benzylchloride1 - 18-3-2009 at 14:13
Pentamethyl cyclopentadiene is a useful ligand in organometallic chemistry that enables the synthesis of cyclopentadienyl complexes of metals for
which normal cyclopentadienyl complexes cannot be synthesized. Chemical suppliers charge very large sums for small quantities; 5g $110 from Aldrich.
Inorganic Syntheses has a method for the synthesis of this compound, but requires very expensive precursers for the starting materials. I am working
on the synthesis of this compound in my home lab from fairly inexpensive compounds. The only hitch is that the synthesis requires expensive
iodomethane and triphenylphosphine for the final Wittig reaction. I started on the synthesis of the 2,3-dibromobutane today. I started by producing
biacetyl monoxime from methyl ethyl ketone. I am currently hydrolyzing the monoxime to produce biacetyl. The biacetyl will be reduced with aluminum
isopropoxide to produce 2,3-butanediol. The diol will be refluxed with concentrated hydrobromic acid to produce 2,3-dibromobutane. The next part of
the synthesis is the synthesis of 3-methyl-2,4-pentadione from iodomethane and 2,4-pentadione. The substituted pentadione will then be reacted with
the triphenylphosphonium salt of 2,3-dibromobutane to produce the desired pentamethyl cyclopentadiene.
Attachment: Proposed Synthesis of pentamethyl cyclopentadiene.doc (22kB)
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not_important - 18-3-2009 at 15:21
Dunno, the only unusual thing in the OS synth is 2-bromo-2-butene, for which there are a number of routes to using pretty common stuff. Ethyl Acetate
and pTsOH are both cheap and easy to make, and the TsOH is just a catalyst.
Why not just purchase diacetyl ? It's an industrial scale chemical, after all.
[Edited on 18-3-2009 by not_important]
Ebao-lu - 18-3-2009 at 15:24
There can be hypothetically some problems with
1) alumitium isopropoxide meervein-pondorff reduction of dione
- the conjugation between two C=O groups can probrbly make them bit inert and less active in the reacion
- are you sure there would be no benzylic rearrangement into (Me)2C(OH)COOH?
otherwise do you have lit on that?
2) are you sure wittig reaction is applicable to 1,2-dihalocompounds and there would not be reduction into alkene(butene)?
If there would be a way of making a,a-methylene-bismethylethylketone CH3COCH(CH3)-CH2-CH(CH3)COCH3 (form MEK + CH2O, or CH2Br2), then it can be
coupled into diol and then probably directly(!) dehydrated with conc HBr refluxing without pinacol rearrangement.
[Edited on 18-3-2009 by Ebao-lu]
Jor - 18-3-2009 at 17:42
Are you sure you need methyl iodide?
Methyl bromide is also very reactive (although less so), and is very easy to make from methanol, KBr and H2SO4. There is a topic on this subject,
where the MeBr gas is used in situ to methylate hydroquinone to 1,4-dimethoxybenzene.
Methyl iodide is much harder to make requiring HI or Red P/I2. HI can be generated from KI and H3PO4, or made by reacting H2S and I2 (dangerous) , but
using MeBr would be much more convenient.
Both of these methylating agent are nasty carcinogens.
I think triphenylphosphine is a chemical wich is very hard to make yourself, and it's something wich you should buy. It's quite cheap, you just need a
source.
I think it's nice to see someone doing such a relatively complicated synthesis at home.
benzylchloride1 - 18-3-2009 at 19:28
I have a small quantity of iodomethane. I am interested in trying the methyl bromide route though because the precursors to the iodomethane are hard
to come by. I have some triphenyl phosphine, 100g. I synthesized triphenylphosphine once, but the yields were very low, 30% using phenylmagnesium and
phosphorus tribromide. If the Meerwein, Pondorf reduction does not work I will use another reducing agent such as sodium borohydride, but would prefer
to use the latter method due to the high cost of borohydride. Benzil which is another 1,2 diketone can be reduced to the diol in high yield with
borohydride. I may end up purchasing the diacetyl, but I am trying to make most of the intermediates from chemicals I currently have access to. The
biacetyl is yellow-green in color. The hydrolysis reaction does not seem to be very high yielding, but the starting materials are cheap. This is a
synthesis that I thought up yesterday and I am starting to work on it. The synthesis of biacetyl monoxime is very easy, but the hydrolysis step is the
problem. I plan on isolating hydroxylamine hydrochloride from the aqueous residue after the hydrolysing and distilling the diketone. The diketone is
steam distilled and extracted from the distillate in which it is highly soluble in with dichloromethane. The dichloromethane is then carefully
fractionally distilled from the biacetyl. If hydrobromic acid does not work for the replacement of the hydroxyl groups, I plan on using phosphorus
tribromide to conduct this reaction. I think that this synthesis will challange my abilities and I will learn much from this exercise. I purchase many
of my chemicals, but I always prepfer to prepare them my self if possible.
