Sciencemadness Discussion Board - Isopropyl iodide attempt

Isopropyl iodide attempt

arkoma - 31-8-2016 at 15:00

Last evening being bored, I put together a reaction that I hoped would give me an alkyl halide. I am hindered by no regular net access, or even access to my literature as am still suffering from broken laptopitis.

250ml two neck RBF.
50ml isopropyl alcohol 91%
I₂ unknown qty (broke scale too) estimation about 3 to 4 gms
Aluminum 2 X 6 inch piece foil rolled into 8 little balls
Refluxed for about an hour. No obvious reaction. Well, I turnt the heat off, but left the condenser water on and went to bed. Today at noon, I looked at my flask and have this pretty light yellow solution. No visible in left and it smells different. Some evidence of reacted aluminum. Using my cuzzins fone, Google isopropyl iodide. Doesn't seem a match. Anyone got a clue?

[Edited on 8-31-2016 by arkoma]

[zts16: changed title for clarity]

[Edited on 9-1-2016 by zts16]

Melgar - 31-8-2016 at 16:20

You probably have http://en.wikipedia.org/wiki/Aluminum_isopropoxide and a bunch of aluminum oxy/hydroxy halides. The yellow is just the elemental iodine that's still in solution. Iodine reacts with aluminum to expose the metal surface, but this would have taken quite a long time because the water in the alcohol would have just oxidized it again. Eventually, the water content would get low enough that the less-reactive isopropyl alcohol could start reacting.

Aluminum isopropoxide is actually a pretty interesting substance. It will set up an equilibrium between the alcohols and the ketones in a solution. So you can oxidize a secondary alcohol to a ketone by dissolving it in a molar excess of acetone, then using the aluminum isopropoxide to establish the equilibrium, which in the process reduces acetone to isopropyl alcohol. The reverse works too; using isopropyl alcohol as a solvent reduces ketones in solution to secondary alcohols. This reaction is also useful, because of how selective it is. Only secondary alcohols and ketones are affected.

Texium - 31-8-2016 at 16:26

I see you're following the ChemPlayer method for ethyl or methyl iodide but with isopropanol instead. While this works for those alcohols, with isopropanol being a secondary alcohol, you could expect it to react differently, if at all. I don't really know though, as this procedure has not been tested on any other alcohols as far as I know. If you just want to make an alkyl halide and it doesn't have to be isopropyl iodide, maybe try making ethyl iodide or methyl iodide instead.

arkoma - 31-8-2016 at 20:35

I knew I seen the iodine/aluminum reaction somewhere. I have broken three laptops in the last year. ATM I just can't spare $65 for ANOTHER 17.3" LCD. Hopefully soon. But it was still fun to actually ATTEMPT some chemistry besides *ahem* ethanol production.

chemplayer... - 3-9-2016 at 05:54

Trying this reaction for IPA is a good idea, but as has been said, no guarantees!

Ethanol works far better than methanol as far as primary alcohols go (in the Al reaction), which is an unusual finding, but nature does what she does and we're left to work out the rest. What is interesting is that in the reaction (and indeed the reaction using phosphorus instead of aluminium), there isn't a huge immediate generation of the alkyl iodide as might be predicted by the theoretical mechanism (PI3 + ROH -> RI etc.). Instead it takes many hours of distilling at higher temperatures than the alkyl iodide boiling point to get all the product off, and during this process there's virtually no original starting alcohol distilled off.

So we wonder whether or not the textbooks have this right (even for the P reaction). Perhaps in the case of the aluminium reaction it's the I which 'initiates' the aluminium, forming the aluminium alkoxide as a first step, and then this reacts more slowly with excess iodine (or HI generated) to form the alkyl iodide.

If true, and it's actually the alkoxide plus iodine / HI reaction which generates the product, then all bets are off in terms of what will and won't work. You may find that one alcohol reacts well and another doesn't at all.

CuReUS - 3-9-2016 at 09:10

what about trying the finkelstein reaction to get isopropyl iodide ?

Melgar - 3-9-2016 at 13:16

Quote: Originally posted by chemplayer...

So we wonder whether or not the textbooks have this right (even for the P reaction). Perhaps in the case of the aluminium reaction it's the I which 'initiates' the aluminium, forming the aluminium alkoxide as a first step, and then this reacts more slowly with excess iodine (or HI generated) to form the alkyl iodide.

If true, and it's actually the alkoxide plus iodine / HI reaction which generates the product, then all bets are off in terms of what will and won't work. You may find that one alcohol reacts well and another doesn't at all.

The reaction would produce some aluminum hydroxyhalides and the like, but the main product would be aluminum isopropoxide for sure, unless water got into the reaction and added some aluminum oxides and hydroxides to the mix. This substance is very well-studied, though not as well understood. If you add water, it'll have a negative pH, although the various aluminum compounds tend to buffer it.

[Edited on 9/3/16 by Melgar]

battoussai114 - 3-9-2016 at 14:16

Are you sure Iodine alone initiates the isopropoxide formation? I was under the impression the method required a mercury salt along with the Iodine.

Also, can some OC guru speculate on why the better results with ethanol vs methanol? I thought maybe the longer chain stabilized a carbocation intermediate in the reaction, assuming of course a mechanism that would have zilch to do with the SN reaction that the P process goes through... but this should mean that IPA would work better right?

