Destroye Ochem Q71

[cuttiedentist]

Advertisement - Members don't see this ad

hi everyne
i am a little confused about Q71 on destroyer. why dosnt butanol rearrange to give 2butene insted of 1 butene when react with heat and acid. doesnt alcohols usuallr rearrange when they go through elimination with acid and heat?
thanks

 

[cuttiedentist]

anyone?

 

[shoombol tallah]

yeah i was wondering the same myself. wouldn't the 2 butene be more stable and therefore the more favored product?

 

[dsony2284]

you mean the tautomeric shift? yea i was also wondering this... i added a post earlier but no one responded

 

[ericdds]

To answer your question (Destroyer Q 71)

The reason there is no rearrangement in the dehydration of butanol is because the formation of the double bond is synchronized with the leaving of the water (protonated hydroxyl group) molecule. Another way to think about this: E2 reaction. First the hydroxyl group is protonated, then, another water molecule comes in and abstracts a proton from carbon #2 as the water is leaving from carbon #1. The only way you could get rearrangements is if you have a cation intermediate, which is not the case here.

Hope that helps.

Eric

 

[shoombol tallah]

i guess whats confusing me about the reaction is that heat is one of the reagents and i thought heat was always synonymous with E1 reactions (most substituted alkene)

 

[ericdds]

i guess whats confusing me about the reaction is that heat is one of the reagents and i thought heat was always synonymous with E1 reactions (most substituted alkene)

That's not the case. Even though you are applying heat, it is a primary alcohol and therefore does not react in an E1 manner to a large extent. The way you need to think about this is: In this dehydration reaction, there will be 2-butene (produced by a E1 mechanism) and 1-butene (produced by an E2 mechanism). However, there will be a majority of 1-butene (more than 80%) because butanol is a primary alcohol and would prefer to react by E2 mechanism because a primary carbocation is not very stable. Therefore, there is a competition between E1 and E2. In this case E2 wins. Most people make a mistake of thinking about organic chemistry reactions in a black and white manner. It's not like you're going to produce 100% of 1-butene here. Dr. Romano is just saying that 1-butene will be the major product produced. So, in explaining that, you are partially right, there will be some 2-butene produced, but not as much as 1-butene.

When you get these rxn's always think about all the factors (you were just thinking about heat). In this case, consider the heat, the substitutions on the alcohol.

Hope that helps

 

[jdoc04]

I thought primary was Sn2 or E2 only and never E1? Since this was a dehydration the result is the E2 product.

 

[Mstoothlady2012]

That's not the case. Even though you are applying heat, it is a primary alcohol and therefore does not react in an E1 manner to a large extent. The way you need to think about this is: In this dehydration reaction, there will be 2-butene (produced by a E1 mechanism) and 1-butene (produced by an E2 mechanism). However, there will be a majority of 1-butene (more than 80%) because butanol is a primary alcohol and would prefer to react by E2 mechanism because a primary carbocation is not very stable. Therefore, there is a competition between E1 and E2. In this case E2 wins. Most people make a mistake of thinking about organic chemistry reactions in a black and white manner. It's not like you're going to produce 100% of 1-butene here. Dr. Romano is just saying that 1-butene will be the major product produced. So, in explaining that, you are partially right, there will be some 2-butene produced, but not as much as 1-butene.

When you get these rxn's always think about all the factors (you were just thinking about heat). In this case, consider the heat, the substitutions on the alcohol.

Hope that helps

nice explanation

 

[cuttiedentist]

thanks for all the explanations.
just to make the point clear. E2 never rearranges. right?

 

[levyusupov]

thanks for all the explanations.
just to make the point clear. E2 never rearranges. right?

E2 never rearenge because it is concerted rxn (one step) and no intermidiate carcocation is formed.

 

[peRsEVERENCE]

To answer your question (Destroyer Q 71)

The reason there is no rearrangement in the dehydration of butanol is because the formation of the double bond is synchronized with the leaving of the water (protonated hydroxyl group) molecule. Another way to think about this: E2 reaction. First the hydroxyl group is protonated, then, another water molecule comes in and abstracts a proton from carbon #2 as the water is leaving from carbon #1. The only way you could get rearrangements is if you have a cation intermediate, which is not the case here.

Eric

wikipedia says primary alcohol will go via E1

"More useful is the E1 elimination reaction of alcohols to produce alkenes. The reaction generally obeys Zaitsev's Rule, which states that the most stable (usually the most substituted) alkene is formed. Tertiary alcohols eliminate easily at just above room temperature, but primary alcohols require a higher temperature."
http://en.wikipedia.org/wiki/Alcohol

 

[peRsEVERENCE]

E1 does not require a strong base while E2 does. Moreover, rearrangements occur instantaneously.

 

[El Sol]

wikipedia says primary alcohol will go via E1

"More useful is the E1 elimination reaction of alcohols to produce alkenes. The reaction generally obeys Zaitsev's Rule, which states that the most stable (usually the most substituted) alkene is formed. Tertiary alcohols eliminate easily at just above room temperature, but primary alcohols require a higher temperature."
http://en.wikipedia.org/wiki/Alcohol

Kaplan also say's E1 for dehydration of alcohol. Their example even has the dehydration of butanol.

 

[ericdds]

wikipedia says primary alcohol will go via E1

"More useful is the E1 elimination reaction of alcohols to produce alkenes. The reaction generally obeys Zaitsev's Rule, which states that the most stable (usually the most substituted) alkene is formed. Tertiary alcohols eliminate easily at just above room temperature, but primary alcohols require a higher temperature."
http://en.wikipedia.org/wiki/Alcohol

My friend, you are a prime example of why people should not be relying on wikipedia. What wikipedia said (the quote that you posted) is semi-true, I said in my previous post: There is a competition between E2 and E1. But the majority of butanol will undergo E2; E1 is a small side reaction. Here is where they are wrong: They are saying that the major product will be from an E1, it should say from and E2. Wikipedia is not contradicting what I said earlier; they're just not too clear. But just in case you don't believe me. I did some reading in my old Organic Chem book "Solomons & Fryhle, Organic Chemistry, 8th Edition" p. 302 section 7.7C. It goes like this, "The dehydration of primary alcohols is believed to proceed through a E2 mechanism because the primary carbocation required for dehydration by an E1 mechanism would be too unstable............." It also gives examples. Please look at this page, it will clear everything up.

Finally, to establish my credentials, I have been a Teacher Assistant for Organic Chemistry in the University of Washington for a year. I know what I am talking about. So you could take it, or leave it. But at least, open your textbook and read about it. For God's sake don't rely on Wikipedia where any ***** can post his two cents.

Eric