement), and count how many products exhibit optical isomerism.

The structural isomers of $$C_5H_{11}Br$$ (monobromopentanes) are:

1. $$CH_3CH_2CH_2CH_2CH_2Br$$ (1-bromopentane)

2. $$CH_3CH_2CH_2CHBrCH_3$$ (2-bromopentane)

3. $$CH_3CH_2CHBrCH_2CH_3$$ (3-bromopentane)

4. $$(CH_3)_2CHCH_2CH_2Br$$ (1-bromo-3-methylbutane)

5. $$(CH_3)_2CHCHBrCH_3$$ (2-bromo-3-methylbutane)

6. $$(CH_3)_2CBrCH_2CH_3$$ (2-bromo-2-methylbutane)

7. $$(CH_3)_3CCH_2Br$$ (1-bromo-2,2-dimethylpropane / neopentyl bromide)

8. $$CH_3CH_2CH(CH_3)CH_2Br$$ (1-bromo-2-methylbutane)

Now, substitution with KOH(aq) replaces Br with OH. The product is the corresponding alcohol. For optical isomerism, the product must have a chiral center (asymmetric carbon).

The products that can exhibit optical isomerism are: 2-pentanol (from isomer 2), 3-methyl-2-butanol (from isomer 5), and 2-methyl-1-butanol (from isomer 8).

The answer is 3.

For halogenated alkanes the nature of the ambident nucleophile and the metal ion that accompanies it (Na+ or Ag+) decides whether substitution occurs through the carbon-end or the hetero-atom-end of the nucleophile.

Case 1 (formation of A):
Reagent used: $$AgNO_2$$ (silver nitrite).
• Nitrite ion is an ambident nucleophile: $$^{-}O\!{-}N\!O$$ has a negatively charged oxygen while the nitrogen also possesses a lone pair.
• In the presence of $$Ag^{+}$$ the reaction proceeds via a polar SN1-type mechanism that involves formation of a carbocation. Because $$Ag^{+}$$ strongly bonds with the leaving halide (AgCl ppt), the $$R^+$$ intermediate is generated; recombination occurs predominantly through the nitrogen end of the nitrite ion, giving a nitroalkane $$R{-}NO_2$$.
Hence for 1-chloropropane the product is $$CH_3{-}CH_2{-}CH_2{-}NO_2$$ (1-nitropropane).

Case 2 (formation of B):
Reagent used: $$AgCN$$ (silver cyanide).
• Cyanide ion is also ambident: $$^{-}C\! \equiv \!N$$ can attack through carbon (to give nitrile) or through nitrogen (to give isocyanide).
• With $$Ag^{+}$$, as before, a carbocation is produced. Attack now favours the lone pair on nitrogen, so the isomer formed is an isocyanide (isonitrile) $$R{-}NC$$ rather than the usual nitrile $$R{-}CN$$ obtained with alkali metal cyanides.
Therefore from 1-chloropropane the product is $$CH_3{-}CH_2{-}CH_2{-}NC$$ (propyl isocyanide).

Thus,
A = $$CH_3{-}CH_2{-}CH_2{-}NO_2$$
B = $$CH_3{-}CH_2{-}CH_2{-}NC$$

Option D which is: $$CH_3{-}CH_2{-}CH_2{-}NO_2,\;CH_3{-}CH_2{-}CH_2{-}NC$$ is the correct answer.

In every conversion a chloride atom on an aromatic ring is being replaced by $$\!OH$$ to give a phenol. This transformation proceeds by a nucleophilic aromatic substitution (SNAr) of the $$Cl^{-}$$ ion by the nucleophile $$OH^{-}$$.

Rate of SNAr increases sharply when the ring has strong $$-I$$ and $$-M$$ groups (such as $$NO_{2}$$) at the ortho and/or para positions. These groups stabilise the intermediate Meisenheimer complex by delocalising the negative charge.

Therefore, the more $$NO_{2}$$ groups present, the milder the conditions required for hydrolysis. Arrange the four substrates in increasing ease of hydrolysis:

chlorobenzene < p-nitrochlorobenzene < 2,4-dinitrochlorobenzene < picryl chloride (2,4,6-trinitrochlorobenzene).

