Question

Which of the following is Lindlar catalyst?

• A. Zinc chloride and $\mathrm{HCl}$
• B. Partially deactivated palladised charcoal
• C. Sodium and Liquid ${\mathrm{NH}}_3$
• D. Cold dilute solution of ${\mathrm{KMnO}}_4$

Answer 1

The correct option is B

Partially deactivated palladised charcoal

The explanation for correct option:

(b) Partially deactivated palladised charcoal

1. Partially deactivated palladised charcoal is known as Lindlar’s catalyst.
2. It is a heterogeneous catalyst.
3. The catalyst is used for the hydrogenation of alkynes to cis-alkenes (i.e. without further reduction into alkanes).
4. It is composed of palladium that is deposited on calcium carbonate (${\mathrm{CaCO}}_3$) and then poisoned by suitable catalytic poisons, the most common of which being lead and sulfur.
5. Composition: $5\%\mathrm{Pd}-{\mathrm{CaCO}}_3+\mathrm{Pb}{\left({\mathrm{OCOCH}}_3\right)}_2+\mathrm{Quinoline}$
6. Reaction:
   $$\begin{gathered} \mathrm{COOH}-\mathrm{C}\equiv \mathrm{C}-\mathrm{COOH}\left(\mathrm{g}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)\stackrel{\mathrm{Lindlar}}{\to }\mathrm{COOH}-\mathrm{C}=\mathrm{C}-\mathrm{COOH}\left(\mathrm{g}\right) \\ \mathrm{Alkyne}\mathrm{Cis}-\mathrm{alkene} \end{gathered}$$
   

The explanation for incorrect option:

(a) Zinc chloride and $\mathrm{HCl}$

1. The solution of anhydrous Zinc chloride in concentrated $\mathrm{HCl}$ is called Lucas reagent.
2. The solution is used to distinguish the alcohols into primary, secondary and tertiary.
3. The reaction is a substitution in which the chloride replaces a hydroxyl group.
4. The best results for this test are observed in tertiary alcohols, as they form the respective alkyl halides fastest due to higher stability of the intermediate tertiary carbocation.
5. Reaction:
   $$\begin{gathered} \mathrm{ROH}\left(\mathrm{aq}\right)+\mathrm{HCl}\left(\mathrm{aq}\right)\stackrel{{\mathrm{ZnCl}}_2}{\to }\mathrm{RCl}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ \mathrm{Alcohol}\mathrm{Alkyl} \\ \mathrm{halide} \end{gathered}$$

(c) Sodium and Liquid ${\mathrm{NH}}_3$

1. It is a strong reducing agent used to reduce alkynes to trans-alkenes.
2. It is also used to reduce aromatic ring to simple alkene (Birch Reduction)
3. Reaction: $$\begin{gathered} -\mathrm{C}\equiv \mathrm{C}-\stackrel{\mathrm{Na}/{\mathrm{NH}}_3}{\to }-\mathrm{C}=\mathrm{C}-+{\mathrm{NaNH}}_2 \\ \mathrm{alkyne}\mathrm{trans}-\mathrm{alkene} \end{gathered}$$

(d) Cold dilute solution of ${\mathrm{KMnO}}_4$

1. It is called Bayer's reagent.
2. It is a strong oxidising agent.
3. It is used for detect the presence of unsaturation in the compound. if the unsaturation or double bond is present the voilet colour of ${\mathrm{KMnO}}_4$ disappears.
4. Reaction:

$$\begin{gathered} -\mathrm{C}=\mathrm{C}-\left(\mathrm{g}\right)+2{\mathrm{KMnO}}_4\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to -\mathrm{C}\left(\mathrm{OH}\right)-\mathrm{C}\left(\mathrm{OH}\right)-+{\mathrm{MnO}}_2\left(\mathrm{aq}\right)+{\mathrm{K}}_2{\mathrm{MnO}}_4\left(\mathrm{aq}\right) \\ \mathrm{alkene}\mathrm{Potassium}\mathrm{water}1,2-\mathrm{diol}\mathrm{Manganese}\mathrm{potassium} \\ \mathrm{permanganate}\mathrm{dioxide}\mathrm{manganate} \\ \left(\mathrm{Violet}\right)\left(\mathrm{Brown}\right) \end{gathered}$$

Therefore, Option (b) is correct, Partially deactivated palladised charcoal is Lindlar's catalyst.