2
$\begingroup$

So today we performed an experiment, where we mixed cyclohexene with Bromine water. The equation was

here

which seems to make sense. The Bromine water we were using was decoloured as evidence that this equation took place, whereas when we mixed cyclohexane with Bromine water nothing happened (to show how double bonds are more 'reactive'). But what I don't understand is why the double bond was broken. I mean, wasn't it happy just being the way it was? Doesn't it still take energy to break the double bond, even if it is less than the energy required to break a single bond? It isn't like it WANTS to take the bromine? (or is it?) The only place where this energy could have come from in my mind is either the vibrational energy due to temperature, or else the kinetic energy from pouring it in? (may be taking too much of a physics approach). If so, doesn't that mean that if I got the cyclohexane, and poured it in from a great height, will it react with the Bromine water? Or is the energy difference between double bonds and single bonds so great that it can only perform under special circumstances? (i.e. under UV light)

$\endgroup$
  • 2
    $\begingroup$ BTW, I've never heard about Br-O-H. When I look up "bromine water," I get descriptions of $\ce{Br2}$ in $\ce{H2O}$ not Br-O-H. Are you sure Br-O-H is reacting? $\endgroup$ – Geoff Hutchison Nov 23 '14 at 1:45
  • 1
    $\begingroup$ A second BTW: what you have drawn is a mono-organylhydroxyboronium ion (it’s lacking a third bonding partner). Bromine is written with a lowercase r, else it becomes organic-residue + boron. $\endgroup$ – Jan Jul 6 '16 at 18:11
6
$\begingroup$

Keep in mind, that the reason reactions occur spontaneously is because there's an overall driving force: $\Delta G = \Delta H - T\Delta S$. Since there are fewer molecules on the right hand, we can guess that $\Delta S$ isn't driving this reaction. ($G$ is Gibbs free energy, $S$ is entropy, $H$ is enthalpy, and $T$ is temperature.)

So then your question boils down to "why is this reaction favorable for $\Delta H$?"

Even before I look up the bond enthalpies for each of the bonds involved, I have an intuition. Yes, we're breaking the double bond in cyclohexene. We're also breaking the $\ce{Br-O}$ bond in $\ce{Br-O-H}$. So that's two bonds that are broken.

You're right -- that costs energy, so there's clearly something more energetically favorable on the other side. Well, in the product, I'm forming a $\ce{C-Br}$ and a $\ce{C-OH}$ bond.

So the net reaction breaks two bonds, and forms two bonds. As you say, the general intuition is that the "double" part of a double bond isn't quite as strong as the standard $\sigma$ single bond.

Let's look up the bond enthalpies also through Wikipedia:

  • $\ce{C=C}$ (vs. $\ce{C-C}$ which remains in the product): ~146 kcal/mol vs. ~84 kcal/mol = ~62 kcal/mol
  • $\ce{Br-O}$ = ~56 kcal/mol (I had to look this one up, so it's a bit iffy)
  • $\ce{C-Br}$ = ~69 kcal/mol
  • $\ce{C-O}$ = ~85 kcal/mol

Net $\Delta H \approx$ -36 kcal/mol.

Even if some of the bond enthalpies are off a bit, it's clearly energetically favorable. It has nothing to do with how the species are mixed, or really the temperature. (Well, my guess is that $\Delta S$ is not favorable here, since two molecules form one, so high temperature might make the reaction less favorable from a $\Delta G$ standpoint.) Nor do you need UV light or anything like that -- the difference in bond strengths will push this reaction.

$\endgroup$
  • $\begingroup$ Could you please clarify the symbols? I'm guessing G is change in energy, S is entropy, H is enthalpy, and T is temperature? $\endgroup$ – Joshua Lin Nov 23 '14 at 1:38
  • $\begingroup$ Yes, that's correct. $\endgroup$ – Geoff Hutchison Nov 23 '14 at 1:43
0
$\begingroup$

Forget about energy due to pouring in from a great height or kinetic energy of mixing, it is not important, all it does is minutely heat the solution.

The thermodynamic answer from @Geoff Hutchinson is a fine explanation but you can look at it in another more molecular way if you want to.

In any reaction there is a potential energy barrier between reactants and products otherwise most stuff would just react instantly and we would not be here :( . The barrier is overcome only slowly in many reactions and this is because it turns out that most barriers are far greater in energy than thermal energy (average 3RT/2 kJ/mol). When the reactant molecules collide in solution most of the times that this happens they just bounce apart again without reacting. Just occasionally by random collisions with the solvent the reactants have enough energy $^*$to surmount the barrier and produce products (your 'vibrational energy due to temperature'). If the products are at lower energy (as in your example) the barrier back to reactants is even larger than the barrier forward to produce products and so the product is produced in large amounts. If the barrier from products to reactants is small then reaction favours the reactants and little product is formed.

($^*$ I'm assuming, for simplicity, that the molecules orientations are unimportant.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.