The given compound is 2-methylbutan-2-ol. $$ \begin{matrix} & & \text{OH} & \\ & & | & \\ \text{CH}_3 & - & \text{C} & - & \text{CH}_2 & - & \text{CH}_3 \\ & & | & \\ & & \text{CH}_3 \end{matrix} $$ Dehydration of an alcohol involves the removal of a hydroxyl (-OH) group from the alpha-carbon (the carbon bearing the -OH group) and a hydrogen atom from an adjacent beta-carbon. This reaction typically occurs in the presence of an acid catalyst (like $\text{H}_2\text{SO}_4$ or $\text{H}_3\text{PO}_4$) and heat, forming an alkene. For tertiary alcohols, the reaction proceeds via an E1 mechanism. According to Zaitsev's rule, the major product of an elimination reaction is the most substituted alkene (the alkene with the most alkyl groups attached to the double-bonded carbons). Let's identify the alpha-carbon and the beta-carbons in 2-methylbutan-2-ol: The alpha-carbon is the carbon bonded to the -OH group. The beta-carbons are the carbons directly bonded to the alpha-carbon. $$ \begin{matrix} & & \text{OH} & \\ & & | & \\ \text{CH}_3^{\beta_1} & - & \text{C}_{\alpha} & - & \text{CH}_2^{\beta_2} & - & \text{CH}_3 \\ & & | & \\ & & \text{CH}_3^{\beta_3} \end{matrix} $$ There are two types of beta-hydrogens that can be removed: 1. Removal of H from $\text{CH}_3^{\beta_1}$ or $\text{CH}_3^{\beta_3}$ (both are equivalent methyl groups): If a hydrogen is removed from one of the methyl groups ($\text{CH}_3$) and the -OH group is removed from the alpha-carbon, a double bond forms between $\text{C}_{\alpha}$ and the methyl carbon. The product formed is: $$ \begin{matrix} & & & \\ & & \text{CH}_3 & \\ & & | & \\ \text{CH}_2 & = & \text{C} & - & \text{CH}_2 & - & \text{CH}_3 \end{matrix} $$ This compound is 2-methylbut-1-ene. The double bond carbons are $\text{C}_1$ and $\text{C}_2$. $\text{C}_1$ has two hydrogens. $\text{C}_2$ has a methyl group and an ethyl group. This is a disubstituted alkene (two alkyl groups on the double bond carbons). This corresponds to option D. 2. Removal of H from $\text{CH}_2^{\beta_2}$ (methylene group): If a hydrogen is removed from the methylene group ($\text{CH}_2$) and the -OH group is removed from the alpha-carbon, a double bond forms between $\text{C}_{\alpha}$ and the methylene carbon. The product formed is: $$ \begin{matrix} & & & \\ & & \text{CH}_3 & \\ & & | & \\ \text{CH}_3 & - & \text{C} & = & \text{CH} & - & \text{CH}_3 \end{matrix} $$ This compound is 2-methylbut-2-ene. The double bond carbons are $\text