## 3.41

Sophie Krylova 2J
Posts: 59
Joined: Fri Sep 29, 2017 7:06 am

### 3.41

How would you approach a formula like this to draw a lewis structure? H2C(NH2)COOH

Posts: 51
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time

### Re: 3.41

So this problem is a doozy.

First, you wanna try and separate the element into groups such as H2C,NH2, and COOH so we get that C and H are bonded such that it is a single bond between C and H giving C four bonded electrons meaning that it connects to two other elements on its sides. So we can place it in the middle since it'll connect to the two other groups.

Then, we take NH2 and we see that N will have 4 bonded electrons with 2/5 of its electrons being bonded so it connects to one side of the H2C element which gives it 6 electrons now and then 2 left over to be a lone pair.

Lastly, we take COOH and from this we get that an O connects to C with a double bond so that O can be a full octet and we can connect the other O with a H to get 7 e and then connect that to the first C to get an octet on the O. The C still has one electron left over for an octet so we place the last electron on the other side of the H2C compound to make it a full octet.

My explanation is most likely super confusing so I suggest you go to this link to get a better picture of what is looks like.

viewtopic.php?t=4167

Jessica Lutz 2E
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am

### Re: 3.41

H2C(NH2)COOH
This was my thought process when doing the problem: start with the two carbons in the center because it has the highest ionization energy. Then, from the way the equation is written, you can tell that NH2 will stem off of one of those carbons (with only N connecting to C). From there, I added the rest of the atoms in a way that would ensure the formal charges of each atom was zero. Looking at the format of this equation, there are hints to suggest that 2 hydrogens will come off the carbon and two oxygens will come of the other carbon, with one oxygen being double bonded and the other connected to a hydrogen (again this stems from each atoms preference to have a formal charge of 0).
This one is definitely harder than the others but I hope that helps you a bit!