[Marcus]

Is it possible to obtain a single product from the dehydration of an alcohol, or will there always be secondary products?

Given the formula C₇H₁₄O, are there any other isomers that could yield a single alkene product?

[stewie griffin]

Weren't you already identified as doing some NMR for a lab class?
Is this question by any chance related to another lab/experiment report?
Shouldn't you make an attempt at the answer yourself (as the forum rules dictate)?

[Marcus]

Indeed, I was "identified" as inquiring about NMR spectra for a lab, and you are correct in assuming that this problem is also related to labwork.

I believe a tertiary alcohol on a methylated cyclic alcochol would readily dehydrate to form an alkene, with the double bond connecting with the methylated carbon, the most substituted position. It seems to me that no matter how the carbocation is formed, the end result will be a single product of 1-methylcycloalkene.

Is that correct?

[democanarchis]

Alcohol dehydration generally follows Zaitsev's rule. There is usually a major and minor product. Is there another product possible from that cyclic alcohol?

[Edit] Didn't see the requirement that the product be an alkene. [/Edit]

Were you given that structure? There are other cyclic tertiary alcohols possible.

[Marcus]

The initial compound is a methylated cyclohexanol.

I was wondering if 1-methylcyclohexanol, being a tertiary alcohol, would dehydrate into simply one product? Basically, is it possible at all for any isomer of the above formula to give anything less than two products?

[stewie griffin]

You can easily answer this question by simply drawing out the mechanism. First off, ask yourself if this is E1 or E2 elimination.

[Arctic-Nation]

The initial compound is a methylated cyclohexanol.

I was wondering if 1-methylcyclohexanol, being a tertiary alcohol, would dehydrate into simply one product? Basically, is it possible at all for any isomer of the above formula to give anything less than two products?

Unless you can draw an isomer that has three identical substituents on the carbon atom bearing the alcohol, no.

[Marcus]

So, the formation of the carbocation (as is typical of E1 reactions) precludes the formation of only one product, since it can receive protons from any adjacent carbons?

Actually, I guess I'm misunderstanding the mechanism, or the definition of isomers, but shouldn't this be the result of dehydration of the tertiary alcohol?



If so, are the two products actually different?

[tmartin]

Here you have only drawn one product...and just rotated it around and drawn it again.  You are however missing something.
If one were to be really nit-picky, which in this case  I would be... your arrows will not get you to the product you have drawn (unless my computer is selectively deleting atoms again).  I understand what you meant to draw, but perhaps if you drew it out correctly, you would give yourself a significant hint.

[Marcus]

Ahh... would there be formation of a terminal alkene between the primary carbon and the methyl group?

Forgot to include deprotonation of carbons via H2O.

[eclecticmess07]

two words: achiral and symmetrical.

[tmartin]

Marcus:  The last comment you made would seem to indicate you're thinking along the lines of Saytzeff and Hoffman products (which I think was one of the points of the lab), so I would follow that line of reasoning.

[stewie griffin]

Perhaps your TA would be better to ask for more in depth one on one assistance.
So you've identified it's E1 and goes through a carbocation. Good. So... draw out EVERY SINGLE HYDROGEN that's alpha to the carbocation. ANY of these H's can be involved in the elimination. Thus to answer your own question, you should go through and draw each product resulting from removal of each H. Are any of the products the same? 
Hint Marcus: your most recent comment shows you are thinking correctly

[darobian]

Sorry for replying to an old thread but I think that a complete answer might be helpful to others now we can be quite sure that the deadline for the original poster's homework has passed so we're not supporting plagiarism.

The initial compound 1-methylcyclohexanol, presumably the product of a Grignard type methylation of cyclohexanone, bears an alcohol group that is vulnerable to dehydration through exposure to heat and/or acidic conditions.

I make the reaction equation and mechanism as follows:



Removal of the protonated hydroxyl group, leaves the carbocation that creates the methylcyclohexene (major product) and methylenecyclohexane (minor) products discussed. Water is of course the final product of a dehydration reaction.

As I'm a total newbie and this is only my first post, I'm not entirely sure if the conventions for denoting reversibility and equilibria have been followed correctly. I'm pretty sure the mechanism is is essentially correct though.

Greetings to everyone on the forum, hope I can help some people.

Cheers,
Andrew

[orgopete]

Wouldn't 4-methylcyclohexanol give a single alkene product?