[Altered State]

Should I try to eliminate the OH from the reactant, creating an alkene?

I don't think so.

I was suggesting something like this:
http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch09/cc-rxmech.gif

Being your alkyne acetylene and your haloalkane a derivative of your starting material.

[rcrandall]

I'm sorry, but I'm still not understanding. Is the oxygen moving carbons, or is that a different oxygen?

[Altered State]

I'm sorry, but I'm still not understanding. Is the oxygen moving carbons, or is that a different oxygen?

Ok, it seems like you are really struggling but you are actually trying, this is what I ment on my first suggestion, probably not covered in your course:




And this was my second route, more steps, but you probably saw all of the reactions (I hope):

[rcrandall]

Oh my goodness, I always forget about how much you can do if you create an alkyne. Thank you!
BTW, how did you make that? Is there a website that helps in drawing these reactions out?

[Altered State]

Oh my goodness, I always forget about how much you can do if you create an alkyne. Thank you!
BTW, how did you make that? Is there a website that helps in drawing these reactions out?

I used a software called ChemDraw. It is not free, but probably your university can provide you with a license.
I believe that there are a few free alternatives, like ChemSketch, although I never tried them.

[kriggy]

Is there a particular reason you reduced the tripple bond? I think better idea is to do the water addition to that tripple bond so the keto-enol tautomerism creates the product in 2 less steps.

chem sketch is fine, I use it a lot  because Im not comfortable with chemdraw.

[Altered State]

Is there a particular reason you reduced the tripple bond? I think better idea is to do the water addition to that tripple bond so the keto-enol tautomerism creates the product in 2 less steps.

chem sketch is fine, I use it a lot  because Im not comfortable with chemdraw.

You are right, but if she/he was not been taught carbonyl reactivity, he/she probably don't know anything about keto-enol tautomerism.

[Babcock_Hall]

One often teaches the keto-enol equilibrium as part of hydration of alkynes.  @OP:  If you hydrate an alkyne to make a methyl ketone, is this Markovnikov or non-Markovnikov addition?  How does this affect your choice of which reagent to use?

[rcrandall]

We covered keto-enol tautomerisms! Did I draw it out correctly?

[Altered State]

We covered keto-enol tautomerisms! Did I draw it out correctly?

Seems good to me.

[rcrandall]

Yay! Thank you so much, that one had a lot of steps.

For the third synthesis, do you think that these steps will work?

[kriggy]

Yea this will work no problem. YOu can also use grignard reagent

[rcrandall]

Great! Thank you!

I'm temporarily skipping problem number 4...
For problem 5, do you think this will work?

[rcrandall]

*For the second to last step, is there a difference between using B(sia)2H vs using BH3? I know I want the product to be antimarkovnikov, and using Boron will get me there...my reagent sheet that lists all of these includes both of those Boron reagents (used with peroxide and NaOH)

[Altered State]

If your professor wants you to make use of radical halogenation of alkanes, it is probably fine. However, keep in mind that that kind of reaction is highly non-selective, and in real life you will end up with a mixture of polihalogenated compounds in different proportions.

Anyway, you can use many kinds of boron reagents to perform a hydroboration which will give you the antimarkovnikov alcohol. BH3 is the simplest one, and it is not very selective. HB(sia)2 or 9-BBN, for example, are big molecules, which can selectively introduce the alcohol group in the less hindered position. This is extremely usefull when it comes to perform a selective hydroboration of a molecule which contains more than one alkene group, or only one but with a less hindered possition, even when they are equally substituted on both sides of the olefin.

In your case, just BH3 would propably yield your product in high proportion, but I suppose it would be even higher if you use a bulkier boron reagent.