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Elimination Reactions – Haloalkanes

by Adam Le Gresley, PhD
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    00:01 is the carbon along. So, let’s talk a little bit more about that.

    00:04 Elimination reactions are, by their very nature, resulting in something coming off. That’s what an elimination reaction actually means. When you’ve got a base, something that doesn’t necessarily react as a nucleophile, it can abstract the proton, the hydrogen, shown in the beta position in this diagram. The abstraction of that proton can result in electrons moving to form a double bond in the center and subsequent loss of a halide - iodide, bromide, chloride.

    00:35 Two possible mechanisms exist for this: E2 and E1. Like the substitution that I was talking about earlier, these are also either bimolecular or unimolecular. Hence, the numbers 2 or 1.

    00:50 So, let’s have a quick look at the E2 mechanism. In this particular case, when we have a base, it is bimolecular. What happens in the first instance is a proton is abstracted by electron donation from the hydrogen. The two electrons in the sigma molecular orbital between the carbon and the hydrogen then form a pi double bond between the two carbons and X, as its halide ion is lost; again, either chloride, bromide or iodide.

    01:24 The concerted mechanism therefore implies that we need to have a large concentration of B- and also a large concentration of the haloalkane for this to be achieved. And as a consequence, the rate kinetics of this, not the thermodynamics, but the rate kinetics, are bimolecular. The higher the concentration of both of those, the faster the rate of the E2 reaction. Now, this is where it also gets interesting.

    01:53 If you recall, from when we were doing the addition reactions of alkenes, a lot of these were... or a lot of the regiochemistry chemistry was dependent upon the stability of the carbocation that was formed. If we actually have a look at this particular scenario where we’re reacting a bromopentane, in this case, with potassium hydroxide, a strong base, in the presence of alcohol, we have two possible outcomes, either where the double bond is formed in a terminal position or where it’s formed in the center or close to the center in the two position, in this case. What was observed, experimentally, is that only 20% of the terminal alkene is formed and the majority of the alkene is formed in the center. The selectivity is not actually proportional to the number of hydrogens on that beta carbon.

    02:46 If, on the other hand, we look at the second example, which is a highly substituted bromopentane, you can see that the terminal alkene forms even less preferably. And indeed, we have the double bond formed at the most substituted possible set of carbons. This is known as Zaitsev’s rule. If we look at the E1 mechanism, on the other hand, what we see is not too dissimilar to the SN1 unimolecular substitution. The first step in this reaction is that we lose the X as the halide. This happens in the absence of the base in question. What then happens is instead of nucleophilic attack on the positively charged carbon, you now have a base abstracting the proton which is nearest to it, on the beta carbon. It results in the same system, which is to say, a double bond.

    03:44 And of course, BH is formed as a consequence, which is the conjugate base B- and the proton that’s been abstracted. In this case, however, as with the SN1, the rate is completely independent upon the concentration of the base because the haloalkane, is this loss of halogen, is the slow step. And the reaction rate is, hence, unimolecular, where the concentration of R-X is the only component that influences the reaction rate.


    About the Lecture

    The lecture Elimination Reactions – Haloalkanes by Adam Le Gresley, PhD is from the course Organic Chemistry.


    Included Quiz Questions

    1. Trans or Entgegen (E)
    2. Cis or Zusammen (Z)
    3. Racemisation
    4. No preference
    5. Mixture of Z and E
    1. …beta (β).
    2. …alpha (α).
    3. …gamma (γ).
    4. …epsilon (ε).
    5. …zeta (ζ).
    1. Propene
    2. Propan-2-ol
    3. Propan-1-ol
    4. Propyne
    5. 1-Bromopropane
    1. …concentrations of both 2-bromobutane and NaOH.
    2. …concentration of 2-bromobutane.
    3. …concentration of NaOH.
    4. …concentration of the solvent.
    5. …concentrations of alcohol.
    1. …formation of a pi double bond between two carbon atoms and releases of halogen as a halide ion.
    2. … formation of a triple bond between two carbon atoms and releases of halogen as a halide ion.
    3. … formation of a triple bond between two carbon atoms and releases of a halogen atom.
    4. … formation of a double bond between two carbon atoms and releases of a halogen atom.
    5. …formation of an alkane with the release of halogen as a halide ion.
    1. …the most stable and highly substituted internal alkene.
    2. …the most stable and highly substituted terminal alkene.
    3. …the most unstable and highly substituted internal alkene.
    4. …the most unstable and highly substituted terminal alkene.
    5. …the most stable and highly substituted terminal alkyne.
    1. …elimination of halide ion takes place in the absence of a base.
    2. …elimination of halide ion depends upon the concentrations of both haloalkane and base.
    3. …elimination of halide ion depends upon the base concentration only.
    4. …formation of terminal alkene dependent upon the temperature of the reaction mixture.
    5. …alcoholic KOH determines the stability of generated carbocation.

    Author of lecture Elimination Reactions – Haloalkanes

     Adam Le Gresley, PhD

    Adam Le Gresley, PhD


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