Sciencemadness Discussion Board

reaction of alkenes

piogun - 17-8-2004 at 07:00

Hello!
Could anyone help me with the following reactions:

C6H5-CH=CH2 ==CHCl3, 50% NaOH in H2O, R4N+Cl- ===> ?

C6H5-CH=CH2 ==CH2I2, Zn(Cu), ether ===> ?

I'd be extremely grateful

[Edited on 17-8-2004 by piogun]

[Edited on 18-8-2004 by piogun]

Blind Angel - 17-8-2004 at 07:36

I don't think anyone here will simple give you the answer, at least proove that you've done some research

piogun - 17-8-2004 at 09:09

AS for the first reaction I read about catalysis in mixture consisting of 2 fases (aqeuos and organic). It said that compounds like R4N+OH- (which might be also true about chlorides) are catalysts for the alkylation of anions in those mixtures. My thoughts are that probably CHCl3 in the first equation is the solvent for the alkene, the solution of natrium hydroxide is the aqeuous fase and the salt catalyst, but I don't know whether alkenes undergo any reaction under the influence of bases. Perhaps, this reaction is the dimerization of the alkene. But I only know about the dimerization. But how would the carbocation be formed?

Geomancer - 17-8-2004 at 13:40

You're right about the phase transfer catalysis, but wrong in assuming that the chloroform is inert. Both of these reagent sets generate an intermediate that then reacts with the alkene (which does not form a carbocation). Both intermediates react by the same process. Is this part of a normal organic course? If so, assuming you looked at the reading, the answer should jump out at you. The reagents are unique and the reaction is unusual--probably the most unusual in the chapter.

piogun - 18-8-2004 at 09:57

Unfortunately, the book that I have at home is quite old and gives only one example of the phase transfer catalysis which is the methylation of C6H5-CH2-Co-CH3 with CH3I and the catalyser.
But when I was looking at the second reaciton it struck me that carbene :CH2 could be generated and then react with the alkene giving C6H5-CH=CH-CH3. Carbene could also be connected to the carbon adjacent to benzene ring but due to sterical hinderness I think the product that I suggested is formed more easily.
You also said that in both reaction the intermediate is the same so perhaps chloroform reacts with natrium hydroxide giving :CHCl (chlorocarben?), NaOCl, NaCl and water. And then the carben reacts with the alkene giving C6H5-CH=CH-CH2Cl. Is this way of thinking correct or not?

JohnWW - 18-8-2004 at 11:56

Quote:
Originally posted by piogun
Hello!
Could anyone help me with the following reactions:

C6H5-CH=CH2 ==CHCl3, 50% NaOH in H2O, R4N+Cl- ===> ?

C6H5-CH=CH2 ==CH2I2, Zn(Cu), ether ===> ?

I'd be extremely grateful
--------------------------------------------------
The formulas given for those conjugated aromatic alkenes, which are styrene cderivatives, appear to be incorrect. Assuming they are supposed to be propene derivates, they should be:

C6H5-CH=CH-CHCl2 and
C6H5-CH=CH-CHI2 .

It is also not stated whether they are the cis or trans isomers, of what would be 1-phenyl-3,3-dihalo-prop-1-enes. They would be produced by direct halogenation in CCl4 solution, the end carbon being preferentially halogenated for steric reasons, although some halogenation of the double bond can be expected. In the presence of a catalyst capable of propagating a free-radical reaction, these would probably polymerize to give substituted polystyrene derivatives.

As regards the reactions with concentrated NaOH with a quaternary ammonium chloride: the latter cannot take part in such a reaction except as a catalyst, as quaternary ammonium salts are usually too stable under such conditions unless the attached functional groups indicate otherwise. The most likely outcome is that the halide atoms would be hydrolysed and replaced by -OH by nucleophilic substitution to give gem-diols, which except in aqueous solution would lose water to become aldehydes:

C6H5-CH=CH-CHCl2 + 2NaOH -> C6H5-CH=CH-CH(OH)2 + 2NaCl

C6H5-CH=CH-CH(OH)2 -> C6H5-CH=CH-CHO + H2O

As regards the reaction with Zn (if alone) in ether, at least of the iodo-compound, the most likely result would be dehalogenation wth migration of an H to form ZnI2 and phenylallene, C6H5-CH=C=CH2. But if we are talking about a Zn-Cu couple, with the Cu present as cuprous chloride, the reaction would be the Simmons-Smith reaction, in which the dihalogenated end carbon is initially dehalogenated to form a diradical plus ZnI2, then adds across the styrene double bond of another molecule of the reagent to form a cyclopropane derivative in about a 90% yield:

C6H5-CH-CH-CHI2
\ /
CH-CH=CH-C6H5

An excess of Zn + CuCl may result in the reaction being repeated on this product, to eventually form a polymeric product containing many cyclopropane rings. In the case of the remaining 10% or so of reagent which forms other products, the most likely other competing product (along with ZnI2) would be that formed by dimerization of the initially-formed diradical to give (both cis and trans isomers) 1,6-diphenyl-hexa-1,3,5-triene:

C6H5-CH=CH-CH=CH-CH=CH-C6H5

A conjugated aromatic-substituted hexatriene like this is likely to be highly colored, and liable to polymerization.

John W.

Geomancer - 18-8-2004 at 17:20

JohnWW: I'm pretty sure that when he wrote "==" he meant "+", that is to say he wanted to know how styrene would react with either basic chloroform or methylene chloride+ZnCu.

piogun: You're right in thinking these reagents will generate carbene type species. In the first case dichlorocarbene, and in the second a species that acts somewhat like free carbene. These will characteristicly add to an alkene in a concerted process to form cyclopropane rings. As JohnWW noted, the second case is referred to as the Simmons-Smith reaction.

JohnWW - 18-8-2004 at 19:25

In that case, i.e. if styrene is the organic reagent, with CHCl3 and CH2I2 being solvents which also participate in the reactions, the major product in the first case would be phenylchlorocyclopropane, and in the second case phenylcyclopropane.

Jhn W.

piogun - 19-8-2004 at 10:24

Thanks a lot for your help.
Do always carbenes react with alkenes to produce cyclopropane ring or do they also react like with the alkanes (the product is by one carbon longer than the substrate)?

JohnWW - 19-8-2004 at 12:03

Carbenes, by which is meant organic diradicals, with an unused pair of electrons on the reactive carbon, e.g. CH2:, as produced ephemerally as reactive intermediates e.g. by gem-dehalogenation (e.g. of CH2I2) with an electropositive metal, react wherever they can find other unused electron pairs to join onto (e.g. CH2: can simply dimerize to ethylene, CH2=CH2), or else add across unsaturated bonds to produce cyclopropane derivatives (e.g. the ethylene produced as above can react with a third carbene to produce cyclopropane).

While the carbene produced from CH2I2 will react with styrene or its derivatives to produce phenylcyclopropanes, a substantial proportion of it would be lost as gaseous ethylene and cyclopropane, hence only about a 90% yield.

John W.