Rep:Mod:xxyte35132

Part 3: Stereochemistry and Reactivity of an Intermediate in the Synthesis of Taxol

Atropisomerism by virtue of a ring flipping barrier

Initial synthesis of a key intermediate in the total synthesis of Taxol, depending on substituents of the cyclohexane ring, gives either a mixture of isomers, or a single isomer, differing in the absolute stereochemistry of the carbonyl group; the group points either up or down, relative to the rest of the ring. On standing, the system isomerises to give a single isomer. We are going to look at an example with no extra substituents on the cyclohexane ring. The two isomers are shown, 9 & 10.

An MM2 optimisation was carried out on both isomers, subject to the stereochemistry shown, specifically with respect to the position of the cyclohexane ring (down to the 9-membered ring). Several energy minima were found, depending upon mostly the conformation of the cyclohexane ring. The lowest energy conformations from both isomers were found when the cyclohexane ring was in the chair conformation. The isomer 10 was found to be lower in energy by 19.3Kjmol^-1. Looking specifically at the bending contribution to the energy, the isomer 10 is lower in energy by 21.7Kjmol^-1. The other contributions therefore change little between the two isomers. In both isomers, the bending contribution to the total is quite high - the bridging carbon atom is highly strained. As the isomer puckers up, the bond angle between the bridge and the carbonyl group increases, from 118.3^o (already quite strained beyond the ideal 109^o) to 122.5^o. This is largely responsible for the large difference in energy between the isomers. Also contributing a lot to the total energy is the torsional component and the 1,4-Van der Waals contribution. These are about equal for the two isomers. These are necessary due to the size of the system, and the geometrical constrains introduced by the stereochemistry and the carbonyl and the double bond.

Taxol Intermediate 9

Taxol Intermediate 10

In order to isomerise, the 9-membered ring of isomer 9 must twist sufficiently to allow the carbonyl group carbon to invert. This is likely to occur exo- to the ring, because the internal dimensions of the ring would likely not permit the rotation to occur within (a higher energy barrier). This would likely increase strain on the already strained C-α to the carbonyl, and on the bridge. Hence, this would likely be quite a high barrier compared to thermal energy. This explains the observation of slow isomerisation to give the most stable isomer, 10, on standing. Hence, the two isomers are atropisomers, separated due to a high rotational barrier compared to thermal energy. Not too high, otherwise the isomers would be separable and would not isomerise. If the solution was cooled, then the two isomers may be separable.

Slow reaction of the bridge-alkene - Hyperstable olefins

It is noted that the alkene reacts slowly on subsequent functionalisation. Why?

Using MM2, we shall now also model the hydrogenated derivative of the more stable isomer, 10, and compare the bond angles between the two large rings in both molecules. On addition we would expect torsional energy to increase, as the number of eclipsing interactions increases, and this is a general rule.

Here is a list of bond angles as we work our way around the large ring system of the most stable of the two rotamers, anticlockwise as we look on it from above, for the alkene and dihydro-derivative.

Alkene 10 angles /o	Dihydro-derivitive angles /o	Bond
95.4	104.2	Bridge atom, sp3 to sp3
123.3	119.5	Alkene sp2 to sp3
123.9	121.9	Alkene sp2 to sp3
109.3	118.2	sp3 to sp3
116.2	116.0	sp3 to sp3
116.5	115.2	sp3 to sp3
120.2	118.4	Ketone sp2 to sp2
118.1	117.3	sp3 to sp3
115.7	117.6	sp3 to sp3

Overall, both ring systems are strained, due to the geometry fixing bridge and the carbonyl group, resulting in the carbon bond angles inside the ring being generally larger than the desired. Except for the alkene. The bond angles are 123.3 and 123.9^o for the alkene, which is close to optimal. On hydrogenation, one reduces to 119.5^o and the other to 121.9^o These now formally sp³ centers are very strained, as they cannot become tetrahedral. Most other bond angles of the ring change little, except for the two carbons alpha to the double bond. The bridging atom increases its angle towards ideal, so reduces its strain, and the other alpha carbon increases its angle beyond ideal, so becomes more strained. Overall, because the alkene reacts relatively slowly, the total energy of the system must rise on reaction, hence the strain introduced by reaction must outweigh the strain reduced from the bridge carbon on reaction. Also, the torsional energy rises, so this also contributes to the overall relative unreactivity of the alkene.

The dihedral angle across the alkene is 11.2^o, which is slightly twisted. Schleyer reported that in alkenes attached to bicyclic systems such as here, that the twist would raise the energy and hence reactivity of the alkene, because of the less effective overlap as the sp² centres twist out of plane. So at first glance, we would expect the alkene to be more reactive. However, the ring is large enough to allow the alkene to sit almost in plane, so it's energy will not be increased too much, relative to a totally planar double bond, and so the increase in strain on reaction must overcome this factor.

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Index