Klute - 19-3-2009 at 00:38
It's a very clever path way, the preparation of the butan-2,3-diol is very clever!
I would directly try NaBH4 for the reduction of the diketone, as you are nearly sure it can work, and Al/Hg might give rise to pinacol coupling..
Cp* is indeed a very usefull ligand, but sold as such a insultingly high price... Any plans on what do to once you get the ligand?
Ebao-lu - 19-3-2009 at 03:03
Home chemists cooking ligands for organometallic chemistry - thats fine))
not_important - 19-3-2009 at 06:22
I'd still go with the OS route, just because I don't have ready access to cheap triphenylphosphine or even cheap PBr3; and because I have doubts
about the Wittig step. (BTW - you might stick with PDF or some open source document format over MS Word)
Methylacetylene + NaNH2 => MeCC-Na + NH3 (use MAPP gas)
Or
MeCCH + Na => MeCCNa + 1/2 H2 (with losses from reduction to propene+propane)
then
MeCC-Na + MeX => MeCCMe + NaX
or
Acetylene + 2 Na => H2 + NaCCNa (with losses from reduction of HCCH)
NaCCNa + 2 MeX => MeCCMe + 2 NaX
MeX can be MeCl, as the sodium acetylides are pretty reactive, or even inorganic esters.
Or, not as low cost or desirable
HCCH + 2 MgXR (say R = nBu) => 2 RH + MgX2 + Mg(CC)
then not waiting very long, as magnesium acetylide changes to the carbide
Mg(CC) + 2 MeBr => MgBr2 + MeCCMe
after that
MeCCMe + HBr => MeC(H)=C(Br)Me and Bob's your uncle.
Ebao-lu - 19-3-2009 at 08:09
And can calcium carbide be alkylated maybe with some special conditions used etc?
Just read a word file. There should be an easier option for ketone nitrosation - with NaNO2 and AcOH(not obligatory glacial) but only for those who
have AcOH.
And probably, you can obtain the 3-methyl-acetylacetone by acylation of MEK with ethylacetate and EtONa.
[Edited on 19-3-2009 by Ebao-lu]
Nicodem - 19-3-2009 at 23:44
1,2-Bis(triphenylphosphonium)alkanes do exist, but as far as I know they can not form the corresponding ylides. Phosphonium ylides are strong bases
and as such decompose if the neighbouring group is reactive toward bases or nucleophiles.
smuv - 20-3-2009 at 12:06
You could try to make/buy sec-butanol and dehydrate it to the alkene in H3PO4/H2SO4 and pass it through bromine water as it is formed.
benzylchloride1 - 23-3-2009 at 20:33
I am trying to avoid the use of hydrogen halide gases, but the method is interesting. I still have to distill my crude biacetyl. Since I only have
around 10mL of crude product, I will use sodium borohydride for the reduction due to the small amount that is theoretically required. Sterics could be
a factor in the final proposed Wittig reaction, which could block the formation of the 4-membered intermediate in the reaction with the ketone. What
percentage of MAPP gas contains methyl acetylene? This is interesting, I did not know that it contained methyl acetylene. Sodium amide is expensive
unless cylinders of liquid ammonia and sodium metal is available. I would be interested in the formation of alkali metal acetylides. I have several
acetylene cylinders that I purchased for my atomic absorbtion spectrometer that I could use for synthetic work.
not_important - 24-3-2009 at 06:54
At one time I was getting MAPP that was 35 to 45 percent methylacetylene, but it looks as if the current stuff is only 1/4 to 1/3 of the total gasses.
Allene(15%), propylene (40-50%), propane(1-5%), and butanes(2-10%) make up the rest. It's also gone up noticeably in cost.