Animated Solution for Chemistry - Organic Chemistry: The major product ($Y$) in the following reaction is ${\text{CH}}_3-\stackrel{\stackrel{{\text{CH}}_3}{|}}{\mathrm{CH}}-\text{C}\equiv \text{CH}\underset{{\text{H}}_2\text{O}}{\overset{{\text{HgSO}}_4,{\text{H}}_2{\text{SO}}_4}{\to }}X\underset{{\text{(ii) Conc. H}}_2{\text{SO}}_4/\Delta }{\overset{{\text{(i) C}}_2{\text{H}}_5{\text{MgBr, H}}_2\text{O}}{\to }}Y$

From Alkyne to Alkene

The beauty of organic synthesis lies in its logical progression. In this problem, we are given a terminal alkyne, specifically 3-methylbut-1-yne, and we are tasked with predicting the final product after a sequence of three distinct chemical transformations. Let's break down this journey step by step.

The Kucherov Hydration

Our first set of reagents is $HgS{O}_4$, $H_{2}S{O}_4$, and $H_{2}O$. This is the classic Kucherov reaction, designed to hydrate alkynes. When water adds across the triple bond, it strictly follows Markovnikov's rule. The hydroxyl ($-OH$) group attaches to the more substituted carbon atom of the alkyne, while the hydrogen attaches to the less substituted terminal carbon.

This initial addition forms an enol intermediate: $C{H}_3-CH(C{H}_3)-C(OH)=C{H}_2$. However, enols are generally unstable and rapidly undergo tautomerization. The double bond shifts, and the proton migrates to the terminal carbon, yielding a highly stable ketone. Thus, our intermediate $X$ is 3-methylbutan-2-one.

The Grignard Addition

Next, we treat our newly formed ketone $X$ with ethyl magnesium bromide ($C_2{H}_{5}MgBr$), a powerful Grignard reagent, followed by an aqueous workup. The Grignard reagent provides an ethyl carbanion ($C_2{H}_5^{-}$), which acts as a strong nucleophile.

This nucleophile attacks the electrophilic carbonyl carbon of the ketone, breaking the carbon-oxygen $\pi$ bond and pushing the electrons onto the oxygen to form an alkoxide ion. Upon hydrolysis with $H_{2}O$, the oxygen is protonated. This transforms our ketone into a tertiary ($3^{\circ }$) alcohol: 2,3-dimethylpentan-3-ol.

Acid-Catalyzed Dehydration

In the final step, we heat this tertiary alcohol with concentrated sulfuric acid ($Conc.H_{2}S{O}_4/\Delta$). This initiates an acid-catalyzed dehydration via an $E1$ mechanism. The strong acid protonates the hydroxyl group, turning it into an excellent leaving group (water). As water departs, it leaves behind a tertiary carbocation: $C{H}_3-CH(C{H}_3)-C^{+}(C_2{H}_5)-C{H}_3$.

To form the final alkene, a proton must be eliminated from a carbon adjacent to the carbocation. Here, we must apply Saytzeff's rule, which dictates that the major product will be the most substituted, and therefore most thermodynamically stable, alkene.

We have three adjacent carbons to choose from. Removing a proton from the $C{H}_3$ group yields a disubstituted alkene. Removing a proton from the $C{H}_2$ of the ethyl group yields a trisubstituted alkene. However, removing the single proton from the adjacent $CH$ group yields a tetrasubstituted alkene.

Following Saytzeff's rule, the elimination of the proton from the $CH$ group is heavily favored. This results in the formation of 2,3-dimethylpent-2-ene, which corresponds to structure $Y$.

Comparing this to our given options, it perfectly matches option (a). This problem is a fantastic demonstration of how fundamental organic reactions—hydration, nucleophilic addition, and elimination—can be chained together to build complex molecular architectures.