Tertiary alcohols

Discussion in 'MCAT Study Question Q&A' started by Deepa100, Jun 14, 2008.

  1. Deepa100

    Deepa100 Junior Member

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    Why do Tertiary alcohols react much more quickly with HCl than do other types of alcohols?
     
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  3. bluemonkey

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    Upon protonation, tertiary alcohols form more stable carbocations than their counterparts. Since carbocation formation the RLS for SN1 & SN2, increased stability means increased rate.
     
  4. Deepa100

    Deepa100 Junior Member

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    But if you are using a strong acid like HCl, what kind of nucleophile is that? The reaction can not be SN1 or SN2, can it? Are we talking an elimination reaction here? Because once the water leaves, Cl- can deprotonate another H+, causing a double bond to form. If that is the case, then how am I to know whether this will result in E1 or E2? E2 requires a strong base. So, can this be E1?
     
  5. 161927

    161927 Guest

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    If you react a tertiary alcohol with HCl, Sn1 and E1 are the two competing reactions that might occur. Sn2 is unfavorable because of steric considerations with a tertiary alcohol. Low temperature favors substitution, while higher temperatures favor elimination. Cl- is a decent nucleophile and can attack the tertiary carbocation, or it can act as a base to eliminate a hydrogen, as you have mentioned, but it is likely through an E1. How do you know it's E1 rather than E2? Because the acidic HCl protonates water, allowing it to leave and form a relatively stable tertiary carbocation.
     
  6. tncekm

    tncekm MS-1

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    A protonated alcohol becomes a good leaving group, that's why #1. Then, through inductive effects (that is electron distribution through sigma bonds) electron donating groups like alkyl groups will donate electrons to help stabilize the positive charge on a carbocation. So, if you've got 1 alkly group you get less inductive stabilization than 2, and 2 gets less inductive stabilization than 3 b/c there are simply less electrons to share. That's why #2.

    Whether you get an elimination or addition in an SN1 or E1 reaction isn't really easy to control b/c both are favored by very similar conditions. I'd imagine being having a slightly basic solution would tend to favor SN1 b/c you'd have nucleophiles that could attack and not be protonated as frequently and induced to leave again, but that's honestly speculation on my part. In addition, if you've got a bulky solvent it would likely have a hard time acting as a nucleophile for a tertiary carbocation so that may favor E1 over SN1.
     
  7. wizenedone

    wizenedone Indeed...

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    Why would it form an alkene? That won't happen because if Cl- takes off one of the H's it will form a strong acid again (HCl) and thus NOT stable. Hence, this won't be "favored"

    Sn1 is favored since it leads to more "stable" product.
     
  8. 161927

    161927 Guest

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    I would not so confidently say that Cl- cannot act as a base to abstract a proton. Cl- is both a decent nucleophile and base. Formation of HCl by acting as a base is not necessarily disfavored energetically, because basicity is a thermodynamic property, and the stabilization of the carbocation may result in a favorable energy change. You cannot readily disentangle Cl-'s behavior as a nucleophile vs. base, but you can consider things like temperature and solvent effects in making a guess about which pathway--substitution or elimination--will be favored.
     
  9. wizenedone

    wizenedone Indeed...

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    See I don't think Cl- would abstract a proton simply due to the fact that HCl will be formed (again) and thus it will be less stable reaction (as opposed to Sn1 reaction). I am not saying Cl- would not act as a base...It might in other reactions but not in this reaction.

    Maybe QofQuimica can chime in?
     
  10. 161927

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    Which is more stable, a carbocation and free H+ and Cl- in solution, or an alkene and HCl, which does, as you noted, dissociate into H+ and Cl- again? I don't know if you can draw a clear-cut answer here. From a thermodynamic perspective, I think abstraction of a proton might be favored, because it stabilizes the carbocation and results in a neutral molecule. Free Cl-, which is what you get after HCl dissociates, can be stabilized by solvent effects. Remember that Cl- is a fairly polarizable nucleophile that can also be stabilized by protic solvents.
     
  11. wizenedone

    wizenedone Indeed...

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    Well a "carbocation" isnt really the final product...the final product will be alkyl halide.

    So, wouldnt you compare stability of tertiary alkyl halide vs alkene + HCl?
     
  12. bluemonkey

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    Sorry I made a typo in my above post. I meant to type carbocation formation is the RLS in SN1 and E1, not SN2. I'm not so good with the late-night posts...
     
  13. QofQuimica

    QofQuimica Seriously, dude, I think you're overreacting....
    Administrator Physician PhD Faculty Lifetime Donor Classifieds Approved

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    This is essentially correct.

    While it is true that Cl- is not a very good base, the carbocation is a stronger (less stable) acid (Lewis acid) than the HCl is. Also, don't forget that your reaction vessel is full of water (where do you think all those protonated hydroxyl leaving groups go???) So, what you will really end up with is Cl- and H3O+. You will have to work the reaction up at the end (extract it and dry it over sodium sulfate) to remove the water.
     
  14. TJames

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    As for the reason for the greater reactivity of tertiary alkyl halides in HCl/H2O as opposed to other alkyl halides, the reason is that a carbocation intermediate will form, and tertiary carbocations are more stable than primary or secondary.

    Reactions like this will give both E1 and Sn1 products. There will be little/no Sn2 or E2 products.
     

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