Sciencemadness Discussion Board

Iodine Cis-Trans Rotation

The Volatile Chemist - 16-9-2015 at 15:33

The mechanism for this reaction is known by many. It makes sense, but I have a few questions about it. Also, it seems like it'd have some application to amateurs. This google book is one place it is described:
here
1. I know this reaction works, but why isn't the alpha,beta - Iodo compound formed? Iodine solution is used to determine the saturation of oils, why doesn't it saturate the double bonds in these reactions?
2. Will this work for any alkyl- groups of reasonable lenght, or is the mechanism based on the fact that the iodine atom is heavier than the side group, and causes rotation that way?

Sorry if this should be in beginnings, but I would like to discuss the catalyst's mechanism. If there's some illicit usage of this mechanism, please, tell me so, as I'd rather not discuss it if such is true.
Nathan

DraconicAcid - 16-9-2015 at 16:27

It probably doesn't give the 2,3-diiodobutane because of sterics- the iodine substituent is huge, so any diiodo product would be likely to eliminate iodine to give the alkene again. This would be favoured at high temperature because the elimination reaction gives an increase in entropy. The addition reaction may be more favourable at low temperatures (thus the iodine addition to oils at low temperature....but isn't iodine bromide used more often?), but the isomerization takes place in the gas phase (presumably at a higher temp, so that the iodine is a gas).

(The google link doesn't work for me, but I have that text. It's on page 696 of the 8th edition.)

UC235 - 16-9-2015 at 21:02

Iodine is NOT directly used to determine the unsaturation in oils. It is done with a known quantity (excess) of ICl or IBr solution which are considerably more electrophilic sources of iodine that would lead to a less hindered product. An iodide salt is then added, which turns excess interhalogen into iodine which is titrated with thiosulfate. The difference between the amount added and the remainder corresponds to the amount of unsaturation.

The Volatile Chemist - 17-9-2015 at 14:03

OK, that makes sense.(^DraconicAcid) Does that mean the iodine isomerisation proceeds when the alkene and iodine are gaseous, or just the iodine? I've heard of gas phase reactions, but I guess most of them aren't feasible to the amateur.

(^UC235) That makes sense, but I've seen 'tincture of iodine' used for determining if a liquid is 'saturated' in middle- and high- school textbooks. I've done it myself. Is it just not accurate for quantitative chemistry?

subsecret - 18-9-2015 at 12:07

Quote: Originally posted by DraconicAcid  


It probably doesn't give the 2,3-diiodobutane because of sterics- the iodine substituent is huge, so any diiodo product would be likely to eliminate iodine to give the alkene again.



Is iodine too big to be added trans because it deflects the methyl groups toward each other? In step 3 of the rotation, why does the methyl group move closer to the iodine atom?

@TheVolatileChemist Using tincture of iodine as an unsaturation test is a great OTC approach if you don't have access to permanganate, bromine water, or chromate/dichromate for this.

DraconicAcid - 18-9-2015 at 12:11

Quote: Originally posted by The Volatile Chemist  
OK, that makes sense.(^DraconicAcid) Does that mean the iodine isomerisation proceeds when the alkene and iodine are gaseous, or just the iodine? I've heard of gas phase reactions, but I guess most of them aren't feasible to the amateur.


In the example given, 2-butene will be gaseous under any conditions that iodine will be gaseous.

PHILOU Zrealone - 18-9-2015 at 16:26

Trying to obtain 1,2,3-triiodopropane from HI/ZnI2 and glycerol; one will only get allyl iodide (CH2=CH-CH2I) because viccinal iodine atoms readily loose I2 and form an unsaturation (C=C).

The Volatile Chemist - 19-9-2015 at 14:06

Quote: Originally posted by DraconicAcid  
Quote: Originally posted by The Volatile Chemist  
OK, that makes sense.(^DraconicAcid) Does that mean the iodine isomerisation proceeds when the alkene and iodine are gaseous, or just the iodine? I've heard of gas phase reactions, but I guess most of them aren't feasible to the amateur.


In the example given, 2-butene will be gaseous under any conditions that iodine will be gaseous.

I wasn't querying for the butene reaction, I was wondering if the reaction mechanism required being in the gas phase to work for other alkenes.