Sciencemadness Discussion Board - Propyl-Iodide Preparation?

Propyl-Iodide Preparation?

smaerd - 28-4-2010 at 12:49

Once again I am a total chem noob(I feel the need to include this with every post). So as a novice, I'm looking for simple experiments to do!

1. I was wondering if this 'write-up' looked right or if I mucked it up?(hopefully the attachment works okay). Edit - also, what kind of reaction is this(by name), just for future knowledge

?

2. If the only by-product of this reaction is water, and according to wiki(not a great source I know), propyl-iodide is soluble in water("0.11 g/100 mL at 20 °C" - http://en.wikipedia.org/wiki/N-Propyl_iodide).

Hypothetically, couldn't you freeze the solution to maybe -17*C so the water would freeze, but the propyl iodide would remain a liquid.(propyl-iodide's melting point = -101c : http://chemicalland21.com/industrialchem/organic/n-PROPYL%20...)?

Or is this an azeotropic relationship(apologies for lack of scientific vernacular)?

Thank you for your time you brilliant minds!

propyl iodide.gif - 3kB

[Edited on 28-4-2010 by smaerd]

Lambda-Eyde - 28-4-2010 at 13:32

What the heck is the phosphine good for? Why do you wanna mess around with that??

Propanol would react with hydrogen iodide without the need for ridiculously toxic and self-igniting phosphine. The HI is often made in situ with KI/NaI + H3PO4.

smaerd - 28-4-2010 at 13:39

Well I for some reason was under the impression that PH3 was red phosphorus! YIKES! What a mistake that was...

so there's not even a need for a phosphorus catalyst?

"It has the chemical formula C3H7I and is prepared by heating n-propyl alcohol with iodine and phosphorus." - from wiki.

thanks for helping me out here jeesh, I mucked that up pretty good(go figure) lol.

hmmm...

[Edited on 28-4-2010 by smaerd]

Sedit - 28-4-2010 at 13:54

Are you hell bent on halogenating with Iodine or would other halogens such as Bromine suffice? If you could I would no doubt go with the Bromo derivative since its easier to aquire H2SO4 and NaBr to perform the reaction. I have posted a writeup on my Bromoethane synthesis using getto equipment on here which should apply just as well here if not better due to the assumed higher Boiling point of Propylbromide over EtBr allowing more time to react before being distilled from the reaction media.

PS: Was the propanol bought from a chem supplier or do you have a clean over the counter source for it? If so PM me please since I would like to have to propanol without distilling assloads of floor stripper.

[Edited on 28-4-2010 by Sedit]

smaerd - 28-4-2010 at 13:57

AH! I see -light-bulb-

so it'd be more like this?

propanol + potassium iodide + phosphoric acid --reflux--> propyl iodide + (still need to work out the products)

Sedit - 28-4-2010 at 13:58

Yes that is correct and you can find a nice writeup on here as well where someone did just that to yeild Methyl iodide.

Heres the Iodomethane writeup using H3PO4,
http://www.sciencemadness.org/talk/viewthread.php?tid=12475#...

[Edited on 28-4-2010 by Sedit]

smaerd - 28-4-2010 at 14:04

I'll have to read your bromoethane work-up, sounds interesting.

I have yet to find the the n-propanol(I have found sources that appear amateur friendly, but it looks pretty expensive), so this is kind of a in future project. However, I could u2u the supplier, though it's kind of annecdotal(lack of better word) because I'd of never used them before. If you were to use floor-stripper would you need to use fractional distillation or would simple distillation suffice?

Thanks so much for the information here, I've got a bit of rearranging to do with my idea in ACD's ChemSketch. I'll certainly post up the final picture just to save a fellow noob the trouble, hehe. I want to make sure I have everything 100% on paper before I even consider it. Which makes sci-mad a necessary resource

.

[Edited on 28-4-2010 by smaerd]

smaerd - 28-4-2010 at 15:55

I know this is more of a "two-step" reaction, but does this look right by means of reactants and products?

