In the Rosenmund's reaction RCOCl+H2 xrightarrowPd/BaSO4RCHO+HCl here BaSO4

Question

In the Rosenmund's reaction RCOCl+H2 xrightarrowPd/BaSO4RCHO+HCl here BaSO4  (A) promotes catalytic activity of Pd  (B) removes the HCl formed in the reaction (C) deactivates Pd  (D) activates Pd

Tardigrade Admin · Accepted Answer

BaSO4 decreases the activity of palladium

Q. In the Rosenmund's reaction $RCOCl + H_{2} P d / B a S O_{4} RC H O + H Cl$ here $B a S O_{4}$

promotes catalytic activity of $P d$

removes the $H Cl$ formed in the reaction

deactivates $P d$

activates $P d$

$B a S O_{4}$ decreases the activity of palladium

Q. In the Rosenmund's reaction RCOCl+H2​Pd/BaSO4​​RCHO+HCl here BaSO4​

promotes catalytic activity of Pd

removes the HCl formed in the reaction

deactivates Pd

activates Pd