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Topic: Having trouble with substitution v. elimination  (Read 2755 times)

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Offline Orgo18

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Having trouble with substitution v. elimination
« on: July 20, 2017, 01:43:49 PM »
Hey everyone,

I'm having some difficulty deciding when a reaction is going to create a substitution or elimination product. From what I understand, this only happens with secondary halides and primary halides in a strong nucleophile. If the nucleophile/base is bulky, it will encourage an elimination reaction. However, some examples seem to not fit with this. Why does the ethoxide ion react with 4-bromo-2,2,3,3-tetramethylpentane in an elimination reaction? I thought that ethoxide ion (a strong nucleophile) would cause the compound to undergo an SN2 reaction because ethoxide isn't bulky. Also, why would the compound want to have the unstable double bond at the end formed by an elimination reaction??

Same thing with compounds attached to Fluorine. Shouldn't 2-fluoropentane have a substitution reaction instead of elimination with a methoxide ion for the above reasons? I'd greatly appreciate it if someone could give me a rule of thumb that would help to understand whether an elimination or substitution reaction will occur. Thank you!

Offline wildfyr

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Re: Having trouble with substitution v. elimination
« Reply #1 on: July 20, 2017, 05:21:00 PM »
The frank answer to this is that you often will get mixed products. Especially in the case of secondary alkyl halides.

The way we generally control this is with heat. Higher temps favor elimination. Cold favors substitution. Also this double bond is not particularly unstable, its not as favorable as the more substituted one.

I really dislike the way these concepts are taught in undergrad ochem, its preposterous to tell people to write the single product of ethoxide with something like 2 bromohexane with no temperature indicated.

Offline Orgo18

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Re: Having trouble with substitution v. elimination
« Reply #2 on: July 20, 2017, 07:49:11 PM »
The frank answer to this is that you often will get mixed products. Especially in the case of secondary alkyl halides.

The way we generally control this is with heat. Higher temps favor elimination. Cold favors substitution. Also this double bond is not particularly unstable, its not as favorable as the more substituted one.

I really dislike the way these concepts are taught in undergrad ochem, its preposterous to tell people to write the single product of ethoxide with something like 2 bromohexane with no temperature indicated.

Thank you for your response! I found a separate commentary in the answer key that said "we know it's E2 because a strong base was used." To clarify, using a strong base doesn't mean that it's more likely to be E2 rather than Sn2, correct?

I appreciate your candid response. What should I do if my professor/textbook don't provide the temperature that the reaction should take place at? Does this mean that the elimination and substitution product are formed equally?

Offline wildfyr

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Re: Having trouble with substitution v. elimination
« Reply #3 on: July 20, 2017, 08:19:43 PM »
Yes that commentary is reasonable. Ethoxide is a stronger base than it is a nucleophile, although it is a decent nucleophile. Alkoxides are quite strong bases in general.

Offline Orgo18

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Re: Having trouble with substitution v. elimination
« Reply #4 on: July 20, 2017, 08:26:55 PM »
Yes that commentary is reasonable. Ethoxide is a stronger base than it is a nucleophile, although it is a decent nucleophile. Alkoxides are quite strong bases in general.

Do you mean that it's reasonable to say it's E2 because we used ethoxide or it's reasonable to say that it could be either Sn2 or E2 because we used ethoxide? If the former, how do you differentiate between the strong bases/nucleophiles that lead to E2 rather than Sn2?

Offline Babcock_Hall

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Re: Having trouble with substitution v. elimination
« Reply #5 on: July 20, 2017, 09:56:32 PM »
"I thought that ethoxide ion (a strong nucleophile) would cause the compound to undergo an SN2 reaction because ethoxide isn't bulky. Also, why would the compound want to have the unstable double bond at the end formed by an elimination reaction??"

It depends on whether the substrate is primary or secondary.  All else held equal, a secondary alkyl halide is more likely to show elimination than a primary one is.

"I found a separate commentary in the answer key that said 'we know it's E2 because a strong base was used." To clarify, using a strong base doesn't mean that it's more likely to be E2 rather than Sn2, correct?'

My interpretation of this passage is that the answer key is that this favors the E2 mechanism over the E1 mechanism; however, that's just my inference, which may not have been what the author intended.

With respect to the question of strong bases and E2 versus SN2, I used to think that bulky bases favored elimination more than small bases.  However, I looked up some data last year that made me think that the story was a bit more complex than that (I don't have the data handy at the moment).  I do think that it's fair to say that there are multiple variables (temperature, solvent, base strength versus nucleophilic strength, perhaps some more) that must all be considered.

Offline wildfyr

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Re: Having trouble with substitution v. elimination
« Reply #6 on: July 21, 2017, 09:11:39 AM »
Babcock,

Your answer perfectly illustrates how poorly thought out textbook questions are for undergrads. Giving them these in between questions using nonhindered strong bases on secondary alkyl halides with no conditions mentioned is just sloppy. Temp and solvent should always be provided to give guidance.

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