# Why is this step exothermic?

From Wade's "Organic Chemistry" [1, p. 1003-1004]:

MECHANISM 21-7 Conversion of an Ester to an Amide (Ammonolysis of an Ester)

Step 2, loss of a poor leaving group is justified by claiming the step is exothermic, hence the TS energy isn't very high (as would be the case w/ $\mathrm{S_N2}$ since it's neither exo or endo)

Why is this step exothermic though? No neutralization of charge and no net additional bonds were formed (1 pi bond formed, 1 sigma bond broken)

### Reference

1. Wade, L. G. Organic Chemistry; Pearson Education, 2012,. ISBN 978-0-321-81139-4.
• The carbonyl pi bond is reformed, compensating the energy used in breaking the bond between the carbon and the -OR leaving group. Thus, it is overall exothermic – Tan Yong Boon Dec 31 '17 at 8:37
• It's just wrong, proton transfer should be written before leaving. – Mithoron Dec 31 '17 at 17:52