Consider the following carbocations:
I. $C l_{3} C^{+}$ II. $C l_{2} C H^{+}$ III. $C l C H_{2}^{+}$ IV. $C H_{3}^{+}$
The stability sequence follows the order:

IV < I < II<III

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I < II < III < IV

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IV < I < III < II

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IV< II < I < III

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Solution

The correct option is B I < II < III < IV
Due to –I effect , Cl destabilises the carbocation. More number of ‘Cl’ atoms lesser is the stability of the carbocation. Thus option (b) is correct.

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Arrange the following in the increasing order of their basicity.

CH₃NH₂(I) ;(CH₃)₂NH(II) ;C₆H₅NH₂(III) ;(CH₃)₃N(IV)

Consider the following carbocations:-
I. II. III. IV.
The stability sequence follows the order:-

(I) H_{2} C = + N = - N

(I I) H_{2} + C - N = - N

(I I I) H_{2} - C - + N \equiv N

(I V) H_{2} - C - N = + N

Arrange the following carbocations in decreasing order of stability

s_{p^{3}}

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