# How will you convert methane to the following: ?

## (a) propane (b) butane (c) methyl butanoate (d) ethanoic acid (e) ethanol

Mar 13, 2016

Methane being as unreactive as any alkane to most reactions (having a pKa of about 50~60), we pretty much have to begin with a radical halogenation to turn it into a reactive species. Specifically we can go with radical bromination because for experimental reasons it gives a better yield than radical chlorination.

From that point, we can get a bit more... original/creative.

MAKING PROPANE

Making propane is not extremely difficult from here; simply use a very strong lewis base like sodium hydride or sodium amide to deprotonate acetylene.

Then you can react that as a nucleophile with the alkyl halide (as an electrophile) you just made to generate methylacetylene, which reduces down to the alkane via ${\text{H}}_{2}$ catalyzed by $\text{Pd}$ over $\text{C}$.

MAKING BUTANE

Simple modification of the above pathway: just replace the acetylene that you bring in from a separate source with methylacetylene, and you have introduced an extra carbon into the semi-final product.

In that case the triple bond is not on the terminal carbon but on carbon-2. Then the reduction to the alkane generates butane.

MAKING METHYL BUTANOATE

This one is quite a bit trickier. I ended up simply making methanol from methane, and making a carboxylic acid from a separate acetylene source.

From acetylene, we used the $\text{NaH}$ and $\text{RBr}$ idea from before. The hydroboration adds anti-Markovnikov onto the alkyne, giving a terminal enol, which tautomerizes to the aldehyde (in this case, in base).

Full oxidation (via chromic acid, for instance) generates the carboxylic acid.

Heating with the alcohol dehydrates the carboxylic acid and alcohol to combine them into the ester in a Fischer esterification.

To be honest, it's much easier to start from acetylene...

MAKING ETHANOIC ACID

With this I invoke the Grignard reaction, which I hope you've learned at this point; it's one of the first labs you do in organic chemistry 2. You wouldn't have prepared the Grignard reagent itself, but you would have used it on an alkyne.

Grignards essentially act as really strong, but water-sensitive nucleophiles (water can donate a proton, deactivating the super strongly-basic nucleophile). i.e. $\text{CH"_3"MgBr}$ is really just ${\text{CH"_3:^((-)) "MgBr}}^{\left(+\right)}$.

Weak acid finishes off the reaction by easily donating a proton to the methyl anion and turning ${\text{MgBr}}^{+}$ eventually into ${\text{Mg}}^{2 +}$, ${\text{Br}}^{-}$, and $\text{H"_2"O}$.

MAKING ETHANOL

Take the ethanoic acid you made, and simply reduce it using ${\text{LiAlH}}_{4}$. It's a dangerous substance, but it is necessary to use it when reducing a carboxylic acid; ${\text{NaBH}}_{4}$ wouldn't be reactive enough, though it can reduce ketones, aldehydes, and the other more reactive carbonyl compounds.

${\text{LiAlH}}_{4}$ essentially acts as a hydride ($\text{H} {:}^{-}$) donor, giving you ethanal, followed by ethanol. It does not stop at the aldehyde.

(A nice alternative that does stop at the aldehyde is lithium tri-tert-butoxyaluminum hydride.)

Mar 27, 2016

#### Explanation:

(a) Propane

${\text{CH"_4 stackrelcolor(blue)("Br"_2//hνcolor(white)(m))(→) "CH"_3"Br" stackrelcolor(blue)("Mg"//"dry ether"color(white)(m))→ "CH"_3"MgBr"stackrelcolor(blue)("ethylene oxide"color(white)(m)) → "CH"_3"CH"_2"CH"_2"OH" stackrelcolor(blue)("HBr"color(white)(m))(→) "CH"_3"CH"_2"CH"_2"Br" stackrelcolor(blue)("LiAlH"_4color(white)(m))(→) "CH"_3"CH"_2"CH}}_{3}$

(b) Butane

${\text{CH"_3"Br" stackrelcolor(blue)("Na"color(white)(m))(→) "CH"_3"CH"_3 stackrelcolor(blue)("Br"_2//hνcolor(white)(m))(→) "CH"_3"CH"_2"Br"stackrelcolor(blue)("Na"color(white)(m))(→) "CH"_3"CH"_2"CH"_2"CH}}_{3}$

(c) Methyl butanoate

$\text{CH"_3"Br" stackrelcolor(blue)("NaOH"color(white)(m))(→) "CH"_3"OH}$

${\text{CH"_3"CH"_2"CH"_2"Br" stackrelcolor(blue)("Mg"//"dry ether"color(white)(m))→ "CH"_3"CH"_2"CH"_2"MgBr" stackrelcolor(blue)("CO"_2color(white)(m))(→) "CH"_3"CH"_2"CH"_2"COOH" stackrelcolor(blue)("SOCl"_2color(white)(m))(→) "CH"_3"CH"_2"CH"_2"COCl" stackrelcolor(blue)("CH"_3"OH"color(white)(m))(→) "CH"_3"CH"_2"CH"_2"COOCH}}_{3}$

(d) Ethanol

$\text{CH"_3"CH"_2"Br" stackrelcolor(blue)("NaOH"color(white)(m))(→) "CH"_3"CH"_2"OH}$