I think I got it this time

. The only thing I'm a bit shakey on is the water forming as the end result. Normally when Hydrogen gas(H2) and Oxygen form to make H2O isn't it a very violent reaction?

revised.gif - 4kB

not_important - 28-4-2010 at 16:38

More like

H3PO4 + KI <=> KH2PO4 + HI

HI + ROH => RI + H2O

The first H on H3PO4 is a fairly strong acid, and can release the halogen acids from their salts. The other 2 H are weaker acids, I doubt that the 2nd plays much of a role and certainly not the 3rd unless you are distilling HI or RI away from the reaction mix. Hmm .. the insolubility of alkyl halides might drive the reaction enough to involve the 2nd hydrogen of the H3PO4, but I'm still doubtful.

H3PO4 works where H2SO4 doesn't because H3PO4 is not an oxidising acid at moderate temperatures such as can be tolerated by lab glassware.

In some cases you can use strong hydrochloric acid and a soluble iodide. HCl reacts noticeably slower with primary alcohols that does HI. The HCl and say NaI give a mixture containing both HCl and HI, even though HI is the stronger acid enough forms that for some alcohols the insolubility of the alkyl iodide keeps removing HI and drives the reaction to completion.

[Edited on 29-4-2010 by not_important]

Pomzazed - 28-4-2010 at 17:26

I doubt you'll get pure 1-halopropane.
acid attacks the alcohol, resulting in a primary carbocation (positive charge at terminal carbon); which in turn rearrange to the more stable secondary carbocation (the charge at the center carbon) and then this combine with the nucleophile (halide; in this case iodide) resulting in 2-iodopropane.

smaerd - 28-4-2010 at 17:58

So how would one theoretically purify the end-results of such a reaction then? simple distillation seems kind of unrealistic.

If HCL would work it'd be a real nice short-cut from ordering phosphoric acid. I just don't want to do anything too 'experimental'. I'm not afraid of failure haha, I am just sure my first chemical creation will already be riddled with problems and bumps along the way as it is.

Another big question I have is how important of a role does purity of the acid play for these kinds of reactions? I know the acids have to be strong. but like 70% strong, or reagent grade?

not_important - 28-4-2010 at 18:00

As I just said in another thread : Org Syn collected vol 1 page 25.

Purification often is just washing and drying for the lower alcohols, for higher alcohols the water solubility is low enough that distillation might be needed. Using HCl it is best to distill the product after wash&dry, as some RCl can be formed; or do the wash-up and dry, then add to a solution of NaI in acetone to convert any RCl to RI, filter off NaCl(s), w&d.

Strength and purity are two different things. Impurities can cause side reactions, or end up in the product and perhaps require distillation to remove. The strength needed is in part going to be determined by the alcohol being converted.

[Edited on 29-4-2010 by not_important]

Nicodem - 29-4-2010 at 01:42

Smaerd, how about learning the basics of chemistry before actually doing any experiment. For someone who thinks HI is iodine and PH3 phosphorous ("phosphane"?) you seem too ambitious, especially by wanting to perform a reaction of which you don't have the slightest clue on how it works. Your reluctance to read the reference not_important gave you, makes you appear quite irresponsible. Such an approach is often a good recipe for disaster. You should start the opposite way, first learn the basics, then do some experiment based on what you learned and learn more from the experience. Having an experience (or mishap) without having the basic tools to interpret it is of little to no use.

Pomzazed, the reaction is a simple SN2 substitution if primary alcohols are used. The acidic media is not for carbocation generation, but just to "activate" the alcohol by protonation (thus making it enough electrophilic for the reaction to proceed at all). The reaction is thus:
R-OH + H+ <=> R-OH2+
R-OH2+ + I- => R-I + H2O

(H+ is just a formal way of describing an acid catalyst without actually describing the whole proton transfer process - otherwise "naked" protons as such do not exist in solutions)

There is no HI involved (if you check the pKa of HI you can see that it is present only in very tiny amounts). This is also consistent with the observation that the ease of formation follows R-I > R-Br > R-Cl as the nucleophilicity of halides in protic solvents drops analogously: I- > Br- > Cl-. If the reaction would be SN1 then the rate of alkyl halide formation would be independent from the halide used. An interesting demonstration of it being an SN2 reaction would therefore be to use HCl instead of H3PO4 as acid catalyst and analyse the R-I vs. R-Cl ratio in the product. Being an SN2 reaction, there should be nearly no R-Cl formed on primary alcohols. With tertiary alcohols which follow the SN1 mechanism, there should form a considerable mixture of both t-alkyl halides.

smaerd - 29-4-2010 at 04:46

"how about learning the basics of chemistry before actually doing any experiment."

