E2 question plz read

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BiomajorPreDent

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From my text and destroyer I know that on an E2 reaction that has a strong bulky base, you get the less substituted product

But I was just watching Chads Ochem videos, and he says that, that is only the case with tertiary halides, and if its a primary or secondary it is still zaitsev product

but the example is destroyer has a secondary halide

its on road map 5 on the right side of the page 2-Bromobutane is reacting

Can someone shed some light on this? I hate the tiny details like this..
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for E2:
Small base gives most substituted alken, large, bulky base gives least substituted.

E1:
major product is always the most substituted.
 
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i asked this same question a few days ago.. and no one could come up with an explanation
 
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Can you look at roadmap 5

where it goes at the top of the page from 2-bromo-2-methylbutane and reacts with -CN/heat

Isn't -CN a strong nucleophile? why would it react E1 instead of E2 makes no sense..
 
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Yeah I have been trying to figure this out..Its confusing but I think I got it.

First thing is..CN- is a weak base (I checked my textbook) We also know its a strong nucleophile. You look at bases for elimination and nucleophiles for substitution

Its a tertiary carbon..and although theres a strong nucleophile, tertiary carbons dont go Sn2. And since we look at CN- as a base now, we see its a weak base. Which means its E1. It cant go SN1 because its a strong nucleophile

The heat further shows E1

It cant go E2, because E2 needs a strong base

Wow I edited this like 10 times, but I got it finally! :)
 
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Haha Chad was right, I answered my own question from my textbook

from my text it shows 2-bromo-2-methylbutane being attacked by tert-butoxide and the least sub. product is formed.

But it then goes on to say:

"If the alkyl halide is not sterically hindered and the base is only moderately hindered the major product will be the more stable product"

It shows Tert-butoxide attacking 2-Iodobutane and it gives 2-butene as the major product. so destroyer was wrong that time I guess. Either that or its another one of those exceptions.

Idk what destroyers deal is: on q 100 (2009 ed) step 2 it does the same thing and gives the less sub product from 2-Chlorobutane. I hope someone here can explain this to me.
 
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yeah you're right CN is a weak base so it would be in fact E1 because heat forces an E1 reaction.. I see thanks !
 
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np

Ok does anyone know the story about what to do with sterically hindered bases reacting with alkyl halides that are not tertiary? Does the solvent have anything to do with more or less sub product?

I still have not figured it out.
 
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BiomajorPreDent said:
np

Ok does anyone know the story about what to do with sterically hindered bases reacting with alkyl halides that are not tertiary? Does the solvent have anything to do with more or less sub product?

I still have not figured it out.
Click to expand...

I just made a thread where I have encouraged others to post usefull charts. I posted a chart all about SN-E2 reactions. I will post it to this message also. Study it for a few minutes and see if it clears anything up for you. If not I will try to explain it... Hope that helps
 

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Philippines03j said:
I just made a thread where I have encouraged others to post usefull charts. I posted a chart all about SN-E2 reactions. I will post it to this message also. Study it for a few minutes and see if it clears anything up for you. If not I will try to explain it... Hope that helps
Click to expand...

hey Philippines thanks for the chart,
I have a quick question if you look at Road map 5 where it has

2-bromo-2-methylbutane reacted with -CN/heat

on your chart you have any charge with a tertiary carbon will react in fact E2 which is what I thought. However it appears CN- is a weak base but a strong nuclophile and destroyer seems to say that this will undergo E1 any ideas?

i thought me and biomajor settled this but your chart just made me more confused... :laugh:
 
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dennisDDS said:
hey Philippines thanks for the chart,
I have a quick question if you look at Road map 5 where it has

2-bromo-2-methylbutane reacted with -CN/heat

on your chart you have any charge with a tertiary carbon will react in fact E2 which is what I thought. However it appears CN- is a weak base but a strong nuclophile and destroyer seems to say that this will undergo E1 any ideas?

i thought me and biomajor settled this but your chart just made me more confused... :laugh:
Click to expand...

