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synthesize acetic acid from alkyl halide

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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)

Can you understand the grinhar compound?
CH3Cl  + Mg ---  CH3MgCl ( in ete)
RMgCl + HA --  RH  +  AMgCl
RMgCl  + R'Cl ---  RR'  +  MgCl2 :'( :-*
 ;D ;D ;D

Can You search the name reaction on:
hi hi:
Are you a student or pupil

Yes you would also end up with the oxidized methyl 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).


--- Quote ---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

--- End quote ---

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.

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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