NCERT Class XI Chemistry Organic Chemistry Some basic principles and Techniques Solutions
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Question : 16
Total: 40
For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
(a)C H 3 O − O C H 3 C H 3
+
C H 3
(a)
Solution:
(a) C H 3 O − O C H 3 C H 3
+
C H 3
This is an example of homolysis which gives rise to free radicals (the word ‘homo’ means same and ‘lysis’ means breakage. The bond breakage which results in species of the same kind e.g., free radicals is called homolysis.).
Electron flow :H 3 C O
O C H 3 2 C H 3 O .
(b)C H 3 −
− C H 3 + O H − C H 3 −
− C H 3 − + H 2 O
This is an example of heterolysis where bond breakage results in formation of two different species.
The electron flow may be depicted as :
Given reaction is an example of heterolysis where reactive intermediate is a carbocation.
Electron flow :
This is heterolysis where intermediate is the arenium (carbocation) ion.
Electron flow :
This is an example of homolysis which gives rise to free radicals (the word ‘homo’ means same and ‘lysis’ means breakage. The bond breakage which results in species of the same kind e.g., free radicals is called homolysis.).
Electron flow :
(b)
This is an example of heterolysis where bond breakage results in formation of two different species.
The electron flow may be depicted as :
Given reaction is an example of heterolysis where reactive intermediate is a carbocation.
Electron flow :
This is heterolysis where intermediate is the arenium (carbocation) ion.
Electron flow :
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