Sapling week 8 #18

(Polar molecules, Non-polar molecules, etc.)

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Joseph Hsing 2C
Posts: 91
Joined: Wed Sep 30, 2020 9:42 pm

Sapling week 8 #18

Postby Joseph Hsing 2C » Sat Nov 28, 2020 3:19 pm

Could someone explain the rule of when x is even or odd in H2C(C)xCH2 that determines if the terminal hydrogens lie in the same plane? More specifically why exactly does this happen? I know from the answer that odd number of carbon atoms will result in pi bonds that are perpendicular and an even number will result in parallel pi bonds.

Lillian Ma 1I
Posts: 104
Joined: Wed Sep 30, 2020 9:58 pm
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Re: Sapling week 8 #18

Postby Lillian Ma 1I » Sat Nov 28, 2020 9:46 pm

The carbons on the end have 3 regions of electron density which means that they have 3 sp2 orbitals and 1 p orbital. The carbons in the middle, whose amount is determined by x only has 2 regions of electron density which means they have 2 sp hybrid orbitals and 2 p orbitals. This means that in H2CCCH2, the sp orbital of the C in the center forms a sigma bond with the other 2 carbons. One of the p orbitals of the center carbon forms a pi bond with the p orbital from the left carbon, which leaves the second p orbital, which is perpendicular to the first, to form a pi bond with the p orbital of the right carbon. This makes the hydrogens on the right lie in a different plane than the ones on the left. However in H2CCCCH2, there are 2 carbons which have 2 sp and 2 p orbitals, and the third carbon forms a pi bond with the perpendicular p- orbital of the second carbon, leaving the original rotated facing p orbital free to form a pi bond with the last carbon. This means that the hydrogen surrounding the last carbon are once again on the same plane as the hydrogens surrounding the first. This pattern repeats for odd then even numbered carbon molecules. Sorry for the long explanation, so here's a picture that might explain it better, but I hope this helped!
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KSV_INORG_CHM_P1_C02_S01_028_S01.png

Joseph Hsing 2C
Posts: 91
Joined: Wed Sep 30, 2020 9:42 pm

Re: Sapling week 8 #18

Postby Joseph Hsing 2C » Sun Nov 29, 2020 12:56 pm

Lillian Ma 1L wrote:The carbons on the end have 3 regions of electron density which means that they have 3 sp2 orbitals and 1 p orbital. The carbons in the middle, whose amount is determined by x only has 2 regions of electron density which means they have 2 sp hybrid orbitals and 2 p orbitals. This means that in H2CCCH2, the sp orbital of the C in the center forms a sigma bond with the other 2 carbons. One of the p orbitals of the center carbon forms a pi bond with the p orbital from the left carbon, which leaves the second p orbital, which is perpendicular to the first, to form a pi bond with the p orbital of the right carbon. This makes the hydrogens on the right lie in a different plane than the ones on the left. However in H2CCCCH2, there are 2 carbons which have 2 sp and 2 p orbitals, and the third carbon forms a pi bond with the perpendicular p- orbital of the second carbon, leaving the original rotated facing p orbital free to form a pi bond with the last carbon. This means that the hydrogen surrounding the last carbon are once again on the same plane as the hydrogens surrounding the first. This pattern repeats for odd then even numbered carbon molecules. Sorry for the long explanation, so here's a picture that might explain it better, but I hope this helped!


Thanks for the explanation it was really helpful!


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