Consider acetaldehyde (ethanal for progressive nomenclaturists). What conformation does it adopt, and why? This question was posed of me by a student at the end of a recent lecture of mine. Surely, an easy answer to give? Read on … There really are only two possibilities, the syn and anti . Well, I have discovered it is useful to start with a search of the Cambridge data base.

Although have dealt with the π-complex formed by protonation of PhNHOPh in several posts, there was one aspect that I had not really answered; what is the most appropriate description of its electronic nature?

Michael Dewar[cite]10.1016/S0040-4039(01)82765-9[/cite] famously implicated a so-called π-complex in the benzidine rearrangement, back in the days when quantum mechanical calculations could not yet provide a quantitatively accurate reality check. Because this π-complex actually remains a relatively unusual species to encounter in day-to-day chemistry, I thought I would try to show in a simple way how it forms.

If you search e.g. Scifinder for N,O-diphenyl hydroxylamine (RN 24928-98-1) there is just one literature citation, to a 1962 patent. Nothing else;

One frequently has to confront the question: will a hydrogen bond form between a suitable donor (lone pair or π) and an acceptor? One of the factors to be taken into consideration for hydrogen bonds which are part of a cycle is the ring size. Here I explore one way of quantifying the effect for the series below, n=1-5 (4-8 membered rings). I will use the NBO approach.

I return to this reaction one more time. Trying to explain why it is enantioselective for the epoxide product poses peculiar difficulties. Most of the substituents can adopt one of several conformations, and some exploration of this conformational space is needed. Amongst the conformational possibilities are the two rotations shown below.

tpap[cite]10.1055%2Fs-1994-25538[/cite], as it is affectionately known, is a ruthenium-based oxidant of primary alcohols to aldehydes discovered by Griffith and Ley.

I noted briefly in discussing why Birch reduction of benzene gives 1,4-cyclohexadiene (diagram below) that the geometry of the end-stage pentadienyl anion was distorted in the presence of the sodium cation to favour this product. This distortion actually has some pedagogic value, and so I elaborate this here.

Birch reduction of benzene itself results in 1,4-cyclohexadiene rather than the more stable (conjugated) 1,3-cyclohexadiene. Why is this?

I promised that the follow-up to on the topic of Birch reduction would focus on the proton transfer reaction between the radical anion of anisole and a proton source, as part of analysing whether the mechanistic pathway proceeds O or M . To add some context, Hammond’s postulate [cite]10.1021/ja01607a027[/cite] states that “ the structure of a transition state resembles that of the species nearest