Hydroboration of asymmetric internal alkynes and alkenes?

[fangrz]

How can you predict which product will be in excess if you have asymmetric hydroboration of internal alkynes and alkenes? For example, if I used 9-BBN or something sterically-hindered like R2BH...

 

[TeethWhitener]

For a bulky borane, the excess product will be the one where addition of the boryl moiety is least sterically hindered. (Again, I haven't done these reactions myself, so I'm not a specialist) For example, if you have 2-methyl-2-butene, the 3 position is less sterically hindered than the 2 position. Therefore the BR₂ moiety will add to the 3 position and the H (much smaller in size compared to the BR₂) will add to the more sterically hindered 2 position. If you're following the hydroboration with a standard oxidation, the major product will ultimately be 3-methyl-2-butanol. Since hydroboration is a syn addition (both substituents add to the same side of the alkene), and the transition state is a 4-membered ring, the addition of the bulky BR₂ group to an already sterically crowded position on an alkene is highly energetically unfavorable. This explains the anti-Markovnikov character of the addition.

 

[fangrz]

For a bulky borane, the excess product will be the one where addition of the boryl moiety is least sterically hindered. (Again, I haven't done these reactions myself, so I'm not a specialist) For example, if you have 2-methyl-2-butene, the 3 position is less sterically hindered than the 2 position. Therefore the BR₂ moiety will add to the 3 position and the H (much smaller in size compared to the BR₂) will add to the more sterically hindered 2 position. If you're following the hydroboration with a standard oxidation, the major product will ultimately be 3-methyl-2-butanol. Since hydroboration is a syn addition (both substituents add to the same side of the alkene), and the transition state is a 4-membered ring, the addition of the bulky BR₂ group to an already sterically crowded position on an alkene is highly energetically unfavorable. This explains the anti-Markovnikov character of the addition.

Thanks! But what if you have, say, 2-pentene? Which side is more sterically-hindered (the one with more carbons?)? And what about if you had 4-methyl-2-pentene? Does the methyl group that's farther away have any effect?

 

[TeethWhitener]

Which side is more sterically-hindered (the one with more carbons?)

My gut says probably. Remember that the ethyl group in 2-pentene can swing 360 degrees around its bond between the 3 and 4 positions, a motion which sweeps out a lot of volume and makes it harder for bulky substituents to attack at the 3 position. Keep in mind, though, that the more similar the groups are, the less difference I'd expect in yield of major vs. minor product. So for something like 3-heptene, where one side has an ethyl group and the other has a propyl group, you probably wouldn't see too much of a difference between major and minor products.

And what about if you had 4-methyl-2-pentene? Does the methyl group that's farther away have any effect?

The placement of the methyl group in this case ends up making the group pretty bulky, so I imagine the effect would be pretty pronounced here. In (for example) something like 10-methyl-2-dodecene, the extra methyl group probably wouldn't have much of an effect. Also, I should point out that there are electronic effects in hydroboration, but this is more often seen with groups that can push electrons around a little more (electronegative species, aryl groups, etc.). The effects from hydrocarbon groups on stereochemistry tend to be much more easily rationalized in terms of steric considerations.

 

[fangrz]

My gut says probably. Remember that the ethyl group in 2-pentene can swing 360 degrees around its bond between the 3 and 4 positions, a motion which sweeps out a lot of volume and makes it harder for bulky substituents to attack at the 3 position. Keep in mind, though, that the more similar the groups are, the less difference I'd expect in yield of major vs. minor product. So for something like 3-heptene, where one side has an ethyl group and the other has a propyl group, you probably wouldn't see too much of a difference between major and minor products.The placement of the methyl group in this case ends up making the group pretty bulky, so I imagine the effect would be pretty pronounced here. In (for example) something like 10-methyl-2-dodecene, the extra methyl group probably wouldn't have much of an effect. Also, I should point out that there are electronic effects in hydroboration, but this is more often seen with groups that can push electrons around a little more (electronegative species, aryl groups, etc.). The effects from hydrocarbon groups on stereochemistry tend to be much more easily rationalized in terms of steric considerations.

Thank you!