Which of the following compounds will give a ketone on oxidation with chromic anhydride ($CrO_3$)? (A) $(CH_3)_2CH-CH_2OH$ (B) $CH_3CH_2CH_2OH$ (C) $(CH_3)_3C-OH$ (D) $CH_3-CH_2-CH(OH)-CH_3$

Which of the following compounds will give a ketone on oxidation with chromic anhydride (CrO₃)? (A) (CH₃)₂CH-C

Question

CrO_3

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★

✓ Free question

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.)
KMnO₄: similar to dichromate, but stronger — can over-oxidize.
Swern oxidation (DMSO + oxalyl chloride): mild, gives aldehydes from 1° alcohols.

Tip

For exams: if you see "mild oxidation" of a primary alcohol, think aldehyde. If you see "strong oxidation" or "acidic dichromate", think carboxylic acid. For secondary alcohols, both mild and strong give ketones.

The Mechanism (Simplified)

In acidic dichromate oxidation, the alcohol oxygen attacks chromium, forming a chromate ester. Then a base (often water) removes a hydrogen from the carbon bearing the –OH, and the C–O bond becomes a C=O. The chromium is reduced from Cr(VI) to Cr(III) — that's the colour change from orange to green.

Note

You don't need to memorise the full mechanism for most Indian board exams (Class 12), but understanding that a hydrogen is removed from the carbon is crucial.

Final Takeaway

Alcohol oxidation = dehydrogenation of the carbon with –OH.

1° → aldehyde (mild) or acid (strong)
2° → ketone
3° → no reaction

That's it. Build your understanding from this single idea, and you'll never confuse the products.

Searches like "oxidation of alcohols primary secondary tertiary" and "alcohols phenols ethers class 12 chemistry reactions" are common, since this is a core reaction covered in the Alcohols, Phenols and Ethers chapter of the NCERT/CBSE Class 12 Chemistry curriculum. Reagent-based questions (PCC vs. acidic dichromate) built on this concept are frequently tested in JEE Main and NEET.

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
Secondary ( $2^\circ$ )	1	Ketone (stops)	Only one hydrogen; ketone has none left
Tertiary ( $3^\circ$ )	0	No reaction	No hydrogen to remove

6. The Mechanism (Simplified for Understanding)

For a primary alcohol with chromic acid ( $H_2CrO_4$ ):

Ester formation: The alcohol oxygen attacks the chromium, forming a chromate ester.
Elimination: A base (water or the solvent) removes a proton from the carbon, while the $C-O$ bond breaks, releasing the aldehyde and reducing $Cr(VI)$ to $Cr(IV)$ .

R-CH_2OH + H_2CrO_4 \rightarrow R-CH_2-O-CrO_3H \xrightarrow{-H^+} R-CHO + Cr(IV) \text{ species}

Why this mechanism? The chromium acts as a leaving group after the ester forms. The carbon–hydrogen bond breaks because the resulting carbocation is stabilized by the adjacent oxygen (resonance).

7. Common Exam Pitfalls to Avoid

Don't say "tertiary alcohols don't oxidize at all" — they do under extreme conditions, but not in standard reactions.
Remember: PCC stops at aldehyde because it's anhydrous — no water for the next step.
For JEE/NEET: Know that $K_2Cr_2O_7$ / $H_2SO_4$ gives carboxylic acid from primary alcohols, while PCC gives aldehyde.

Final Takeaway

The number of hydrogens on the carbon bearing the –OH group is the single most important factor. It determines:

Whether oxidation occurs
What product forms
Whether the reaction stops or continues

This is why the formulas and products are not arbitrary — they follow directly from the structure of the alcohol.

Chromic anhydride oxidizes secondary alcohols to ketones; only the secondary alcohol $(CH_3)_3CH-CH(OH)-CH_3$ in option (D) will yield a ketone.

The key to this question lies in understanding how different classes of alcohols respond to oxidation. Chromic anhydride ( $CrO_3$ ) is a strong oxidizing agent that behaves predictably based on the structure of the alcohol it encounters.

Alcohols are classified by the number of carbon atoms attached to the carbon bearing the hydroxyl group:

Primary alcohols ( $1°$ ): one carbon attached → oxidize to aldehydes, then further to carboxylic acids
Secondary alcohols ( $2°$ ): two carbons attached → oxidize to ketones (and stop there)
Tertiary alcohols ( $3°$ ): three carbons attached → resist oxidation under normal conditions (no $C-H$ bond on the carbinol carbon)

The reason secondary alcohols stop at ketones is simple: once the $C-OH$ becomes $C=O$ , there's no hydrogen left on that carbon to remove. Ketones are stable to further oxidation because breaking a $C-C$ bond requires much harsher conditions.

