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I have recently been trying to come up with a method to synthesize 2-methyl-2-butanol or tert-amyl alcohol. I wish to avoid complex/expensive steps since as a home chemist/student, I don't have access to much money or equipment.

What I have done so far: I noticed that methyl ethyl ketone has pretty much the correct structure, minus a methyl group.

I then realized this meant I could use a methylating agent to get the correct molecule. What I needed was to break the $\ce{C=O}$ bond from the carbonyl group and add a methyl group.

Long story short, I found that either Grignard reagents or something like methyl lithium would do it since it would donate an electron pair to the carbon, and form a bond with the methyl. I would then have the 2-methyl-2-butanol as a conjugate base with $\ce{Li}$ ions present. I could then react them out with $\ce{HCl}$ to precipitate out the $\ce{Li}$ salt and add a hydrogen to the negatively charged oxygen.

My problem: I don't know if this would work at all. I didn't find any documentation. Also, methyl lithium is very expensive and difficult to make so I can't use it.

Does anyone have any other route suggestions? Would it be possible to make methyl sodium and if so, would it work the same? Is it possible to make Grignard reagents easily at home?


I'm a hobby chemist, so please be detailed with how you predicted the organic reaction.

Also, I don't necessarily care about what chemicals I use for the reaction, so if anyone can suggest a different starting chemical, I'd be glad to hear it. I was even thinking of using an alkene group of some kind and methylating the $\ce{C=C}$ bond, but I don't know for sure, as I'm pretty new to organic chem.

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    $\begingroup$ Yes it would work excellently in theory with either Grignard reagents or alkyllithiums. I can't say about in practice although both reagents can be dangerous to handle so this is something to think about. However, if you're only interested in the product then buying it is the easiest option. $\endgroup$ – bon Aug 18 '15 at 15:37
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    $\begingroup$ Grignards are not the kind of thing that should be worked with without a hood. $\endgroup$ – jerepierre Aug 18 '15 at 16:31
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    $\begingroup$ Acid-catalyzed hydration of 2-methyl-2-butene should also work. $\endgroup$ – Jannis Andreska Mar 26 '16 at 18:57
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Isopropyl lithium and ethyl bromide would be my go-to if I were to try to prepare this stuff (which I will call TAA). This requires a solid fume hood, and the resulting mixture of salt, TAA, and isopropyl lithium solvent (usually n-hexane) should be easy to separate. The hexane can be stripped from the TAA via fractional distillation, with more distillation passes resulting in a more pure product. TAA boils at $\pu{102 ^\circ C}$ (Sigma Aldrich), and n-hexane boils at around $\pu{70 ^\circ C}$ - this $\pu{30 ^\circ C}$ gap in BP passes the rule-of-thumb "can I separate with distillation?" question.

All that said, even pricey Sigma Aldrich can get you 99% for $115 a liter.

I am aware that TAA is an experimental recreational drug, and a possible replacement for ethanol. I advise caution in sourcing this material for that purpose, as the impurities present are likely to be more toxic than TAA itself. I believe the industrial synth is from 2-methyl-2-butene (2m2b) with the addition of water (this would be cheapest); TAA prepared this way can be purified further by refluxing with an open condenser, enabling any unreacted alkene to evaporate (Boiling point = $\pu{39 ^\circ C}$). Acetone and ethyl Grignard would also form TAA, but with much more hassle and hazard, so the remnant ethyl chloride and acetone from such a synthesis are not likely impurities in any cheap sample.

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Or, cheaper alternative: I suspect that the acid dehydration of isoamyl alcohol would yield a carbocation with a subsequent 1,2 hydride shift, thus giving (to a yield that I don't know, and with reaction conditions that I don't know) 2-Methyl-2-butene (caution: flammable and with a bp of 40°C).

Further treatment with aqueous acid could yield, via Markovnikov hydration of the alkene, the desired product.

Any trace of a secondary alcohol product (derived from a elimination without hydride shift), if isolable, could be added to a reaction environment described in the first step, as it would easily convert to a trisubstituted alkene, that via Markovnikov hydration, again, gives 2-Methyl-2-butanol.

Please don't perform these reactions without a proper fume hood, anyway.

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