Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: polly on March 31, 2004, 05:13:47 PM
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Hi! How do you synthesize acetic acid starting with any alkyl halide and acetylene. Would it just be KMnO4/H3O+ ?? Also, when synthesizing a (well I can't name this structure so I will do my best to describe it- it is trans with 2 benzene groups attached by a 2 propene in the middle)- hope that makes sense. The question is how do you snthesize this using any alkyl halide and acetylene? Can you use NaH2,NH3 twice as 2 separate steps and from there use Li/NH3??
Thanks!
Edit: edited title for better indexing. Mitch
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React the acetylene with NaNH2 (or Na) to produce an acetylide anion, alkylate using an methyl halide (chloride), then oxidize completely.
I'm not quite sure what you're referring to in your second question. Perhaps end substituded propene by two benzenes...if so I have an idea what you'll need.
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Thanks for your input. I will try and explain the second material better. There is a double bond in the middle. Coming off both sides of the double bond in a trans manner is an ethyl benzene on each side. The ethyl part is atached to the double bond. My question is can you form this by using acetylene and then adding an ethyl benzene with NaNH2,NH3, THF on two separate occasions and then use Li/NH3 to cut the tripple bond to double and form a trans product?? If not, how would you make it? I hope that is more clear. I wish I could draw it and show you but I have no idea how to do that on the computer.
Thanks
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PS- I goofed on the first email- I meant 2 butene not propene- Sorry!
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Can you post a picture? Just use Paint and save it as a JPeg?
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I hope this works...
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When reacting acetylene with NaNH2/CH2Cl, then oxidize to make acetic acid (CH3CO2), would you end up with too many carbons?
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I don't suppose you have learned the Wettig reaction yet? Is there anything in particular that you have to start with?
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No, I have not learned the wettig reaction yet, but would love to learn it. I need to start both reactions with any alkyl halide (if needed) and acetylene.
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You can do :
CH3Cl + Mg -- CH3MgCl (grinhar)
acetylene + 2CH3MgCl -- ClMg-CC-MgCl + 2CH4
ClMg-CC-MgCl + H2 --- ClMg-CH=CH-MgCl (cat: Pd)
2C6H5CH2Cl + ClMg-CH=CH-MgCl --- The (cis- product) + by product
::)
;D
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CH3Cl + Mg -- CH3MgCl (grinhar)
acetylene + 2CH3MgCl -- ClMg-CC-MgCl + 2CH4
2C6H5CH2Cl + ClMg-CC-MgCl --- C6H5CH2-CC-CH2C6H5(A) + MgCl2
A + NaNH2(NH3) --- trans(the desire product)
:o
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Can you understand the grinhar compound?
CH3Cl + Mg --- CH3MgCl ( in ete)
RMgCl + HA -- RH + AMgCl
RMgCl + R'Cl --- RR' + MgCl2 :'( :-*
;D ;D ;D
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Can You search the name reaction on:
http://www.monomerchem.com/display4.html
hi hi:
Are you a student or pupil
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Yes you would also end up with the oxidized methyl group...to the acid. You can separate this from the acetic acid though.
Reactions are usually very specific and even if alternatives were to work there is usually only few reactions which are realistically plausible given the product one desires. This is why I doubt that the wittig reaction would be a good choice (more input later).
Also
CH3Cl + Mg -- CH3MgCl (grinhar)
acetylene + 2CH3MgCl -- ClMg-CC-MgCl + 2CH4
ClMg-CC-MgCl + H2 --- ClMg-CH=CH-MgCl (cat: Pd)
2C6H5CH2Cl + ClMg-CH=CH-MgCl --- The (cis- product) + by product
This simply will not work, there are many flaws. First of all, the hydrogens of acetylene are acidic, the reaction is too costly and too long, some of the products are incorrect.
I hope I did not sound condenscending.
I'll reply to Polly's question on the next post.
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Polly, for your second question, this is what I can think of for now...
Use two equivalents of a toulene halide (Ar-CH3I) and two equivalents of the base NaNH2 as reagents to react with the acetylene. This will yield an di-anionic acetylene and thus will react with two of the halides to for your desired compound...use a suitable reducing agent to go from reduce to the double bond...depending on which reducing agent you use you can form either the trans or cis products. If you still need help feel free to ask.
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Hi GCT! I appreciate your input. However, I am not sure I understand fully what you mean. Can you show me by writing out the reaction?? Also, will reacting acetylene with KMnO4 give you acetic acid?? If not what would you get?
Thanks
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Hi GCT! I appreciate your input. However, I am not sure I understand fully what you mean. Can you show me by writing out the reaction?? Also, will reacting acetylene with KMnO4 give you acetic acid?? If not what would you get?
Reacting acetylene with KmnO4...I doubt will give you acetic acid. Because it will cleave the triple bond to create two methyl acid, formic acid...or it might create different products. Acetylene is very reactive.
What you need to do is to react acetylene with NaNH2, this will ionize in water to create the base NH2 anion which will react through a rather non-reversible acid base reaction. This will create the acetylide anion. Use a methyl halide...CH3I to react through SN2 reaction with the anion and thus the actylide anion will attack the halide, alkylateing it. You are left with a three carbon species with triple bond at one end. Using Kmno4 (you might need to use a stronger oxidant) will cleave the triple bond, substituting one hydroxyl group for each triple bond, which will then rearrange itself to form the carboxlic acid...acetic acid, and a formic acid I imagine. You can separate these two through various methods.
As for the second equation, the simplest way (although other ways might be safer) is to react the aceytlene with two equivalents of the NaNH2, and two equivalents of the toulene halide (Ar-CH2I). This will substitute at the ends of the aceytlene...reduce this...depending whether you want trans or cis product you will need to use the proper reagent (tell me if you do not know which agent to use).
Hope this helps. Feel free to ask more questions.
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Thank you so much GCT! That helped a great deal. For the last one I am pretty sure you use lithium to get trans. :)