Q.CH3CH2OH can be converted into CH3CHO by ____________.
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🔒 Start your 14-day free trial to unlock the full solution →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) | — |
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−CH2OH): Has two hydrogens on the carbon attached to –OH.
- Secondary alcohol (R2CHOH): Has one hydrogen on that carbon.
- Tertiary alcohol (R3COH): 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−CH2OH) is oxidized, the first product is an aldehyde (R−CHO).
Why? The oxidizing agent (like K2Cr2O7 / H2SO4 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−CH2OH[O]R−CHO+H2O
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., K2Cr2O7 / H2SO4, heat), the aldehyde is further oxidized to a carboxylic acid (R−COOH).
R−CHO[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 (R2CHOH) has only one hydrogen on the carbon with –OH. Oxidation removes:
- One hydrogen from –OH
- One hydrogen from the carbon
This forms a ketone (R2C=O).
R2CHOH[O]R2C=O+H2O
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 (R3COH) 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 K2Cr2O7 / H2SO4, 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∘) | 2 | Aldehyde → Carboxylic acid | Two hydrogens available; aldehyde still has one more |
The key idea is Alcohol Oxidation — specifically, the controlled oxidation of a primary alcohol to an aldehyde without over-oxidising to the carboxylic acid.
- CH3CH2OH (ethanol) is a primary alcohol. To get CH3CHO (acetaldehyde), we need an oxidising agent that stops at the aldehyde stage.
- Catalytic hydrogenation and LiAlH4 are reducing agents — they would convert ethanol to ethane or keep it as alcohol, not oxidise it.
- KMnO4 is a strong oxidant — it would directly oxidise ethanol all the way to acetic acid (CH3COOH), not stop at the aldehyde. …
The conversion of ethanol (CH3CH2OH) to acetaldehyde (CH3CHO) is a controlled oxidation reaction. The correct reagent is pyridinium chlorochromate (PCC), which selectively oxidizes primary alcohols to aldehydes without over-oxidation to carboxylic acids. The answer is (iii).
The question asks for a method to convert ethanol into acetaldehyde. This is a classic example of partial oxidation of a primary alcohol. The key challenge is to stop the oxidation at the aldehyde stage, not push it all the way to the carboxylic acid (acetic acid, CH3COOH).
Let's think about what each option does.
-
Catalytic hydrogenation (Option (i)) adds hydrogen across double or triple bonds. Ethanol has no such bonds — it's already saturated. Hydrogenation would not change it. This is a reduction process, not an oxidation. So this is wrong.
-
Treatment with LiAlH4 (Option (ii)) is a powerful reducing agent. It reduces carbonyl compounds (aldehydes, ketones, acids) back to alcohols. Using it on ethanol would do nothing — ethanol is already at the alcohol oxidation level. This is the opposite of what we need. So this is wrong.
-
Treatment with pyridinium chlorochromate (PCC) (Option (iii)) is a mild oxidizing agent specifically designed for this job. PCC in anhydrous conditions (usually in dichloromethane, CH2Cl2) oxidizes primary alcohols to aldehydes and stops there. It does not have water present to hydrate the aldehyde, so no further oxidation to the carboxylic acid occurs. This is the correct choice. …
Method: Controlled Oxidation of Primary Alcohols
The concept here is that a primary alcohol (CH3CH2OH, ethanol) can be oxidised to an aldehyde (CH3CHO, ethanal) only if the oxidising agent is mild and the reaction is stopped before further oxidation to a carboxylic acid occurs.
Steps to identify the correct reagent:
-
Recognise the functional group change
- Starting material: Primary alcohol (−OH on a terminal carbon).
- Product: Aldehyde (−CHO).
- This is a two-electron oxidation (loss of two hydrogen atoms).
-
Eliminate reducing agents
- Catalytic hydrogenation (A) and LiAlH4 (B) are reducing agents — they would convert an aldehyde back to an alcohol, not oxidise it.
- ✗ (A) and (B) are incorrect.
