
For \[C{H_3} - CH = CH - CHO\,\,\,\,\xrightarrow{X}\,\,\,C{H_3} - CH = CH - COOH\], ‘\[X\]’ can be
A.Tollen’s reagent
B.\[KMn{O_4}/KOH\],hot
C.\[KMn{O_4}/{H_2}S{O_4}\], hot
D.\[{K_2}C{r_2}{O_7}/{H_2}S{O_4}\](conc.), hot
Answer
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Hint: In the above question, the oxidation of an aldehyde occurs and it gives carboxylic acid. So, ‘\[X\]’ must be an oxidizing agent as an oxygen atom is added. We need to find which oxidizing agent is used for the above reaction.
Complete answer:
There are four reagents given in the question. These are all oxidizing agents.
We know that tollen’s reagent is considered to be the best reagent for the oxidation of aldehyde group into carboxylic group. The chemical formula Of Tollen’s reagent is \[(Ag{(N{H_3})_2}OH)\] .Tollen’s reagent is an alkaline solution of ammoniacal silver nitrate which means it contains silver nitrate \[(AgN{O_3})\] and ammonia \[(N{H_3})\] and sodium hydroxide \[(NaOH)\]. This is used to test for aldehydes. The aldehyde is oxidized by a Tollen reagent and forms carboxylic acids. It acts as a mild reagent. The silver ions present in the Tollen’s reagent are reduced into metallic silver and form a silver mirror. That’s why this is also called the silver mirror test.
The other three options are also oxidizing agents which means that they can also oxidize aldehydes. Now the question arises why they are not used for the oxidation of aldehyde.
This is so because the oxidizing agent such as \[KMn{O_4}/KOH\], \[KMn{O_4}/{H_2}S{O_4}\], and \[{K_2}C{r_2}{O_7}/{H_2}S{O_4}\] is a strong oxidizing agents, so it can also oxidizes the double bond present between the carbon atom. So, to protect the double bond, a mild oxidizing agent is used like Tollen’s reagent. It oxidizes the \[CHO\] to \[COOH\] without affecting \[(C = C)\] and silver mirror\[Ag(0)\] is precipitated.
So, ‘\[X\]’ will be Tollen’s reagent.
Hence, the correct answer is option (A).
Note:
Tollen’s reagent test, also called the silver mirror test, is used to distinguish between aldehyde and ketone. Remember ketone does not give Tollen’s reagent test as they do not have hydrogen to remove attached to the functional group.
Complete answer:
There are four reagents given in the question. These are all oxidizing agents.
We know that tollen’s reagent is considered to be the best reagent for the oxidation of aldehyde group into carboxylic group. The chemical formula Of Tollen’s reagent is \[(Ag{(N{H_3})_2}OH)\] .Tollen’s reagent is an alkaline solution of ammoniacal silver nitrate which means it contains silver nitrate \[(AgN{O_3})\] and ammonia \[(N{H_3})\] and sodium hydroxide \[(NaOH)\]. This is used to test for aldehydes. The aldehyde is oxidized by a Tollen reagent and forms carboxylic acids. It acts as a mild reagent. The silver ions present in the Tollen’s reagent are reduced into metallic silver and form a silver mirror. That’s why this is also called the silver mirror test.
The other three options are also oxidizing agents which means that they can also oxidize aldehydes. Now the question arises why they are not used for the oxidation of aldehyde.
This is so because the oxidizing agent such as \[KMn{O_4}/KOH\], \[KMn{O_4}/{H_2}S{O_4}\], and \[{K_2}C{r_2}{O_7}/{H_2}S{O_4}\] is a strong oxidizing agents, so it can also oxidizes the double bond present between the carbon atom. So, to protect the double bond, a mild oxidizing agent is used like Tollen’s reagent. It oxidizes the \[CHO\] to \[COOH\] without affecting \[(C = C)\] and silver mirror\[Ag(0)\] is precipitated.
So, ‘\[X\]’ will be Tollen’s reagent.
Hence, the correct answer is option (A).
Note:
Tollen’s reagent test, also called the silver mirror test, is used to distinguish between aldehyde and ketone. Remember ketone does not give Tollen’s reagent test as they do not have hydrogen to remove attached to the functional group.
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