Q.The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.

The key to this problem is working backwards from the ozonolysis product. Ozonolysis cleaves a carbon-carbon double bond and replaces it with carbonyl groups. If the only product is ethanol ($\ce{CH3CH2OH}$), that means the original alkene $\ce{C4H8}$ must have been symmetrical in a very specific way — each half of the double bond, after cleavage, becomes an aldehyde or ketone that is then reduced to an alcohol. Ethanol has two carbons, so the alkene must be but-1-ene ($\ce{CH3CH2CH=CH2}$). Let’s verify each statement against this structure.

1. Statement I: "It does not exhibit cis/trans isomerism"

   But-1-ene has the double bond at the terminal position (C1–C2). One of the carbons in the double bond (C1) has two identical hydrogen atoms attached. For cis/trans (geometric) isomerism, each carbon of the double bond must have two different substituents. Here, C1 has two H’s, so no geometric isomers exist.

   Watch out

   This statement is true for but-1-ene, but the question asks for correct statements about the reactant (X). We must check all statements — a true statement about the wrong alkene is still wrong if the alkene is misidentified. But here, but-1-ene indeed lacks cis/trans isomerism. However, read on — the trick is that statement I is true for but-1-ene, but the question’s answer choices will reveal if it’s considered correct. We’ll evaluate after checking all.

2. Statement II: "It adds water in the presence of dil. $\ce{H2SO4}$ to give secondary alcohol"

   Hydration of but-1-ene follows Markovnikov’s rule: the hydrogen adds to the less substituted carbon (C1), and the OH adds to the more substituted carbon (C2). The product is butan-2-ol ($\ce{CH3CH2CH(OH)CH3}$), which is a secondary alcohol. So this statement is also true for but-1-ene.

   Tip

   Markovnikov addition: $\ce{H}$ goes to the carbon with more H’s already; $\ce{OH}$ goes to the other carbon.

3. Statement III: "It is obtained as major product in the reaction of secondary butylchloride with alc. $\ce{KOH}$"

   Secondary butyl chloride is $\ce{CH3CH2CHClCH3}$ (2-chlorobutane). Alcoholic $\ce{KOH}$ favours elimination (dehydrohalogenation) over substitution. The major product is the more substituted alkene (Saytzeff’s rule): but-2-ene ($\ce{CH3CH=CHCH3}$), not but-1-ene. But-1-ene is the minor product. So this statement is false for but-1-ene.

Now, we have a conflict: statements I and II are true for but-1-ene, but statement III is false. However, look at the answer choices — none says “I, II only” is correct? Wait, option (B) is “I, II only”. But we must double-check: is but-1-ene really the only $\ce{C4H8}$ that gives only ethanol on ozonolysis? Let’s confirm.

The key is to first identify the alkene. The formula C₄H₈ could be either 1‑butene or 2‑butene (or isobutylene). The problem says it does not exhibit cis/trans isomerism.

• 2‑Butene (CH₃–CH=CH–CH₃) does have cis/trans isomers because each doubly‑bonded carbon has two different groups (H and CH₃).
• 1‑Butene (CH₂=CH–CH₂–CH₃) has one doubly‑bonded carbon with two H’s, so no cis/trans.
• Isobutylene (CH₂=C(CH₃)₂) also has no cis/trans, but its bromination product would lead to a different final alkyne.

So the alkene X is 1‑butene.

Now follow the reaction sequence:

1. Bromination: Br₂ in CCl₄ adds across the double bond.

   CH₂=CH–CH₂–CH₃ + Br₂ → CH₂Br–CHBr–CH₂–CH₃ (1,2‑dibromobutane).

2. First dehydrohalogenation: Alcoholic KOH (alc. KOH) removes one HBr to give an alkene. The more substituted alkene (Saytzeff product) is 2‑bromo‑2‑butene, but here we want to eventually get a terminal alkyne, so the elimination will occur to give 1‑bromo‑1‑butene (or 2‑bromo‑1‑butene). In practice, alc. KOH gives a mixture, but the next step pushes toward the terminal alkyne. …

The concept first

Cannizzaro reaction: an aldehyde that has no $\alpha$-hydrogen (benzaldehyde, formaldehyde, trimethylacetaldehyde …) cannot enolise, so it cannot do an aldol condensation. Treated with concentrated alkali, two molecules instead disproportionate:

• one molecule is reduced to the primary alcohol,
• the other is oxidised to the carboxylate salt.

Mechanism in one line: $\ce{OH-}$ adds to the carbonyl carbon to give an alkoxide; this transfers a hydride ion to a second aldehyde molecule; the donor becomes the acid (then the salt in the alkaline medium), the acceptor becomes the alkoxide of the alcohol.

A second trap in the same question: does NaOH displace the ring chlorine? No. Aryl halides are extremely unreactive towards nucleophilic substitution because the $\ce{C-Cl}$ bond has partial double-bond character and the ring carbon is sp$^2$. (Chlorobenzene → phenol needs $\ce{NaOH}$ at $\sim$623 K and 300 atm, the Dow process.) So the chlorine survives, and every option containing an $\ce{-OH}$ in place of $\ce{-Cl}$ is wrong.

