Preparation of aldehydes

1. Controlled oxidation of a primary alcohol. A primary alcohol has two H atoms on the carbinol carbon; removing them in a controlled way leaves exactly one H behind on a new $\text{C=O}$:

where: R is the alkyl/aryl group, PCC is pyridinium chlorochromate (a mild, anhydrous $\text{Cr(VI)}$ oxidant), and dilute $K_2Cr_2O_7$ under gentle, distillation-controlled heating is the aqueous alternative. Strong aqueous oxidants (acidic $\text{KMnO}_4$, excess $K_2Cr_2O_7$) do not stop here — they carry straight through to $\text{R–COOH}$.

2. Ozonolysis of an alkene. The $\text{C=C}$ bond is cleaved cleanly, and each carbon that used to be part of the double bond becomes its own carbonyl carbon:

where: the ozonide intermediate formed in step (i) is reductively cleaved by zinc dust in water in step (ii) (the $Zn$ prevents the $H_2O_2$ by-product from over-oxidising the aldehydes to acids). A terminal alkene $\text{R–CH=CH}_2$ therefore gives $\text{R–CHO}$ plus formaldehyde, $\text{HCHO}$.

3. Markovnikov hydration of acetylene. $Hg^{2+}/H_2SO_4$-catalysed addition of water to a terminal alkyne places $\text{–OH}$ on the more substituted carbon, so the unstable enol formed tautomerises to a ketone for every terminal alkyne except the very first one — acetylene has no substituent to be "more substituted," so its enol tautomerises to an aldehyde:

4. Rosenmund's reduction of an acid chloride. Catalytic hydrogenation over palladium, deliberately poisoned, replaces the chlorine with hydrogen and stops there:

where: BaSO₄ is the catalyst support and the sulphur–quinoline mixture is the poison that partially deactivates the palladium surface, dialled down just enough to stop before the aldehyde is reduced further to the alcohol.

5. Stephen reaction from a nitrile. Tin(II) chloride and HCl reduce the nitrile to an imine-hydrochloride salt, which is then hydrolysed:

Notice the pattern across all five: an aldehyde is a metastable stopping point on the way to either a fully reduced alcohol or a fully oxidised acid, and the reagent/catalyst is always chosen or tuned specifically to arrest the reaction there.

There is no single derivable numeric formula behind aldehyde preparation — each route is a named organic transformation — so the useful thing to nail down is why the control step in each method actually holds the reaction at the aldehyde and does not let it slide further. In the alcohol oxidation, the danger is that once $\text{R–CHO}$ exists, its carbonyl carbon still carries one H, and any oxidant capable of abstracting that H (via the aldehyde's hydrate, $\text{R–CH(OH)}_2$, in aqueous medium) will push on to $\text{R–COOH}$. PCC works because it is used in an anhydrous, non-aqueous solvent (dichloromethane): without water present, the aldehyde cannot form the hydrate that further oxidation needs, so the reaction is chemically incapable of proceeding past $\text{R–CHO}$, not merely "gentle" by luck. In Rosenmund's reduction, the danger is the opposite: Pd is an excellent hydrogenation catalyst and would happily reduce the aldehyde on to the alcohol once it forms. Poisoning the catalyst surface with sulphur/quinoline lowers its activity just enough that it can still activate the more electrophilic acid-chloride carbon (fast reaction) but is too sluggish to touch the less reactive aldehyde carbonyl (slow reaction) within the reaction time — a kinetic window, not a thermodynamic wall. In ozonolysis, the control is structural rather than kinetic: the ozonide intermediate is cleaved by zinc in water, and zinc's role is specifically to reduce away the hydrogen peroxide that the cleavage would otherwise generate, because $H_2O_2$ is itself an oxidant that would carry the fresh aldehydes on to acids. In the acetylene hydration, the control is purely structural, decided before any catalyst acts: Markovnikov's rule places $\text{–OH}$ on the carbon that can best stabilise the intermediate carbocation, which is the more substituted carbon. Acetylene is the one terminal alkyne with no substituent asymmetry between its two carbons, so the "more substituted" carbon is not defined and the addition lands symmetrically, tautomerising to $\text{CH}_3\text{CHO}$ rather than a ketone.

Problem. Propanal, $\text{CH}_3\text{CH}_2\text{CHO}$, is to be prepared from propanoyl chloride, $\text{CH}_3\text{CH}_2\text{COCl}$, by Rosenmund's reduction. Write the equation, then explain what would happen if an unpoisoned Pd/C catalyst were used instead, and contrast this with the alternative route from 1-propanol.

Step 1. Set up the Rosenmund reduction: hydrogenate the acid chloride over palladium supported on barium sulphate, with a trace of sulphur–quinoline poison.

Step 2. Consider the unpoisoned case. Without the poison, the Pd surface retains full hydrogenation activity. It still reduces the acid chloride to the aldehyde first (that step is unavoidable and fast), but it does not stop there — it goes on to reduce the aldehyde's $\text{C=O}$ to $\text{C–OH}$, giving 1-propanol, $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$, as the major product instead of the desired propanal. The poison is therefore not optional decoration; removing it changes the isolated product.

Step 3. Contrast with the alcohol route. Starting from 1-propanol and using PCC in dichloromethane gives the same target, $\text{CH}_3\text{CH}_2\text{CHO}$, by controlled oxidation rather than controlled reduction:

Both routes reach propanal, but they approach it from opposite oxidation-state directions — one holding an oxidation back, the other holding a reduction back — which is exactly why the choice of route in synthesis depends on which starting material (alcohol, acid chloride, alkene, alkyne or nitrile) is already available in the synthetic sequence.

Propanal, $\text{CH}_3\text{CH}_2\text{CHO}$, is obtained either by poisoned catalytic reduction of propanoyl chloride or by PCC oxidation of 1-propanol; an unpoisoned catalyst in the first route over-reduces to 1-propanol instead.

• Writing "$\text{KMnO}_4$" as the reagent for alcohol to aldehyde without specifying controlled, mild, non-aqueous (PCC) conditions — under ordinary aqueous strong-oxidant conditions the acid forms instead.
• Forgetting that ozonolysis cleaves the double bond symmetrically: each carbon of the original $\text{C=C}$ becomes its own carbonyl carbon, so a disubstituted alkene gives two separate carbonyl products, not one.
• Assuming acid-catalysed hydration of any terminal alkyne gives an aldehyde — only acetylene does; every other terminal alkyne gives a methyl ketone because Markovnikov addition places $\text{–OH}$ on the more substituted carbon.
• Omitting the catalyst poison in Rosenmund's reduction, or naming the wrong support (it is $\text{BaSO}_4$, with sulphur/quinoline as the poison, not the support itself).

Ask the tutor: if you want to practise choosing the right route for a specific target aldehyde or want the Stephen reaction mechanism walked through step by step, ask the AI tutor to work through it with you.