Which major product is formed in the acid-catalyzed hydration of ${\mathrm{CH}}_3$-${\mathrm{CH}}_2$-${\mathrm{CH}}_2$=${\mathrm{CH}}_2$ (3-methyl-1-butene)?

• A. ${\mathrm{CH}}_3$CH(OH)CH$2_2$CH$3_3$ (2-methyl-1-butanol)
• B. ${\mathrm{CH}}_3$CH$2_2$C(OH)(CH$3_3$)$2_2$ (tert-amyl alcohol)
• C. ${\mathrm{CH}}_3$CH$2_2$CH(OH)CH$3_3$ (2-methyl-2-butanol)
• D. ${\mathrm{CH}}_3$CH$2_2$CH$2_2$CH$2_2$OH (1-butanol)

Answer: (C)

Verified step by step guidance

1. Identify the structure of the starting alkene, 3-methyl-1-butene. Draw the molecule to clearly see the double bond location and substituents. The double bond is between C1 and C2, with a methyl group on C3.
2. Recall that acid-catalyzed hydration of alkenes proceeds via Markovnikov's rule, where the proton (H⁺) adds to the carbon with more hydrogens, and the OH group attaches to the more substituted carbon.
3. Determine the more substituted carbon of the double bond. Between C1 and C2, C2 is more substituted because it is connected to more alkyl groups compared to C1.
4. Predict the carbocation intermediate formed after protonation of the double bond. Protonation will occur at C1, generating a carbocation at C2. Consider possible carbocation rearrangements (like hydride or alkyl shifts) to form a more stable carbocation if applicable.
5. Attach the OH group to the carbocation center (C2) to form the major product. This results in 2-methyl-2-butanol, where the OH is on the more substituted carbon, consistent with Markovnikov addition.