Primary Alcohol Dehydration

[baronen26]

Is it e1 or e2? I am getting conflicting information when about when rearrangement is/isn't possible.

This site says its e2

http://chemwiki.ucdavis.edu/Organic...of_Alkenes/Alkenes_by_Dehydration_of_Alcohols

but then also provides an example where rearrangement occurs.

Chad say's its e2 (according to my notes my subscription is expired) and khan academy says rearrangement always occurs.

What does the dat expect on this subject?

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[Cool Beans]

Your question is really getting into the minutiae of the topic and the DAT is much more concerned with general principles. But if this gives you some peace, the original mechanism is E2, but once the alkene forms it will be in equilibrium with possible carbocations due to the addition of H+ in the acidic environment (think of the reaction of an alkene with H3O+). If a carbocation that forms could once again undergo elimination to form an even more substituted alkene than the initial one formed, then it will. This second elimination would be analogous to the last step of an E1 elimination.

To sum it up, dehydration of a primary alcohol to form an alkene will proceed via the E2 mechanism, but the rearrangement (if it occurs) would more resemble the E1 mechanism (at least the last step of it).

Hope this helps!

 

[FeralisExtremum]

Here's what I have on my notes from Chad's video on alcohol dehydration:

"Recap: We can also dehydrate alcohols (elimination reaction) (4-12). Follows Zaitsev’s rule: take the H from the more substituted adjacent carbon (less H’s available) to form the more substituted/stable alkene. Done with H2SO4, H3PO4, or mix of both: highly protic so E1 is preferred which goes through a carbocation intermediate (on 3º and 2 º alcohols), but 1º alcohols must go through E2. Mechanism not important, the basics: protonate the OH to make it good leaving group, then proceed through elimination."

So it looks like primary alcohols go by E2, secondary and tertiary by E1. Any minutiae beyond these basic rules are unlikely to be tested within the scope and depth of the DAT.

 

[baronen26]

yeah okay thank you that all makes sense.

Another specific question - LiAlH4 reduces anyhrides to two primary alcohols, but would NaBH4 only be able to reduce one side to a primary alcohol and the other is cleaved off as a carboxylate?

 

[FeralisExtremum]

NaBH4 wouldn't work on an anhydride at all, it's too weak (remember that it will only reduce ketones, aldehydes, and acid chlorides effectively). LiAlH4 is much stronger and would reduce it down to two primary alcohols.

 

[baronen26]

http://www.chem.wisc.edu/areas/reich/chem547/2-redox{06}.htm

also chad says nabh work on anything until an ester (in terms of reactivity) so i am assuming this means anhydride

 

[FeralisExtremum]

The link above does suggest it will work, but Chad says otherwise (verbatim):

"The key is this: this is what NaBH4 does - reacts with ketones, aldehydes, and acid chlorides (and you’re much more likely to see ketones and aldehydes). LAH can do everything that NaBH4 can do and more, LAH can also react with all the rest of the carboxacid derivatives as well."

Page 14 here also states "Although NaBH4 is a good source of creating a new C—H bond, it does not react with anything that is less reactive than an aldehyde or a ketone", and Mastering Organic Chemistry says: "What it’s used for: Sodium borohydride is a good reducing agent. Although not as powerful as lithium aluminum hydride (LiAlH4), it is very effective for the reduction of aldehydes and ketones to alcohols. By itself, it will generally not reduce esters, carboxylic acids, or amides (although it will reduce acyl chlorides to alcohols)." Which is interesting because it seems anhydride was intentionally omitted, but both sources seem to confirm what Chad says.

I'm guessing NaBH4 + anhydride is something of a grey area in terms of reactivity, which means it's unlikely to show up on the DAT. When in doubt, I'd stick with the basic level taught by Chad.

 

[baronen26]

thanks a lot for the responses... one last question - I'm getting super paranoid before my exam

I just saw on destroyer where cl2 and h20 added to benzamine to both ortho positions (para is blocked). I don't think i've seen an eas where it adds twice in one reaction.

 

[DAT Destroyer]

thanks a lot for the responses... one last question - I'm getting super paranoid before my exam

I just saw on destroyer where cl2 and h20 added to benzamine to both ortho positions (para is blocked). I don't think i've seen an eas where it adds twice in one reaction.

I think you mean Aniline,,,, The NH2 group is HIGHLY activating........thus is very difficult to stop in a single step.......thus both ortho positions are hit. In real life.......we have to protect the NH2 group to convert it into a less activating group.

BTW......this is indeed an Electrophilic Aromatic Substitution.......but as I stated.......Highly reactive !!!!!!

Hope this helps.

 

[DAT Destroyer]

Is it e1 or e2? I am getting conflicting information when about when rearrangement is/isn't possible.

