2) $C H_{2} = C H - ⨁ C H_{2}$
3) $C_{6} H_{5} - ⨁ C H_{2}$
4) $C H_{3} - ⨁ C H - C H_{3}$

Decreasing order of stability of the given above carbocations is:

3 > 2 > 4 > 1

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1 > 3 > 4 > 2

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1 > 3 > 2 > 4

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3 > 2 > 1 > 4

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Solution

The correct option is C $1 > 3 > 2 > 4$
The carbocation 4 is least stable as it is not resonance stabilized. Carbocation 1 is most stable as the delocalisation of positive charge over the ring will not break aromaticity. Carbocation 3 is more stable than carbocation 2 as the extent of delocalisation carbocation 3 is more than that in carbocation 2.

So, the decreasing order of stability of given carbocations is $1 > 3 > 2 > 4$ .

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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

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}}^{-}

H_{2} C = C H - C H_{2} - + C H - C H_{3}

H_{3} C - C H_{3} | C | C H_{3} - + C H_{2}

Write these carbocations in decreasing order of their stabilities.

Q. Given below are isomers of

C_{4} H_{8}

. Which of the following statements is correct?

1.

1 C H_{2} = 2 C H - 3 C H_{2} - 4 C H_{3} But-1-ene

1 C H_{3} - 2 C H = 3 C H - 4 C H_{3} But-2-ene

1 C H_{2} = 2 C (C H_{3}) - 3 C H_{3} 2-Methylprop-1-ene

Q. The IUPAC name of compound :

7 C H_{3} - 6 C H_{2} - 5 C | H C H_{3} - 4 C H = 3 C H - 2 C H_{2} - 1 C H O

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2)CH2=CH−⨁CH23)C6H5−⨁CH24)CH3−⨁CH−CH3Decreasing order of stability of the given above carbocations is:

2) $C H_{2} = C H - ⨁ C H_{2}$
3) $C_{6} H_{5} - ⨁ C H_{2}$
4) $C H_{3} - ⨁ C H - C H_{3}$

Decreasing order of stability of the given above carbocations is: