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प्रश्न
For the following bond cleavages, use curved-arrows to show the electron flow and classify as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
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उत्तर
The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as
It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the bromine ion. The reaction intermediate formed is a carbocation.
संबंधित प्रश्न
For the following bond cleavages, use curved arrows to show the electron flow and classify homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
\[\ce{CH3O - OCH3 -> CH3\overset\bullet{\text{O}} + \overset\bullet{\text{O}}CH3}\]
For the following bond cleavages, use curved-arrows to show the electron flow and classify as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
For the following bond cleavages, use curved-arrows to show the electron flow and classify as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
Which of the following carbocation is most stable?
What is the correct order of decreasing stability of the following cations.
| I. | \[\ce{CH3 - \overset{⊕}{C}H - CH3}\] |
| II. | \[\ce{CH3 - \overset{⊕}{C}H - OCH3}\] |
| III. | \[\ce{CH3 - \overset{⊕}{C}H - CH2 - OCH3}\] |
Covalent bond can undergo fission in two different ways. The correct representation involving a heterolytic fission of CH3 – Br is:
Write structures of various carbocations that can be obtained from 2-methylbutane. Arrange these carbocations in order of increasing stability.
Which of the following carbocation's is most stable?
 |
 |
 |
 |
| (I) | (II) | (III) | (IV) |
Among the given species the resonance stabilised carbocations are:
Which of the following is the most stable carbocation?
The increasing order of stability of the following free radicals is ______.
