[Cooper]

Hi,

I've attached the problem below (scroll down for it, 2nd post). Here's what I think (I've added a pic. of my synthesis),

In retrosynthesis, before adding sulfuric acid in water, you would have had an alkene. The double bond would have been terminal because after it plucked the proton from sulfuric acid, there would have been a carboncation on carbon-2, and then a hydride shift to put the carboncation on carbon-3.

Then before adding H2 in Lindlar's catalyst, you would have had hexyne. And before that (on the "?" stage) you would be adding 1-bromobutane to an ethyne ion with a lone pair.

The other possibilities are that after the first retrosynthetic step, there would be a carbocation on the 2- or 3-carbon, but this would give you more than one answer, because it would proceed in both directions.

But the answers say that the double bond was on the 2-carbon, why?

Thanks!
~Cooper

EDIT: The triple bonded carbons should have linear confirmation, sorry

[Cooper]

Here is the Q:

~Cooper

[discodermolide]

Well if you use mixture f you get hex-3-yne as the product which gives (Z)-hex-3-ene upon Lindlar hydrogenation. This alkene is symmetrical. Acid catalysed hydration gives hexan-3-ol.

[Cooper]

Well if you use mixture f you get hex-3-yne as the product which gives (Z)-hex-3-ene upon Lindlar hydrogenation. This alkene is symmetrical. Acid catalysed hydration gives hexan-3-ol.

Ah, I see. And I didn't realize that it was symmetrical so you get hexan-3-ol no matter where the OH goes. But you also get that with D. Is F the "better" choice, because there is no terminal alkene involved - F's alkene is more stable?

Thanks!
~Cooper

[Dan]

I think you have made a mistake in your retrosynthesis. How do you envisage formation of hexan-3-ol from hex-1-ene (the double bond is between C1 and C2)?

[Cooper]

I think you have made a mistake in your retrosynthesis. How do you envisage formation of hexan-3-ol from hex-1-ene (the double bond is between C1 and C2)?

Well when the alkene takes a proton from sulfuric acid, a carbocation will appear on the 2-carbon. I was thinking that a hydride would shift from the 3-carbon to the 2-carbon and give you a carbocation on the 3-carbon, which would lead to the alcohol being there.

Would the shift not happen because the carbocations are both secondary?

[Dan]

Would the shift not happen because the carbocations are both secondary?

Exactly. You need a thermodynamic driving force for the hydride shift (such as sec/pri carbocation tert carbocation). As there is a relatively small difference in stability between the two secondary carbocations you propose, I very much doubt it would happen - and certainly not to a synthetically useful extent.

[Cooper]

Thanks