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NCERT Exemplar · Q9

Q.What is the correct order of decreasing stability of the following cations?
I. CH3—CH^+—CH3
II. CH3—CH^+—OCH3
III. CH3—CH^+—CH2—OCH3

(i) II > I > III
(ii) II > III > I
(iii) III > I > II
(iv) I > II > III
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Carbocation stability depends on how well the positive centre is fed electrons. The

methoxy group (−OCHX3-\ce{OCH3}) directly on the cationic carbon (II) donates its oxygen

lone pair by resonance — powerful stabilisation. One carbon further away (III) the

oxygen can no longer donate by resonance and its electronegative −I-I pull now

destabilises the cation, dropping it below the plain secondary ion (I).

Order: II > I > III.

Two effects compete for an oxygen-substituted carbocation: the oxygen lone pair

(stabilising, but only when it can overlap the empty p-orbital, i.e. when oxygen is

directly attached) and the oxygen's electronegativity (an electron-withdrawing −I-I

pull that always destabilises a cation).

  1. Cation I: CHX3−CHX+−CHX3\ce{CH3-CH^+-CH3} (isopropyl cation). A plain secondary carbocation,

    stabilised only by the +I+I effect and hyperconjugation of two methyl groups. This is

    our reference.

  2. Cation II: CHX3−CHX+−OCHX3\ce{CH3-CH^+-OCH3}. The methoxy oxygen is bonded directly to the

    cationic carbon, so its lone pair overlaps the empty p-orbital and delocalises the

    charge onto oxygen:

CHX3−CHX+−OCHX3↔CHX3−CH=O+CHX3\ce{CH3-CH^+-OCH3 <-> CH3-CH=\overset{+}{O}CH3}

Every atom gains an octet; this resonance donation far outweighs oxygen's −I-I pull,

making II the most stable. …

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