Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: RCL circuit on November 24, 2008, 08:43:07 PM
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I have been working on this synthesis problem for two weeks but keep running into dead ends. Any help at all would be appreciated.
Synthesize the following molecule using organic compounds with one or two carbon atoms only (not more) and any inorganic compounds. Must be major products!
CH3
l
C2H5--C--CH2Cl
l
OH
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First things, change the OH group. What ways do you know of making a 3° alcohol? You can have any two pieces that you want to react and make that OH group. Post that structure, and we'll talk from there.
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I am in Organic I right now so this might be a little frustrating for you. Im sorry!
I would use oxymercuration to yield the markinov product on an alkene. but I also considered using an epoxide in H3O+ to open it up to yield a 1,2 diol than change one of the OH groups by SOCl2 and maybe use an alkyl substitution rxn. Im sorry if I didnt really understand the question. I appreciate the help you are offering.
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Hmm...ok, fair enough, those are legitimate answers. Have you learned about the grignard reaction?
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We have studied them a little bit are you suggesting using MgX on a carbonyl compound?
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only 1 to 2 carbon containing compounds can be used at a time Im not sure how this would effect the carbonyl compound
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Correct. So, change the alcohol into a carbonyl, and then you need to break off one of the groups. Can you see one of the groups that you could break off and form a symmetric ketone?
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Heres my try.. I'm sure I've made errors.. just let me know
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Good question. It stumped me for a while, too.
Macman, b/c of the primary chloride, I don't think Grignard as the last step will work. I think halohydrin formation is a more plausible last step.
RCL, I got there in 6 steps starting from acetylene.
refid, I don't see a scheme for your post.
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azmanam, can you tell where I went wrong in my reaction?
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I don't think your enamine chlorination will be selective, and you might see unwanted byproducts when you subject the choride to acidic conditions in your final step. Creative, though.
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azmanam, what if you start with the alpha-chloride? The Cl shouldn't interfere with a grignard if you start with ClCH2CHO. Add, MeMgBr, oxidize, add EtMgBr, no?
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Sure, the carbonyl carbon is electrophilic, probably the most electrophilic, but the chloride-bearing carbon is also electrophilic. I think you might see competition with SN2 substitution.
(edit: although Scifinder says this JOC paper from 1981 says you can react allyl Grignard with chloroacetaldehyde to get the halohydrin in 80% yield, so your way might work as well: http://dx.doi.org/10.1021/jo00318a019)
4,5-Epoxy-1-pentene. Allylmagnesium bromide was reacted with chloroacetaldehyde, and the resulting chlorohydrin, bp 65-70 "C (16-17 mm) (lit.27b p 163 "C), was obtained in 80% yield. Cyclization with NaOH gave 4,5-epoxy-1-pentene, bp 92-94 "C (lit.27b p 93.5 "C), in 71% yield (99% purity).
I did acetylene -> propyne -> propene -> methyloxirane --(H+, EtMgBr)--> 2-methyl-1-butanol --(SOCl2)--> 1-chloro-2-methylbutane --(NaOEt, EtOH, heat)--> 2-methyl-1-butene --(Cl2, H2O)--> pdt.
Your way's shorter.
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What about addition of ClOH across a double bond?
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What about addition of ClOH across a double bond?
2-methyl-1-butene --(Cl2, H2O)--> pdt.
that's what this last step does. Cl2 and H2O make ClOH in solution.
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that's what this last step does. Cl2 and H2O make ClOH in solution.
Oh, wonderful, I didn't realize that.
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methyloxirane --(H+, EtMgBr)--> 2-methyl-1-butanol
hmm... That doesn't look right, now does it...
I reworked it. It's attached, but very clunky...
I think the Grignard into the carbonyl is probably the way to go.
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*chalks one up for himself* ;)
*brush dirt off shoulder*
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azmanam, your molecule has a chlorine, wouldn't the Grignard step (Mg(s) + R-Cl) also would react?
And how can you ensure that the Grignard reactions doesn't happen twice (i.e. aldehyde into tert. alcohol)?
Anyways, the Non-Grignard way looks correct. :)
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azmanam, your molecule has a chlorine, wouldn't the Grignard step (Mg(s) + R-Cl) also would react?
You aren't doing Mg(s) + R-Cl, you're doing Mg(s) + R1-Br, and adding that to R(C=O)-Cl.And how can you ensure that the Grignard reactions doesn't happen twice (i.e. aldehyde into tert. alcohol)?
You are reacting an aldehyde to make a ketone. Ketones are less reactive than aldehydes by a degree large enough that you can be pretty sure the aldehyde will always react first. Also, you can help control it by dripping the grignard into a large pot of aldehyde, thus ensuring that the aldehyde's are in excess and that there isn't lots of grignard sitting around waiting to react.
