# Possible Products in the Addition To a Double Bond in a Ring

I don't understand why both these products should be different. I know hydrogenation is syn addition and both the hydrogens add from the same side, but why should the plane in which both the hydrogen atoms are present even matter?

Also, bond rotation is possible around sigma bond which makes both these products identical.

Please explain why the two products are different and where am I wrong with this logic?

• You are right, these two are identical. May 8, 2016 at 15:27
• Those two are identical but the rotation of that sigma bond is not possible. May 8, 2016 at 15:32
• @orthocresol what is your logic of calling them identical?
– user23923
May 8, 2016 at 15:39
• Just flip the one on the left upside-down, it is the same May 8, 2016 at 15:40
• Of course, if you really don't like flipping the molecule you can always flip yourself and look at it from the bottom up instead of looking at it top down. Why don't you try using a molecular modelling set and make the two molecules. That will also show you why that bond can't be rotated. May 8, 2016 at 15:44

The two molecules are not different, they are identical. You can prove this in a number of ways:

1. Build both with a molecular model kit and see if you can orient them in the same way (spoiler: you can).

2. Use your imagination and 3D-vision to rotate one of the two structures until it looks like the other one.

3. Look for decisive symmetry elements. In this case, there is a plane of symmetry traversing the molecule through the $\ce{C-Br}$ bond. Therefore, the molecule cannot be chiral. Therefore, two versions of it that differ only in stereochemical orientation must be identical.

I can only speculate why your source drew it that way. One possibility would be to stress that the hydrogenation is not only syn-selective but can also happen from both sides (even though it gives the same product here). That doesn’t change that they’re identical, though.

why should the plane in which both the hydrogen atoms are present even matter?

In this case it doesn't. The author of the example is laying the groundwork for the future.

The area above and below the double bond is often referred to as a "face". In your molecule, the two faces are not equivalent. If you build two models of your compound and in both color the top of the double bond red and the bottom blue, you'll find there is no way to superimpose one model onto the other where the red face is superimposed on the blue face. These two faces are therefore not equivalent, they are said to be enantiotopic (mirror images) in this molecule.

In this case, addition of (an achiral reagent such as) hydrogen to the two faces produces the same molecule. In more complex molecules, or when using chiral reagents this will not be the case.

Consider the case where there is a methyl group attached to the methylene group adjacent to the $\ce{C-Br}$ carbon in your molecule. Now the faces are different (diastereotopic) and addition of hydrogen to the top and bottom faces will produce two different products (methyl and bromine on the same side or opposite sides of the molecule).

The concept of faces is important in chemistry. Always consider if the faces are equivalent, enantiotopic or diastereotopic. Depending on what you find, the products from attack at the two faces may be the same, enantiomers or diastereomers. The concept of faces is introduced in your problem.