Indian Forest Service (Mains), Chemistry optional · 2023 · Paper II

Q1(g). A double oxy-Cope rearrangement

Question

Identify A and B when the drawn vicinal dienediol is heated to 190 °C. [structure of the dienediol given in the paper]

A is the rearranged dienol MeC(OH)=CHCH₂CH₂CH=C(OH)Me; tautomerization gives B, octane-2,7-dione.

The original six-atom array is C1=C2–C3(OH)(Me)–C4(OH)(Me)–C5=C6.
A thermal six-electron [3,3] rearrangement breaks C3–C4, forms C1–C6 and shifts the two pi bonds to C2=C3 and C4=C5.
The two OH-bearing carbons are now enolic carbon atoms. Each terminus tautomerizes to a methyl ketone.
The final connectivity is CH₃CO(CH₂)₄COCH₃. Strong carbonyl formation makes the oxy-Cope sequence favourable; no carbon is lost.

The full model answer

As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.

Coversoxy-Cope rearrangement[3,3] sigmatropic shiftketo–enol tautomerismoctane-2,7-dione1,5-diene