Answer
Verified
471k+ 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
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
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
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Who was the Governor general of India at the time of class 11 social science CBSE
Distinguish between the following Ferrous and nonferrous class 9 social science CBSE
Name five important trees found in the tropical evergreen class 10 social studies CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE