Why are Tertiary Alkyl Halides More Reactive Than Secondary and Primary Ones?

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Discussion Overview

The discussion centers around the reactivity of tertiary, secondary, and primary alkyl halides in the context of halogenation, substitution, and elimination reactions. Participants explore the underlying reasons for the observed trends in reactivity and the role of the inductive effect, carbocation stability, and leaving groups.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions why tertiary hydrogen is more reactive than secondary and primary hydrogen, leading to a discussion on the reactivity of alkyl halides.
  • Another participant asks for clarification on the context of reactivity, suggesting the inductive effect as a relevant concept.
  • A participant expresses confusion about the application of the inductive effect, stating that their conclusion contradicts the expected reactivity trend of 1°<2°<3°.
  • There is a suggestion that the stability of intermediates, such as carbocations and carbanions, may influence reactivity, with a focus on the inductive effect.
  • One participant mentions that halogenation reactions utilize radical intermediates, which may follow trends similar to carbocation stability, but admits uncertainty about the underlying reasons.
  • Another participant discusses the stability of carbocations and carbanions, noting that tertiary carbocations are relatively stable while tertiary carbanions are unstable, leading to different reactivity patterns.
  • Concerns are raised about the role of good and bad leaving groups in elimination and substitution reactions, with a participant questioning the nature of bond cleavage and the resulting species.
  • There is a consensus that the stability of intermediates is relative and that various factors, including solvent properties and steric effects, influence reaction outcomes.

Areas of Agreement / Disagreement

Participants express differing views on the application of the inductive effect and the stability of intermediates, indicating that multiple competing perspectives remain. The discussion does not reach a consensus on the reasons behind the reactivity trends.

Contextual Notes

Participants acknowledge the complexity of the reactions involved, noting that factors such as solvent properties, steric effects, and nucleophilicity may influence the outcomes, which are not fully resolved in the discussion.

Who May Find This Useful

This discussion may be of interest to students and professionals in chemistry, particularly those exploring organic reaction mechanisms and the factors influencing reactivity in alkyl halides.

andyrk
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Why is tertiary hydrogen the most reactive then secondary hydrogen then primary hydrogen?
Similarly, why are tertiary alkyl halides much more reactive than secondary alkyl halides which are in turn more reactive than primary alkyl halides?
 
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Reactive in what context? You may want to look up the Inductive Effect and let us know if you still have any problems.
 
Reactive in Halogenation, Substitution and Elimination reactions. And I applied the inductive effect concept here but then I come up with the conclusion that 1°<2°<3° and not the other way round as it is like. So I was confused here as to what to do.
 
andyrk said:
Reactive in Halogenation, Substitution and Elimination reactions. And I applied the inductive effect concept here but then I come up with the conclusion that 1°<2°<3° and not the other way round as it is like. So I was confused here as to what to do.

Does that trend follow for all intermediates formed during a given reaction? Would it not changed based on the charge of the carbon intermediate formed? IOW, apply the inductive effect to a primary through tertiary carbanion and carbocation. Don't worry about how to make any of the given intermediates, just draw out any primary through tertiary carbon structure and remove the atoms/electrons necessary to create that given intermediate, then analyze them according to the inductive effect.
 
Halogenation reactions use radical intermediates which tend to follow the trends of carbocation stability. Its just a rule of thumb that I have been taught and don't really know the deep reason for such behavior, maybe someone can enlighten us all about that.
 
Yanick said:
Does that trend follow for all intermediates formed during a given reaction? Would it not changed based on the charge of the carbon intermediate formed? IOW, apply the inductive effect to a primary through tertiary carbanion and carbocation. Don't worry about how to make any of the given intermediates, just draw out any primary through tertiary carbon structure and remove the atoms/electrons necessary to create that given intermediate, then analyze them according to the inductive effect.


Hmm. So if we remove a tertiary hydrogen from an alkyl halide, a 3° carbocation is formed. As it is 3° carbocation so it gets +I effect from 3 other alkyl groups that it is connected with. So this carbocation becomes relatively stable and wouldn't want to react now. But why does it still react and that too vigorously? Similarly if a 3° carbanion is formed it is very very unstable due to more +I effect on the already negative carbanion. So it would react quickly. Is this it then?

I think that the role of a good leaving group and bad leaving group is also played in the elimination and substitution reactions. That is, if the group is a good leaving group, its leaving is highly feasible and so it leaves as an negative radical making the alkyl halide from which it left a positive radical. Is this true?

Does the same explanation which I gave also fit for substitution and elimination reactions? I am confused. :/
 
Last edited:
Is anyone there?
 
Sorry, busy times at the moment for myself.

andyrk said:
Hmm. So if we remove a tertiary hydrogen from an alkyl halide, a 3° carbocation is formed.

Don't confuse the situation with an alkyl halide right now. You should look at, for example, tert-butane (for the tertiary carbon case) and n-butane (for secondary and primary carbon cases).

andyrk said:
As it is 3° carbocation so it gets +I effect from 3 other alkyl groups that it is connected with. So this carbocation becomes relatively stable and wouldn't want to react now. But why does it still react and that too vigorously?

That sounds okay but I always confuse the notation using +I or -I. I prefer to just say electron donating or electron withdrawing to avoid confusion. Any substituent with electron donating potential will make a cation less positive (or an anion more negative). Since charge plays a vital role in energy of a system (think coulomb potential), this situation results in a more stable charged intermediate relative to a similar situation without such substituents.

To answer your last question up there, these intermediates are not stable in an absolute sense. Relative to methane, for example, any given carbocation will be less stable. Its all relative as these are rules of thumb which are useful when you don't have the time to do ab initio QM type calculations. Its nice to be able to just look at a given molecule and take educated guesses about what and how things would happen. So when we say stable, its just that a tertiary carbocation is more stable than a secondary. No absolutes, just trends and rules of thumb.

andyrk said:
Similarly if a 3° carbanion is formed it is very very unstable due to more +I effect on the already negative carbanion. So it would react quickly. Is this it then?

It may never form because it is so unstable! Check the pKa's of primary through tertiary carbons, they are all near 50. Pretty tough to find a base to deprotonate an alkane, regardless of neighboring group inductive effects.

andyrk said:
I think that the role of a good leaving group and bad leaving group is also played in the elimination and substitution reactions. That is, if the group is a good leaving group, its leaving is highly feasible and so it leaves as an negative radical making the alkyl halide from which it left a positive radical. Is this true?

Homolytic bond cleavage usually does not result in charged species. Draw this out and see for yourself.

Also when doing this type of analysis you need to look at the system as a whole. A good leaving group will be stable when it leaves with its own electrons but you also need a relatively stable cation on your molecule. You always need to think about everything that is in the flask.

andyrk said:
Does the same explanation which I gave also fit for substitution and elimination reactions? I am confused. :/

Yes except with the caveat that there are a multitude of factors involved in any given reaction (solvent properties, steric effects, nucleophilicity, basicity etc).
 

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