Which type of bond has a higher molar absorptivity C=O or C=C, why?

Question

Which type of bond has a higher molar absorptivity C=O or C=C, why?

Answer 1

Let's assume we're talking about a molecule with just the one chromophore, either a C=C or a C=O, and there's no funny business with conjugation or a benzene ring or anything.

Previous answer said:

"From Beer's Law, we know that A= εcl

So the more it absorbs, the higher the ε."

So far so good.  After that it's all wrong, though.

The implication seems to be that because an isolated C=C tends to absorb below 200nm, that it doesn't absorb at all, and so has an ε of zero.  Which is bollocks.  You might not be able to measure the transition with the spectrometer sitting in your typical undergraduate lab, but it's there.   There's a range of extinction coefficients for the π-->π* of a double bond depending on the substituents around it, and although most people quote ethylene as an example, it's a bad one, because its ε is abnormally high.

H2C=CH2 lambda ~165 nm, ε ~ 15000 M-1cm-1

3-hexene lambda ~185 nm, ε ~ 8000

C=O is trickier, because there are two bands.  The previous poster has noted the n-->π*, but a C=O also has a π-->π* just like a C=C does, albeit at slightly lower energy.  The n-->π* comes around 270 to 300nm, the π-->π* around 170 to 200.

As it happens, the π-->π* of a C=O is comparable to that of a C=C.  I'd have to compare two specific compounds to be definitive.  But as examples:

formaldehyde 175 nm, ε ~ 18000

acetaldehyde 182 nm, ε ~ 10000

acetone 195 nm, ε ~ 9000

This is complicated by the fact that these are gas phase numbers, while the hexene value above was in solution.  You get different values, and those differences can be dramatic.  Bottom line: the molar absorptivities of the π-->π* transition of C=C and C=O both span a range depending on compound and conditions, and those ranges overlap.

The previous poster's conclusion, that the n-->π* is the most intense, is simply wrong.  The C=O n-->π* bands have ε around 10 to 20, about 1000x *less intense* than the π-->π* .

So, first part of your question: depends on the compounds and conditions, actually, because C=C and C=O have an overlapping range of ε values for the π-->π* transitions.  The C=O has a second n-->π* at lower energy and far weaker intensity.

As to the last part of your question, which I guess we can now formulate as: why is n-->π* weaker in intensity than π-->π* ...? Well, I guess there are lousy handwavy rationales involving orthogonality of ground state vs excited state orbitals or something, but to be honest if the answer doesn't involve quantum mechanics and hardcore overlap integrals and the underlying physical basis for selection rules, it's not really an answer.  Unless you've done your third year quantum course, I'd just go with "empirically, n-->π* is weaker than π-->π* " and leave it there.