# Why does this compound only have one pair of enantiomers?

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?

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.