-
Distinguish between strong and mild oxidants
- KMnO4 (D) is a strong oxidant. It will over-oxidise a primary alcohol all the way to a carboxylic acid (CH3COOH), not stop at the aldehyde.
- ✗ (D) is incorrect.
-
Identify the mild, selective oxidant …
Here are the common mistakes students make on this question, along with how to avoid each.
Mistake 1: Confusing Oxidation with Reduction
- The Error: Choosing (A) catalytic hydrogenation or (B) treatment with LiAlH4.
- Why it happens: Students see a conversion and forget to check whether the carbon is gaining or losing oxygen/hydrogen. They might recall that LiAlH4 is a strong "reagent" without remembering its function.
- How to avoid: Always check the oxidation state of the carbon attached to the -OH group.
- In CH3CH2OH (ethanol), the carbon has one bond to oxygen.
- In CH3CHO (acetaldehyde), the carbon has a double bond to oxygen.
- Moving from a single bond to a double bond with oxygen is oxidation (loss of hydrogen).
- Catalytic hydrogenation (H2/catalyst) and LiAlH4 are reducing agents. They add hydrogen, which would turn an aldehyde back into an alcohol. They cannot perform the oxidation needed here.
Mistake 2: Overlooking the "Mild vs. Strong" Oxidizer
- The Error: Choosing (D) treatment with KMnO4.
- Why it happens: Students know KMnO4 is an oxidising agent, so they assume it will work. They forget that strong oxidisers do not stop at the aldehyde stage.
- How to avoid: Memorise the selectivity of common oxidising agents.
- Mild oxidisers (like PCC – pyridinium chlorochromate) stop at the aldehyde.
- Strong oxidisers (like acidic KMnO4 or K2Cr2O7) will over-oxidise a primary alcohol all the way to a carboxylic acid (CH3COOH).
- Therefore, KMnO4 would give CH3COOH, not CH3CHO.
Mistake 3: Not Knowing the Specific Reagent (PCC)
- The Error: Not recognising option (C) as the correct answer. …
Showing the 12 most recent of 14 on this concept.
- GUJCET 2025Set 031 markMCQQ.Identify R', R'' and R''' for the following reaction. [FIGURE: a ketone R'R''C=O reacting via(i) R'''MgX(ii) H2O to give 2-methylbutane-2-ol.] (A) R′=C2H5,R′′=C2H5,R′′′=CH3 (B) R′=CH3,R′′=C2H5,R′′′=CH3 (C) R′=C2H5,R′′=CH3,R′′′=C2H5 (D) R′=CH3,R′′=CH3,R′′′=CH3
›Reveal solutionSolution
Ketone R′R′′C=O + R′′′MgX→R′R′′R′′′C−OH; the product's three alkyls are CH3, CH3, C2H5.
Concept — Grignard synthesis of tertiary alcohols. A ketone gives a tertiary alcohol whose carbinol carbon bears the ketone's two groups plus the Grignard's group.
Steps.
- 2-methylbutan-2-ol: CH3−C(OH)(CH3)−CH2CH3. The C–OH carbon carries CH3, CH3, C2H5.
- So {R′,R′′,R′′′}={CH3,CH3,C2H5}. …
- GUJCET 2025Set 031 markMCQQ.For the given reaction, identify the proper reagent. [FIGURE: (hydroxymethyl)cyclohexane (cyclohexane ring bearing a CH2OH group) converted to cyclohexanecarbaldehyde (cyclohexane ring bearing a CHO group).] (A) KMnO4/H2SO4 (B) O3/H2O−Zn dust (C) C5H5NH+CrO3Cl− (D) CrO3+(CH3CO)2O
›Reveal solutionSolution
[!TLDR]
Oxidising a primary alcohol to an aldehyde requires the mild, selective reagent PCC (C5H5NH+CrO3Cl−).
Concept
Primary alcohols are oxidised to aldehydes and can be further oxidised to carboxylic acids by strong oxidants. To stop cleanly at the aldehyde, a mild oxidant such as PCC (in anhydrous dichloromethane) is used.