Let me examine each option systematically:

Option (A): $(CH_3)_2CH-CH_2OH$ (isobutanol)

The hydroxyl group is on a $-CH_2OH$ carbon, which has two hydrogens and is bonded to only one other carbon. This is a primary alcohol. Oxidation with $CrO_3$ would first give an aldehyde $(CH_3)_2CH-CHO$ , which would immediately oxidize further to the carboxylic acid $(CH_3)_2CH-COOH$ . No ketone forms.
Option (B): $CH_3CH_2CH_2OH$ (1-propanol)

Again, the $-OH$ sits on a $-CH_2OH$ group at the end of the chain. This is a primary alcohol. It oxidizes to propanoic acid $CH_3CH_2COOH$ via the aldehyde intermediate. No ketone.
Option (C): $(CH_3)_3C-OH$ (tert-butanol)

The central carbon bearing the hydroxyl group is attached to three methyl groups—no hydrogen atoms remain on the carbinol carbon. This is a tertiary alcohol. Tertiary alcohols do not undergo oxidation with $CrO_3$ under normal conditions because oxidation requires a $C-H$ bond adjacent to the $C-OH$ . No reaction occurs.
Option (D): $CH_3-CH_2-CH(OH)-CH_3$ (2-butanol)

The hydroxyl group is on the second carbon, which carries one hydrogen and is bonded to two other carbons (an ethyl group and a methyl group). This is a secondary alcohol. Oxidation removes the hydrogen from the $CHOH$ group, converting it cleanly to a ketone:

CH_3-CH_2-CH(OH)-CH_3 \xrightarrow{CrO_3} CH_3-CH_2-CO-CH_3

The product is 2-butanone (methyl ethyl ketone).

Watch out

A common mistake is confusing the position of the $-OH$ group. In option (A), the branching (two methyls) is on the adjacent carbon, not the one bearing the hydroxyl—so it remains a primary alcohol despite the branching nearby.

✓Final answer

The correct option is (D), as it is the only secondary alcohol that oxidizes to a ketone.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 22 on this concept.

CBSE 2026Set A1 markMCQ
$Q. When vapours of an alcohol are passed over hot reduced copper, it gives an alkene. The alcohol is$
(a) Primary
(b) Secondary
(c) Tertiary
(d) None of these
›Reveal solutionSolution
Over hot reduced copper (573 K), a primary alcohol gives an aldehyde, a secondary gives a ketone, and a tertiary gives an alkene.

When alcohol vapours are passed over hot reduced copper the behaviour depends on the class of alcohol:

Primary alcohol -> dehydrogenation -> aldehyde

Secondary alcohol -> dehydrogenation -> ketone

Tertiary alcohol -> dehydration -> alkene (there is no H on the carbinol carbon to remove, so it loses water instead)

Since the product here is an alkene, the alcohol must be tertiary.

✓Final answer
$(c) Tertiary.$
CBSE 2026Set ANNUAL1 mark
$Q. Write the name of product obtained when vapour of ethyl alcohol are passed over heated Copper at 573 K.$
›Reveal solutionSolution

Passing alcohol vapours over heated copper catalyses either dehydrogenation (for 1° and 2° alcohols) or dehydration (for 3° alcohols), depending on alcohol type.

Ethyl alcohol (a primary alcohol) undergoes dehydrogenation over copper at 573 K:

CH3CH2OH --Cu, 573K--> CH3CHO (acetaldehyde) + H2

✓Final answer
$Acetaldehyde (CH3CHO).$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Dehydration of tertiary alcohols with copper at 573 K gives:$
(a) Aldehyde
(b) Ketone
(c) Alkene
(d) None of these
›Reveal solutionSolution
Passing alcohol vapours over heated copper at 573 K is a classification test: 1° alcohols → aldehydes, 2° alcohols → ketones, but 3° alcohols (no α-H on the carbinol carbon available for dehydrogenation) undergo dehydration to give an alkene.

When vapours of an alcohol are passed over copper catalyst at 573 K:

Primary alcohols are dehydrogenated (lose $H_2$ ) to aldehydes: $RCH_2OH \xrightarrow{Cu, 573K} RCHO + H_2$

Secondary alcohols are dehydrogenated to ketones: $R_2CHOH \xrightarrow{Cu, 573K} R_2C=O + H_2$

Tertiary alcohols have no hydrogen on the carbinol (C–OH) carbon itself, so dehydrogenation to a carbonyl is not possible. Instead, they undergo dehydration (loss of water) over the hot copper surface to form an alkene: $R_3C-OH \xrightarrow{Cu, 573K} \text{alkene} + H_2O$

This contrast (dehydrogenation vs dehydration) is a classic way to distinguish primary/secondary alcohols from tertiary alcohols.

