Question 3.95

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Maria1E
Posts: 61
Joined: Sat Jul 22, 2017 3:01 am

Question 3.95

Postby Maria1E » Wed Nov 01, 2017 5:46 pm

Draw resonance structures for the trimethylenemethane anion C(CH2)3^2- in which a central carbon atom is attached to three CH2 groups.

Why does the solution manual show each CH2 group as only having 4 valence electrons? I'm confused as to how that was calculated. Thank you!

Alex Kashou
Posts: 38
Joined: Fri Sep 29, 2017 7:07 am

Re: Question 3.95

Postby Alex Kashou » Wed Nov 01, 2017 6:36 pm

Its the C in the CH2 that has only 4 valence electrons because you do not calculate formal charges on hydrogen atoms. Thus, the solution manual is correct in finding the lowest energy form of this molecule. A better way to look at this is to actually draw out the CH2's that are connect to the central C and then calculate the formal charges of all of them without the double bond and then look to see how you can fix it with formal charges.

I hope this helped.

Maria1E
Posts: 61
Joined: Sat Jul 22, 2017 3:01 am

Re: Question 3.95

Postby Maria1E » Wed Nov 01, 2017 7:05 pm

I'm still a little confused as to why CH2 wouldn't form an octet. Carbon on its own only has four valence electrons but it usually forms an octet, so why wouldn't CH2?

Alex Kashou
Posts: 38
Joined: Fri Sep 29, 2017 7:07 am

Re: Question 3.95

Postby Alex Kashou » Thu Nov 02, 2017 9:16 am

It is forming an octet because each Carbon has 2 single bonds to each H and then 1 single bond to the central carbon and last 2 loose electrons. In total that is 8. The central Carbon however has 2 single bonds to 2 of the carbons and 1 double bond which is also 8. Therefore, they all satisfy the octet rule.

I really encourage you to draw out the structure rather than look at the answer sheet because they didnt draw it out fully.


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