Single bonds generally experience free rotation at room temperature (as will be discussed in more detail in Chapter 4):
Alright guys. So, for this question it's saying that the above molecule at the top in 1 has a free barrier of rotation to get to our compound over here, this product, but in the bottom over in 2, it says that now this compound, notice it's rotational along another single bond has a much greater barrier of rotation hence it uses a lot more energy to achieve this transformation and it wants to know why is that, right? So, let's take a look, well, we look at compound 2, notice that it has a lot of double bond character, right? That bond that's been rotated, because take a look, what we can do is we can actually show that there's a resonance structure, right? This is a possible resonance structure for this compound, so notice that all the carbon atoms in here exhibit a little bit of double bond character, right? All these single bonds and notice that in double bonds we have a p orbital and our p orbital actually needs to overlap, right? That's going to give us that resonance stabilization but here notice that in order to rotate here, we are going to have to use a lot of energy, because what's going to happen when we rotate is we are going to now destroy that overlapping p orbital character of this molecule, so essentially, look at it as this compound in 2, this molecule contain a lot of double bond character and for that reason the barrier of rotation is going to be a lot larger when you have a molecule like this, that resonance stabilized, so rotating the molecule destroys that resonance stabilization which is very hard to achieve. Alright guys, hopefully this make sense and just let me know if you have any questions.