Preparation of Aldehydes and Ketones

1. Oxidation of alcohols — the main route. A primary alcohol has two hydrogens available for removal at its oxidation-bearing carbon: one on the $-\text{OH}$ and one on the $\alpha$-carbon. Removing both gives an aldehyde, which still has one $\alpha$-hydrogen left and can be oxidised again to a carboxylic acid. A secondary alcohol has only one $\alpha$-hydrogen to begin with, so removing it gives a ketone with no hydrogen left on that carbon — oxidation simply cannot continue. This is why ketones are "safe" from over-oxidation but aldehydes are not.

where: $\text{PCC}$ is pyridinium chlorochromate, a mild, non-aqueous oxidant that supplies exactly enough oxidising power to remove two hydrogens and no more; $[\text{O}]$ denotes any oxidant (mild or strong) since a secondary alcohol cannot over-oxidise regardless of strength.

2. Dehydrogenation over hot copper. Passing alcohol vapour over copper at $573\ \text{K}$ pulls off $\text{H}_2$ directly (no oxidant needed): primary alcohols give aldehydes, secondary give ketones. Tertiary alcohols have no $\alpha$-hydrogen at all, so instead of dehydrogenating they dehydrate to an alkene under the same conditions — a fact examiners like to test as a trap.

3. From acid chlorides — Rosenmund reduction. An acid chloride is already one oxidation level above the aldehyde, so this is a controlled reduction rather than an oxidation.

where: $\text{Pd/BaSO}_4$ is palladium metal whose catalytic surface is partially "poisoned" by precipitated $\text{BaSO}_4$ — this deliberately weakens the catalyst just enough that it can add one $\text{H}_2$ (acid chloride to aldehyde) but not a second (aldehyde to alcohol).

4. From nitriles and esters. $\text{DIBAL-H}$ reduces a nitrile part-way, and hydrolysis of the resulting imine intermediate gives an aldehyde; a Grignard reagent adds once to a nitrile and hydrolysis of the imine salt gives a ketone. An ester reacts with two equivalents of Grignard reagent — the first addition displaces the alkoxide to give a ketone intermediate, which is more electrophilic and reacts with the second equivalent — so in practice esters go straight to a tertiary alcohol unless the Grignard is limited; the ketone stage is the syllabus's simplified teaching picture.

5. Friedel-Crafts acylation — the only route to an aromatic ketone.

where: anhydrous $\text{AlCl}_3$ is a Lewis acid catalyst that pulls $\text{Cl}^-$ off the acid chloride to generate the electrophilic acylium ion $\text{CH}_3\text{CO}^+$, which the benzene ring attacks.

Every named method in this topic is a control problem: how do you park a reaction exactly at the carbonyl oxidation state, when the "obvious" reagent would either overshoot (past the aldehyde to the acid) or undershoot (stop before losing enough hydrogen)? Three separate control tricks solve this, and CBSE tests all three.

Trick one — pick a reagent that runs out of oxidising power at the right point. PCC removes exactly two hydrogens (the $\text{O–H}$ and one $\alpha\text{–H}$) and forms $\text{C=O}$; it has no mechanism to insert a further oxygen atom into the aldehyde, so it simply cannot make the acid. Strong aqueous oxidants such as acidic $\text{KMnO}_4$ or $\text{K}_2\text{Cr}_2\text{O}_7$ can, because they supply "nascent oxygen" that keeps attacking the remaining $\alpha\text{–H}$ on the aldehyde. The choice of oxidant, not the alcohol, decides whether you land on the aldehyde or the acid.

Trick two — remove the extra hydrogen so over-reaction is structurally impossible. A ketone carbon has no hydrogen attached once it is formed, so there is nothing left for any oxidant, mild or strong, to remove. This is why acidic $\text{K}_2\text{Cr}_2\text{O}_7$ — normally too harsh for a primary-alcohol-to-aldehyde conversion — is perfectly fine for oxidising a secondary alcohol to a ketone: the reaction has nowhere further to go.

Trick three — deliberately cripple the catalyst. In the Rosenmund reduction, plain palladium would reduce the acid chloride all the way to the primary alcohol (two additions of $\text{H}_2$). Precipitating $\text{BaSO}_4$ onto the palladium surface blocks enough active sites that the catalyst can just barely manage the first, easier reduction (acid chloride to aldehyde) but not the second, harder one (aldehyde to alcohol). This "poisoned catalyst" idea — using a partially deactivated catalyst to stop a reduction halfway — reappears elsewhere in organic chemistry (for example, Lindlar's catalyst for alkynes) and is worth remembering as a general strategy, not just a fact about Rosenmund's method.

Extend the Pass-1 case: starting from propan-1-ol, show both what PCC gives and what excess acidic $\text{KMnO}_4$ would give instead, and separately confirm why propan-2-ol cannot be pushed past acetone.

Step 1. Propan-1-ol, $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$, is primary. With PCC it stops cleanly at the aldehyde: $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{PCC}} \text{CH}_3\text{CH}_2\text{CHO}$ (propanal).

Step 2. The same alcohol treated with excess hot acidic $\text{KMnO}_4$ does not stop at propanal. Because propanal still has an $\alpha\text{–H}$, the strong oxidant removes it and inserts oxygen: $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow[\text{KMnO}_4]{\text{H}^+} \text{CH}_3\text{CH}_2\text{COOH}$ (propanoic acid). This confirms the Pass-1 result that PCC, not acidic $\text{KMnO}_4$, is the correct choice whenever the target is the aldehyde.

Step 3. Now take propan-2-ol, $(\text{CH}_3)_2\text{CHOH}$, a secondary alcohol, and treat it with the same strong oxidant, acidic $\text{K}_2\text{Cr}_2\text{O}_7$: $(\text{CH}_3)_2\text{CHOH} \xrightarrow[\text{acidic}]{\text{K}_2\text{Cr}_2\text{O}_7} (\text{CH}_3)_2\text{C=O}$ (acetone). Even though this oxidant is "strong," the product acetone has no hydrogen on its carbonyl carbon, so there is no further reaction possible — acetone is the final answer regardless of how strong or how much excess oxidant is used, unlike the primary-alcohol case in Steps 1–2 where the choice of oxidant changed the product entirely.

• Assuming any oxidant will do for a primary-alcohol-to-aldehyde conversion — only mild, non-aqueous oxidants like PCC stop at the aldehyde; acidic $\text{KMnO}_4$/$\text{K}_2\text{Cr}_2\text{O}_7$ push through to the acid.
• Thinking ketones resist over-oxidation because they are "unreactive" — the real reason is structural: no $\alpha$-hydrogen remains on the carbonyl carbon to remove.
• Forgetting that tertiary alcohols dehydrate (lose water, form an alkene) rather than dehydrogenate over hot copper, since they have no $\alpha$-hydrogen to pair with the $\text{O–H}$ hydrogen.
• Mixing up which reagent poisons the Rosenmund catalyst — it is $\text{BaSO}_4$ on Pd, and using unpoisoned Pd would over-reduce the acid chloride to the alcohol.
• Treating Friedel-Crafts alkylation and acylation as interchangeable for making aryl ketones — only acylation (acid chloride/anhydride plus $\text{AlCl}_3$) gives the ketone directly; alkylation risks rearrangement and over-alkylation instead.

Ask the tutor: not sure which oxidising agent or route fits a given starting alcohol, acid chloride, nitrile or arene — ask the AI tutor to walk through your specific structure step by step.