Nicodem that's what I'm doing right now

. Like I said above, I'm not doing anything until I know 100% what I'm doing. I'm not there yet at all, obviously. That's why I'm here, to pick up information, and find out more of what I need to know. You're prolly thinking "hit the books", and your right, I should, and I will! Right after I get back from class.

Also I've been reading and researching every resource given to me. Please don't think I haven't! Whether or not I understand it all, is another story. However, he gave me that reference at the very end of this thread and I fell asleep before checking back here:
http://www.orgsyn.org/orgsyn/default.asp?dbname=orgsyn&d...

It looks like an incredibly valuable resource for what I'm trying to understand. Thanks not_important

.

I know I'm new here and new to chemistry in general, but I'm not without any commonsense. I'm a new chemistry major, I understand the scientific method, and I'd never do anything I'm unsure of. Hence why I haven't even considered aquiring anything for this idea, cuz I just don't know it(which in turn is also why I am here)! Sorry for such a long drawn out post of little value.

[Edited on 29-4-2010 by smaerd]

smaerd - 29-4-2010 at 09:52

On the link provided by not_important, it shows

and it says it'll work for n-pr, assumedly that's n-propanol.

Now that is for the ROH-Br. couldn't the same be done but with I instead of Br?

Pomzazed - 29-4-2010 at 10:01

@Nicodem
Yes I know that. But that's why i said i "doubt" he will get "pure" 1-halopropane. I did get fraction of a rearranged isomer the attemp of pentanol->halopentane from fractional distillation, though pentyl is larger thus having more steric hindrance than a propyl. I did not say he wont get the product, but "pure" product.

@smaerd
The equation shown above use the different reagent (phosphorous bromide) and have kinda different mechanism.

smaerd - 29-4-2010 at 10:26

well I talked to someone who is a biochem major(years ahead of me) and he looked over it and said it would work. So I don't know what to think now... Wiki's statement also agrees that this would work. So who should I trust, you guys, or my friend and wiki?

PI3 would be the reagent instead of PBr3(obviously).

[Edited on 29-4-2010 by smaerd]

DJF90 - 29-4-2010 at 11:38

Nobody said it wouldn't work, just that it was a different mechanism; Using PI3 is slightly different to using KI and H3PO4 , though it would work, as would other reactions like the Appel reaction, using CX4 and PPh3:

R-OH + CX4 + PPh3 => R-X + CHX3 + Ph3P=O (IIRC)

where X = Cl, Br, I.

smaerd - 29-4-2010 at 14:32

Hopefully, I finally got this right.

final.gif - 4kB

Nicodem - 2-5-2010 at 11:31

Smaerd, that is different reaction from the one discussed upthread and it is (still) not clear if you are aware of this.

Quote: Originally posted by Pomzazed

@Nicodem
Yes I know that. But that's why i said i "doubt" he will get "pure" 1-halopropane. I did get fraction of a rearranged isomer the attemp of pentanol->halopentane from fractional distillation, though pentyl is larger thus having more steric hindrance than a propyl. I did not say he wont get the product, but "pure" product.

Are you sure? Have you analysed that fraction? By what means?
I don't have much first hand experience in preparing alkyl halides by acid catalysed halogenation of alcohols (except for isopropyl bromide and isobutyl bromide). I did however once made 1-bromo-3-phenylpropane from 3-phenylpropanol using 48% aq. HBr (the product is otherwise dirt cheap, but I did this out of curiosity more than need). This was a biphasic reaction so it is not exactly the same. Anyway, at room temperature there was no reaction (by HPLC). It did go well at reflux though. There were no unaccounted side products on the HPLC chromatogram and the product was pure by 1H NMR (no isomers as far as I remember).
Also, keep in mind that the potential presence of isomeric alkyl halides in the nucleophilic acid catalysed halogenation of primary alkyl alcohols is not proof enough that the reaction proceeds (or partially proceed) via SN1 substitution (that is via carbocations - with the exception of electron rich benzylic or allylic alcohols). The origin of the sec-alkyl halides could still be accounted by implying a more probable elimination-addition pathway: E2 elimination of the primary alcohol to the terminal alkene and electrophilic addition of HX to this which gives a sec-alkyl halide. To complicate things even further, some alkenes give regioisomeric side products upon electrophilic addition of HX resulting from the hydride shifts in the intermediate secondary carbocations. But primary carbocations from simple primary alcohols are highly unlikely and require usually more than just H3PO4 or HX (once again with the exception of benzylic and allylic alcohols, etc.). These acids are certainly no superacids.

