Will 2-Chloroethanol react with aqeous ammonia to produce ethanolamine?
thanks,Rich_Insane - 24-7-2009 at 13:56
Hm. I'm not sure, but if this is a nucleophilic substitution, I'd think
Here, the NH3 is the nucleophile and 2-chloroethanol is the reagent. The leaving group is the Cl. I think you'd have to use anhydrous NH3. I might be
wrong. There might be competition between water and NH3
the reaction would be 2-chloroethanol + NH3 --> 2-ethanolamine + HCl
With water you might get a diol in small quantities, and you may get the amine? Tell me if I'm wrong.
You might want to put this in Beginnings or Short Answer thread.kclo4 - 24-7-2009 at 14:26
I wouldn't be surprised if it did. Ethyl bromide, Dichloromethane, etc all react with ammonia to form the corresponding amines. I don't know how the
alcohol part of it would affect it though, I'd try it if you have the reagents. It would be easy to determine if it formed, wouldn't it?Picric-A - 24-7-2009 at 14:43
It would be easy to determine if it formed, wouldn't it?
It would? maybe with an IR or NMR scan, but other than that i cant think how it would be easy...
I was just worried the alcohol would be attacked by the nucleophile NH3... guess the is a better leaving group...Sedit - 24-7-2009 at 14:48
The amine would more then likely form as the hydrochloride salt and allow you to evaporate leaving the crystals which could be set free with NaOH. No
NMR needed anyware there.DJF90 - 24-7-2009 at 15:09
I would assume you to form the desired product. The reaction may go by direct Sn2 to displace chloride, or the reaction may proceed by forming the
epoxide intermediate (chlorohydrin in base...), then nucleophilic opening by ammonia to yield the ethanolamine. As both products are the same then it
makes no difference. 1-chloropropan-2-ol might not give 1-aminopropan-2-ol though. And you should probably expect some di- and tri- ethanolamine to
be formed, as if you have ever read an organic chemistry book you would know that halides form a mixture of products when reacted with ammonia/an
amine.
Obviously there are ways to circumvent this, but you will have to read about those yourself. I wont spoonfeed you. Being that this is a short question
bearing no reasoning on your part, I feel it would be more appropriately placed in the short questions thread.Picric-A - 25-7-2009 at 01:43
I thought that if there was a chance the alcohol group woiuld be attacked i could protect it with an ester group however i am not sure how resistant
ester groups are to hydrolysis by ammonia.
@DJF90 - you are right, as i have read in basic organic chemistry books, a mixture is formed, however, it is possible to produce predominantly the
mono amine via changing the ratio's of reactants. DJF90 - 25-7-2009 at 09:00
You have to use a large excess of the ammonia/amine, and prefeably add the electrophile to the ammonia/amine solution. But yes it is theoretically
possible to get predominantly the monoamine. But I expect it will be far from quantitative.Rich_Insane - 25-7-2009 at 09:40
Quote:
It would? maybe with an IR or NMR scan, but other than that i cant think how it would be easy... I was just worried the alcohol would be attacked by
the nucleophile NH3... guess the is a better leaving group...
I thought Cl is a far better leaving group? Sometimes OH- is used itself as a nucleophile. How good of a nucleophile is NaNH2? That way, you don't get
your hands on ammonia. Anhydrous ammonia would work the best, because you may see competition between water and ammonia, even though ammonia is a far
better nucleophile.
Indeed if this worked you would get the 1 isomer, because it is a secondary intermediate (as opposed to the 2-isomer, which is primary). That would
mean you would get a small quantity of 2-ethanolamine. I'm sure there's something you can do to increase the yield of 2-ethanolamine. I believe a
little heat improves yield, but that's bad, because you are handling ammonia. If you did a substitiution and started with 2-chloroethanol that is.
Check out this for amine detection. Some of the reagents may be out of reach, but you never know:
Ammonia/amine solution actually sounds great. But you would need anhydrous ammonia.
Ozone - 26-7-2009 at 07:51
The hydroxyl is not sufficiently electron withdrawing as to make the a-hydrogen acidic and thus ammenable to abstraction by an alkali as weak as
ammonia (or as strong as NaOH, for that matter). No enolate, no go; it should be OK.
The ester, on the other hand, would probably transamidate to some extent giving the amide. Avoid this.
Remember that the amine you have made is still a decent nucleophile, and so also is the second, etc.