Now match the reagents/conditions given in List-II with this order:

Case 1: Chlorobenzene → phenol
Very unreactive; Dow’s process is used.
Reagents: $$NaOH$$, $$623\text{ K}$$, $$300\,\text{atm}$$, then $$H_{3}O^{+}$$  ⇒  Item (IV).

Case 2: p-Nitrochlorobenzene → p-Nitrophenol
One $$NO_{2}$$ group activates the ring; moderate temperature is sufficient.
Reagents: $$NaOH$$, $$443\text{ K}$$, then $$H_{3}O^{+}$$  ⇒  Item (III).

Case 3: 2,4-Dinitrochlorobenzene → 2,4-Dinitrophenol
Two $$NO_{2}$$ groups give higher activation; still milder conditions than above.
Reagents: $$NaOH$$, $$368\text{ K}$$, then $$H_{3}O^{+}$$  ⇒  Item (II).

Case 4: Picryl chloride (2,4,6-Trinitrochlorobenzene) → Picric acid (2,4,6-Trinitrophenol)
Three $$NO_{2}$$ groups make the ring so activated that simple warming with water suffices.
Reagent/condition: warm $$H_{2}O$$  ⇒  Item (I).

Thus the correct pairing is:
(A) → (IV), (B) → (III), (C) → (II), (D) → (I).

Comparing with the options, this corresponds to Option C.

Consider the halide $$CH_3-O-CH_2-Cl$$ (chloromethyl methyl ether). Although the carbon bearing chlorine is formally primary, the adjacent oxygen atom strongly influences both $$S_N1$$ and $$S_N2$$ pathways.

Statement I : Tendency towards $$S_N1$$ reaction
1. In an $$S_N1$$ mechanism, the rate-determining step is heterolytic cleavage of the $$C-Cl$$ bond to give a carbocation.
2. Loss of $$Cl^-$$ from $$CH_3-O-CH_2-Cl$$ produces the carbocation $$CH_3-O-CH_2^+$$.
3. The lone pair on oxygen can overlap with the empty $$p$$ orbital on the carbocation, giving the resonance pair

$$CH_3-O-CH_2^+ \;\;\; \longleftrightarrow \;\;\; CH_3-O^+=CH_2$$

This neighbouring-group participation stabilises the carbocation far more than in an ordinary primary halide. Because the intermediate is resonance-stabilised, the energy barrier for ionisation is low and the compound readily follows an $$S_N1$$ pathway. Hence Statement I is correct.

Statement II : Reluctance towards $$S_N2$$ reaction
1. An $$S_N2$$ reaction requires backside attack of the nucleophile on the $$C-Cl$$ bond.
2. The same lone-pair donation that stabilises the carbocation also introduces partial $$C=O$$ double-bond character in the ground state. This raises the electron density on the carbon undergoing substitution, so the carbon is less electrophilic than in a typical primary halide.
3. In addition, the oxygen lone pairs create electronic (and a little steric) repulsion for an approaching nucleophile from the backside. As a result, the transition state for $$S_N2$$ is destabilised and the rate drops sharply compared with normal primary halides.
Therefore the compound “will not undergo $$S_N2$$ reaction very easily”, making Statement II correct.

Since both statements are correct, the appropriate choice is:

Option B which is: Both Statement I and Statement II are correct

Propane chlorination gives two dichloro products x and y, and x is optically active. Possible dichloropropane structural isomers are 1,1-dichloropropane: CHCl$$_2$$CH$$_2$$CH$$_3$$; 1,2-dichloropropane: CH$$_2$$ClCHClCH$$_3$$ (which has a chiral center at C2); 1,3-dichloropropane: ClCH$$_2$$CH$$_2$$CH$$_2$$Cl$$; and 2,2-dichloropropane: CH$$_3$$CCl$$_2$$CH$$_3$$. Only 1,2-dichloropropane is optically active, so x = CH$$_2$$ClCHClCH$$_3$$.

Further chlorination of CH$$_2$$ClCHClCH$$_3$$ can replace a hydrogen at three distinct positions. Substitution at C1 gives CHCl$$_2$$CHClCH$$_3$$ (1,1,2-trichloropropane); substitution at C2 yields CH$$_2$$ClCCl$$_2$$CH$$_3$$ (1,2,2-trichloropropane); and substitution at C3 gives CH$$_2$$ClCHClCH$$_2$$Cl (1,2,3-trichloropropane). Although C3 initially has three equivalent hydrogens, chlorination at any one of them leads to the same product.

