In which of the following reactions the final product is neither acid nor an acid salt?
A. $Ph-CHO\xrightarrow{Tollen's\text{ }reagent}$
B. ${{H}_{3}}C-CH=C{{H}_{2}}\xrightarrow{KMn{{O}_{4}}/O{{H}^{-}}}$
C. $Ph-COH\xrightarrow{\text{Fehling }reagent}$
D. $Ph-C{{H}_{2}}-OH\xrightarrow{{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}^{+}}}$
Answer
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Hint: As we know that acid is able to donate a proton, and acid salts are those that produce an acidic solution after being dissolved in a solvent. To solve this question, we will write each of the given reactions individually.
Complete Solution :
- We will first discuss about the first option that is $Ph-CHO\xrightarrow{Tollen's\text{ }reagent}$
Here, in this reaction, we can see that when benzaldehyde is reacted with the tollen’s reagent, there is a formation of benzoic acid.
$Ph-CHO+2{{\left[ Ag{{\left( N{{H}_{3}} \right)}_{2}} \right]}^{+}}O{{H}^{-}}+{{H}_{2}}O\to Ph-COOH+2Ag+4N{{H}_{3}}+2{{H}_{2}}O$
- We will discuss about the next ${{H}_{3}}C-CH=C{{H}_{2}}\xrightarrow{KMn{{O}_{4}}/O{{H}^{-}}}$
$2C{{H}_{3}}CHC{{H}_{2}}+4KMn{{O}_{4}}\to 3C{{H}_{3}}COOK+4Mn{{O}_{2}}+KOH+{{H}_{2}}O$
Here, we can see that when propene reacts with $KMn{{O}_{4}}$ there is a formation of acid salt takes place.
- We will discuss about the next option that is $Ph-COH\xrightarrow{\text{Fehling }reagent}$
As we know that aldehydes that lack alpha hydrogen cannot form the enolate like benzaldehyde and hence, doesn’t give Fehling’ test under normal conditions.
- We will first discuss about the next option that is: $Ph-C{{H}_{2}}-OH\xrightarrow{{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}^{+}}}Ph-COOH$
Here, we can see that formation of acid takes place.
- Hence, we can conclude that the correct option is (c), that benzaldehyde can’t give Fehling test and the product is neither acid nor an acid salt.
So, the correct answer is “Option C”.
Note: - We should note here that as Fehling solution is a very weak oxidising agent, which fails to oxidise aromatic aldehydes, whereas it oxidises aliphatic aldehydes .
- And also aliphatic aldehydes will show Fehling test only if they will contain an alpha hydrogen atom.
Complete Solution :
- We will first discuss about the first option that is $Ph-CHO\xrightarrow{Tollen's\text{ }reagent}$
Here, in this reaction, we can see that when benzaldehyde is reacted with the tollen’s reagent, there is a formation of benzoic acid.
$Ph-CHO+2{{\left[ Ag{{\left( N{{H}_{3}} \right)}_{2}} \right]}^{+}}O{{H}^{-}}+{{H}_{2}}O\to Ph-COOH+2Ag+4N{{H}_{3}}+2{{H}_{2}}O$
- We will discuss about the next ${{H}_{3}}C-CH=C{{H}_{2}}\xrightarrow{KMn{{O}_{4}}/O{{H}^{-}}}$
$2C{{H}_{3}}CHC{{H}_{2}}+4KMn{{O}_{4}}\to 3C{{H}_{3}}COOK+4Mn{{O}_{2}}+KOH+{{H}_{2}}O$
Here, we can see that when propene reacts with $KMn{{O}_{4}}$ there is a formation of acid salt takes place.
- We will discuss about the next option that is $Ph-COH\xrightarrow{\text{Fehling }reagent}$
As we know that aldehydes that lack alpha hydrogen cannot form the enolate like benzaldehyde and hence, doesn’t give Fehling’ test under normal conditions.
- We will first discuss about the next option that is: $Ph-C{{H}_{2}}-OH\xrightarrow{{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}^{+}}}Ph-COOH$
Here, we can see that formation of acid takes place.
- Hence, we can conclude that the correct option is (c), that benzaldehyde can’t give Fehling test and the product is neither acid nor an acid salt.
So, the correct answer is “Option C”.
Note: - We should note here that as Fehling solution is a very weak oxidising agent, which fails to oxidise aromatic aldehydes, whereas it oxidises aliphatic aldehydes .
- And also aliphatic aldehydes will show Fehling test only if they will contain an alpha hydrogen atom.
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