Question

Oxidation of the alkene 1 with osmium tetroxide gives a mixture of two isomeric triols.

Oxidation of the alkene 1 with osmium tetroxide gives a mixture of two isomeric triols. Он Он ОН OH OsO4 Ho 'Bu ОН 'Bu 'Bu он НО The coupling constants for the major isomer are: J1,2 3.2 Hz J23 9.3 Hz Assuming that the 'Butyl group occupies an equatorial position on the cyclohexane ring, deduce which triol product is the major isomer. Draw this isomer in a chair conformation showing the equatorial/axial positions of the protons at H-1, H-2 and H-3 and explain your assignment of coupling constants in terms of angular dependence.

Answer 1

Major isomer:

он ОН 2 "Вu НО

Explanation:

он Он Он МОН Bu Он Bu он н н ОН ОН 2 Bul ОН ОН Bu Н ОН 3 н Но н н т

In the chair conformer of the isomer on the right, the H-1 and H-3 are situated at same angle with respect to the H-2. This implies that J_1,2 = J_2,3.

But, given, J_1,2 (3.2 Hz) $\neq$J_2,3 (9.3 Hz).

Hence, the major isomer is the isomer on the left.

H-1 and H-2 are cis-oriented and they are at a dihedral angle of 50^o to 70^o. H-2 and H-3 are trans-oriented and they are at a dihedral angle of about 180^o with respect to each-other. Hence, J_1,2 > J_2,3.