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Question
CH≡CH+H2quinoline−−−−−−−→Pd−BaSO4ACl2/H2O−−−−−→B
The compound B is :
The compound B is :
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Solution
The correct option is C Cl−CH2CH2−OH
Quinoline + Pd−BdSO4 is the Lindlar's catalyst which decreases the site activity of the substrate at which the hydration of alkyne occurs.
So, the hydration of alkyne will be stopped till the formation of alkene. No further hydration of alkene takes place, so the molecule A is ethene.
After that, there is an addition of Cl2 to the ethene to give cyclic halonium ion and H2O nucleophile attacks on one of the carbon which results in the formation of Cl−CH2CH2−OH.
CH≡CH+H2quinoline−−−−−−−→Pd−BaSO4H2C=CH2Cl2/H2O−−−−−→Cl−CH2CH2−OH
Option C is correct.
Quinoline + Pd−BdSO4 is the Lindlar's catalyst which decreases the site activity of the substrate at which the hydration of alkyne occurs.
So, the hydration of alkyne will be stopped till the formation of alkene. No further hydration of alkene takes place, so the molecule A is ethene.
After that, there is an addition of Cl2 to the ethene to give cyclic halonium ion and H2O nucleophile attacks on one of the carbon which results in the formation of Cl−CH2CH2−OH.
CH≡CH+H2quinoline−−−−−−−→Pd−BaSO4H2C=CH2Cl2/H2O−−−−−→Cl−CH2CH2−OH
Option C is correct.
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