Electrophilic Attack – Alkanes and Alkenes

by Adam Le Gresley, PhD

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    00:01 So, let’s have a look in a little bit more detail as to the exact mechanism of electrophilic attack on our alkene from a couple of slides ago. Here, we have the pi bond which is an electron rich. Bearing in mind the pi bond, remember from Module I, contains two electrons.

    00:18 The underlying sigma bond also contains two electrons. So, that’s quite an electron rich kind of area. Remember also, when we talked about double headed arrows and their significance. A double headed arrow imparts that two electrons or an electron pair are moved from one end to the other where the actual end of the arrow is. In this scenario, what’s happening is that the electron ridge, alkene group, is having its electrons transferred on to the delta+ of the H+ or if it were in solution, the H+ cation in the presence of the Br- anion.

    00:57 What happens there in the first place is we have generated now a new sigma bond between a hydrogen and the carbon on that alkene. But, of course, charges conserved and where we have moved the pair of electrons from that bond, we must, by necessity, be creating a positive charge on the other carbon. This carbocation intermediate then reacts with the bromide ion that results and you have now the addition of hydrogen bromide over that double bond. This is a two step mechanism. The carbocation happens to be the intermediate. Strong acids add to a double bond, but weak acids, such as ascetic acid and water, don’t. Water, though, can add to a double bond in the presence of an acid, strong acid. Right. So, as we saw before, it’s possible for hydrogen bromide to add over an alkene. However, when we have more than two carbons involved such as, as we see here in the case of propene; of course, we have three carbons in our alkene here, there are two possible products depending on where the H+ or delta H is attacked. So, in the case of this propene molecule, as we can see here, it’s possible for either the hydrogen or the H+ to be added on to either C-1 resulted… resulting in a carbocation with a charge in the center. So, this is shown at the top or at the bottom where, for example, the hydrogen adds to the carbon in the center, thus resulting in a carbocation which has its positive charge on a terminal, CH2 group, as you can see. The bromide ion can then add either to the center, in the case of the first situation, or it can add to the terminal position, in the case of the second. This gives us two different regioisomers. Remember, we talked about stereochemistry back in Module I. And this is an example of how we can achieve, from a synthetic perspective, two particular regioisomers. So, two different products.

    About the Lecture

    The lecture Electrophilic Attack – Alkanes and Alkenes by Adam Le Gresley, PhD is from the course Organic Chemistry.

    Included Quiz Questions

    1. CH3-CH=CH-CH3
    2. CH3-CH2-CH2-CH2-CH3
    3. C6H12
    4. CH3¬-CH2-CHCH3-CH3
    5. CH3-CHCH3-CHCH3-CH3
    1. …nucleophile.
    2. …electrophile.
    3. …carbocation.
    4. …carbanion.
    5. …a negatively charged ion.
    1. H2SO4
    2. NaOH
    3. Acetic acid
    4. CH3NH2
    5. HCOOH

    Author of lecture Electrophilic Attack – Alkanes and Alkenes

     Adam Le Gresley, PhD

    Adam Le Gresley, PhD

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