Expert: Dr. Jeffery Raymond - 11/10/2011

Question
QUESTION: How many valance electrons in Methyl salicylate C7H8O3 are used to make sigma bonds?

that's how it looks like

http://usermeds.com/static/dd9680d634351eb004f9d7c8619ff5f6.gif

ANSWER: In organic and biochemistry, sigma bonds are the overlap of two s, sp, sp2, or sp3 orbitals. Pi bonds are the overlap of two p orbitals.  In terms of a method, it is easiest to just count the connections between atoms, and ignore all of the second bonds in the double bonds (which are p-p/pi bonds).  So, starting from the terminal methyl, lets count pairs of electrons in the molecule attributed to sigma bonds:
3xCH
CO (methyl to O)
OC (O to C)
CO (C to O ignoring the pi bond)
C to C (at the first benzene carbon we run into)
going clock wise we have C-C, C-H, C-C, C-H, C-C, C-H, C-C, C-H, C-C.. and the last C-C of the ring - again ignoring the instances where there is a second bond because it is a pi bond.
Now, lets get the C-O from the ring to the hydroxyl group, and the O-H on the hydroxyl group

All together this gives 19 sigma bonds, with two electrons in each bond, giving 38 electrons in sigma bonds all together.

The word valence in the question is not important, as only valance electrons typically form bonds.

I hope this helps!


---------- FOLLOW-UP ----------

QUESTION: Thank you so much
so when i am trying to find the number of pi bonds i would subtract total valance electrons from the electrons of the sigma bonds?

Answer
Short answer: No.  
Long answer:  Valence electrons can be typically thought of as having three behaviors in most molecules - either they lie directly between atoms and form a sigma bond, they lie out of plane between atoms and form a pi bond, or they sit as lone pairs on atoms (form no bond).

So
Valence e- = e- in sigma + e- in pi + e- in lone pairs

So, in your molecule, you want to look for double bonds, specifically to the ring (3 double bonds) and the O double bonded to the C in the ester group.  This gives us 4 double bonds.  Out of the eight bonds that this represents, 4 of those bonds have already been noted as sigma bonds, meaning the remaining bonds are pi bonds.  4 pi bonds = 8 e- total in pi bonds.

To double check these numbers, let's finally count our lone pairs: We have 3 O's with two pairs each and no lone pairs on any of the Cs or Hs.  So 6 pairs are unbonded, giving 12 e- unbonded.

So using V= s + p + lp above, we get 38+8+12= 58 e total.... lets check if this makes sense.
In this molecule we have  8C, 3O, 8H... looking to the columns these are in on the periodic table we see that each neutral H comes with 1 Ve-, 4 Ve- from C, and 6 Ve- from O.
So from Cs we have 4*8= 32, Os we have 6*3=18, and from Hs we have 8*1=8.
32+18+8= 58.

Everything checks out... so we are good.

This algorithm should be good for almost all neutral organic molecules where you are dealing with things like C, N, O, H, F, Cl, Br, P, S and B.  

I hope this helps.
Take care.