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The treatment of H-C\equiv C-H with 2H_2 in presence of Pd produces

Option: 1

Alkane


Option: 2

gem-dihalide


Option: 3

Alcohol


Option: 4

Aldehyde


Answers (1)

best_answer

Dihydrogen gas adds to alkenes and alkynes in the presence of finely divided catalysts like platinum, palladium or nickel to form alkanes. This process is called hydrogenation. These metals adsorb dihydrogen gas on their surfaces and activate the hydrogen–hydrogen bond. Platinum and palladium catalyse the reaction at room temperature but relatively higher temperature and pressure are required with nickel catalysts.

\\\mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{H}_{2} \stackrel{\mathrm{Pt} / \mathrm{Pd} / \mathrm{Ni}}{\rightarrow} \mathrm{CH}_{3}-\mathrm{CH}_{3}\\\mathrm{Ethene\quad\quad\quad\quad\quad\quad\quad \quad \: Ethane}

\\\mathrm{CH}_{3}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}+2 \mathrm{H}_{2} \overset{\mathrm{Pt/Pd/Ni}}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\\\mathrm{Propyne \quad\quad\quad\quad\quad\quad\quad\quad \quad \quad \quad Propane}

Therefore, option (1) is correct.

Posted by

rishi.raj

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