1
$\begingroup$

I am to turn 1-octyne to 1-hexene.

Method 1

enter image description here

1) Reduction using sodium metal in pure ammonia to an alkene.

2) Halohydrin formation; the -OH installs at the more substituted carbon due to its greater ability to stabilize positive charge (this always confused me; the more substituted carbon is both more positive in character and has more electron-donating methyl groups?)

3) Conversion of OH to a good leaving group (-OTs) using a sulfonate ester and a weak base, pyridine.

4) E2 elimination using NaOH and heat. Because of resonance and sheer bulk and negative induction the OTs(-) should make for a better leaving group than Br-. So the internal alkene is formed.

5) Two-part one step ozonolysis; creation of aldehyde.

6) Reduction of carbonyl with hydrogen and transition metal to terminal alcohol.

7) Conversion of terminal alcohol to a better leaving group using tosyl chloride and pyridine again.

8) Use of LDA to deprotonate and create the terminal alkene.

Method 2

Also we can take another route with the brominated alkene; we can use OsO4 and basic hydrogen peroxide to create a 1,2 (vic) diol, which can be cleaved by periodic acid, and this creates a primary alcohol, which we convert to a good leaving group using tosyl chloride and pyridine, and we can eliminate. Same number of steps, but just something different that I thought of. Well, perhaps slightly simpler since there isn't a need for a two-step process (ozonolysis).

enter image description here

Improve this question
$\endgroup$
11
  • 1
    $\begingroup$ "Oxidative cleavage of alkynes proceeds via cycloaddition to metal oxides. Most famously, potassium permanganate converts alkynes to a pair of carboxylic acids." | en.wikipedia.org/wiki/Hofmann_rearrangement | en.wikipedia.org/wiki/Hofmann_elimination $\endgroup$
    – permeakra
    Sep 5, 2014 at 5:25
  • $\begingroup$ So I have to make an amine first @permeakra? Seems like a good route but we I'm only employing methods I know already. I will however keep what you mentioned in mind for future stuff. $\endgroup$
    – Dissenter
    Sep 5, 2014 at 5:28
  • $\begingroup$ Is oxidative cleavage of alkynes and carboxilic acid reductions already known? $\endgroup$
    – permeakra
    Sep 5, 2014 at 5:30
  • $\begingroup$ Yes to oxidative cleavage; no to COOH reductions. $\endgroup$
    – Dissenter
    Sep 5, 2014 at 5:48
  • 1
    $\begingroup$ No it did not. Reread your writing. $\endgroup$
    – permeakra
    Sep 5, 2014 at 5:58

3 Answers 3

Reset to default
5
$\begingroup$

In principle, this can be realized in two steps:

  1. Oxidative cleavage using catalytic amounts of $\ce{RuO2}$ and excess of oxone converts 1-octyne to heptanoic acid. (DOI)

  2. Heating a mixture of acetic anhydride and a carboxylic acid in the presence of $\ce{Pd}$ catalyst turns the acid to a terminal alkene lacking one carbon atom (DOI). Admittedly, the high temperatures needed might be a problem here and may force you to perform the reaction in a sealed tube in an autoclave.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ +1, interesting techniques and nice answer. I was thinking along the same lines: oxidative cleavage of the alkyne to give heptanoic acid (with, e.g., acidified potassium permanganate) followed by decarboxylation. I had some version of the Hunsdiecker reaction in mind for the latter step, followed by elimination of the halogen. $\endgroup$
    – Greg E.
    Sep 5, 2014 at 8:01
5
$\begingroup$

Your second step will not work as written. Reaction of an alkene with aqueous HBr will produce the bromide (not the bromohydrin) with Markovnikov regioselectivity and the possibility for carbocation rearrangement. To get the bromohydrin, se $\ce{Br2/H2O}$

Your fourth step will be challenging also. There are two leaving groups, and LDA is a very strong base. There are regioselectivity and chemoselectivity problems, and unless the reaction is controlled well, you will also get 1-octyne.

Better to install just one functional group in step 2 (oxymercuriation, while old-fashioned, gives you just 2-octanol with little to no rearrangement), and then your elimination in step 4 will behave better.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Why if I created a fluorohydrin? $\endgroup$
    – Dissenter
    Sep 5, 2014 at 12:33
2
$\begingroup$

This might be a shorter route that employs common synthetic techniques you're probably already familiar with.

enter image description here

The route begins by forming the acetylide anion, followed by alkylation with pentyl bromide. Oxidation of the resultant acetylene will produce the two possible carboxylic acids. These two acids should be readily separable by distillation (C6 b.p. 206C; C7 b.p. 223C). Further reaction (reduction, tosylation, elimination) of the hexanoic acid fraction would produce 1-hexene.

Non-Subverting Route

enter image description here

I think you've studied radical allylic oxidation (maybe it was radical allylic bromination that you posted about? Well, same idea.). In any case, it is a well-known reaction (here are some examples). You might get a mixture of the propargylic alcohol and ketone - no matter, carry them both along to the next oxidation step. In this second step we oxidize the triple bond and any remaining propargylic alcohol to the alpha-ketoacid, which will decarboxylate (see the first few sentences in this reference) under the reaction conditions. This yields hexanoic acid which can then be converted to the desired olefin.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I think this is a good approach, but it subverts the spirit of the question, as you throw away most of the carbons from the starting material. $\endgroup$
    – jerepierre
    Sep 5, 2014 at 13:59
  • $\begingroup$ Regarding your comment, it's perfectly fine, it doesn't upset me, but I'd like to share my thoughts. There's a fine line between subverting the spirit of a question and thinking outside the box. I used the chemical properties of A to get to B. If a student ever correctly answered one of my exam questions in a way I hadn't anticipated, the student would receive full credit. I never thought of it as subverting my intent. Either I hadn't written the question in a definitive enough manner, or (more likely) the student was smarter than me. In any case, I appended a "non-subverting" route :) $\endgroup$
    – ron
    Sep 6, 2014 at 0:17
  • $\begingroup$ I totally see where you are coming from, and if I was in the position as the grader, I would have to view it the way that you do. I only pointed that out because (1) it's a practical limitation and (2) I could see other instructors either wording the question in such a way to avoid this method (hopefully) or not receiving the answer in the same way you do. Thanks for adding the alternative route. Another thoughtful answer, as I've come to expect from you. $\endgroup$
    – jerepierre
    Sep 6, 2014 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.