# Why are the hydrogen-carbon bonds bent in a graphical depiction of an alkene, but are straight horizontally and vertically in an alkane?

In propane, for example, the $$\ce{C-H}$$ bonds are depicted by straight vertical and horizontal lines. But in propene, the same bonds are bent.

Is it a rule that $$\ce{C-H}$$ bonds are always bent if the carbon atoms are double-bonded? The image below illustrates what I am asking.

• It is a matter of taste how you project a 3D structure onto paper. Further on you will only use skeletal formulas with implicit carbon and hydrogen atoms. There is one major rule for ACS publications regarding angles in hydrocarbons: the chain angle should be 120° (which both your structures fail to comply with). May 23 '21 at 7:53
• Different level of information drawn. As a rule, chemical drawing should not convey information which are not really established. But at this level all is simplified, randomly applied but still intelligible. Note that the left structure doesn't adopt the same criterion, half is "bent" half is not. It is even not a projection. But vertical bond to H at the CH2 terminal of an alkene would look even more strange and ugly May 23 '21 at 8:51
• Chemicals are 3D structures so no 2D drawing represents their real structure well. Very simple drawings {like propane above) do a very bad job and many chemists would never draw them like that, preferring a 2D picture that uses ~120° angles for all the carbon and hydrogen bonds. You should infer no significance at all in the angles in the structure you show which is just an overly simple version few chemists would use. May 23 '21 at 22:54
• I think it's because of hybridization. Alkanes are sp3 hybridized and thier geometry is tetrahedral, which would make drawing their 3d structures with wedged and dashed lines extremely time consuming; alkenes are sp2 hybridized and their structure is trigonal planar which means there's a 120° bond angle. The middle carbon should also have 120° bond angles as it is also sp2 hybridized. May 23 '21 at 23:23

Fundamentally, it depends on how you put a three-dimensional structure onto a 2D surface.

The Valence Shell Electron Repulsion (VSEPR) model can predict the structure of most molecules. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsion. So with the first example (propane), each bond is equally 90 degrees from each other, and in the second (propene), the bonds are 120 degrees from each other. If the bonds were placed 90 degrees away from one another, at least one gap between electron groups would be 180 degrees, thus a less stable arrangement of electrons, regardless of if it is a double bond (c=c) a or a single bond (c-c)

• The first lines are correct but do not apply to the specific case. The rest is not much relevant to the discussion which should be about molecules representation and not about prediction or energy. I suggest OP to not overthink, at least not on the bases of the example s/he gave. In there, the only message is connectivity. When necessary or confusion might rise, then yes, there are rules for chemical drawing. May 23 '21 at 8:58