# Predicting product ratio by rearrangement in Dehydration Reactions

I came across this question recently, which would have a clear major product. The OP has given satisfactory explanation for the product ratio, but wasn't sure about the major product. When I was looking for reference for better explanation, I show the following question in undergraduate textbook (Ref.1):

Dehydration of 2,2,4-trimethyl-3-pentanol with acid gives a complex mixture of the alkenes in the indicated percentages. Write a mechanism that accounts for each product.

• I: 2,3,4-trimethyl-1-pentene, 29%
• II: 2,4,4-trimethyl-1-pentene, 24%
• III: 3,3,4-trimethyl-1-pentene, 2%
• IV: 2,4,4-trimethyl-2-pentene, 24%
• V: 2,3,4-trimethyl-2-pentene, 18%
• VI: 2-isopropyl-3-methyl-1-butene, 3%

I thought it would be a good practice for our readers to predict the mechanism for the obtained product ratio and give explanation for the major products. Can anybody give a reasonable mechanism for this product ratio and explanation for why $$\bf{I} \gt \bf{II} \ge \bf{IV} \gt \bf{V}$$, and why $$\bf{III}$$ and $$\bf{VI}$$ are in such small amounts?

Late edit: It is evident that the textbook print had an error (Thanks Nisarg Bhavsar for the finding). The compound V is actually 2,3,4-trimethyl-2-pentene, not as 3,3,4-trimethyl-2-pentene in the print. I have corrected it now. Nonetheless, the question is still interesting.

References:

1. Robert J. Ouellette and J. David Rawn, “Chapter 9: Haloalkanes and alcohols Nucleophilic Substitution and Elimination Reactions,” In Organic Chemistry: Structure, Mechanism, Synthesis; Second Edition; Academic Press (an imprint of Elsevier): London, United Kingdom, 2019, pp. 255-298 (ISBN: 978-0-12-812838-1).

The reaction pathway I would propose for the obtained products is as below:

Coming to the percentage of each product obtained, I would defer from the data provided in the question. I would believe that V would be the major product because it is both thermodynamically and kinetically the best possible product.

I would be the next best as it is formed from the best intermediate.

Than II and IV would be the next major products as they are formed from the second best intermediate. Actually I would assume that IV would have a slight higher percentage due to the fact that it is thermodynamically more stable.

III and VI are produced in very less quantities because to form the carbocation required to form these products it has to undergo a very unfavourable rearrangement which would happen only in very less proportion.

So according to me the order should be V>I>IV≥II and very minor quantities of III and VI.

• You are correct about the compound V. You can do the editing to your answer now. – Mathew Mahindaratne May 23 at 18:54
• @MathewMahindaratne I have edited the answer in response to the edit to the question. – Nisarg Bhavsar May 23 at 19:07
• @ Nisarg Bhavsar: The double headed arrows are for resonance structures. In your scheme, they are not resonances but intermediate carbocations. Hydride or methide shifts did not imply resonance. – Mathew Mahindaratne May 23 at 20:27
• @MathewMahindaratne Opps! They were supposed to be equilibrium signs. I am editing them – Nisarg Bhavsar May 23 at 20:41
• @ Nisarg Bhavsar: Actually, $\bf{IV} \gt \bf{V}$ simply because hydride shift is faster than methide shift. In addition, $\bf{IV}$ can be achieved from initial secondary carbocation as well. – Mathew Mahindaratne May 23 at 21:44

Although this question is from textbook written by a professor in well established institute, it lacks completeness. For example, the question is written such a poor way that it even didn't give any conditions or even in which solvent the reaction has performed. So, it's safe to assume that the reaction has performed in thermodynamic control. We'll say it is an acid catalyzed dehydration reaction in a refluxing condition in a protic solvent. Thus, it'd be a $$\mathrm{E1}$$ elimination reaction with carbocation intermediate(s). Let's see the products and their yields:

The amount of products with significant yields suggest that the reaction has gone through few relatively stable intermediates. Let's look at these possible intermediates:

The original dehydration of the substrate gives secondary carbocation, intermediate 1. This intermediate can gives only one product, 2,4,4-trimethyl-2-pentene $$(\bf{IV})$$, which is Zaitsev product (there are no possibility to form a Hofmann product from this intermediate). Since $$\bf{IV}$$ is not the only product detected, it is fair to say rate of this product formation is slower than the carbocation (intermediate 1) rearrangement to give more stable tertiary carbocation(s). Intermediate 1 can be stabilized $$(2^\circ \rightarrow 3^\circ)$$ by either hydride shift (reaction path $$b$$) to give intermediate 2 or methide shift (reaction path $$a$$) to give intermediate 3. Intermediate 3 can be further rearranged by another hydride shift (reaction path $$c$$) to give intermediate 4, which can be less favorably $$(3^\circ \rightarrow 2^\circ)$$ rearranged to intermediate 5 by a methide shift (reaction path $$d$$). Note that this secondary carbocation, intermediate 5, can gives only one product, 3,3,4-trimethyl-1-pentene $$(\bf{III})$$ with the least yield percentage (2%) to justify.

Let's see how would the products would formed by these five intermediate carbocations:

Intermediates 2, 3, and 4 are tertiary carbocations with possibility to form both Zaitsev and Hofmann products. In one glance, one might think all would give favorable Zaitsev products under the conditions. However, in reality, they have given almost equal ratio of both Zaitsev and Hofmann products except for intermediate 5, which gave only 3% of the Hofmann product, 2-isopropyl-3-methyl-1-butene $$(\bf{VI})$$. This result is also justified by less steric hindrance on proton abstraction from either of two isopropyl groups to form the Zaitsev product, 2,3,4-trimethyl-2-pentene $$(\bf{V})$$, compared to other two intermediates:

• In intermediate 2, a proton must be abstracted from $$\ce{C}$$3, which has an enormous steric hindrance created by nearby tertiary-butyl group. Thus, Hofmann product, 2,4,4-trimethyl-1-pentene $$(\bf{II})$$ would be the major product from this intermediate (there are two methyl groups to give this product). However, its Zaitsev product, 2,4,4-trimethyl-2-pentene $$(\bf{IV})$$, is also produced by intermediate 1, and hence it could be expected that $$\bf{IV}$$ may have significant yield as well.
• In intermediate 3, a proton must be abstracted again from $$\ce{C}$$3, which has almost equally enormous steric hindrance created by nearby iso-propyl group in addition to methyl group on $$\ce{C}$$3. The two methyl groups on positively charged carbon will play a role as well. Thus, Hofmann product, 2,3,4-trimethyl-1-pentene $$(\bf{I})$$ would be the major product from this intermediate (there are two methyl groups to give this product). However, its Zaitsev product, 2,3,4-trimethyl-2-pentene $$(\bf{V})$$, is also produced by intermediate 4 as its major product, and hence it could be expected that $$\bf{V}$$ may have significant yield as well.

It's worth noting that according to the product ratios, it is safe to say that intermediate 3 is the major contributor during this reaction. Even though hydride shift is faster to form intermediate 2, its relatively slow rate of double bond formation to give Zaitsev product due to the steric hindrance by tert-Butyl group may have coursed the slower methide shift to dominate at the end to get intermediate 3.