Q.How do you account for the formation of ethane during chlorination of methane ?

The question asks which compound is not primarily responsible for stratospheric ozone depletion. To answer, we need to understand the chemistry of ozone destruction — specifically, the catalytic cycles that break down ozone (O₃) in the stratosphere.

Concept and Intuition:

Ozone depletion is not caused by stable molecules reacting directly with ozone in a single step. Instead, it happens through catalytic cycles where a free radical (like chlorine atom, Cl•, or nitric oxide, NO•) reacts with ozone, forming an intermediate that then reacts with another ozone molecule, regenerating the original radical. This allows one radical to destroy thousands of ozone molecules before being removed. The main culprits are compounds that release such radicals in the stratosphere — especially chlorofluorocarbons (CFCs) and nitrogen oxides. Methane (CH₄) is different: it is a sink for chlorine radicals (it reacts with Cl• to form HCl, which is harmless to ozone), so it actually reduces ozone depletion.

Let’s examine each option step by step.

1. Option (A): NO (nitric oxide) NO is a key player in the NOₓ cycle of ozone destruction. In the stratosphere, NO reacts with ozone:

Then the NO₂ produced reacts with an oxygen atom (O) to regenerate NO:

Net: $\text{O}_3 + \text{O} \rightarrow 2\text{O}_2$. NO is a catalyst and is regenerated, so it is primarily responsible for ozone depletion. (Sources: supersonic aircraft, lightning, etc.)

2. Option (B): CF₂Cl₂ (dichlorodifluoromethane, a CFC) CFCs are stable in the troposphere but rise to the stratosphere, where UV light breaks them down, releasing chlorine atoms:

The chlorine atom then destroys ozone catalytically:

Net: $\text{O}_3 + \text{O} \rightarrow 2\text{O}_2$. This is the infamous CFC-ozone link. So CF₂Cl₂ is primarily responsible.

3. Option (C): CH₄ (methane) Methane does not directly destroy ozone. In fact, it reacts with chlorine radicals in the stratosphere:

HCl is a stable reservoir that does not attack ozone. This reaction removes active chlorine, slowing ozone depletion. Methane also reacts with hydroxyl radicals (OH•) but that does not deplete ozone. So CH₄ is not primarily responsible for ozone depletion — it is actually protective.

4. Option (D): Cl₂ (chlorine gas) Cl₂ can be photolyzed in the stratosphere by UV light:

The chlorine atoms then enter the same catalytic cycle as from CFCs. While Cl₂ is less abundant than CFCs, it is still a direct source of ozone-destroying radicals. So it is primarily responsible (though less common).

Note on the source

The two reaction schemes this question refers to are not fully captured in the text available here (only the answer options are legible). The identifications below follow the official key and standard reagent chemistry.

X — decarboxylative halogenation

A silver carboxylate treated with a halogen undergoes Hunsdiecker-type decarboxylation, replacing $-\text{COOAg}$ by halogen on the residual alkyl group:

From an acetate this leaves the one-carbon methyl halide, i.e. $X=\text{CH}_3\text{Cl}$.

Y — acyl chloride formation

A carboxylic acid on treatment with a chlorinating agent such as $\text{SOCl}_2$ (or $\text{PCl}_5$) gives its acyl chloride:

So $Y=\text{CH}_3\text{COCl}$ (acetyl chloride).

The Wurtz reaction couples two alkyl halides using sodium metal in a solvent. The key is that the solvent must be dry (water-free) and inert toward sodium. Water reacts violently with sodium, producing hydrogen gas and destroying the reagent. So any solvent must be completely anhydrous.

But beyond dryness, the solvent must also be aprotic — it cannot have an acidic hydrogen that could react with the strongly basic alkyl sodium intermediate formed during the reaction. Protic solvents like alcohols or water would quench the carbanion-like species and kill the coupling.

1. Dry ether — Diethyl ether is the classic solvent for the Wurtz reaction. It is non-polar, aprotic, and completely inert toward sodium. It also dissolves the alkyl halide and allows the sodium surface to remain exposed for the reaction. This is the standard choice in every textbook.

2. Dry ethanol — Ethanol has an -OH group with an acidic hydrogen. Even though it's "dry," the alcohol proton will react with the alkyl sodium intermediate, giving an alkane instead of the coupled product. So this is unsuitable.

3. Dry acetonitrile — Acetonitrile ($\ce{CH3CN}$) has a slightly acidic $\alpha$-hydrogen (pKa ~25) that can be abstracted by the strongly basic alkyl sodium. It also has a polar aprotic nature that can coordinate sodium ions, but the risk of side reactions makes it a poor choice.

4. Dry dichloromethane — Dichloromethane ($\ce{CH2Cl2}$) has chlorine atoms that can react with sodium metal in a Wurtz-type side reaction, producing polymers or other byproducts. It is not inert under these conditions.

The four routes are presented as reaction schemes (structures/equations shown as figures in the paper). Comparing them with the standard preparations of methane:

• Genuine methane routes include hydrolysis of aluminium carbide, $\text{Al}_4\text{C}_3 + 12\,\text{H}_2\text{O} \rightarrow 3\,\text{CH}_4 + 4\,\text{Al(OH)}_3$, and decarboxylation of sodium acetate with soda lime, $\text{CH}_3\text{COONa} + \text{NaOH}\xrightarrow{\text{CaO},\ \Delta}\text{CH}_4 + \text{Na}_2\text{CO}_3$.
• Routes that do not deliver methane are the ones that either give a higher/different hydrocarbon, an oxygenated product, or no C–H bond of the required kind.

Against the official key, the schemes labelled A, C and D are the ones by which methane cannot be prepared.