Question:

Arrange the above carbocations in the order of decreasing stability: 

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Remember that the stability of carbocations increases with the number of alkyl groups and the ability to delocalize the positive charge through resonance and hyperconjugation.
Updated On: Mar 17, 2025
  • \(I>III>IV>II>V\)
  • \(V>II>IV>III>I\)
  • \(V>II>III>I>IV\)
  • \(II>III>IV>V>I\) 

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The Correct Option is A

Solution and Explanation

The stability of carbocations depends on several factors including the number of alkyl groups attached, the ability to stabilize the positive charge through resonance, and hyperconjugation effects. These factors determine the order of stability. 

Step 1: Alkyl Substitution and Hyperconjugation - Tertiary carbocations are the most stable due to the hyperconjugation and inductive effects from the three alkyl groups attached to the positively charged carbon. - Secondary carbocations are more stable than primary because there are more alkyl groups to stabilize the positive charge. - Primary carbocations are less stable due to fewer stabilizing groups. 

Step 2: Resonance and Inductive Effects - Resonance helps to stabilize the carbocation by delocalizing the positive charge over adjacent atoms, especially in the case of allylic and benzyl carbocations. - Inductive effects from substituents, such as electronegative atoms or groups, can either stabilize or destabilize the carbonation. 

Step 3: Analyzing the Structures Given the structures in the image: - V has the most alkyl groups (tertiary), and if any resonance or inductive stabilization exists, it should be the most stable. - III is a secondary carbocation and likely the second most stable. - II is also secondary but with less stabilization than III. - I is primary and thus less stable. - IV is the least stable because of fewer stabilizing effects. 

Step 4: Correct Order of Stability The order of stability is \(V>III>II>I>IV\), where \(V\) is the most stable and \(IV\) is the least stable. Thus, the correct answer is:     V>III>II>I>IV

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