Question

The major product obtained when 2 -phenylpropene is treated with water containing acid is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{OH}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CHOHCH}_{3}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHOH} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{OH}\)

Short answer

Answer: (a) C₆H₅CH(CH₃)CH₂OH

Step-by-step solution

Identify the reactant and the type of reaction

We are given 2-phenylpropene as the reactant and the reaction is an electrophilic addition between an alkene (2-phenylpropene) and water in the presence of an acid.

Draw the structure of 2-phenylpropene

The structure of 2-phenylpropene is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\) with a double bond between the \(\mathrm{CH}\) and \(\mathrm{CH}\) of the propene part.

Addition of water across the double bond

When electrophilic addition occurs across the double bond, the hydrogen from the acid will add to one of the carbons while the hydroxyl group from water will add to the other carbon. This reaction follows Markovnikov's rule, which states that the hydrogen will add to the carbon of the double bond that has more hydrogen atoms, and the hydroxyl group will add to the other carbon atom. Applying Markovnikov's rule to this reaction, we get that hydrogen will add to \(\mathrm{CH}\) (second carbon) and hydroxyl group to \(\mathrm{CH}\) of \(\mathrm{CH}_{3}\) (third carbon).

Determine the product

After the reaction, we get the product \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{OH}\).

Compare the product with the given options

The product obtained in step 4 is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{OH}\), which matches with option (a). Hence, the major product obtained when 2-phenylpropene is treated with water containing acid is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{OH}\).

Key concepts

Markovnikov's Rule

Understanding Markovnikov's rule is key to predicting the outcomes of many organic chemistry reactions involving alkenes. This rule guides us in electrophilic addition reactions, where an electrophile and a nucleophile are added across a double bond. It states that in such reactions, the electrophile will attach to the less substituted carbon atom -- the carbon with more hydrogen atoms -- while the nucleophile will bond with the more substituted carbon. This rule is a direct result of the stability of carbocation intermediates formed during the reaction.

For instance, in the hydration of alkenes, like the treatment of 2-phenylpropene with water in the presence of an acid, a proton (H⁺) first adds to the less substituted carbon of the double bond. Following this, the hydroxyl group (OH^-) attacks the more substituted carbocation, leading to the final product. The application of Markovnikov's rule to such a reaction results in the major product being more highly substituted alcohol.

Alkene Reactions

Alkenes are characterized by the presence of a carbon-carbon double bond and are highly reactive in electrophilic addition reactions. These reactions are pivotal in forming a variety of compounds in organic synthesis. Alkenes can undergo reactions with different electrophiles, such as halogens, hydrogen halides, and water, in the presence of an acid catalyst.

The reactivity of alkenes comes from the electrons in the pi bond of the double bond that are available for attack by electrophiles. In a typical alkene reaction, the pi bond breaks, allowing the formation of two new sigma bonds. As seen in hydration reactions, where water is added across the double bond, the outcome often depends on the structure of the alkene and the reaction conditions. By understanding these reactions, we grasp how complex molecules in pharmaceuticals and natural products are constructed.

Organic Chemistry Mechanisms

The mechanism of a reaction in organic chemistry unravels the step-by-step process by which reactants are converted into products. These mechanisms provide insights into the movement of electrons, the formation and breaking of bonds, and the intermediate states such as carbocations. Grasping this concept is foundational for predicting reaction outcomes and for designing new synthetic pathways.

Mechanisms are depicted using arrow notation to indicate electron flow, and they typically involve several stages: initiation, propagation, and termination. By studying these mechanisms, chemists can optimize reaction conditions and yields, and they can understand the behavior of molecules under different circumstances. A well-documented organic mechanism is a tool that allows for the prediction and explanation of the outcome of the large array of reactions that organic molecules can undergo.