
ASSERTION: Ethyl acetoacetate will give the iodoform test.
REASON: It contains the\[C{H_3}CO\] group linked to a Carbon atom.
A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
C. Assertion is incorrect but Reason is correct.
D. Both Assertion and Reason are incorrect.
Answer
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Hint:We need to see that ethyl acetoacetate gives iodoform test. We can find out by looking for $C{H_3}CO$ group at the end of the chain which is necessary for an iodoform test. Acetoxy (\[C{H_3}CO\]) group as well as \[C{H_3}CH(OH) - \] gives the iodoform test.
Complete step by step solution:
Carbonyl compounds (aldehydes and ketones having methyl group \[C{H_3}\] group at position) give the iodoform test.
Ethyl acetoacetate \[C{H_3}COC{H_2}COOC{H_2}C{H_3}\] is a carboxylic acid derivative, more specifically an ester.
Carboxylic acid derivatives do not give the iodoform test. This is due to resonance the does not remain freely available which is needed for the reaction to take place.
Even though ethyl acetoacetate contains \[C{H_3}CO\], an acetoxy group yet it does not give the iodoform test. This is because the -atoms of the \[C{H_2}\] group are attached to the electron-withdrawing group making them more acidic than the - atom of \[C{H_3}\] group. As a result, the iodination will take place at the more acidic position, which is at \[C{H_2}\] group and thus \[C{H_3}I\] will not be formed.
Thus, the assertion is incorrect.
Ethyl acetoacetate \[C{H_3}COC{H_2}COOC{H_2}C{H_3}\] does have a \[C{H_3}CO\] attached to a Carbon atom. So, from the structure the reason is correct.
Therefore, option C is correct.
Additional information:
The iodoform test is used to detect carbonyl compounds. Some alcohols having the structure \[R - CH(OH) - C{H_3}\] always form precipitate, giving a positive iodoform test. Ethyl alcohol gives a positive iodoform test.
Note:
\[C{H_3}I\] precipitates as a yellow solid and confirms the iodoform test. The students need to pay attention that the mere presence of \[C{H_3}CO\] group in Ethyl acetoacetate is not a confirmatory parameter to decide whether or not the compound gives the iodoform test.
In reaction with Iodine in an alkaline medium, the \[C{H_2}\] group reacts more rapidly than the methyl group of \[C{H_3}CO\], preventing the formation of iodoform.
Complete step by step solution:
Carbonyl compounds (aldehydes and ketones having methyl group \[C{H_3}\] group at position) give the iodoform test.
Ethyl acetoacetate \[C{H_3}COC{H_2}COOC{H_2}C{H_3}\] is a carboxylic acid derivative, more specifically an ester.
Carboxylic acid derivatives do not give the iodoform test. This is due to resonance the does not remain freely available which is needed for the reaction to take place.
Even though ethyl acetoacetate contains \[C{H_3}CO\], an acetoxy group yet it does not give the iodoform test. This is because the -atoms of the \[C{H_2}\] group are attached to the electron-withdrawing group making them more acidic than the - atom of \[C{H_3}\] group. As a result, the iodination will take place at the more acidic position, which is at \[C{H_2}\] group and thus \[C{H_3}I\] will not be formed.
Thus, the assertion is incorrect.
Ethyl acetoacetate \[C{H_3}COC{H_2}COOC{H_2}C{H_3}\] does have a \[C{H_3}CO\] attached to a Carbon atom. So, from the structure the reason is correct.
Therefore, option C is correct.
Additional information:
The iodoform test is used to detect carbonyl compounds. Some alcohols having the structure \[R - CH(OH) - C{H_3}\] always form precipitate, giving a positive iodoform test. Ethyl alcohol gives a positive iodoform test.
Note:
\[C{H_3}I\] precipitates as a yellow solid and confirms the iodoform test. The students need to pay attention that the mere presence of \[C{H_3}CO\] group in Ethyl acetoacetate is not a confirmatory parameter to decide whether or not the compound gives the iodoform test.
In reaction with Iodine in an alkaline medium, the \[C{H_2}\] group reacts more rapidly than the methyl group of \[C{H_3}CO\], preventing the formation of iodoform.
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