Solution
The substrate (hydroxymethyl)cyclohexane has a −CH2OH group that must become −CHO (cyclohexanecarbaldehyde) — a controlled oxidation to the aldehyde.
- (A) KMnO4/H2SO4: strong oxidant, over-oxidises to the carboxylic acid.
- (B) O3/H2O–Zn: ozonolysis, cleaves C=C double bonds — not applicable to an alcohol. …
- GUJCET 2023Set 091 markMCQQ.Which of the following alcohol undergo dehydration reaction with Cu (Copper) metal at 573 K temperature? (A) Secondary and Tertiary (B) Primary & Secondary (C) Primary and Tertiary (D) Only Tertiary
›Reveal solutionSolution
With Cu at 573 K: 1° and 2° alcohols dehydrogenate (→ aldehyde/ketone), while only tertiary alcohols dehydrate (→ alkene).
Concept. Passing alcohol vapour over heated copper at 573 K:
- Primary → aldehyde (dehydrogenation, loss of H2)
- Secondary → ketone (dehydrogenation)
- Tertiary → has no α-H on the carbinol carbon to lose as H2, so it instead loses water and dehydrates to an alkene. …
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.R'-X --Na/ether--> 2,3-dimethylbutane. Identify R'.(a) (CH3)2CH-(b) (C2H5)2CH-(c) (CH3CH2)3C-(d) (CH3)3C-
›Reveal solutionSolution
2,3-dimethylbutane is symmetric, made by Wurtz coupling of two isopropyl (2-propyl) groups.
Wurtz reaction: 2 R'-X + 2 Na --dry ether--> R'-R' + 2 NaX. It couples two alkyl groups to give a symmetrical alkane.
…
- GUJCET 2022Set 171 markMCQQ.Which product is obtained from following reaction? [FIGURE: a cyclohexanone ring (C=O on the ring) bearing a −CH2−CO−OCH3 substituent at the 2-position] NaBH4 (A) Cyclohexanol ring (ring bearing OH) with a −CH2−CH2−OCH3 substituent (B) Cyclohexane ring with a −CH2−CO−OCH3 substituent (no ring OH) (C) Cyclohexenol ring (ring bearing OH and a ring double bond) with a −CH2−CO−OCH3 substituent (D) Cyclohexanone ring (ring C=O) with a −CH2−CH2−OCH3 substituent
›Reveal solutionSolution
NaBH4 is a mild reducing agent — it reduces aldehydes/ketones to alcohols but does NOT reduce esters.
Concept: Sodium borohydride selectively reduces the cyclohexanone carbonyl (C=O) to a secondary alcohol (CH−OH), converting the ring ketone into a ring alcohol. The methyl ester group −CH2−CO−OCH3 is unreactive toward NaBH4 and is retained unchanged. Among the options, only the choic …
- GUJCET 2021Set 151 markMCQQ.Which Grignard reagent gives 2-methylpropan-1-ol with reaction with methanal? (A) CH3−CH2−CH2−Mg−X (B) CH3−CH(CH3)−Mg−X (C) CH3−CH=CH−Mg−X (D) CH3−CH(CH3)−CH2−Mg−X
›Reveal solutionSolution
Grignard + methanal → primary alcohol R−CH2OH; work backwards to find R.
Concept: R−MgX+HCHO→R−CH2−OMgXH2OR−CH2OH. Methanal always adds one carbon and gives a primary alcohol. …
- GUJCET 2021Set 151 markMCQQ.Which reagent is used to convert Allyl alcohol to propenal? (A) PCC (B) O3/H2O - Zn (Powder) (C) DIBAL-H (D) All above
›Reveal solutionSolution
PCC cleanly oxidises 1° alcohol → aldehyde and leaves the double bond intact.
Concept: Pyridinium chlorochromate (PCC) is a mild oxidant that stops at the aldehyde stage and does not attack C=C.
