Question

In the following table, match the reactants given in List-I with the correct product in.List-II as per the reaction of hydration of alkene under acidic conditions.

• (A) - (I), (B) - (III), (C) - (II), (D) - (IV)
• (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
• (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
• (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

Answer 1

The Correct Option is A

The hydration of alkenes under acidic conditions typically follows Markovnikov's rule, where the hydrogen from the water adds to the less substituted carbon, while the OH group attaches to the more substituted carbon. Below is the detailed matching of the given alkenes and their corresponding products:

List-I (Reactants)	List-II (Products)
(A) Propene	(I) Isopropanol
(B) 2-Methylpropene	(III) 2-Methyl-2-propanol
(C) 1-Butene	(II) 2-Butanol
(D) 2-Butene	(IV) 2-Butanol

By understanding the application of Markovnikov's rule and examining each of the given reactions, we can correctly match the reactants to their corresponding products: 

• Propene, when hydrated, produces Isopropanol. Thus, (A) matches with (I).
• 2-Methylpropene reacts to form 2-Methyl-2-propanol, so (B) matches with (III).
• 1-Butene yields 2-Butanol due to rearrangement fitting Markovnikov's product, matching (C) with (II).
• 2-Butene also gives 2-Butanol indicating that (D) matches with (IV).

Hence, the correct answer is (A) - (I), (B) - (III), (C) - (II), (D) - (IV).

Answer 2

To solve the problem of matching reactants with products for the hydration of alkenes under acidic conditions, we need to understand the general mechanism of the reaction:

• Alkenes, when hydrated in the presence of an acid, follow Markovnikov's rule. This means the hydrogen from water will attach to the carbon with more hydrogens, and the hydroxyl group (OH) will attach to the carbon with fewer hydrogens.
• The intermediate carbocation formed should be the most stable one possible. Stability order is tertiary > secondary > primary due to hyperconjugation and the inductive effect.

(A) Propene: Following Markovnikov's rule, the OH group attaches to the second carbon. Thus, the product is Propan-2-ol (I).

(B) 2-Methylpropene: The double bond breaks to stabilize around the tertiary carbon, resulting in 2-Methylpropan-2-ol (III).

(C) But-1-ene: Following the rule, OH attaches to the second carbon, forming Butan-2-ol (II).

(D) Pent-2-ene: The most stable carbocation forms by attaching OH to the third carbon, leading to Pentan-3-ol (IV).

Thus, the correct answer is: (A) - (I), (B) - (III), (C) - (II), (D) - (IV)