Ok, I read the conversation again and again... I realized at the begining we were talking about secondary halides and elimination and then it switched to tertiary. So after looking in my text book and examining my chart I like you started to second guess my self. First off I dont have road map #5 I have the 2007 edition. So I dont see exactly how it is telling you it is an E1. Does it say it right above/at the reaction "E1"? If so I am not sure what He is talking about. I looked up some charts similar to mine on other web sites. Here they are
1. http://www.personal.psu.edu/the1/e2.htm Look at the bottom of the page it has a chart saying that E1 is only in competition with SN1. E2 occurs when a base is present.
2. http://www.learnchem.net/orgo/sn1.shtml bottom of this page also
3.http://faculty.northseattle.edu/jpatterson/chem238/pdf/235sn1sn2.pdf and on this chart it kind of explains the heat thing with E1 and it competing with SN1.

Sorry that I confused you more than before... I just know that E1 is usually a minor product and it competes with SN1.
 
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No offense, but your chart is confusing me

this one seems to be okay though:

http://faculty.northseattle.edu/jpatterson/chem238/pdf/235sn1sn2.pdf

The reaction listed in the roadmap shows 2-bromo-2-methylbutane reacting with cyanide anion and heat forming 2-methyl-2-Butene


I am sticking to my knowledge of these reactions to solve this

CN- is a weak base. and also a strong nucleophile

Eliminations need bases, Substitutions need nucleophiles

So since its a weak base, right there it cant be E2. Now it cant be SN2 either because its a tertiary halide.

Now between E1 and Sn1.

Heat promotes Elimination over substitution, and we already stated that it is a strong nucleophile, so E1 is the only option left.

I hope that cleared it up.

Yeah Philippines03j sorry, my original question was relating to E2 reactions with substituted bases, like t-butoxide.

Destroyer shows any alkyl halide (usually theyre secondary in the examples) reacting with a bulky base to form the less sub (hoffman) product.

But Chad's videos says this is only the case if you have a tertiary alkyl halide. If you have anything like 2ndary or primary it is still the zaitsev product.

my textbook seems to say the same thing. But is not too specific. I am trying to figure out what exactly you need to have the less sub product
 
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haha o man.. yeah I sorta hi-jacked Biomajor thread only because I too asked the same question he did and got no response.. sorry about that..


but yeah.. Chad says most people confuse the tertbutoxide doing the least subsitutted alkene. And it takes a lot for someone to talk about it for like 2 mins so I guess I have to agree with chad.. lol..
 
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I have a question on Roadmap 5 (2009 edition)...

On the right side of the page, they show a 2-bromobutane undergoing elimination....but the LEAST substituted double bond is formed to give a butene with the double bond on the first carbon.

I thought elimination favors the MOST substituted double bond?? Or are they just showing a minor product??

Please help...I feel like this is a fundamental concept I should know. Thanks!
 
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hahaha too funny. I even searched the forums before posting, and that thread didn't come up!

At least I'm not the only one confused....man, that thread was hard to follow. From what I can tell, no one really gave a straight answer to the question that BioMajorPreDent and I have, right?

Also, just want to be sure...the reaction would probably be an E2 since it uses a bulky base?
 
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predental1187 said:
hahaha too funny. I even searched the forums before posting, and that thread didn't come up!

At least I'm not the only one confused....man, that thread was hard to follow. From what I can tell, no one really gave a straight answer to the question that BioMajorPreDent and I have, right?

Also, just want to be sure...the reaction would probably be an E2 since it uses a bulky base?
Click to expand...

yeah.. I think it got confusing because i asked him another question.. shouldn't have done that..

but Chad said that in fact it does do the more subsitutted alkene while destroyer said no.. biomajor looked it up in the book and said it does do least subsituted but changes to the more sub when there is no steric hinderance so to sum we didn't get a definite answer..

I would vote for E2 too...
 
