The equilibrium constant Kc for the gas-phase thermal decom-position of cyclopropane to propene is 1.0 * 105 at 500 K:

(e) Why is cyclopropane so reactive? (Hint: Consider the hybrid orbitals used by the C atoms.)

Video transcript

Hello. Everyone in this video, we're giving this three D diagram of cyclo butane and saying that cycle butane in the unstable cycle out. Kane that undergoes thermal decomposition at low pressures. So the question is specifically asking us, why is cyclo butane unstable? And you think about hybridization of each of these carbon atoms. So it's kind of taking this three D. Structure and translating into a two D. Structure or a Lewis structure. We have these four carbons connected in a ring to make a square shape and each of those carbons also has to hire a gens. So just draw those in. All right now take a look at every single one of these carbon atoms. We can see that. So let's actually write this down to all carbon atoms Have four electron groups because we have two of them being the carbon atoms and two of them being hydrogen. Each carbon has the same four different electron groups. And because we have four different groups, then the hybridization on that is going to be S P three hybridization. I remember that the ideal bonding angle for our sp three hybridization is going to be one oh 9.5. So ideal angle is one oh 9.5 degrees. We can see here that in square here in basic um math classes that we've taken the past. We know that square, Our angles are 90 degrees. So actual angle Is 90°. So because we have this molecule enclosed in a um ring of a sort there's a very high ring strain here and because of ring strain, everything is constructed meaning that the entire molecule is going to be very, very unstable because it wants to have this much room. But we all kind of squishing everything into just 90°. So the firm answer, then, is unstable because of high ring strain, so that right, there is going to be my final answer for this problem.