Chemistry · Aldehydes, Ketones And Carboxylic Acid · NEET
It gives an ALDEHYDE, not an alcohol. This is the whole point of the reaction. Normally, hydrogen gas over palladium would keep reducing the acyl chloride past the aldehyde and turn it into an alcohol. To stop this over-reduction, the palladium catalyst is weakened (poisoned) with BaSO4. So R-COCl becomes R-CHO and the reaction stops there. If a NEET option says 'gives alcohol', it is wrong.
BaSO4 is used to PARTLY deactivate the palladium catalyst. A fully active Pd catalyst is too strong and would reduce the acyl chloride all the way to a primary alcohol. Adding BaSO4 makes the catalyst weaker (this is called poisoning the catalyst), so it only reduces up to the aldehyde stage. Sometimes sulphur or quinoline is also added for the same reason. Remember: BaSO4 = brake pedal that stops at aldehyde.
Both make aldehydes, but the starting material and reagent are different. Rosenmund reduction: starts from an ACYL CHLORIDE (R-COCl), uses H2 with Pd-BaSO4. Stephen reaction: starts from a NITRILE (R-CN), uses SnCl2 (stannous chloride) with HCl, then water. NEET often gives you the reagent and asks the name, or gives the name and asks the reagent, so memorise the pairs.
Stannous chloride (SnCl2) in the presence of hydrochloric acid (HCl). The nitrile R-CN is first reduced to an imine (R-CH=NH), and this imine is then hydrolysed (broken by water) to give the aldehyde R-CHO. So the reagent chain is: SnCl2 / HCl, then H2O. Do not confuse SnCl2 with LiAlH4 or NaBH4.
Yes. NCERT gives a second method: DIBAL-H (diisobutylaluminium hydride). DIBAL-H reduces a nitrile only up to the imine stage, and after hydrolysis you get the aldehyde. DIBAL-H is a mild, selective reducing agent. The same DIBAL-H can also reduce esters to aldehydes. So for nitrile to aldehyde you can quote either SnCl2/HCl (Stephen) OR DIBAL-H.
Because special mild or poisoned reagents are used on purpose. In Rosenmund, the Pd catalyst is poisoned with BaSO4 so it is too weak to reduce the aldehyde further. In Stephen and DIBAL-H methods, the reagent only adds enough hydrogen to reach the imine, which then gives the aldehyde after water is added. If you used a strong reducing agent like LiAlH4 instead, you would over-reduce and get an amine or alcohol.
The following conversion is known as: C6H5-COCl + H2 (Pd-BaSO4) -> C6H5-CHO
For toluene --(i) CrO2Cl2, CS2--> --(ii) H3O+--> P, choose the correct statement about P.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. Any acyl chloride R-COCl (aromatic like benzoyl chloride, or aliphatic like acetyl chloride) can be reduced to its aldehyde by H2 over Pd-BaSO4. Benzoyl chloride to benzaldehyde is the most common NEET example.
An IMINE (R-CH=NH). SnCl2/HCl reduces the nitrile R-CN to this imine, and then water hydrolyses the imine to the aldehyde R-CHO.
BaSO4 (barium sulphate) is the usual poison. It partly deactivates palladium so the reaction stops at the aldehyde. Sulphur or quinoline can also be used as poisons.
Yes. DIBAL-H reduces nitriles to aldehydes (via an imine) and also reduces esters to aldehydes. It is a mild, selective hydride that stops at the aldehyde stage, unlike LiAlH4.
Aldehyde preparation is a repeat favourite in NEET. Questions usually give a reagent and ask the reaction name, or give the name and ask the product. Knowing Rosenmund (acyl chloride) and Stephen (nitrile) as a pair helps you answer them in seconds.
