The stability order of the following carbocation is:

I I > I V > I I I > I

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I V > I I > I I I > I

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I I > I I I > I > I V

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I > I I I > I I > I V

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Solution

The correct option is C $I I > I I I > I > I V$
$I V$ has an anti-aromatic ring and hence is the least stable. $I I$ has the carbocation in conjugation with $2$ double- bonds whereas $I I I$ has it in conjugation with only $1$ double bond.

Hence, $I I$ will be more stable than $I I I$ . There is no such conjugation in $I$ .

Therefore, the order of stability is $I I > I I I > I > I V$

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C l_{3} C^{+}

C l_{2} C H^{+}

C l C H_{2}^{+}

C H_{3}^{+}

Arrange the following carbocations in decreasing order of stability

The stability of carbanions in the following

i) $R C \equiv ⊖ C$

ii)

iii) $R_{2} C = ⊖ C H$

iv) $R_{3} C - ⊖ C H_{2}$

is in the order of

The correct order for decreasing stability of the following free radicals is

K_{2} C O_{3} (I), M g C O_{3} (I I), C a C O_{3} (I I I), B e C O_{3} (I V)

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