[spirochete]

For example: In a reaction of an acyl chloride with ammonia, why can't the addition of ammonia be concerted with loss of chloride, forming an amide in one step? 

This reaction looks really nice on paper but doesn't happen in reality.  Is it even known why the reaction can't proceed like this? 

[macman104]

For example: In a reaction of an acyl chloride with ammonia, why can't the addition of ammonia be concerted with loss of chloride, forming an amide in one step?

This reaction looks really nice on paper but doesn't happen in reality.  Is it even known why the reaction can't proceed like this?

Did you find somewhere that said this, or were you told this?  If no, why do you think it wouldn't?  It sounded nice to me to, so I googled it to see if I could find an explanation, and I come up with numerous places and organic websites that provide the mechanism for this.  I can't imagine why they would all provide the mechanism, but not a mention of the reaction not occurring...

[kryptoniitti]

Hmmm, here's a quote from Grossman's book that might be on the subject:

C(sp2) electrophiles never, ever, ever undergo substitution by the SN2 mechanism. The two-step mechanism for substitution at carbonyl C is much more reasonable than a one-step, SN2 mechanism for several reasons. The C=O pi bond is higher in energy than the C(sp2)–X bond, so it’s easier to break. Moreover, addition of Nu to C=O can occur along a trajectory that is out of the plane of the carbonyl group, whereas an SN2 trajectory must be in the crowded plane of the carbonyl group. In addition to these theoretical considerations, plenty of experimental evidence suggests that the two-step mechanism is always operative. (See any physical organic chemistry textbook for details.)

[spirochete]

I looked it up some more and found this explanation on some lecture notes:

"The pi* LUMO is lower in energy and more accessible to the homo of the nucleophile"

http://users.ox.ac.uk/~magd1571/Teaching/carbonylA.pdf

Is this just another way of saying that the C=O pi bond is weaker, or is this a different argument?

[kryptoniitti]

I looked it up some more and found this explanation on some lecture notes:

"The pi* LUMO is lower in energy and more accessible to the homo of the nucleophile"

http://users.ox.ac.uk/~magd1571/Teaching/carbonylA.pdf

Is this just another way of saying that the C=O pi bond is weaker, or is this a different argument?

Hello spirochete, sorry about the late answer.
Yes, I believe it is the same argument: when you look at MO electron diagrams, the HOMO for sigma bond should usually be lower in energy than the HOMO for pi bond and it shows in the bond dissociation energies. However, the LUMOs are the opposite. Lower for energy for the pi* LUMO than the sigma* LUMO. Try looking for some diagrams if you are still puzzled.