Answer
64.8k+ views
Hint: Iodoform reaction of acetone will lead to the products - acetic acid (${ CH }_{ 3 }{ CO }{ O }{ H }$) (sodium acetate) and iodoform (${ CH }I_{ 3 }$). On writing the mechanism of the reaction, we can find certain stable intermediates are also formed. None of these intermediates contains a C=C bond. So, among the options given, the compound with C=C is the answer.
Complete Step by Step Solution:
-Structure of acetone is given below:
![](https://www.vedantu.com/question-sets/56135059-b8a8-470c-945c-b1c4ee62fe222544487898822686034.png)
-Iodoform reaction is used to identify the presence of carbonyl compounds in compounds.
-Here in the iodoform reaction of acetone, the acetone will be reacted by ${ I }_{ 2 }$ in the presence of NaOH to form acetic acid (sodium acetate) and iodoform.
-When we look at the mechanism of this reaction, we can see many intermediates formed in between.
-Given below is the diagrammatic representation of the mechanism of iodoform reaction of acetone.
-From the mechanism, we can see that the iodoform reaction of acetone leads to the products - iodoform (${ CH }I_{ 3 }$) and acetic acid (${ CH }_{ 3 }{ CO }{ O }{ H }$ ).
-Apart from the final products, there are certain stable intermediates formed in the reaction - i.e, on addition of iodine on acetate ion, it leads to the formation of 1-iodoacetic acid (${ CH }_{ 3 }{ CO }{ CH }_{ 2 }{ I }$) and on successive addition of iodine , two more compounds are formed which are 1,1-di iodoacetic acid (${ CH }_{ 3 }{ CO }{ CH }{ { I }_{ 2 } }$ ) and 1,1,1- tri iodoacetic acid (${ CH }_{ 3 }{ CO }{ C }{ { I }_{ 3 } }$)
-So, these compounds ${ CH }_{ 3 }{ CO }{ CH }_{ 2 }{ I }$,${ CH }_{ 3 }{ CO }{ CH }{ { I }_{ 2 } }$ , ${ CH }_{ 3 }{ CO }{ C }{ { I }_{ 3 } }$are formed in the iodoform reaction of acetone, which are satisfied by Option (A), option (C) and option (D).
-In the question, it has been asked which among the options are not formed in the iodoform reaction of acetone and that compound is Option (B)${ ICH }_{ 2 }{ CO }{ C }{ H }_{ 2 }$.
Therefore, the correct answer here is Option (B)${ ICH }_{ 2 }{ CO }{ C }{ H }_{ 2 }$.
Note: Iodoform reaction occurs when there is an $\alpha $- hydrogen in that compound, and this $\alpha $ - hydrogen will be substituted with iodine leading to the formation of iodoform. In the case of acetone, there are three $\alpha $ - hydrogens present in the compound, so all the three $\alpha $ - hydrogens are replaced by iodine successively and thereby these stable intermediates are formed.
Complete Step by Step Solution:
-Structure of acetone is given below:
![](https://www.vedantu.com/question-sets/56135059-b8a8-470c-945c-b1c4ee62fe222544487898822686034.png)
-Iodoform reaction is used to identify the presence of carbonyl compounds in compounds.
-Here in the iodoform reaction of acetone, the acetone will be reacted by ${ I }_{ 2 }$ in the presence of NaOH to form acetic acid (sodium acetate) and iodoform.
-When we look at the mechanism of this reaction, we can see many intermediates formed in between.
-Given below is the diagrammatic representation of the mechanism of iodoform reaction of acetone.
![](https://www.vedantu.com/question-sets/880cb099-e3e8-42e1-ae04-208210ef7b9b5275882117592146694.png)
-Apart from the final products, there are certain stable intermediates formed in the reaction - i.e, on addition of iodine on acetate ion, it leads to the formation of 1-iodoacetic acid (${ CH }_{ 3 }{ CO }{ CH }_{ 2 }{ I }$) and on successive addition of iodine , two more compounds are formed which are 1,1-di iodoacetic acid (${ CH }_{ 3 }{ CO }{ CH }{ { I }_{ 2 } }$ ) and 1,1,1- tri iodoacetic acid (${ CH }_{ 3 }{ CO }{ C }{ { I }_{ 3 } }$)
-So, these compounds ${ CH }_{ 3 }{ CO }{ CH }_{ 2 }{ I }$,${ CH }_{ 3 }{ CO }{ CH }{ { I }_{ 2 } }$ , ${ CH }_{ 3 }{ CO }{ C }{ { I }_{ 3 } }$are formed in the iodoform reaction of acetone, which are satisfied by Option (A), option (C) and option (D).
-In the question, it has been asked which among the options are not formed in the iodoform reaction of acetone and that compound is Option (B)${ ICH }_{ 2 }{ CO }{ C }{ H }_{ 2 }$.
Therefore, the correct answer here is Option (B)${ ICH }_{ 2 }{ CO }{ C }{ H }_{ 2 }$.
Note: Iodoform reaction occurs when there is an $\alpha $- hydrogen in that compound, and this $\alpha $ - hydrogen will be substituted with iodine leading to the formation of iodoform. In the case of acetone, there are three $\alpha $ - hydrogens present in the compound, so all the three $\alpha $ - hydrogens are replaced by iodine successively and thereby these stable intermediates are formed.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)