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Arrange, the following carbocation on increasing order of stability:enter image description here

What have I though is that in second it will be most unstable, as $\ce{-OH}$ group will create most inductive effect (negative), further destabilizing carbocation, followed by (3) compound because, fluorine is at a distance, so inductive effect would be effectively weaker than (2). So $(2)<(3)<(1)$

But in my book, answer is given:$(3)<(1)<(2)$, which I couldn't understand. Any, help is massively appreciated, and point out any mistakes in my explanation if any.

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    $\begingroup$ You have to consider resonance structures. Can you draw one for (1) where an electron pair from the oxygen atom is donated to make an extra bond and give all atoms a filled valence shell? What about (2), or (3) with the fluorine atom? $\endgroup$ Oct 2 at 10:01
  • $\begingroup$ @Oscar Lanzi, resonance happens when lone pair and cation or anions are alternate/adjacent? I might be wrong so can you show resonance in proper answer? $\endgroup$ Oct 2 at 10:31
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enter image description here

Actually, you just need to know $1$ point that Mesomeric effect or M effect has a greater effect than Inductive effect or I effect. In this case, the -OH group shows +M (donating electron density) effect and the -F group shows the -I (withdrawing electron density) effect.

In compound $2$, -OH group shows the +M effect which stabilizes the positive charge.

Compound $1$ also has -OH group but due to a farther distance shows a weaker +M effect.

Compound $3$ has -F attached to it showing -I effect destabilizing the positive charge.

Therefore, final order becomes: $(2) > (1) > (3)$

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The fact that 1) is more stable than 3) can be explained by the concept of Neighboring Group Participation (NGP)

Oxygen being more polarisable and less electronegative than fluorine tends to delocalise its lone pairs to the C⁺ atom. (Nucleophilicity of oxygen atom is more due to greater polarisability which makes oh group capable of NGP).

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