Stability of carbonation in decreasing order is :

3 > 2 > 1

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1 < 2 < 3

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3 < 2 < 1

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None of these

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Solution

The correct option is A $3 > 2 > 1$
As $3^{o}$ carbocation is most stable, the correct decreasing order is $3^{o} > 2^{o} > 1^{o}$ .
Hence, option A is correct.

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Similar questions

Consider the following carbocations

(1) $C H_{3} - \oplus C H_{2}$ (2) $C H_{2} = \oplus C H$ (3) $C H_{2} = C H - \oplus C H_{2}$ (4) $C_{6} H_{5} - \oplus C H_{2}$ stability of these carbocations in decreasing order is

The stability of given free radicals in decreasing order is

$1) C H_{3} - ˙ C H_{2} 2) C H_{3} - C ˙ H - C H_{3} 3) C H_{3} - ˙ C - C H_{3} | C H_{3} 4) C_{3}$

Q. The order of stability of the carbon radicals

3^{\circ}, 2^{\circ}, 1^{\circ}

and

˙ C H_{3}

is:

Q. The order of stability of the carbon carbocations

1^{\circ}, 2^{\circ}, 3^{\circ}

and

^{⨁} C H_{3}

Q. The order of decreasing stability of the carbanions is:
1)

(C H_{3})_{3} C^{-}

(C H_{3})_{2} {C H}^{-}

C H_{3} {C H_{2}}^{-}

C_{6} H_{5} {C H_{2}}^{-}

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Stability of carbonation in decreasing order is :

The correct option is A 3>2>1As 3o carbocation is most stable, the correct decreasing order is 3o>2o>1o.Hence, option A is correct.

The correct option is A $3 > 2 > 1$
As $3^{o}$ carbocation is most stable, the correct decreasing order is $3^{o} > 2^{o} > 1^{o}$ .
Hence, option A is correct.