Some oxidation reactions of methane are given below. Which of them is/are controlled oxidation reactions? (Not

NCERT Exemplar · Q11

Q. Some oxidation reactions of methane are given below. Which of them is/are controlled oxidation reactions? (Note: more than one of the given options may be correct.)

(i) CH4

(g) + 2O2

(g) --> CO2

(g) + 2H2O (l)

(ii) CH4

(g) + O2

(g) --> C (s) + 2H2O (l)

(iii) CH4

(g) + O2

(g) --Mo2O3--> HCHO + H2O

(iv) 2CH4

(g) + O2

(g) --Cu/523/100 atm--> 2CH3OH

Chhattisgarh CgbseMCQ· 1mImportance★★★★★

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Concept understanding — Alcohol Oxidation

Alcohol Oxidation: The Intuition First

Imagine you have a molecule of ethanol — the alcohol in your hand sanitizer or a drink. It has a carbon atom bonded to an –OH group. Now picture that –OH group as a "handle" that can be transformed. Oxidation, in organic chemistry, doesn't always mean adding oxygen — it often means removing hydrogen from a carbon that already has a bond to oxygen. For alcohols, oxidation is like "stripping away" hydrogen atoms from the carbon that holds the –OH, turning the alcohol into a more oxidized functional group.

Think of it this way: a primary alcohol (R–CH₂–OH) has two hydrogens on the carbon with the –OH. If you remove one hydrogen and the hydrogen from the –OH, you get an aldehyde (R–CHO). Remove both hydrogens (and the –OH hydrogen), and you get a carboxylic acid (R–COOH). A secondary alcohol (R–CHOH–R') has only one hydrogen on that carbon — remove it, and you get a ketone (R–CO–R'). A tertiary alcohol has no hydrogen on that carbon — so it cannot be oxidized without breaking the carbon skeleton.

That's the core intuition: oxidation of an alcohol is about removing hydrogens from the carbon bearing the –OH group. The more hydrogens you can remove, the more oxidized the product.

The Precise Statement

Alcohol oxidation is the process in which an alcohol loses hydrogen atoms (dehydrogenation) from the carbon bonded to the -OH group, increasing the number of C-O bonds (or decreasing C-H bonds). The outcome depends on the class of the alcohol:

Alcohol Class	Structure	Product after oxidation	Reagent example
Primary (1°)	R–CH₂–OH	Aldehyde (R–CHO) then Carboxylic acid (R–COOH)	PCC (stops at aldehyde); K₂Cr₂O₇/H⁺ (goes to acid)
Secondary (2°)	R–CHOH–R'	Ketone (R–CO–R')	K₂Cr₂O₇/H⁺, CrO₃, etc.
Tertiary (3°)	R₃C–OH	No reaction (under normal conditions)	—

Watch out

A common mistake: students think "oxidation" always adds oxygen. For alcohols, it's removal of hydrogen from the carbon with the –OH. The oxygen from the –OH stays — it's the hydrogens that leave.

Why Does Tertiary Alcohol Not Oxidize?

Look at the carbon with the –OH in a tertiary alcohol: it has three carbon groups attached and no hydrogen. To form a C=O bond, you'd need to remove a hydrogen from that carbon — but there is none. The only way to oxidize a tertiary alcohol is to break a C–C bond (strong and difficult), which is not typical oxidation. So in standard organic chemistry, tertiary alcohols are inert to mild oxidizing agents.

A Real-World Analogy

Think of the alcohol carbon as a "parking spot" with a certain number of hydrogen "cars." Primary alcohol has two cars parked. Oxidation is like towing away one car (→ aldehyde) or both cars (→ carboxylic acid). Secondary alcohol has one car — tow it away, and you get a ketone. Tertiary alcohol has zero cars — nothing to tow, so no reaction.

Key Reagents to Remember (for exams)

PCC (pyridinium chlorochromate): oxidizes 1° alcohols to aldehydes only — stops there.
K₂Cr₂O₇ / H₂SO₄ (acidified potassium dichromate): oxidizes 1° alcohols all the way to carboxylic acids; 2° alcohols to ketones. (Not to be confused with Jones reagent, which is specifically CrO₃ dissolved in dilute aqueous H₂SO₄, often used in acetone — a related but distinct oxidant with the same general 1°→acid / 2°→ketone outcome.) …

Why this formula?

Alcohol Oxidation: Why the Reactions Work the Way They Do

Alcohol oxidation is a fundamental reaction in organic chemistry, and understanding why it proceeds as it does is crucial for Indian board exams (Class 12, JEE, NEET). Let's break it down step-by-step.

1. The Core Idea: Loss of Hydrogen

Oxidation in organic chemistry means loss of hydrogen (or gain of oxygen). For alcohols, this happens at the carbon bearing the –OH group.

Primary alcohol ( $R-CH_2OH$ ): Has two hydrogens on the carbon attached to –OH.
Secondary alcohol ( $R_2CHOH$ ): Has one hydrogen on that carbon.
Tertiary alcohol ( $R_3COH$ ): Has zero hydrogens on that carbon.

Key insight: The number of hydrogens on the carbon with –OH determines if and how far oxidation can go.

2. Why Primary Alcohols Give Aldehydes (Then Carboxylic Acids)

Step 1: Aldehyde formation

When a primary alcohol ( $R-CH_2OH$ ) is oxidized, the first product is an aldehyde ( $R-CHO$ ).

