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.
