
Clemmensen reduction of the ketone is carried in the presence of:
A. $\text{LiAl}{{\text{H}}_{4}}$
B. $\text{Zn - Hg with HCl}$
C. $\text{Glycol with KOH}$
D. ${{\text{H}}_{2}}\text{ with Pt as catalyst}$
Answer
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Hint: The Clemmensen reduction was named after scientist Clemmensen. The Clemmensen reduction is used to yield alkanes from ketone or aldehyde with the help of the amalgam whose atomic number is 30. The reduction is the process in which the oxygen is removed and hydrogen is added to the element.
Complete step by step answer:
-As we know that the common formula of the ketone is R-CO-R, so it has a double bond between oxygen and carbon.
-So, the reduction of oxygen will take place by removing it from the carbon and also the double bond will break down and form alkanes i.e. saturated hydrocarbon.
-Saturated hydrocarbon is the compound which consists of a single bond only and also known as paraffin.
-Here, the zinc amalgam is the reducing agent which is an alloy of the mercury with metals such as zinc, sodium, etc.
-So, here when zinc amalgam with HCl reacts with ketone it removes the oxygen atom from the ketone and adds the hydrogen group in place of oxygen.
-So, the product formed after the reduction will be alkane i.e.
$\text{R-CO-R }\xrightarrow[\text{HCl}]{\text{Zn-Hg}}\text{ R-C-R}$
-Lithium aluminium hydride is also a reducing agent but it is not as considered as Clemmensen reduction because Clemmesen described the reduction of ketone by using zinc amalgam with hydrochloric acid.
-So, this option is incorrect.
-Moreover, glycol with potassium hydroxide is a reagent which can be used to convert ketone into alkane and the reaction is known as Wolf- Kishner reaction.
-So, this option is also incorrect.
-Hydrogen with platinum as the catalyst is commonly known as Lindlar's catalyst which can be used to reduce alkynes partially.
-So, this option will be incorrect because it is not a Clemmensen reduction.
So, the correct answer is “Option A”.
Note: Clemmenson's reduction is only suitable for those compounds which do not react with acid. But Wolf-Kishner reagent can be used for the compound which is sensitive in the acidic medium. Moreover, the carboxylic group (-COOH) cannot be reduced using Clemmesem's reduction.
Complete step by step answer:
-As we know that the common formula of the ketone is R-CO-R, so it has a double bond between oxygen and carbon.
-So, the reduction of oxygen will take place by removing it from the carbon and also the double bond will break down and form alkanes i.e. saturated hydrocarbon.
-Saturated hydrocarbon is the compound which consists of a single bond only and also known as paraffin.
-Here, the zinc amalgam is the reducing agent which is an alloy of the mercury with metals such as zinc, sodium, etc.
-So, here when zinc amalgam with HCl reacts with ketone it removes the oxygen atom from the ketone and adds the hydrogen group in place of oxygen.
-So, the product formed after the reduction will be alkane i.e.
$\text{R-CO-R }\xrightarrow[\text{HCl}]{\text{Zn-Hg}}\text{ R-C-R}$
-Lithium aluminium hydride is also a reducing agent but it is not as considered as Clemmensen reduction because Clemmesen described the reduction of ketone by using zinc amalgam with hydrochloric acid.
-So, this option is incorrect.
-Moreover, glycol with potassium hydroxide is a reagent which can be used to convert ketone into alkane and the reaction is known as Wolf- Kishner reaction.
-So, this option is also incorrect.
-Hydrogen with platinum as the catalyst is commonly known as Lindlar's catalyst which can be used to reduce alkynes partially.
-So, this option will be incorrect because it is not a Clemmensen reduction.
So, the correct answer is “Option A”.
Note: Clemmenson's reduction is only suitable for those compounds which do not react with acid. But Wolf-Kishner reagent can be used for the compound which is sensitive in the acidic medium. Moreover, the carboxylic group (-COOH) cannot be reduced using Clemmesem's reduction.
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