✓Final answer
$Alkene.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. When vapour's of a compound X are passed over heated copper, the major product obtained is the acetone. The compound X is:$
(a) n-Propyl alcohol
(b) Iso-propyl alcohol
(c) Acetaldehyde
(d) Propane
›Reveal solutionSolution
Vapours passed over heated copper dehydrogenate 2° alcohols to ketones; since the product is acetone (a ketone), X must be a secondary alcohol — isopropyl alcohol.

Over heated copper (573 K), alcohols are catalytically dehydrogenated based on their class:

1° alcohol → aldehyde

2° alcohol → ketone

3° alcohol → alkene (dehydration)

The product here is acetone, $CH_3COCH_3$ , which is a ketone. A ketone can only arise from a secondary alcohol via dehydrogenation:

$(CH_3)_2CHOH \xrightarrow{Cu,\ 573K} (CH_3)_2C=O + H_2$

So X = isopropyl alcohol (propan-2-ol), $(CH_3)_2CHOH$ .

✓Final answer
$X is isopropyl alcohol (propan-2-ol).$
CBSE 2026Set ANNUAL1 markMCQ
$RCH_2OH \rightarrow RCHO$
(a) $KMnO_4$
(b) $K_2Cr_2O_7$
(c) $LiAlH_4$
(d) PCC (Pyridinium Chlorochromate)
›Reveal solutionSolution

Selective oxidation of a 1° alcohol to an aldehyde (without over-oxidation to the acid) requires an anhydrous, mild oxidant — PCC — rather than a strong aqueous oxidant.

Why $KMnO_4$ and $K_2Cr_2O_7$ (a, b) fail: these are strong oxidising agents used in aqueous, typically acidified medium. The initially formed aldehyde reacts with water to form a geminal diol (aldehyde hydrate), $RCH(OH)_2$ , which is itself readily oxidised further by these strong oxidants to the carboxylic acid, $RCOOH$ . So the reaction cannot be stopped cleanly at the aldehyde stage.

Why $LiAlH_4$ (c) fails: this is a powerful reducing agent (it reduces esters, acids and other carbonyls down to alcohols) — the wrong direction entirely for an oxidation.

Why PCC (d) works: PCC, $C_5H_5NH^+CrO_3Cl^-$ , is used in an anhydrous solvent such as dichloromethane. With no water present, the aldehyde cannot form the hydrate that would otherwise be further oxidised, so the reaction is cleanly controlled:

$RCH_2OH \xrightarrow{PCC,\ CH_2Cl_2} RCHO$

✓Final answer
$(d) PCC (Pyridinium Chlorochromate)$
CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: In addition of bromine in CCl4 to an alkene resulting in disappearance of reddish brown colour of bromine constitutes, an important method for the detection of ______ in a molecule.$
›Reveal solutionSolution

Decolourisation of bromine in CCl4 detects unsaturation (C=C double bond).

An alkene readily adds bromine across its carbon-carbon double bond to form a colourless dibromide:

C=C + Br2 -> Br-C-C-Br

As the bromine is consumed, its characteristic reddish-brown colour disappears. Therefore, the disappearance of the colour of bromine (in carbon tetrachloride) on adding it to a compound is an important test for the presence of unsaturation (a double bond) in the molecule.

✓Final answer
$Unsaturation (carbon-carbon double bond).$
CBSE 2026Set ANNUAL1 mark
$Q. Write True or False: Bromo, iodo and polychloro derivatives of hydrocarbons are heavier than water.$
›Reveal solutionSolution

True - these halogen derivatives are denser than water.

The heavy halogen atoms (Br, I) and multiple chlorine atoms greatly increase the molar mass and density of the molecule. As a result, bromo, iodo and polychloro derivatives of hydrocarbons (e.g. bromoform, iodoform, chloroform, carbon tetrachloride) have densities greater than 1 g/mL and therefore sink in and are heavier than water.

✓Final answer
$True.$
CBSE 2025Set 56/5/11 markMCQ
$CH_3CH_2OH$
›Reveal solutionSolution
The key idea is that converting ethanol ( $CH_3CH_2OH$ ) to ethanal ( $CH_3CHO$ ) is a controlled oxidation of a primary alcohol to an aldehyde. The correct reagent is PCC (pyridinium chlorochromate), which stops at the aldehyde stage without over-oxidizing to a carboxylic acid.

