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In Rosenmund reduction, which of the following poison the catalyst Pd?
A.$BaS\mathop O\nolimits_4 $
B.S
C.Quinoline
D.Xylene

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Last updated date: 24th Jun 2024
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Answer
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Hint:
Rosenmund reduction:- The process by which acid chlorides are converted in aldehyde on catalytic reduction with Pd and BaSO4 in S or quinoline.
Lindlar’s catalyst produces alkenes and Rosenmund catalyst produces aldehyde. Methanal cannot be produced by this method.
$C\mathop H\nolimits_3 COCL + \mathop H\nolimits_2 \xrightarrow{{Pd,BaS\mathop O\nolimits_4 }}C\mathop H\nolimits_3 CHO + HCL$

Complete answer:
$BaS\mathop O\nolimits_4 $:- It, when added with Pd, reduces the activity of it to avoid the over reduction of acid. Barium sulphate has low surface area. So, it has low reducing power that’s why it reduces the activity of palladium hence acts as a poison . So, this option is correct

S:- when aldehyde is formed sulphur helps to prevent its hydrogenation immediately. Hence it does not act as a poison. So, this option is not correct.

Quinoline:- It is a constituent of Lindlar’s catalyst. It does not play any role in poisoning because they are used to form alkenes but not aldehydes hence this option is not correct.

Xylene:- it has no role in Rosenmund reaction. Hence this option is totally incorrect.

Our correct Answer is ‘A’ that is barium sulphate ($BaS\mathop O\nolimits_4 $)

Note: Formaldehyde is not prepared by this method because formaldehyde is unstable at room temperature.
This reaction is used for the preparation of alkyl halide
Used to prepare aryl halides
Used to prepare saturated and fatty aldehydes.
Lindlar’s catalyst used to covert alkynes to cis-alkenes and is composed of palladium in support of $BaC\mathop O\nolimits_3 $