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Organic Chemistry


EvoN1020v

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yes, though there will be slight competition with SN1 depending on the concentration of the acid- the more concentrated the more E1 will be favored. with dilute acid, there would be a bit more competition with SN1 because of the abundance of water and the fact that the hydroxyl group is at the end of the alkane

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What's "E1" and "SN1"? I apologize, but I only know basic organic chemistry.

 

I also have another question:

 

What is this chemical structure called?

 

@@@H@@@H@@@H@@@

@@@|@@@|@@@|@@@

H----C-----C-----C----H

@@@|@@@|@@@|@@@

@@@OH@@OH@@OH@@

 

Something like 1,2,3 trihydropropane?

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In the presence of acid catalyst, 1-butanol would form primarily (E)-2-butene. The acid catalyst would promote the formation of a primary carbocation through an E1 mechanism, which would then undergo an intramolecular rearrangement to a secondary carbocation. The secondary carbocation would then preferentially form the internal, trans alkene, although minor amounts of the terminal alkene (1-butene) and the cis alkene ((Z)-2-butene) would be formed.

 

A biochemist would refer to the second structure as glycerol, although insane_alien's names are more common among organic chemists.

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What's "E1" and "SN1"? I apologize' date=' but I only know basic organic chemistry.

[/quote']

E1 refers to elimination, unimolecular (as opposed to E2, elimination bimolecular). It occurs when you have a good leaving group on a molecule that will form a stable (i.e. tertiary) carbocation. A proton is then abstracted and a double bond thus forms (hence elimination).

 

Sn1 refers to substitution, nucleophilic, unimolecular (as opposed to Sn2, which would be bimolecular). It is when the molecule in question (again) has a good leaving group, but this time, is relatively unhindered. This leaves it open for a nucleophilic attack from behind.

 

For more information, perhaps you could look through Wikipedia, or an organic chemistry textbook :)

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so, to apply all this:

 

the E1 elimination that would occur would have a lone pair of electrons from the oxygen from the hydroxy group grab a proton from H2SO4 just like water does when H2SO4 is diluted. so, like H2SO4+H2O-->H3O+ + HSO4-, in this particular case we get H3CH2CH2CH2OH2+ and HSO4-. eventually, a water molecule or HSO4- anion (though it will most likely be water because HSO4- is a dreadful base) will take a proton from the carbon next to the carbon that has the OH2+ group on it. the electrons from the C-H bond are then transferred to form a second bond between the last two carbon atoms. this pushes the OH2+ group off the end carbon and it leaves as water. the push occurs because we would see 10 electrons around the last carbon if the OH2+ group were not to leave. another important note is that if the hydroxyl group weren't protonated to form OH2+, we wouldn't see this reaction occur because OH-, due to its basicity, is a terrible leaving group.

 

the sN1 we would see has the same first step- the OH group gets protonated. but instead of having a base take away a hydrogen, the OH2+ just leaves, resulting in the formation of H3CH2CH2CH2C+ and water. then later on, a nucleophile attacks the carbocation. but anyway, the sN1 wouldn't occur in any significant degree because a terminal carbocation is reaaaaaaaaally unstable

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