Ask A Scientist Chemistry Archive

(Created prior to 1993)

Question: Most undergraduate chemistry texts state that the higher the bond 
order (single, double, triple) the higher the bond energy for covalent bonds.  
Furthermore, the bond energy is defined as the amount of energy required to 
break the bond.  However, it is experimentally observed that the doubly-bonded 
ethylene and the triply-bonded acetylene (C2H2) are much less stable tan the 
singly-bonded ethane (saturated:C2H6).  In fact, welding bottles of acetylene 
are kept at relatively low pressures and are filled with a clay matrix and 
acetone to keep them from exploding!  So, what gives?  Are the multiple bonds 
actually stronger or do they just tie up more energy.  Why do they appear to 
act like a hair-trigger?  Your response would be much appreciated.
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The bond energy of a C=C double bond is about 145 kcal/mole, 
compared with the bond energy of a C-C single bond (about 80 kcal/mole).  As 
you said, these numbers seem to suggest that alkenes should be more stable 
than alkanes, since the bond energy of C=C is larger than C-C.  However, 
molecular orbital theory says that this is not the case, according to the 
following:  The double bond consists of a sigma bond and a pi bond.  By 
subtracting 80 kcal/mole (the s-bond energy) from 145 kcal/mole (the C=C bond 
energy) we calculate that the pi bond has a dissociation energy of about 65 
kcal/mole.  This is muchless than the s-bond energy, and therefore pi bonds 
are more reactive than sigma bonds.  This means that, all things being equal, 
the pi bond of an alkene is more likely to undergo a reaction than the sigma 
bond of an alkane.  This is a fairly deep concept and will not make much sense 
if you are uninitiated into the mysteries of molecular orbital theory.  PS: my 
source was L.G. Wade Jr., "Organic Chemistry".
Topper
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