Step-by-step

1. Check for $\alpha$-hydrogens. In $\ce{m-Cl-C6H4-CHO}$, the carbon next to $\ce{CHO}$ is an aromatic ring carbon carrying no H that can be removed as an enolisable proton ⇒ no $\alpha$-H ⇒ Cannizzaro pathway. …

The key here is understanding carbocation rearrangement in alkane isomerization. When an n-alkane is treated with a Lewis acid (like anhydrous AlCl₃) and a proton source (HCl), the reaction proceeds via carbocation intermediates. These carbocations undergo hydride shifts and alkyl shifts to form more stable (tertiary > secondary > primary) carbocations, which then lead to branched alkanes. The driving force is thermodynamic stability: branched alkanes are more stable than straight-chain ones due to lower steric strain and better hyperconjugation. So the major products are branched, not minor.

Let’s break it down:

1. Assertion (A) is true.

   n-Alkanes (e.g., n-butane, n-pentane) do undergo isomerization when heated with anhydrous AlCl₃ and HCl gas. The AlCl₃ acts as a Lewis acid, and HCl provides protons, generating carbocations. These carbocations rearrange via 1,2-hydride shifts or 1,2-alkyl shifts to more stable carbocations, which then abstract a hydride from another alkane molecule to form branched alkanes. This is a well-known industrial process (e.g., isomerization of n-butane to isobutane).

2. Reason (R) is false.

   The reason states that “branched isomers are formed as minor isomers.” In reality, branched isomers are the major products because they are thermodynamically more stable. For example, isomerization of n-pentane yields a mixture where isopentane (2-methylbutane) and neopentane (2,2-dimethylpropane) dominate over n-pentane at equilibrium. The equilibrium constant favors branched isomers.

3. Evaluating the options: …

The concept first

Why is chlorobenzene so stubborn?

• The chlorine lone pair conjugates with the ring, giving the C–Cl bond partial double-bond character — it is shorter and stronger than in an alkyl halide.
• The carbon bearing Cl is $sp^2$ (more electronegative, holds its electrons tightly).
• The $\pi$ cloud repels an approaching nucleophile.

So ordinary hydroxide bounces off. Substitution becomes possible only through the $S_NAr$ (bimolecular addition–elimination) route:

The rate depends on both the arene and $\mathrm{OH^-}$ — hence bimolecular. The slow step creates a carbanion, and the reaction is fast only if that negative charge can be delocalised away.

Step-by-step

1. Where does the negative charge go? In the intermediate, the charge is spread over the carbons ortho and para to the point of attack.
2. What helps? A group sitting at those very positions that can absorb negative charge. A nitro group is superb: its resonance structure places the charge on the electronegative oxygens,

3. Test each option.
   • (A) Chlorobenzene — no activating group; needs $623\ \mathrm{K}$ and $300\ \mathrm{atm}$ (Dow process). Not "readily". …

The concept first

Why chlorobenzene is so unreactive:

• Resonance donation from Cl's lone pair gives the C–Cl bond partial double-bond character — shorter and stronger than a C–Cl bond in an alkyl halide.
• The carbon holding Cl is $sp^2$-hybridised, hence more electronegative and less willing to release the electron pair.
• The aromatic $\pi$ cloud electrostatically repels an incoming nucleophile.
• An $S_N1$ route is hopeless too: the phenyl cation is enormously unstable.

The only viable path is $S_NAr$ — bimolecular addition–elimination:

Rate $\propto [\mathrm{ArCl}][\mathrm{OH^-}]$ — genuinely bimolecular.

Step-by-step

1. Locate the negative charge. In the Meisenheimer intermediate the carbanion charge is delocalised onto the carbons ortho and para to the site of attack. So a substituent at those positions is the one that matters.
2. What stabilises a carbanion? An electron-withdrawing group. A nitro group is the champion, because resonance can push the charge right onto its oxygens:

3. Screen each option.
   • (A) Chlorobenzene — unactivated. Hydrolysis needs $623\ \mathrm{K}$ and $300\ \mathrm{atm}$ (the Dow process). Not "readily". …

The question asks for a single-step conversion of ethanol ($\text{CH}_3\text{CH}_2\text{OH}$) to ethane ($\text{CH}_3\text{CH}_3$). This is a reduction: we need to replace the –OH group with a hydrogen atom. The key is to recognize which reagent can do this in one step without first converting the alcohol to something else.

Let’s examine each option carefully.

1. Option (A): Na – Hg + water

   Sodium amalgam in water is a mild reducing agent. It is typically used for reducing carbonyl groups (like aldehydes and ketones) to alcohols, not for removing an –OH group from an alcohol. Ethanol already has an –OH; this reagent won’t replace it with hydrogen. So this cannot give ethane.

2. Option (B): Zn – Hg + conc. HCl

   This is the Clemmensen reduction, a classic method for reducing a carbonyl group ($\text{C=O}$) to a methylene group ($\text{CH}_2$). But here we have an alcohol, not a carbonyl. However, under the strongly acidic conditions of conc. HCl, ethanol can first undergo dehydration to form ethene ($\text{CH}_2=\text{CH}_2$), and then the Zn–Hg (amalgamated zinc) in the presence of HCl can reduce the double bond to ethane. In practice, this reagent is known to reduce alcohols to alkanes in one step when the alcohol can be dehydrated in situ. For ethanol, this works: $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Zn–Hg, conc. HCl}} \text{CH}_3\text{CH}_3$. This is a valid one-step method.

3. Option (C): Aluminium isopropoxide and isopropyl alcohol

   This is the Meerwein–Ponndorf–Verley (MPV) reduction, used to reduce aldehydes and ketones to alcohols. It does the opposite of what we need — it adds hydrogen to a carbonyl, not removes an –OH. Ethanol is already an alcohol; this reagent won’t convert it to ethane.

4. Option (D): LiAlH$_4$ + ether …