Answer
Verified
461.7k+ views
HintHunsdiecker reaction is a decarboxylation and halogenation reaction in which silver salts of carboxylic acids are used to form alkyl halides by reacting them with a halogen.
Complete step by step answer
-Hunsdiecker reaction is also known as Borodin reaction.
-It is a name reaction where the silver salts of carboxylic acids are reacted with a halogen to produce an organic halide. In this reaction both decarboxylation (removal of $C{O_2}$) and halogenations (addition of halogen) takes place at the same time.
This reaction is as follows:
$RCO{O^ - }A{g^ + }\xrightarrow[{CC{l_4}}]{{B{r_2}}}R - Br$
-In 1861, Alexander Borodin was the first one to demonstrate this reaction, when he prepared methyl bromide ($C{H_3}Br$) using silver acetate ($C{H_3}CO{O^ - }A{g^ + }$).
$C{H_3}COOAg + B{r_2} \to C{H_3}Br + C{O_2} + AgBr$
But now it is known as Hunsdiecker reaction because this process was developed into a general method for preparation of organic halide by Clare Hunsdiecker and her husband Heinz Hunsdiecker.
-We can say that this is a decarboxylation reaction because the product has 1 less carbon in it as compared to the parent molecules and a carbon dioxide molecule is released. Since a halogen atom is added to it, it is a halogenations reaction.
-In this reaction radical intermediates are formed. When the silver salt of carboxylic acid reacts with bromine an acyl hypohalite intermediate is formed. Then the O and Br in this hypohalite intermediate leave each other keeping their own electrons to themselves to form a diradical pair. This is followed by decarboxylation resulting in the formation of R free radical (${R^ \bullet }$) and Br free radical ($B{r^ \bullet }$), which combine quickly with each other to form alkyl halide.
This mechanism is shown below:
So, the correct option is: (A) Borodin.
Note:
Although the reaction is named after Hunsdiecker, they were not the first one to demonstrate this reaction. They made it a general reaction but it was first demonstrated by Borodin. Also if we take carboxylate to iodine (as a halide) ratio of 1:1 then we obtain alkyl halide as product, but if we take them in a ratio of 2:1 we obtain an ester product and for a 3:1 ratio we obtain a mixture of both alkyl halide and an ester.
Complete step by step answer
-Hunsdiecker reaction is also known as Borodin reaction.
-It is a name reaction where the silver salts of carboxylic acids are reacted with a halogen to produce an organic halide. In this reaction both decarboxylation (removal of $C{O_2}$) and halogenations (addition of halogen) takes place at the same time.
This reaction is as follows:
$RCO{O^ - }A{g^ + }\xrightarrow[{CC{l_4}}]{{B{r_2}}}R - Br$
-In 1861, Alexander Borodin was the first one to demonstrate this reaction, when he prepared methyl bromide ($C{H_3}Br$) using silver acetate ($C{H_3}CO{O^ - }A{g^ + }$).
$C{H_3}COOAg + B{r_2} \to C{H_3}Br + C{O_2} + AgBr$
But now it is known as Hunsdiecker reaction because this process was developed into a general method for preparation of organic halide by Clare Hunsdiecker and her husband Heinz Hunsdiecker.
-We can say that this is a decarboxylation reaction because the product has 1 less carbon in it as compared to the parent molecules and a carbon dioxide molecule is released. Since a halogen atom is added to it, it is a halogenations reaction.
-In this reaction radical intermediates are formed. When the silver salt of carboxylic acid reacts with bromine an acyl hypohalite intermediate is formed. Then the O and Br in this hypohalite intermediate leave each other keeping their own electrons to themselves to form a diradical pair. This is followed by decarboxylation resulting in the formation of R free radical (${R^ \bullet }$) and Br free radical ($B{r^ \bullet }$), which combine quickly with each other to form alkyl halide.
This mechanism is shown below:
So, the correct option is: (A) Borodin.
Note:
Although the reaction is named after Hunsdiecker, they were not the first one to demonstrate this reaction. They made it a general reaction but it was first demonstrated by Borodin. Also if we take carboxylate to iodine (as a halide) ratio of 1:1 then we obtain alkyl halide as product, but if we take them in a ratio of 2:1 we obtain an ester product and for a 3:1 ratio we obtain a mixture of both alkyl halide and an ester.
Recently Updated Pages
Who among the following was the religious guru of class 7 social science CBSE
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Trending doubts
A rainbow has circular shape because A The earth is class 11 physics CBSE
Which are the Top 10 Largest Countries of the World?
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is BLO What is the full form of BLO class 8 social science CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE