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Exercises · 8.16

Q.For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify the reactive intermediate produced as free radical, carbocation and carbanion. [Refer to the bond cleavages printed in the NCERT textbook.]

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Reading the four drawn cleavages: (a) CHX3O−OCHX3\ce{CH3O-OCH3} splits homolytically into

two methoxy free radicals; (b) the ketone + OHX−\ce{OH-} cleavage is

heterolytic, giving a carbanion (enolate); (c) (CHX3)X3C−Br\ce{(CH3)3C-Br} ionises

heterolytically to a carbocation (tert-butyl) + BrX−\ce{Br-}; (d) the

electrophile EX+\ce{E+} adds to benzene to give a ring carbocation (arenium ion).

Homolysis (fish-hook arrows) shares the bonding pair one electron to each fragment,

giving neutral radicals; heterolysis (a full curved arrow) gives both electrons to one

fragment, producing ions — a carbocation if carbon is left electron-deficient, a

carbanion if carbon keeps the pair.

(a) CHX3O−OCHX3→CHX3OX ∙ +X ∙ X22 ∙ OCHX3\ce{CH3O-OCH3 -> CH3O^. + ^.OCH3} — the weak O–O bond breaks with one electron to

each oxygen: homolysis, giving two methoxy free radicals.

(b) The base OHX−\ce{OH-} removes an α\alpha-hydrogen from the ketone; the C–H pair stays

on carbon, forming the enolate: heterolysis, producing a carbanion (stabilised as

the enolate) and water. …

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