What happens in dehydrations of primary alcohols (in presence of acid catalyst at high temperatures) which don't have a β-hydrogen?

For example, in the case of neopentyl alcohol, will it dehydrate as it doesn't have a β-hydrogen? Obviously, E2 reaction will not be possible, but can it react via an E1 reaction, with a rearrangement in the carbocation step?

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    $\begingroup$ Under what dehydration conditions? $\endgroup$ – Waylander Jul 18 at 6:03


Most dehydration reactions proceed via E1 method in (acidic conditions because that is the most common way of proceeding with it)

So the the protons in the acidic medium attack the lone pair of oxygen. And this in turn leads to the cleavage of the C-O bond, as it is weaker than C-H bond.

So this leads to the formation of a carbocation on the carbon atom.

As you said, this undergoes rearrangements before a proton is released and thus a pi bond forms.

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  • $\begingroup$ Is ring expansion also possible in case of primary cyclo-alcohol? $\endgroup$ – Apurvium Aug 11 at 5:24
  • $\begingroup$ @Apurvium Yes it is. Any carbocation rearrangement happens provided the stability increases. $\endgroup$ – Vamsi Krishna Aug 11 at 6:16

In this case, $\mathrm{E2}$ elimination is impossible regardless of condition used, because of lack of $\beta$-hydrogens. However, in acidic conditions, it is possible to have elimination reaction. Since 2,2-dimethylpropanol is a $1^\circ$-alcohol, initial carbocation formation is difficult. However, this formation of carbocations is accompanied by a structural rearrangement, which is the driving force for the $1^\circ$-carbocation formation, which is otherwise not favored:

Product formation from neo-pentanol

Such rearrangements take place by a 1,2-shift of a neighboring alkyl group or aryl group or hydrogen, and are favored when the rearranged carbocation is more stable than the initial cation. In this case, migrating group is methyl group and carbocation changed from $1^\circ$ to $3^\circ$, which is very stable compared to initial carbocation and favored. Once $1^\circ$-carbocation formed, it quickly rearranged to $3^\circ$-carbocation and as a result, the reaction make forward progress. It can also be considered a concomitant elimination of water and methide migration where methide group acting as an intramolecular nucleophile (See the mechanism shown in bottom box of the scheme). Either way, the initial step is very slow because of very high activation energy:

Dehydration of 2,2-dimethylpropanol

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    $\begingroup$ Source of the pictures? $\endgroup$ – Mithoron Jul 18 at 15:51
  • $\begingroup$ @Mithoron: Thanks for reminding. I included the source. I answered this pass midnight, so probably sleeping to forget that. :-) $\endgroup$ – Mathew Mahindaratne Jul 18 at 16:02

If the reaction is being done in acidic conditions, dehydration of neopentyl alcohol can take place with E1 Mechanism. A 1,2 Methyl shift will happen and the major product will be 2-methylbut-2-ene.

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  • $\begingroup$ You can also get ether formation by reaction with a second molecule of alcohol $\endgroup$ – Waylander Jul 18 at 6:41
  • $\begingroup$ But the yield will be poor since formation of ether will take by SN2 mechanism. 2-methylbut-1-ene can be another minor product. $\endgroup$ – the.eleventh.letter Jul 18 at 7:05
  • $\begingroup$ that rather depends on the reaction condtions. 2-Me-2-butene can also be protonated to give fairly stable cation (see the excellent answer above) which act as an electrophile for the alcohol. $\endgroup$ – Waylander Jul 18 at 14:10
  • $\begingroup$ Yes, it can act as an excellent electrophile but I doubt weather a 3° alcohol will be able to act as a good nucelophile. $\endgroup$ – the.eleventh.letter Jul 18 at 16:07

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