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macman104, I'm just wondering how would I prevent the remainder of the Mg(s) coming in contact R-Cl forming R-MgCl.
And instead of Grignard, is it possible that we use Organozinc (Reformatsky) to react with the aldehyde once?
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I was actually wondering about that. But also, like I said, you would drip the grignard into the aldehyde, and minimize such things. The reformatsky is a alpha-haloester reaction, I think it'd be more work than it's worth.
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And how can you ensure that the Grignard reactions doesn't happen twice (i.e. aldehyde into tert. alcohol)?
You are reacting an aldehyde to make a ketone. Ketones are less reactive than aldehydes
It's actually simpler than that. The Grignard attacks the carbonyl carbon, and the electrons in the C=O double bond kick up onto oxygen to create an O-. For the Grignard to react again, the electrons on O- need to collapse back down and kick out a good leaving group. The choices are R-, R-, or H-. None of these are good leaving groups. The reaction stops at the secondary alcohol. You need a separate re-oxidation to the ketone before you can make the tertiary alcohol you desire.
This is why adding Grignard to aldehydes is different than adding Grignard to esters. When the O- attempts to collapse back down, it still needs to kick out a good leaving group, and now its choices are R-, R-, and OR-. OR- is now an ok leaving group. Now we've generated a ketone in situ and we will get double addition to form the tertiary alcohol.
how would I prevent the remainder of the Mg(s) coming in contact R-Cl forming R-MgCl.
Well, simple Grignards can be bought without excess Mg(s). If you do make the Grignard, they are made in a separate flask at elevated temperatures (often with I2 as a catalyst). The insertion of Mg(s) into R-X requires a bit of heat and needs a catalyst to go. It doesn't just happen spontaneously. This allows you to control the amount of Mg and R-X.
Additionally, the actual Grignard reaction (where the RMgX is added to the electrophile) is usually done at low temperatures (below 0 oC). This means that even if there is unreacted Mg(s), it will not insert into any other R-X bonds.
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Also, I thought about it a bit more this morning, and managed to shave a few steps off my route. It's now 7 steps instead of 10... No one's going to comment on the alcohol -> bromide transformation where I used the often-undesired carbocation rearrangement to my benefit? I thought that was a pretty awesome step. Let this be a lesson that there is often more than one way to make a molecule.
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azmanam, thank you for your informative post. You've answered my questions with detail and I learned something new.
+1
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im just starting organic this year. i havent taken the calss yet but i have gone throuhg the text book to see how well i could do on my own and i wanted to try something more challenging than what was in the book. i could be way wrong but i would appreciate any feed back.
basicly my solution was alkylation of an acetylide anion, followed by addition of HBr. then that product would be added to another acetylide anion. then halohydration to produce the alcohol. and finally a catalytic reduction of the alkyne.
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Considering you haven't taken the class yet, not bad (but not correct either). No SN2 on sp2 carbons and questionable addition of Cl2/H2O. Again, I understand your thinking though.
With the other posts, here is another. It uses different reactions. Organic synthesis problems are always difficult to guess what reactions a student knows.
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thanks for the feedback. i didnt even consider the SN2 problem. and is the addition of Cl2 and H20 only questionalbe because it would attack the weaker triple bond, or is there something else im missing?
and again thanks for the help
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The reaction of an ene-yne is difficult to predict (without looking some precedent). If the same reaction were performed on a diene, then 1,2 and 1,4 addition products are frequently found with 1,2 addition the kinetic products and 1,4 the thermodynamic products. However, I am uncertain how that reaction changes in the presence of water or an acetylene group. I would expect the electrons of a double or triple bond to participate in the reaction whether they alter the product distribution or not.
All in all, your scheme isn't bad considering you haven't taken the class yet. There are other ways to synthesize this compound even along the lines you have chosen.
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MwAhA i know the answer you fools.
this problem was so f&#^$*@ easy, stop acting like you know your s#*$, your answers are all wrong.
I have created this molecule out of naught but gringards reagents and carbonyl groups.
give me something thats actually a challenge mortal.
if you want the answer, youll have to beat me in a sythesis battle...try it organic bitches
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all your answers are seriously wrong. your reagents have more than 2 carbons. that PPh3 have 3 phenyl rings, thats 36 carbons. duh lolz. and the gilman reagents has the same problem fools.
try again
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MwAhA i know the answer you fools.
this problem was so f&#^$*@ easy, stop acting like you know your s#*$, your answers are all wrong.
Ban.
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I have thought about this synthesis for a long time, and at last designed a route as below. I want to use 1eq two Grignard agent to react with the substrate in tandem. Can the synthetic route be applicable? Thank you!!