According to this forum post, there should be no problem with the first step. The methyl iodide would not destroy the double bond.
There is also no hydrogen available to perform an addition reaction with the iodide ion to the N-methylethenamine to destroy the double bond.
I believe that the second step is also viable, also because of the lack of a good electrophile (silver doesn't count because silver can't react with carbon-carbon double bonds).
The third step should be slower because eliminiation reactions that result in alkynes are slower than elimination reactions that result in alkenes in general.
Therefore, the product would be ethyne $\ce{HC#CH}$ instead of ethylene $\ce{HC=CH}$.
According to this forum post, there should be no problem with the first step. The methyl iodide would not destroy the double bond.
There is also no hydrogen available to perform an addition reaction with the iodide ion to the N-methylethenamine to destroy the double bond.
I believe that the second step is also viable, also because of the lack of a good electrophile (silver doesn't count because silver can't react with carbon-carbon double bonds).
The third step should be slower because eliminiation reactions that result in alkynes are slower than elimination reactions that result in alkenes in general.
Therefore, the product would be ethyne $\ce{HC#CH}$ instead of ethylene $\ce{HC=CH}$.
Yeah the supposed enamine product doesnt exist. Furthermore even if you could isolate the enamine it is more likely to react with MeI on carbon not nitrogen.More
The reaction has three steps.
I shall use N-methylethenamine, $\ce{CH2=CH-NH-CH3}$ (the fourth option), as an example.
Step $1$:
$$\ce{CH2=CH-NH-CH3 + 2CH3I -> [CH2=CH-N(CH3)3]+ + I-}$$
This step is an $\mathrm{S_N2}$ step.
The carbon in the methyl iodide is the electrophile while the nitrogen in the N-methylethenamine is the nucleophile.
As the comments have pointed out, this step is not viable. The following tautomerism happens:
$$\ce{CH2=CH-NH-CH3 <=> CH3-CH=N-CH3}$$
Also, the electrophile methyl iodide would attack on the carbon-carbon double bond instead of the nitrogen.
Step $2$:
$$\ce{Ag2O + I- -> AgI + AgO-}$$ $$\ce{AgO- + H2O -> AgOH + OH-}$$
Step $3$:
$$\ce{[CH2=CH-N(CH3)3]+ + OH- ->[\Delta] CH#CH + CH3-N(CH3)-CH3 + H2O}$$
This resembles an $\mathrm{E1}$ step.
According to this forum post, there should be no problem with the first step. The methyl iodide would not destroy the double bond.
There is also no hydrogen available to perform an addition reaction with the iodide ion to the N-methylethenamine to destroy the double bond.
I believe that the second step is also viable, also because of the lack of a good electrophile (silver doesn't count because silver can't react with carbon-carbon double bonds).
The third step should be slower because eliminiation reactions that result in alkynes are slower than elimination reactions that result in alkenes in general.
Therefore, the product would be ethyne $\ce{HC#CH}$ instead of ethylene $\ce{HC=CH}$.
The reaction has three steps.
I shall use N-methylethenamine, $\ce{CH2=CH-NH-CH3}$ (the fourth option), as an example.
Step $1$:
$$\ce{CH2=CH-NH-CH3 + 2CH3I -> [CH2=CH-N(CH3)3]+ + I-}$$
This step is an $\mathrm{S_N2}$ step.
The carbon in the methyl iodide is the electrophile while the nitrogen in the N-methylethenamine is the nucleophile.
As the comments have pointed out, this step is not viable. The following tautomerism happens:
$$\ce{CH2=CH-NH-CH3 <=> CH3-CH=N-CH3}$$
Also, the electrophile methyl iodide would attack on the carbon-carbon double bond instead of the nitrogen.
Step $2$:
$$\ce{Ag2O + I- -> AgI + AgO-}$$$$\ce{AgO- + H2O -> AgOH + OH-}$$
Step $3$:
$$\ce{[CH2=CH-N(CH3)3]+ + OH- ->[\Delta] CH#CH + CH3-N(CH3)-CH3 + H2O}$$
This resembles an $\mathrm{E1}$ step.
According to this forum post, there should be no problem with the first step. The methyl iodide would not destroy the double bond.
There is also no hydrogen available to perform an addition reaction with the iodide ion to the N-methylethenamine to destroy the double bond.
I believe that the second step is also viable, also because of the lack of a good electrophile (silver doesn't count because silver can't react with carbon-carbon double bonds).
The third step should be slower because eliminiation reactions that result in alkynes are slower than elimination reactions that result in alkenes in general.
Therefore, the product would be ethyne $\ce{HC#CH}$ instead of ethylene $\ce{HC=CH}$.
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