## sapling #18

Lucy Wang 2J
Posts: 101
Joined: Wed Sep 30, 2020 10:09 pm
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### sapling #18

Select the correct statement about the relative positions of the hydrogen atoms in the three structures.

->The hydrogen atoms of H2CCH2 and H2CCCCH2 lie in the same plane.
The hydrogen atoms of H2CCCH2 and H2CCCCH2 lie in the same plane.
The hydrogen atoms of H2CCH2 and H2CCCH2 lie in the same plane.

Why does an even number of C atoms mean that the hydrogen atoms lie in the same plane. If the pi bonds between the C=C is perpendicular, wouldn't you want an even number of perpendicular bonds to make them in the same plane?

Emily Vu 1L
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Joined: Wed Sep 30, 2020 9:40 pm
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### Re: sapling #18

Hi Lucy! So I'm not sure if this fully answers your question but what helped me answer this question was visually imagining how the rigid, non-rotational pi bonding affects the locations of the H atoms. So, to minimize electron repulsion between neighboring double bonds in the case of all of the molecules in this question, you would want sequential pi bonds to be perpendicular from each other. If we take the first molecule H2CCH2, we can imagine that because there is only one pi bond with p-orbitals oriented side by side, then the planes of the H atoms should also be side by side, or coplanar. But in the case of H2CCCH2, if we imagine the left-most pi bond to be on the y-axis for example, to minimize repulsion and keep the p-orbitals side-by-side, the right-most (or second) pi bond, should be located on the z-axis - or coming out of the page. So, the central C atom divides the molecule into two different planes in which the H atoms could be located.

So I was able to answer the last question of #18 by just observing these trends. The EVEN number of C ATOMS and thus, an ODD number of perpendicular bonds (in reference to your last question), results in coplanar H atoms.

I hope this helps!