Predict the products of the following reactions.
(b) 1-phenylethanol + excess I₂ in base

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Step 1: Recognize the reactants and reaction conditions. The reactants are 1-phenylethanol (Ph-CH(OH)-CH3) and excess iodine (I2) in a basic medium (OH⁻). This is a classic example of the haloform reaction.

Step 2: Understand the mechanism of the haloform reaction. In the presence of excess iodine and base, the methyl group attached to the alcohol is oxidized to a ketone (acetophenone, Ph-CO-CH3). The methyl group is then further converted into a trihalomethyl group (CCl3 or CI3) and subsequently cleaved to form a haloform (CHI3, iodoform).

Step 3: Analyze the intermediate steps. The alcohol group (-OH) is first oxidized to a ketone (acetophenone). Then, the methyl group adjacent to the carbonyl undergoes successive halogenation to form a trihalomethyl group (-CI3).

Step 4: Predict the cleavage products. The trihalomethyl group is cleaved in the presence of base, yielding iodoform (CHI3) and a carboxylate ion (Ph-COO⁻).

Step 5: Summarize the final products. The reaction produces iodoform (CHI3), which is a yellow precipitate, and sodium benzoate (Ph-COO⁻Na⁺) as the carboxylate salt in the basic medium.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Nucleophilic Substitution

Nucleophilic substitution is a fundamental reaction mechanism in organic chemistry where a nucleophile attacks an electrophile, resulting in the replacement of a leaving group. In the context of 1-phenylethanol reacting with iodine in a basic medium, the hydroxyl group (-OH) can be converted into a better leaving group, facilitating the substitution reaction with iodine.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions involve the transfer of electrons between species, leading to changes in oxidation states. In the reaction of 1-phenylethanol with iodine, the alcohol can undergo oxidation to form a carbonyl compound, while iodine is reduced. Understanding these changes is crucial for predicting the products of the reaction.

Iodination of Alcohols

Iodination of alcohols is a specific reaction where iodine reacts with alcohols in the presence of a base, often leading to the formation of iodoalkanes. In this case, the excess iodine reacts with the hydroxyl group of 1-phenylethanol, resulting in the formation of an iodo derivative. Recognizing the conditions and mechanisms of iodination is essential for predicting the final products.

Predict the products of the following reactions.(b) 1-phenylethanol + excess I2 in base