See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is 3-methylbutan-2-ol (Me2CH–CH(OH)–Me), a secondary alcohol. Step 2 - Reaction with Al2O3 at 350°C (dehydration): Al2O3 at high temperature causes dehydration of the alcohol to give an alkene. The alcohol Me2CH–CH(OH)–Me loses water. The major product (A) by Zaitsev's rule would be the more substituted alkene. Removing H from the Me2CH side gives 2-methylbut-2-ene: Me2C=CH–Me. However, there can also be a carbocation rearrangement: the secondary carbocation (Me2CH–CH+–Me) can rearrange via hydride shift to give tertiary carbocation (Me2C+–CH2–Me), which then loses a proton to give 2-methylbut-2-ene (Me2C=CHMe) or 2-methylbut-1-ene. The predominant alkene (A) formed is 2-methylbut-2-ene (Me2C=CH–Me) via the more stable tertiary carbocation after hydride shift. Step 3 - Reaction with HI (Markovnikov addition): HI adds to 2-methylbut-2-ene (Me2C=CH–Me) following Markovnikov's rule. The H adds to the less substituted carbon (=CH–Me side, i.e., the CH), and I– adds to the more substituted carbon (the CMe2 side). This gives Me2C(I)–CH2–Me. Step 4 - Reaction with AgOH: AgOH reacts with the alkyl iodide in a nucleophilic substitution (SN1 or SN2), replacing I with OH. Me2C(I)–CH2–Me + AgOH → Me2C(OH)–CH2–Me + AgI. Step 5 - Product B: Me2C(OH)CH2Me, which is 2-methylbutan-2-ol, a tertiary alcohol. This corresponds to option (a). Why other options fail: - Option (b) is the same as the starting alcohol — no net change, which is not the result of this sequence. - Option (c) Me–CH(OH)–CMe3 would require a different carbon skeleton rearrangement not consistent with this pathway. - Option (d) HO–CH2–(CH2)Me is a primary alcohol not consistent with Markovnikov addition to 2-methylbut-2-ene. Therefore, the correct answer is A.