
Differentiate between acetaldehyde and acetone with suitable reactions.
Answer
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Hint: We know that the difference between the Aldehyde and Ketone is the presence of a hydrogen atom. The presence of the hydrogen atom makes it very easy to oxidize aldehydes, they are fast reduction agents.
Complete step by step answer:
As we know that aldehydes act as a reducing agent. They reduce Fehling's solution, Tollen’s reagent and Schiff’s reagent while ketones don’t.
Fehling’s Test:
Fehling’s solution appears in the form of two different solutions as Fehling A and Fehling B. Fehling A contains copper sulphate and it is a blue colored solution while Fehling B is a colorless liquid containing potassium sodium tartrate. These two solutions are consequently combined in equivalent amounts to produce the deep color Fehling’s solution.
Aldehydes such as acetaldehyde give reddish brown precipitate while ketones don’t.
$
C{H_3}CHO + 2C{u^{2 + }} + 5O{H^ - }\xrightarrow{\Delta }C{H_3}CO{O^ - } + C{u_2}O \downarrow + 3{H_2}O \\
\\
$
$C{H_3}COC{H_3} + 2C{u^{2 + }} + 5O{H^ - }\xrightarrow{\Delta }no\,reaction$
Tollen’s reagent test:
This reagent is used in the identification of an Aldehyde functional group or a functional alpha hydroxy Ketone group in a given material. Silver nitrate and ammonia are named as Tollens reagents. This test makes use of the Tollen’s reagent contains ${\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]^ + }$ complex which is easily prepared by combining the ammonia and silver nitrate. If an Aldehyde is heated with the reagent the silver metal gets reduced and forms a coat of metallic silver while ketones don’t.
From these two tests we can easily differentiate the acetaldehyde and acetone.
$
C{H_3}CHO + 2{\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]^ + } + 3O{H^ - }\xrightarrow{\Delta }C{H_3}CO{O^ - } + 2Ag \downarrow + 2{H_2}O + 4N{H_3} \\
\\
$
$C{H_3}COC{H_3} + 2{\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]^ + } + 3O{H^ - }\xrightarrow{\Delta }no\,reaction$
Note: We also remember that the Brady’s reagent is used for the qualitative identification of Ketone or Aldehyde functional group carbonyl functionality. Aldehydes and ketones react to yellow, orange, or reddish-orange precipitates while alcohols do not react. This is a successful method for checking for the existence of a drug or representing its absence.
Note:
Note: We also remember that the Brady’s reagent is used for the qualitative identification of Ketone or Aldehyde functional group carbonyl functionality. Aldehydes and ketones react to yellow, orange, or reddish-orange precipitates while alcohols do not react. This is a successful method for checking for the existence of a drug or representing its absence.
Complete step by step answer:
As we know that aldehydes act as a reducing agent. They reduce Fehling's solution, Tollen’s reagent and Schiff’s reagent while ketones don’t.
Fehling’s Test:
Fehling’s solution appears in the form of two different solutions as Fehling A and Fehling B. Fehling A contains copper sulphate and it is a blue colored solution while Fehling B is a colorless liquid containing potassium sodium tartrate. These two solutions are consequently combined in equivalent amounts to produce the deep color Fehling’s solution.
Aldehydes such as acetaldehyde give reddish brown precipitate while ketones don’t.
$
C{H_3}CHO + 2C{u^{2 + }} + 5O{H^ - }\xrightarrow{\Delta }C{H_3}CO{O^ - } + C{u_2}O \downarrow + 3{H_2}O \\
\\
$
$C{H_3}COC{H_3} + 2C{u^{2 + }} + 5O{H^ - }\xrightarrow{\Delta }no\,reaction$
Tollen’s reagent test:
This reagent is used in the identification of an Aldehyde functional group or a functional alpha hydroxy Ketone group in a given material. Silver nitrate and ammonia are named as Tollens reagents. This test makes use of the Tollen’s reagent contains ${\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]^ + }$ complex which is easily prepared by combining the ammonia and silver nitrate. If an Aldehyde is heated with the reagent the silver metal gets reduced and forms a coat of metallic silver while ketones don’t.
From these two tests we can easily differentiate the acetaldehyde and acetone.
$
C{H_3}CHO + 2{\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]^ + } + 3O{H^ - }\xrightarrow{\Delta }C{H_3}CO{O^ - } + 2Ag \downarrow + 2{H_2}O + 4N{H_3} \\
\\
$
$C{H_3}COC{H_3} + 2{\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]^ + } + 3O{H^ - }\xrightarrow{\Delta }no\,reaction$
Note: We also remember that the Brady’s reagent is used for the qualitative identification of Ketone or Aldehyde functional group carbonyl functionality. Aldehydes and ketones react to yellow, orange, or reddish-orange precipitates while alcohols do not react. This is a successful method for checking for the existence of a drug or representing its absence.
Note:
Note: We also remember that the Brady’s reagent is used for the qualitative identification of Ketone or Aldehyde functional group carbonyl functionality. Aldehydes and ketones react to yellow, orange, or reddish-orange precipitates while alcohols do not react. This is a successful method for checking for the existence of a drug or representing its absence.
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