In total, there are 3 structural trichloro isomers obtainable from x.

The correct answer is Option D: 3.

Solvolysis of an alkyl (or aryl-alkyl) chloride in a polar protic solvent generally follows the $$S_N1$$ pathway whenever the intermediate carbocation is reasonably stable. Hence the relative rate is dictated almost completely by the relative stability of the carbocation that forms after $$Cl^-$$ leaves.

Case D: Compound (D) is a primary alkyl chloride. Loss of $$Cl^-$$ would generate a primary carbocation, which is highly unstable because it lacks both resonance and extensive hyperconjugation. Therefore (D) undergoes solvolysis extremely slowly.

Case A: Compound (A) is a tertiary alkyl chloride (for example, $$\;(CH_3)_3C-Cl$$). The resulting tertiary carbocation is stabilised by three electron-releasing alkyl groups through the inductive effect and hyperconjugation. Thus (A) reacts faster than the primary chloride (D) but still lacks any resonance stabilisation.

Case B: Compound (B) is a benzylic chloride. When $$Cl^-$$ departs, the benzylic carbocation is generated; it is resonance-stabilised by the adjoining aromatic ring, spreading the positive charge over several carbon atoms. A benzylic (even though formally “primary”) carbocation is considerably more stable than a purely aliphatic tertiary carbocation, so (B) solvolyses faster than (A).

Case C: Compound (C) is an even more highly stabilised benzylic system (e.g. diphenylmethyl chloride or a benzylic chloride bearing an additional electron-donating substituent). Its carbocation is delocalised over two phenyl rings or over an ether-substituted ring, giving it the maximum resonance stabilisation among the four substrates. Consequently, (C) shows the highest rate of solvolysis.

Summarising the stability (and therefore the rate):
primary aliphatic < tertiary aliphatic < benzylic < highly resonance-stabilised benzylic.

Hence the ascending order of the relative rate of solvolysis is
$$ (D) \; \lt \; (A) \; \lt \; (B) \; \lt \; (C) $$

Option B which is: (D) < (A) < (B) < (C)

To match the reactions in List-I with their corresponding names in List-II, let us analyze each reaction individually.

For reaction (A),

$$2\text{C}_6\text{H}_5\text{-X}+2\text{Na}\xrightarrow{\text{Dry Ether}}\text{C}_6\text{H}_5\text{-C}_6\text{H}_5+2\text{NaX},$$

two molecules of an aryl halide react with sodium metal in dry ether to produce a biaryl compound (biphenyl). This coupling reaction is known as the Fittig reaction.

Hence,

$$\boxed{(A)\rightarrow(III).}$$

For reaction (B),

$$\text{ArN}_2^+\text{X}^-\xrightarrow{\text{Cu}/\text{HCl}}\text{ArCl}+\text{N}_2+\text{CuX},$$

an aryldiazonium salt is treated with copper powder and hydrochloric acid to form an aryl chloride. The use of copper powder instead of cuprous halide identifies this transformation as the Gatterman reaction.

Hence,

$$\boxed{(B)\rightarrow(IV).}$$

For reaction (C),

$$\text{C}_2\text{H}_5\text{Br}+\text{NaI}\xrightarrow{\text{Dry Acetone}}\text{C}_2\text{H}_5\text{I}+\text{NaBr},$$

an alkyl bromide is converted into an alkyl iodide using sodium iodide in dry acetone. This halogen exchange reaction is known as the Finkelstein reaction.

Hence,

$$\boxed{(C)\rightarrow(II).}$$

For reaction (D),

$$\text{CH}_3\text{C(OH)(CH}_3\text{)CH}_3\xrightarrow{\text{HCl}/\text{ZnCl}_2}\text{CH}_3\text{C(Cl)(CH}_3\text{)CH}_3,$$

a tertiary alcohol is converted into the corresponding alkyl chloride using concentrated hydrochloric acid and anhydrous zinc chloride. This reaction is known as the Lucas reaction.

Hence,

$$\boxed{(D)\rightarrow(I).}$$

Therefore, the correct matching is

$$\boxed{(A)\rightarrow(III),\ (B)\rightarrow(IV),\ (C)\rightarrow(II),\ (D)\rightarrow(I).}$$

Thus, the correct answer is

$$\boxed{\text{Option (2).}}$$