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Alkyl Halide Reactivity

  1. Jan 23, 2014 #1
    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?
  2. jcsd
  3. Jan 23, 2014 #2
    Reactive in what context? You may want to look up the Inductive Effect and let us know if you still have any problems.
  4. Jan 23, 2014 #3
    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.
  5. Jan 23, 2014 #4
    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.
  6. Jan 23, 2014 #5
    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.
  7. Jan 23, 2014 #6

    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: Jan 23, 2014
  8. Jan 23, 2014 #7
    Is anyone there?
  9. Jan 24, 2014 #8
    Sorry, busy times at the moment for myself.

    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).

    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.

    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.

    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.

    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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