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:
Hence, the correct answer is (A) - (I), (B) - (III), (C) - (II), (D) - (IV).
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:
(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)
Complete the following reactions by writing the structure of the main products: