Why does this compound only have one pair of enantiomers?

Question

I've been doing some reading about enantiomers, which are apparently chiral molecules that are non-superimposable mirror images of each other. An example I found of this in some problems that I've been attempting seems to have stumped me somewhat:

Take the compound $\ce{CH2=CH-CHD2}$, where $\ce{D}$ is deuterium, or hydrogen-2, and react it with deuterium gas. A pair of enantiomers is formed, both with the condensed formula $\ce{DCH2-CHD-CHD2}$. The answers provided say that these enantiomers are as follows:

...where $\ce{A}$ is standing in for deuterium.

Why don't you get enantiomers where the orientation of the $\ce{C-D}$ bonds at both ends of the molecule varies? (ie. where those $\ce{C-D}$ bonds are also in varying planes, represented by different combinations of hashed and solid wedges on paper.) Here's an example of what I mean:

Why aren't these two structures considered to be another pair of enantiomers also?

Is this - the orientation of bonds at the end of carbon chains counting effectively for nothing - the case with all stereoisomerism in alkanes?

Answer 1

Usually it doesn't as rotation around single bonds is generally very quick in room temperature. However, sometimes it matters because rotation can be blocked by bulky substituents, or other parts of the molecule. Even double bonds usually connected with E/Z isomerism, can in certain situations cause optical activity.

These are all examples of so called axial chirality. In the first case (rotation blocked by substituents) it's called atropoisomerism. Second (blocking by other parts of molecule) is present for example in helicenes. Third case (cumulated double bonds) occurs in allens.

Answer 2

You should draw the molecule showing the $\ce{H}$ atoms as well as the D (or A) atoms.

In the hydrogenated/deuterogenated molecule, all the $\ce{C-C}$ bonds are single bonds, meaning there is free rotation about all of them.

The left (as you have drawn it) end of the molecule is $\ce{DH2C-~}$ and the right end is $\ce{~-CHD2}$. Since each end-cap carbon has two identical substituents, there is no chirality. Flipping the whole molecule about a left-to-right axis parallel to the plane of the screen, and then rotating the $\ce{ C-C}$ bond going to a particular endcap carbon, would transform the "chiral" representation you drew into its "enantiomer". That wouldn't / can't happen with the truly chiral central carbon atom.