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lol Dennis its ok

Yeah OP, I have no clue what to do in this situation. I am hoping to god that if I see a reaction like this that they want the less sub product for they are going to make it a tertiary alkyl halide with a clearly bulky base

The exceptions on the website you pointed out are in my text.

The reason I think destroyer is wrong is b/c in my text it shows two reactions

one has 2-bromo-2-methylbutane (which is tertiary) and reacting with tert-butoxide it gives less sub product

then right under it they show 2-iodobutane with tertbutoxide and it gives the more sub product

the 2-X-butane molecule is in destroyer going to a less sub product in like 4 places atleast. (road map 5 on the right, and question 100 step 2, question 151, and question 158)

So I think destroyer is wrong

NOTE: in my text..they show a solvent of (CH3)3COH in the reaction with 2-bromo-2-methylbutane and they dont show a solvent with the reaction with 2-iodobutane.

But I dont see how the solvent can be involved and it doesnt say anything about the solvent. In every instance in destroyer, there is this same solvent.

Does anyone know if it plays a role in the product?
 
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Ugh, I definitely see the discrepancy now. The reason I thought that website sided with Destroyer was because it said that as the bases get bulkier, less substituted alkenes can form. Therefore, I thought it was saying that when the base is very bulky, either less or more substituted alkenes are possible. And I figured Destroyer may have chosen the less substituted possibility for the purpose of the roadmap sequence.

What textbook do you have? I couldn't find any similar reactions in mine :( I think I will trust your text though, over destroyer/the ambiguous statement on this website.


Hopefully, like you said, it would be pretty obvious on the test. Looks like you've taken it already...do you remember any ambigious-type reactions like this being on there or was it usually pretty straight-forward ones whre only one product was possible?
 
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from what i got from chad's lectures...

Big Bulky Base+Tertiary-->Least Sub(used for antimarkonikov)
Big Bulky Base + primary or secondary-->Most Sub/with NO rearrangment, many times this no rearrangment thing causes the double bond to occur in the least subed area.(used to ensure E2 hence no rearrangment)

correct me if i'm wrong...
 
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BiomajorPreDent said:
Destroyer shows any alkyl halide (usually theyre secondary in the examples) reacting with a bulky base to form the less sub (hoffman) product.

But Chad's videos says this is only the case if you have a tertiary alkyl halide. If you have anything like 2ndary or primary it is still the zaitsev product.

my textbook seems to say the same thing. But is not too specific. I am trying to figure out what exactly you need to have the less sub product
Click to expand...

In my opinion Destroyer is wrong on question 100 (2009 Ed). The product of the second step should be the more sub'd product 2-butene, rather than 1-butene. What confuses me is that this is an E2 reaction (E2 reactions favor polar aprotic solvents) and we are using (CH3)COH, a polar protic solvent. If polar aprotic solvents increase the rate of E2 reactions, maybe it's possible that the polar protic solvent in this reaction is what causes the formation of the less sub'd product... That question is definitely confusing.
 
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I have no clue what to do in this situation. I am hoping to god that if I see a reaction like this that they want the less sub product for they are going to make it a tertiary alkyl halide with a clearly bulky base

The reason I think destroyer is wrong is b/c in my text it shows two reactions (my textbook is Organic Chemistry 5thEd by Bruice)

one reaction is: 2-bromo-2-methylbutane (which is tertiary) and reacting with tert-butoxide it gives less sub product

then right under it they show: 2-iodobutane with tertbutoxide and it gives the more sub product

the 2-X-butane molecule is in destroyer going to a less sub product in like 4 places atleast. (road map 5 on the right, and question 100 step 2, question 151, and question 158)

So I think destroyer is wrong

NOTE: in my text..they show a solvent of (CH3)3COH in the reaction with 2-bromo-2-methylbutane and they dont show a solvent with the reaction with 2-iodobutane.

But I dont see how the solvent can be involved and it doesnt say anything about the solvent. In every instance in destroyer, there is this same solvent. Which is protic like you pointed out..so that makes even LESS sense..

Does anyone know if it plays a role in the product?
 
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