This question tests your understanding of alcohol oxidation — a core reaction in organic chemistry. Ethanol is a primary alcohol. To get an aldehyde, you need to oxidize it partially. The challenge is that many strong oxidizers will push the reaction all the way to the carboxylic acid (acetic acid, $CH_3COOH$ ). So the trick is choosing a reagent that is mild enough to stop at the aldehyde.

Let’s examine each option.

Option (A): Catalytic hydrogenation

Hydrogenation ( $H_2$ with a metal catalyst like Pd, Pt, or Ni) is a reduction process. It adds hydrogen across double or triple bonds. Ethanol has no multiple bonds to reduce — it’s already saturated. This would do nothing. So this is wrong.

Option (B): Treatment with $LiAlH_4$

Lithium aluminium hydride is a powerful reducing agent. It reduces carbonyl compounds (aldehydes, ketones, acids, esters) to alcohols. Using it on ethanol would be pointless — ethanol is already an alcohol. It cannot oxidize anything. So this is also wrong.

Option (C): Treatment with PCC

PCC (pyridinium chlorochromate, $C_5H_5NH^+CrO_3Cl^-$ ) is a mild oxidizing agent specifically designed for the conversion of primary alcohols to aldehydes. It works in anhydrous conditions (typically in dichloromethane) and stops cleanly at the aldehyde stage.

The reaction:

$CH_3CH_2OH \xrightarrow{PCC} CH_3CHO$
This is the textbook method. So this is correct.

Option (D): Treatment with $KMnO_4$ Potassium permanganate is a strong oxidizing agent. In neutral or acidic conditions, it will oxidize a primary alcohol all the way to the carboxylic acid.

$CH_3CH_2OH \xrightarrow{KMnO_4} CH_3COOH$
It does not stop at the aldehyde. So this is wrong for the given target.

Watch out

A common mistake is to think that any oxidizer will work. Strong oxidizers like $KMnO_4$ or $K_2Cr_2O_7$ (acidic) give the acid, not the aldehyde. Only mild, anhydrous oxidizers like PCC or the Swern reagent give the aldehyde.

Tip

Remember the mnemonic: PCC = Primary to Carbonyl, Controlled. For a primary alcohol, PCC gives an aldehyde; for a secondary alcohol, it gives a ketone. Strong oxidizers like $KMnO_4$ or $Na_2Cr_2O_7/H_2SO_4$ give carboxylic acids from primary alcohols.

✓Final answer
$The correct option is (C) — treatment with PCC.$
CBSE 2025Set 56/6/11 markMCQ
$HNO_2$
›Reveal solutionSolution
The key idea is that primary aliphatic amines react with nitrous acid ( $HNO_2$ ) to give alcohols via a diazonium intermediate that decomposes. Among the options, only $C_2H_5NH_2$ (ethylamine) is a primary aliphatic amine, so it yields ethanol. The correct option is (B).

The reaction of an amine with nitrous acid ( $HNO_2$ ) is a classic test to distinguish between primary, secondary, and tertiary amines. Nitrous acid is unstable and is prepared in situ by reacting sodium nitrite ( $NaNO_2$ ) with a mineral acid like HCl or $H_2SO_4$ . The outcome depends entirely on the class of the amine.

For primary aliphatic amines (like ethylamine), the reaction proceeds through an unstable alkyldiazonium salt. This salt spontaneously decomposes to give a carbocation, which then reacts with water to form an alcohol. This is the only case where an alcohol is the major product.

For primary aromatic amines (like aniline), the diazonium salt formed is stable at low temperatures (0–5°C) and does not give an alcohol with water — it gives phenol only upon heating or under specific conditions. At room temperature, aniline reacts with $HNO_2$ to give a diazonium salt that can couple or decompose to other products, but not ethanol.

For secondary amines (like diethylamine), the reaction yields a yellow, oily N-nitrosamine — no alcohol is formed.

For tertiary amines (like triethylamine), the reaction gives a nitrosamine salt or simply dissolves, again no alcohol.

So the only amine that reliably gives an alcohol under standard conditions is a primary aliphatic amine.

