CHEMISTRY 1b
SPECIAL BONDING PROBLEM SET
Cytosine is one of the four bases used in the genetic code. It has the molecular formula C
4
H
5
N
3
O. The molecular structure of cytosine calculated using Spartan with a 3-21G* basis set is displayed below.
1) Draw an electron-dot (valence bond) structure of cytosine. Show all the lone pairs. Using your electrondot structure, predict the structure and hybridization of the heavy atoms. Use your structure to explain why cytosine is classified as a base.
2) Calculations and experiments such as X-ray crystallography show that cytosine possesses a plane of symmetry. That is, all the atoms lie in the same plane. a) What does the molecular planarity imply about the hybridization of the heavy atoms? b) Where are the lone pairs on the N and O atoms, in the plane of the molecule or perpendicular to the plane? b) Is the hybridization sufficient to explain global planarity? Explain.
3) Draw a resonance structure for cytosine. Which of the two resonance structures that you have drawn makes the greater contribution to the wave function? Explain why.
4) The following bond lengths in Å were measured in compounds in which there is only one electron-dot structure, i.e. no resonance: C-C, 1.54; C=C, 1.32; C

C, 1.20; C-N, 1.47; C=N, 1.26; C

N, 1.16; N-N,
1.47; N=N, 1.25; N

N, 1.10; C-O, 1.43; C=O, 1.20. The following bond lengths in Ångstrom were determined from the Spartan structure of cytosine: C1-N1, 1.37; N1-C2, 1.30; C2-C4, 1.41; C4-C3, 1.37;
C3-N2, 1.35; N2-C1, 1.42; C3-N3, 1.35; C1-O1, 1.21. Interpret these cytosine data using your resonance structures.
5) Cytosine has several isomers. Draw the structures of those isomers with the same ring structure, i.e. the carbon atoms and two of the nitrogen atoms lie in a 6-membered ring. Consideration of these isomers was important in the elucidation of the structure of DNA by Watson and Crick.