Could you please rephrase your question. In theory you can prepare easily acetic-hexanoic anhydride. It most probably will behave like Ac2O (i.e., acetylate alcohols).

Thank you for the quick answer. I was interested if the behavior of mixed anhydrides was random (e.g. both acetic and hexanoic esters are formed) or if it is dependent on any properties of parent acid (e.g. transesterification occurs with strong acids vs weak acids(or any other mechanism) yielding only one product).

The reaction you mention is random in the sense that molecules move about under the influence of Brownian motion and sweep through an infinite number of energy landscapes over the course of reaction. The outcome of the reaction will most definitely reflect the weight averaged contributions of these random motions. In particular, the major pathway chosen (meaning major product) will be the one, whose energy landscape depicts the lowest energy barrier. In the case of nucleophilic acyl substitution, one can imagine that the competition between acetylation and benzoylation will depend on the weight-averaged contributions of energy terms relating to electrophilicity of the carbonyl carbon, which functions to attract a nucleophile, and steric bulk, which functions to impede approach of the nucleophile. Even for this single reaction mechanism, one can imagine a continuum of different landscapes and energy barriers, with the optimum landscape reflecting the best orientation of the approaching nucleophile-electrophile pair. In the case of the benzoyl side, the carbonyl carbon is less accessible than that of the acetyl carbon due to the steric bulk of the aromatic group. Also, benzoyl is likely less electropositive because ring electrons can partially negate the positive charge of the carbonyl carbon through resonance and induction of electron clouds coming from out of the ring. Comparatively speaking, the acetyl side only experiences partial negation through hyperconjugation of the neighboring methyl group, which is a small effect. So in terms of accessibility (steric), and charge-to-charge attraction between electrophilic carbon and a nucleophile (electronic), the acetyl side is stereo-electronically more open to attack and in my opinion, it should be highly, if not exclusively transformed by a nucleophile compared to the benzoyl side. Please note that your argument about comparing acid strength of the different leaving groups also comes into play. However, I prefer to discuss acid strength when comparing ground states to ground states, and put more generally, I prefer to predict the outcome of reactions on the basis of the initial and final ground state energy levels ONLY when the forward activation barrier involves energy changes on the order of kT (i.e., the thermal energy of the system). When comparing ground state to activation state (i.e., situations where Ea >> kT), I find the above discussion more productive. Please note that any contribution of structures, which renders a leaving group more or less acidic will also influence the stereo-electronic character of the transition state species by directly influencing electronic terms via electron donation, induction and/or hyperconjugation. Hence, we are not contradicting one another; we are merely discussing the same influential factors using a slightly different language. Lastly, please note that all of the above discussion emphasizes the activation enthalpy. I completely excluded any discussion relating to the contribution of the activation entropy. In chosing to ignore delta S double dagger, I made the sweeping assumption that as nucleophile approaches electrophilic carbon, two free molecules become one rigid system; more importantly, I made the assumption (a reasonable one) that the relative loss of freedom in comparing the approach of Nu to either carbonyl carbon is about the same. Hence, the activation entropies for attack of acetyl or benzoyl carbonyl carbon were deemed equivalent, and I focused my discussion on comparing the enthalpic terms to ascertain which pathway would be favored. Wishing you all the best, AT

Hi, I saw this in Apay's link. The answer to your question is that it depends on the reaction mode. If it's a standard Lewis acid catalysed reaction then the most stable Lewis acid base complex is formed (even although its a kinetic process of course). If you read through George Olah's work you'll find your answer and many others. He was one on my chemistry idols when I was working on acylations!