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Chapter 7: Problem 12

Glycine, an essential amino acid, has the formula \(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\). Its skeleton structure has \(\mathrm{C}-\mathrm{C}\) and \(\mathrm{C}-\mathrm{N}\) bonds but \(\mathrm{no}\) N-O bonds. Write its Lewis structure.

Short Answer

Expert verified
Answer: The Lewis structure of glycine is drawn by counting the total number of valence electrons (23), arranging the atoms with the lowest electronegative atom (carbon) at the center, connecting atoms with single bonds, and distributing the remaining electrons to satisfy the octet rule. The final structure is: H O | || H-N-C-C-O-H | H

Step by step solution

01

Count the valence electrons

To do this, we find the valence electrons of each atom in the molecule and sum them up. - N has 5 valence electrons - H has 1 valence electron (there are 4 H atoms, total 4 valence electrons) - C has 4 valence electrons (there are 2 C atoms, total 8 valence electrons) - O has 6 valence electrons Summing up, we have a total of 23 valence electrons in this molecule.
02

Arrange the atoms

As carbon has a lower electronegativity in comparison to nitrogen and oxygen, we start with a carbon atom at the center. Place the second C and N adjacent to central the C, and the O adjacent to the second C. Connect the hydrogen atoms to the first carbon atom and the oxygen atom. The preliminary arrangement can be written as \(\mathrm{H-N-C-C-O-H}\).
03

Connect the atoms with single bonds

Connect the atoms using single bonds to construct the molecule's backbone. Now, we have \(\mathrm{H-N-C-C-O-H}\). Each single bond contains 2 electrons, and in this arrangement, we've used 10 of the total 23 valence electrons.
04

Distribute the remaining electrons

With 13 valence electrons remaining, we distribute them to satisfy the octet rule for each atom, except for hydrogen which only needs 2 electrons in total. Assign the remaining electrons as lone pairs, starting with the most electronegative atoms (O and N). After doing this, we find that there are still 3 electrons remaining. We assign these to nitrogen, thus forming a lone electron or free radical on N, which is commonly seen in amino acid structures.
05

Verify the formal charges

Lastly, we verify the formal charges for each atom to make sure the structure is correct. In our case, the formal charges are: - N: -1 - C: 0 (for both carbon atoms) - O: 0 - H: 0 (for all hydrogen atoms) Since the sum of the formal charges is zero, which equals the net charge of the molecule, we can conclude that the Lewis structure is drawn correctly. The final Lewis structure of glycine is: H O | || H-N-C-C-O-H | H

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