Arrange the following carbocations in the increasing order of stability with respect to their labels: Label Carbocation $1$ $CH_3-CH^+-CH_3$ $2$ $CH_3^+$ $3$ $(CH_3)_3C^+$ $4$ $CH_3-CH_2^+$
| Label | Carbocation |
| $1$ | $CH_3-CH^+-CH_3$ |
| $2$ | $CH_3^+$ |
| $3$ | $(CH_3)_3C^+$ |
| $4$ | $CH_3-CH_2^+$ |
- A$2 < 4 < 1 < 3$
- B$1 < 2 < 3 < 4$
- C$4 < 3 < 2 < 1$
- D$2 < 4 < 3 < 1$
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Similar Questions
The decreasing order of hydride affinity for the following carbocations is:
$A: CH_2=CH-C^+(CH_3)_2$
$B: (C_6H_5)_3C^+$
$C: (CH_3)_3C^+$
$D: (Cyclopropyl)_3C^+$
Choose the correct answer from the options given below:
$A: CH_2=CH-C^+(CH_3)_2$
$B: (C_6H_5)_3C^+$
$C: (CH_3)_3C^+$
$D: (Cyclopropyl)_3C^+$
Choose the correct answer from the options given below:
The following carbocation is stabilized by the interaction of the empty $p$ orbital with:
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View SolutionSome pairs of ions are given below. In which pair is the first ion more stable than the second?
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View SolutionWhich of the following has an unstable enol form?
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View SolutionIntermediates formed in the reactions $P$ and $Q$ are:
$P: CH_3-CH=CH_2 \xrightarrow{HCl}$
$Q: CH_3-CH=CH_2 \xrightarrow[\text{Peroxide}]{HCl}$
$P: CH_3-CH=CH_2 \xrightarrow{HCl}$
$Q: CH_3-CH=CH_2 \xrightarrow[\text{Peroxide}]{HCl}$
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