Pomzazed - 2-5-2010 at 11:47

Quote:

Also, keep in mind that the potential presence of isomeric alkyl halides in the nucleophilic acid catalysed halogenation of primary alkyl alcohols is not proof enough that the reaction proceeds (or partially proceed) via SN1 substitution (that is via carbocations - with the exception of electron rich benzylic or allylic alcohols). The origin of the sec-alkyl halides could still be accounted by implying a more probable elimination-addition pathway: E2 elimination of the primary alcohol to the terminal alkene and electrophilic addition of HX to this which gives a sec-alkyl halide.

Agreed here.

Quote:

Have you analysed that fraction? By what means?

The fraction (minor) from distilling after the reflux gives a colorless liquid out at 115.5C plus it has the smell of haloalkane. So it's likeliy to be the 2-bromopentane. Though there is no further analysis of this fraction because it was such a small amount comparing to the scale, so it was just then disposed away. (not the product i want anyway). The next major fraction comes out next at 129.5C which is 1-bromopentane.

aonomus - 2-5-2010 at 14:18

Can this method be extended to do

phosphoric acid + KBr + EtOH -> EtBr + KH2PO4 + H2O

Nicodem - 3-5-2010 at 13:38

Smaerd, if you want to do some experiment on preparative synthesis, please give us a list of chemicals you have and we'll give you a proposal and guide you, first trough the learning process and then trough the practical aspect. I think this would be a much wiser approach for someone in your situation and you would certainly learn a lot more.

Quote:

Have you analysed that fraction? By what means?

Maybe if you used H2SO4 and did not add enough water. Depends on the conditions you used, but already in the first place I would be sceptical that this fraction was truly 2-bromopentane without proper characterisation. Besides, your experience can not be generalized to all similar reactions, even those using iodides with H3PO4 on other primary alcohols. If this was a serious and common problem then the Org. Synth. entry would have surely mentioned it somewhere among the so many examples given (some of which use quite harsh HBr/H2SO4(aq) mixtures with heating).

smaerd - 6-5-2010 at 19:45

I have been distracted by finals this week so I haven't had much time to finish my work-up. However, finals are over, and I am back to researching, and writing. I have none of the chemicals required at this point in time, so this is very much a dream still. I figure tomorrow I'll put a few hours into it and hopefully I'll have the theoretical all down on paper/electronic document for your review/consultation.

Thanks a bunch guys

. It means a lot to me that I can get some guidance and some extracurricular studying done before my next semester.

smaerd - 7-5-2010 at 11:04

Alright, I think this is everything. I rounded some decimals so I know it's off by a little bit(according to the law of conservation of mass), and I wouldn't do it with a batch as big as this obviously haha.

Does everything look right to you guys?

Propyl Total.png - 41kB

smaerd - 14-5-2010 at 09:21

it's been about a week or so since I posted this. I figure it's safe to bump it. If that's against forum rules I apologize in advance.

Does my last procedure out-line look alright?

S.C. Wack - 14-5-2010 at 10:39

What is the point? We can read OS and Vogel, you can read OS and Vogel, and you're not going to improve their directions.

Nicodem - 14-5-2010 at 10:48

Smaerd, you still do things in the opposite order. You were already told to start with literature first. Have you even bothered checking the preparation of n-propyl iodide in Vogel's which uses I2 and P? I bet you didn't or else you would not ask such stuff. Now do your homework and study that procedure!

PS: Since I know you would never search yourself, here is the location in the 5th edition: experiment 5.59, page 568-569. Now you have no more excuses!

smaerd - 14-5-2010 at 12:06

Sorry I guess as a noob I didn't know what resources to have aside from google and a couple text-books. Now I have a vogels book. Thanks for letting me know where to look.