[Edited on 3-9-2016 by battoussai114]

AJKOER - 6-9-2016 at 08:17

I may have come up with an explanation on how Hg could serve as a 'catalyst' (actually, I contend it is an inert anode in an aqueous battery cell) for the Aluminum/l2(aq) reaction here. The postulated mechanism also suggest alternatives to the use of Mercury metal and a refinement to methodology by adding some, say, NaI (to act as an electrolyte) as well. My logic starts by paralleling the construction of the so-called bleach battery (that is, a galvanic cell capable of producing electricity based on hypochlorous acid) which I have previously constructed and have generally outlined in the following steps:

1. Make HOCl by the action of a weak acid (dilute acetic, carbonic, or a very diluted mineral acid ) on NaOCl.

2. Add a piece of copper metal which will function as the cathode.

3. Add an Aluminum source to act as the anode or anodic zone (like finely cut up Al foil, for example).

4. Add a touch of sea salt (better than NaCl) to act as the electrolyte to get things started.

My take on the underlying electrochemical related reactions:

3 H2O <--> 3 H+ + 3 OH-

At the anode (Aluminum):

Al + 3OH- ⇒ Al(OH)3 + 3e- Eo = -2.3

At the cathode (copper):

3 HOCl + 3 H+ + 3 e- ⇒ 3/2 Cl2(g) + 3 H2O Eo = 1.63

for an implied net cell reaction of:

3 HOCl + Al --> Al(OH)3 (s) + 3/2 Cl2 (g) Eo net = 3.93 V

So, the battery cell based on HOCl /Al is theoretically capable of generating 3.93 volts. Reference: see http://www.exo.net/~pauld/saltwater/ and also pages 6 and 7 at: http://www.dtic.mil/dtic/tr/fulltext/u2/d019917.pdf

Now, in the current context with Iodine, water, Aluminum, alcohol and NaI (the electrolyte), we note that by the slow action of iodine with water,

I2 + H2O <--> HOI + H+ + I-

we have halogen analog of HOCl cited in Step 1 above. Assuming a similar reaction chain as above:

3 H2O <--> 3 H+ + 3 OH-

At the anode (Aluminum): Al + 3OH- ⇒ Al(OH)3 + 3e- Eo = -2.3

At the cathode (copper?): 3 HOI + 3 H+ + 3 e- ⇒ 3/2 I2(g) + 3 H2O Eo = 1.44

Net cell: 3 HOI + Al --> Al(OH)3 (s) + 3/2 I2 Eo = 3.74

Interestingly, the theoretical cell voltage for HOI / Al cell is only less than the bleach battery by under 1/5 of a volt. Note, unlike chlorine chemistry which does not have significant chlorate formation, I am ignoring the presence of HIO3, as in sufficiently acid conditions in the presence of HI, it is broken down back to elemental iodine:

IO3- + 6 H+ + 5 I- ---> 3 I2 + 3 H20

There may also be an electrochemical decomposition of the HIO3 as well. For example (see https://en.m.wikipedia.org/wiki/Standard_electrode_potential_(data_page) ):

IO3− + 5 H+ + 4 e− ⇌ HOI(aq) + 2 H2O Eo = +1.13

But, being less preferred to HOI + H+ with an Eo of +1.44, only a significant concentration presence may allow the reaction to proceed.

I am similarly recommending jump starting the reaction (that is, as in the case of the bleach battery, to reduce the induction period) by employing Aluminum foil that has been heated to redness (forming a weakened gamma Al2O3 protective coating) and briefly heating the solution mix until visible signs of a reaction using a microwave.

As:

3/2 I2 + 3/2 H2O <--> 3/2 HOI + 3/2 H+ + 3/2 I-

in a repeating cycle in the galvanic cell, it is apparent that the reaction likely consume I2, removing water and HOI, thereby creating mainly HI. The latter may then act on an alcohol forming the iodide of the alcohol if iodine is in excess, otherwise an aluminum alcoholate may be created.

For an interesting patent using a mercury salt and isopropyl alcohol, see
https://www.google.com/patents/US3083218 .

The remaining issue, before deciding to use Cu in place of Hg, is whether the copper electrode as compared to the well known inert mercury electrode, is sufficiently passive so as to not present an unacceptable level of copper ion impurities in the targeted alcohol iodide.

[Edited on 7-9-2016 by AJKOER]

Melgar - 9-9-2016 at 20:02

Quote: Originally posted by battoussai114

Are you sure Iodine alone initiates the isopropoxide formation? I was under the impression the method required a mercury salt along with the Iodine.

Also, can some OC guru speculate on why the better results with ethanol vs methanol? I thought maybe the longer chain stabilized a carbocation intermediate in the reaction, assuming of course a mechanism that would have zilch to do with the SN reaction that the P process goes through... but this should mean that IPA would work better right?

[Edited on 3-9-2016 by battoussai114]

It's a much slower reaction than with an amalgam, and also a much lower, less pure yield, because it's a solid that's reacting, and because iodine is a weird element in general, but assuming that the iodine catalyzed the reaction of all the water in the IPA with aluminum, then you'd be left with I2 and I- species, both of which react with aluminum to form aluminum triiodide. And aluminum triiodide would react almost instantaneously with isopropyl alcohol to form the isopropoxide and HI, which would go to work on the rest of the aluminum. Again though, very slow reaction, which is why you can see a small amount of elemental iodine dissolved in there.

There could never be any isopropyl iodide though, because it would react with aluminum so quickly. The only way to get that from these starting materials would be to form aluminum triiodide in an organic solvent, hydrolyze with water (though phosphoric acid would be better) then distill off the HI and react that with isopropyl alcohol.

Edit: that may not be the only way; there may be others, but this is at least one way.

[Edited on 9/10/16 by Melgar]