How many pi and sigma bonds are in caffeine?

1 Answer
Feb 21, 2016

Caffeine has the following structure:

https://upload.wikimedia.org/

Suppose the molecule only had single bonds. That is, suppose we drew the skeletal structure, with no double bonds whatsoever. We can count each single bond in that case as one #sigma# bond.

That means there are these #sigma# bonds, followed by the count in parentheses:

  • #"N"-"C"# (3; #"N"-"CH"_3# connection)
  • #"C"-"N"# (8; all in the rings)
  • #"C"-"O"# (2)
  • #"C"-"C"# (2)
  • #"C"-"H"# (10, from each #"CH"_3# and one #"C"-"H"# between the #"N"-"C"="N"#)

That's #3+8+2+2+10=25#.

And now we count each double bond as having one #sigma# bond and one #pi# bond. Since we have already accounted for all #sigma# bonds, each double bond contributes one #pi# bond.

The number of #pi# bonds is therefore:

  • #"C"="O"# (2)
  • #"C"="C"# (1)
  • #"C"="N"# (1)

That's #2+1+1 = 4#.

Hence, there are #25# #sigma# bonds and #4# #pi# bonds.