This site says its e2

http://chemwiki.ucdavis.edu/Organic...of_Alkenes/Alkenes_by_Dehydration_of_Alcohols

but then also provides an example where rearrangement occurs.

Chad say's its e2 (according to my notes my subscription is expired) and khan academy says rearrangement always occurs.

What does the dat expect on this subject?

baronen26 this is a great question!

Normally a primary alcohol will dehydrate by E2......HOWEVER,,,,,,,if we have at least 4 carbons or more BOTH can OCCUR !!!! LOL...... Why ? A normal E2 certainly occurs......but it is possible for a simultaneous shift and attack......resembling the E1. We call this a mixed mechanism.. As a rule of thumb..... Secondary and tertiary alcohols,,,,,do E1 to dehydrate.....if under 3 carbons do E2. If over 3 carbons on a primary.....both occur.....with preference for the E1-like product.

Hope this helps.

 

[DAT Destroyer]

The link above does suggest it will work, but Chad says otherwise (verbatim):

"The key is this: this is what NaBH4 does - reacts with ketones, aldehydes, and acid chlorides (and you’re much more likely to see ketones and aldehydes). LAH can do everything that NaBH4 can do and more, LAH can also react with all the rest of the carboxacid derivatives as well."

Page 14 here also states "Although NaBH4 is a good source of creating a new C—H bond, it does not react with anything that is less reactive than an aldehyde or a ketone", and Mastering Organic Chemistry says: "What it’s used for: Sodium borohydride is a good reducing agent. Although not as powerful as lithium aluminum hydride (LiAlH4), it is very effective for the reduction of aldehydes and ketones to alcohols. By itself, it will generally not reduce esters, carboxylic acids, or amides (although it will reduce acyl chlorides to alcohols)." Which is interesting because it seems anhydride was intentionally omitted, but both sources seem to confirm what Chad says.

I'm guessing NaBH4 + anhydride is something of a grey area in terms of reactivity, which means it's unlikely to show up on the DAT. When in doubt, I'd stick with the basic level taught by Chad.

I see such misinformation here.....I would like to address the LiAlH4 question. I have done these reactions for more than 25 years .... Using LiAlH4 is best for any anhydride if your goal is to get the primary alcohol. Things get tricky with NaBH4.....the less powerful reducing agent. Saying they will not work is incorrect......I have used this reagent many times on cyclic anhydrides that reduced to lactones. Phthalic anhydride is a great example. Normally, however avoid NaBH4 with anhydrides. For the DAT......use LiAlH4 for a clean....sure -bet Reduction Kill.

Hope this helps.... Dr. Jim Romano

 

[baronen26]

okay so to recap
primary alcohols- rearranged product would be expected even though the mechanism most resembles e2
highly donating group can cause a double addition

and NaBH4 and LiAlH4 can reduce anhdrides. Both turn cyclic anhydrides into lactones?? What would NaBH4 do to a noncyclic anhydride?

 

[DAT Destroyer]

Usually non-cyclic anhydrides would not react......or very slowly with NaBH4...LiAlH4 would break the ring apart and make primary alcohols cleanly. As I stated.....cyclic anhydrides do react to form lactones with NaBH4.....the ring is destroyed with LiAlH4 to the primary alcohols. If you really want to pursue this....The Advanced Organic Chem book by Francis Carey is a classic.....but way beyond what you need for the DAT exam......For your first question......dehydrating a primary alcohol of say 4 carbons gives 2-butene as a major product. Other theories besides rearrangement have been proposed through the years. One states that you first form 1-butene,,,,,,,as normal,,,,by the usual E2...then.....the acidic environment protonates the 1-butene via a normal Markovnikov addition to give 2-butanol......which then forms your product !!!!!! Not easy to tell things apart......recall.....MECHANISMS are not laws...and can be modified or even changed. For the DAT......we will say 4 or more carbons does E1 like reaction......smaller ones like 2 or 3 carbons E2. I hope this helps,,,,,your question has been debated by us organic chemists for years, don't feel so bad about being confused.

Regards, Dr. Romano

 

[Artdent97]

Thanks for the detailed explanation.

 

[baronen26]

thanks!

 

[DAT Destroyer]

thanks!

You are welcome, please feel free to post other questions and I will do my best to answer.

 

[baronen26]

I was doing some math destroyer and I have a unit conversion question.
You are told that there are 231 cubic inches in a gallon...how many gallons are in two cubic yards.

My train of thought was 12*3*2= 72 inches in two yards, then, (72^3)/231. the answer is (36^3)*2/231. I understand why the second one is correct, but I guess I don't get what I'm missing. I would have quickly chosen my answer if it was an option.

wow okay nvm that would be 8 cubic yards...