Why? The oxidizing agent (like $K_2Cr_2O_7$ / $H_2SO_4$ or $PCC$ ) removes two hydrogens:

One from the –OH group
One from the carbon atom

The carbon–oxygen bond becomes a double bond ( $C=O$ ), forming the aldehyde.

R-CH_2OH \xrightarrow{[O]} R-CHO + H_2O

But why stop here? The aldehyde still has one hydrogen on the carbonyl carbon. If a strong oxidant is present, it can remove that hydrogen too.

Step 2: Carboxylic acid formation

With excess strong oxidant (e.g., $K_2Cr_2O_7$ / $H_2SO_4$ , heat), the aldehyde is further oxidized to a carboxylic acid ( $R-COOH$ ).

R-CHO \xrightarrow{[O]} R-COOH

Why does this happen? The aldehyde's carbonyl carbon is electrophilic (partially positive). Water (from the reaction medium) adds to it, forming a gem-diol intermediate. The oxidant then removes two more hydrogens, giving the acid.

Exam tip: To stop at the aldehyde, use a mild oxidant like PCC (pyridinium chlorochromate) in anhydrous conditions — no water means no gem-diol formation.

3. Why Secondary Alcohols Give Ketones (and Stop)

A secondary alcohol ( $R_2CHOH$ ) has only one hydrogen on the carbon with –OH. Oxidation removes:

One hydrogen from –OH
One hydrogen from the carbon

This forms a ketone ( $R_2C=O$ ).

R_2CHOH \xrightarrow{[O]} R_2C=O + H_2O

Why does it stop here? The ketone has no hydrogen on the carbonyl carbon. Without that hydrogen, further oxidation (under normal conditions) is impossible — you'd need to break a $C-C$ bond, which requires much harsher conditions.

Key result: Secondary alcohols cannot be oxidized further than ketones under standard conditions.

4. Why Tertiary Alcohols Do NOT Oxidize

A tertiary alcohol ( $R_3COH$ ) has zero hydrogens on the carbon bearing –OH.

What happens if you try? The oxidant cannot remove any hydrogen from that carbon. The only possible reaction would be breaking a $C-C$ bond, which doesn't happen under normal oxidation conditions.

Result: Tertiary alcohols are resistant to oxidation under mild to moderate conditions. They require strong heating with powerful oxidants (like $K_2Cr_2O_7$ / $H_2SO_4$ , heat) to break carbon–carbon bonds — this is destructive oxidation, not useful for synthesis.

5. The "Why" in One Table

Alcohol Type	Hydrogens on C–OH	Product	Why?
Primary ( $1^\circ$ )	2	Aldehyde → Carboxylic acid	Two hydrogens available; aldehyde still has one more

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Controlled oxidation reactions selectively convert a reactant into a specific intermediate product, often requiring specific catalysts and conditions. Options (iii) and (iv) are controlled oxidation reactions because they convert methane into formaldehyde and methanol, respectively, using specific catalysts and conditions.

Understanding Controlled Oxidation

Oxidation, in organic chemistry, generally refers to reactions that increase the number of bonds to oxygen or decrease the number of bonds to hydrogen. For hydrocarbons like methane, complete oxidation (combustion) leads to carbon dioxide and water. However, it's often desirable to obtain intermediate oxygen-containing organic compounds (like alcohols, aldehydes, or carboxylic acids) from hydrocarbons.

Controlled oxidation is a process where a hydrocarbon is partially oxidised under specific conditions (catalyst, temperature, pressure, reactant ratio) to yield a particular desired product, rather than undergoing complete combustion or forming a mixture of uncontrolled products (like soot). The key is selectivity and specificity of the product.

Let's analyze each option:

Option (i): $\text{CH}_4 \text{(g)} + 2\text{O}_2 \text{(g)} \rightarrow \text{CO}_2 \text{(g)} + 2\text{H}_2\text{O (l)}$
- This reaction represents the complete combustion of methane. Methane reacts with sufficient oxygen to produce carbon dioxide and water. All the carbon is oxidised to its highest oxidation state ( $+4$ in $\text{CO}_2$ ).
- This is an uncontrolled, highly exothermic reaction that releases a large amount of energy, typically used as a fuel source. It does not yield an intermediate organic product.
- Therefore, this is not a controlled oxidation reaction in the sense of producing a specific organic intermediate.
Option (ii): $\text{CH}_4 \text{(g)} + \text{O}_2 \text{(g)} \rightarrow \text{C (s)} + 2\text{H}_2\text{O (l)}$
- This reaction represents incomplete combustion of methane. When there is insufficient oxygen, methane burns to produce carbon (soot) and water.
- This is an uncontrolled partial oxidation, leading to an undesirable product (soot) rather than a useful organic compound.
- Therefore, this is not a controlled oxidation reaction.
Option (iii): $\text{CH}_4 \text{(g)} + \text{O}_2 \text{(g)} \xrightarrow{\text{Mo}_2\text{O}_3} \text{HCHO} + \text{H}_2\text{O}$
- In this reaction, methane ( $\text{CH}_4$ ) is oxidised to formaldehyde ( $\text{HCHO}$ , also known as methanal).
- The use of a specific catalyst, molybdenum oxide ( $\text{Mo}_2\text{O}_3$ ), is crucial here. This catalyst directs the oxidation specifically towards the formation of formaldehyde, preventing further oxidation to carbon dioxide or uncontrolled side reactions. …

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Q.Some oxidation reactions of methane are given below. Which of them is/are controlled oxidation reactions? (Note: more than one of the given options may be correct.)

Understanding Controlled Oxidation

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