Question

A molecule (" P ") on treatment with acid undergoes rearrangement and gives ("Q"). ("Q") on ozonolysis followed by reflux under alkaline condition gives (" R "). The structure of (" R ") is given below.

The structure of (" P ") is

Answer 1

P = 1,2–dihydroxy–2–methyl–5–methylbicyclo[3.2.0]heptane.

Answer 2

We will show that one acceptable answer is to “reverse‐engineer” the rearrangement–ozonolysis sequence. (Any answer equivalent to the answer below is correct.) In what follows we describe a very schematic solution aimed at the JEE/NEET level.

The idea of the reaction sequence

1. Under acidic conditions a vicinal diol (which we call “P”) undergoes a pinacol rearrangement. In such rearrangements one of the –OH groups is protonated and –OH2⁺ leaves to give a carbocation; then a neighbouring alkyl (or aryl) group migrates and the resulting cation is trapped by water to give a keto‐alcohol (or after tautomerization, a ketone). In our problem the acid treatment converts P into a rearranged molecule (“Q”) in which one of the two originally –OH–substituted carbons becomes a carbonyl.

2. Next the molecule “Q” is subjected to ozonolysis followed by alkaline (reflux) work–up. Ozonolysis cleaves a C=C bond so that after further reaction the fragments recombine (often via an aldol–like process) to give a bicyclic ketone “R”. The structure of R (shown in the question) has a five–membered ring bearing a C=O group; a methyl group is attached on the carbon next to the carbonyl in that ring and a second methyl group is present on the four–membered ring that is fused to the five–membered ring. (A C-atom count shows that the R–system is a bicyclo[3.2.0]heptanone derivative.)

3. Working “back–wards” one sees that the positions of the –CH₃ groups in R force the conclusion that in the starting diol (P) one of the groups attached to a –OH bearing carbon was a –CH₃ which subsequently migrated (this group must have been the one that “moves” in the pinacol rearrangement). Also, the –CH₃ that ends up on the four–membered ring is not altered by the acid treatment.

A possible answer

One acceptable answer is to show that

P is the bicyclic vicinal diol

1,2-dihydroxy-2-methyl-5-methylbicyclo[3.2.0]heptane

i.e. a bicyclic compound having a seven–carbon skeleton which may be “drawn” as two fused rings (a five–membered and a four–membered ring sharing two adjacent carbons) in which the two –OH groups are located on the two ring–fusing carbons. One of these two carbons bears a –CH₃ group; this is the group that migrates during the acid–catalysed pinacol rearrangement so that in the product Q (and finally in R) the –CH₃ appears on the five–membered ring (adjacent to the carbonyl). The other –CH₃ group is present on one of the carbons of the four–membered ring and remains unaltered.

A representative (skeletal) structure is shown below.

Answer (structure of P):

An acceptable answer is:

P = 1,2–dihydroxy–2–methyl–5–methylbicyclo[3.2.0]heptane

This may be drawn as

       HO        │      C*–––C(–CH₃)      /
(–CH₃)C C–H      \ /      C––––––C        │        OH

where the two rings (a five–membered and a four–membered ring sharing two adjacent carbons) are obviously present. (Any clear drawing which shows the diol at the ring–junction with the appropriate –CH₃ substituents is correct.)

Summary (minimal explanation):

1. P is a bicyclic diol that on acid treatment undergoes a pinacol rearrangement (with migration of the CH₃ group on the diol carbon) to give Q.
2. Ozonolysis of Q (cleaving a double bond) under subsequent alkaline conditions gives the bicyclic ketone R whose structure shows a five–membered ring with a carbonyl and a CH₃ at the α–position and a CH₃ on the four–membered ring.
3. Thus one acceptable answer is to identify P as 1,2–dihydroxy–2–methyl–5–methylbicyclo[3.2.0]heptane.

Final Answer:

P = 1,2–dihydroxy–2–methyl–5–methylbicyclo[3.2.0]heptane.

Any answer which shows a bicyclic diol having the two –OH groups at the ring–junction with a –CH₃ on the shifting carbon (leading to the CH₃ next to the C=O in R) and the other –CH₃ on the four–membered ring is correct.