CH2=CH−CH2OHPCCCH2=CH−CHO (propenal) …
- GUJCET 2020Set 071 markMCQQ.Cyclohexanone bearing a −CH2−C(=O)−OCH3 (methyl ester) substituent at the alpha position NaBH4 "X". What is "X" in the reaction? [FIGURE: structures shown for the substrate and each option] (A) The corresponding cyclohexanol (ring C=O reduced to CH-OH) still bearing the −CH2−C(=O)−OCH3 ester group (B) Cyclohexanone (ring C=O intact) bearing a −CH2−CH(OH)−CH3 group (C) Cyclohexanol bearing a −CH2−CH2−CH2−OH group (D) Cyclohexanol bearing a −CH2−CH2−CH3 group
›Reveal solutionSolution
NaBH₄ reduces the ketone (→ cyclohexanol) and does not touch the ester. …
- GUJCET 2020Set 071 markMCQQ.Which reagent is required to convert cyclohexanol to cyclohexanone? (A) Anhydrous CrO3 (B) O3/H2O - Zn dust (C) PCC (D) DIBAL-H
›Reveal solutionSolution
[!TLDR] Secondary alcohol → ketone needs a mild oxidant; PCC cleanly gives cyclohexanone.
Concept
Secondary alcohols are oxidised to ketones. PCC (pyridinium chlorochromate) is a mild, selective, non-aqueous oxidant that converts secondary alcohols to ketones without further oxidation. Ozonolysis reagents and DIBAL-H are not alcohol-oxidation reagents.
Solution
- (A) Anhydrous CrO3: a strong Cr(VI) oxidant, generally used with acid; not the selective mild reagent intended here. …
- GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQQ.Substance A, on reaction with Cu at 573 K, gives Isobutylene. Which is the structural formula of substance A in this reaction?(a) CH3-CH(OH)-CH2-CH3(b) CH3-CH2-CH2-CH2-OH(c) CH3-CH(CH3)-CH2-OH(d) CH3-C(CH3)(CH3)-OH (i.e. (CH3)3C-OH)
›Reveal solutionSolution
Passed over hot copper at 573 K, a TERTIARY alcohol cannot dehydrogenate (it has no H on the carbinol carbon to lose alongside the O-H), so it instead undergoes dehydration to an alkene; a primary or secondary alcohol would dehydrogenate to an aldehyde or ketone instead.
Alcohols passed over copper catalyst at 573 K behave differently by class:
- Primary alcohols dehydrogenate to aldehydes (R-CH2-OH -> R-CHO + H2).
- Secondary alcohols dehydrogenate to ketones. …
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.Give the IUPAC name of the product obtained when Phenol is oxidized by chromic acid. (Na2Cr2O7 + Conc. H2SO4).(a) Cyclohexa-2,5-diene-1,4-dione(b) Cyclohexa-1,4-dione(c) Cyclohexanone(d) Cyclohexa-1,4-diene-2,5-dione
›Reveal solutionSolution
Phenol + Na2Cr2O7/conc. H2SO4 -> benzoquinone = cyclohexa-2,5-diene-1,4-dione.
Phenol on oxidation with chromic acid (from Na2Cr2O7 + conc. H2SO4) is converted to para-benzoquinone. Its IUPAC name is cyclohexa-2,5-diene-1,4-dione: a six-membered ring with C= …
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.Identify Pyridinium chlorochromate from the following.(a) pyridine ring, N+ - H . CrO3Cl^-(b) pyridine ring, N+ - H . CrO2Cl^-(c) pyridine ring, N+ - CrO3Cl^-(d) pyridine ring, N+ - H2 . CrO3Cl^-
›Reveal solutionSolution
PCC = pyridinium (C5H5N-H+) chlorochromate (CrO3Cl-), i.e. option (a).
Pyridinium chlorochromate (PCC) is a mild oxidising reagent (C5H5NH+ ClCrO3-) used to oxidise primary alcohols to aldehydes (without over-oxidation to acids). It consists of:
- the pyridinium cation: pyridine protonated at nitrogen, so N carries a positive charge and an N-H bond, and …
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