Let’s check each option:

Option (A): $C_6H_5NH_2$ (aniline) — This is a primary aromatic amine. With $HNO_2$ at 0–5°C, it forms a stable benzenediazonium salt. This salt does not decompose to give an alcohol at low temperature; it requires heating with water to yield phenol. Under the usual conditions of the reaction (room temperature or slightly above), aniline gives a diazonium salt that may undergo coupling or other reactions, but not an alcohol. So this is not the answer.

Option (B): $C_2H_5NH_2$ (ethylamine) — This is a primary aliphatic amine. The reaction with $HNO_2$ proceeds as:

$C_2H_5NH_2 + HNO_2 \rightarrow [C_2H_5N_2^+] \xrightarrow{H_2O} C_2H_5OH + N_2 + H^+$
The intermediate ethyldiazonium ion is unstable and immediately loses $N_2$ to form an ethyl carbocation, which then reacts with water to give ethanol. This is the classic case where an alcohol is produced. So this is the correct option.

Option (C): $(C_2H_5)_2NH$ (diethylamine) — This is a secondary amine. With $HNO_2$ , it forms a yellow, oily N-nitrosamine:

$(C_2H_5)_2NH + HNO_2 \rightarrow (C_2H_5)_2N-N=O + H_2O$
No alcohol is formed. So this is not the answer.

Option (D): $(C_2H_5)_3N$ (triethylamine) — This is a tertiary amine. With $HNO_2$ , it forms a nitrosamine salt (if at all) or simply dissolves. No alcohol is produced. So this is not the answer.

Watch out

A common mistake is to think that aniline (option A) gives an alcohol because it is a primary amine. But primary aromatic amines form stable diazonium salts that do not decompose to alcohols under standard conditions — they give phenols only on heating with water. The alcohol-forming reaction is specific to primary aliphatic amines.

Tip

The reaction of primary aliphatic amines with $HNO_2$ is a reliable way to convert an $NH_2$ group to an $OH$ group, but it comes with a catch: the carbocation intermediate can rearrange or eliminate, so the alcohol product may not always be the same as the original alkyl group. For ethylamine, however, no rearrangement is possible, so ethanol is the clean product.

✓Final answer
$C_2H_5NH_2$
CBSE 2025Set D1 markMCQ
$Q. Methyl alcohol on oxidation with acidified K2Cr2O7 gives$
(a) CH3COCH3
(b) CH3CHO
(c) HCOOH
(d) CH3COOH
›Reveal solutionSolution

A primary alcohol is oxidised to an aldehyde and then to a carboxylic acid; methanol → HCHO → HCOOH with acidified K2Cr2O7.

Methyl alcohol (methanol, CH3OH) is a primary alcohol. Oxidation with a strong oxidising agent such as acidified potassium dichromate (K2Cr2O7/H2SO4) proceeds:

CH3OH → (oxidation) HCHO (formaldehyde) → (further oxidation) HCOOH (formic acid)

With excess of the strong oxidiser the final product is the carboxylic acid, formic acid (HCOOH). Acetone (CH3COCH3) and acetic acid (CH3COOH) contain two or more carbons and cannot form from one-carbon methanol.

✓Final answer
$(C) HCOOH (formic acid) is obtained.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Ethyl alcohol --Cu/573K--> A, 'A' is$
(a) Acetaldehyde
(b) Propionaldehyde
(c) Acetone
(d) Ethanoic acid
›Reveal solutionSolution

Copper catalyses dehydrogenation (not dehydration) of a primary alcohol at 573 K, converting ethanol to the corresponding aldehyde.

$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow[\text{573 K}]{\text{Cu}} \text{CH}_3\text{CHO} + \text{H}_2$

Copper at 573 K acts as a dehydrogenation catalyst for primary alcohols, removing two hydrogen atoms to give an aldehyde (a secondary alcohol under the same conditions would give a ketone, and a tertiary alcohol would give an alkene instead, since it has no alpha-hydrogen to lose in this way).

✓Final answer
$(a) Acetaldehyde.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Oxidation of propan-1-ol with alkaline KMnO4 solution gives$
(a) propanoic acid
(b) propane
(c) ethane
(d) propanal
›Reveal solutionSolution

Alkaline KMnO4 is a strong oxidising agent, so it oxidises primary alcohols completely to carboxylic acids (via the aldehyde stage, which cannot be isolated).

$\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{alkaline KMnO}_4} \text{CH}_3\text{CH}_2\text{COOH (propanoic acid)}$

With a mild oxidant (like PCC), a primary alcohol stops at the aldehyde stage, but strong oxidants such as acidified or alkaline KMnO4/K2Cr2O7 push the oxidation further to the carboxylic acid.

✓Final answer
$(a) propanoic acid.$

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