Question

The rate of reduction of following with DIBAL-H O=Cc1ccccc1 O=C(Cl)c1ccccc1 CC(=O)c1ccccc1 (I) (II) (III) (A) III \(>\mathrm{I}>\mathrm{II}\) (B) \(\mathrm{III}>\mathrm{II}>\mathrm{I}\) (C) II > III > I (D) II > I > III

Short answer

The short answer based on the Step-by-Step solution is: The rate of reduction of the given compounds with DIBAL-H is (A): III > I > II.

Step-by-step solution

Identify the functional groups in each compound

First, let's identify the functional groups in each compound: (I) An ester functional group, -COO-: \(\mathrm{O=Cc1ccccc1}\) (II) An acyl chloride functional group, -COCl: \(\mathrm{O=C(Cl)c1ccccc1}\) (III) A ketone functional group, -C(O)R: \(\mathrm{CC(=O)c1ccccc1}\) Now that we've identified the functional groups, we can proceed to discuss their reactivity towards DIBAL-H.

Analyze the reactivity of each functional group

Now let's analyze the reactivity of each functional group towards DIBAL-H: 1. Esters (I) are known to be reduced by DIBAL-H to aldehydes relatively easily. 2. Acyl chlorides (II) are highly reactive towards DIBAL-H, and their reactivity may be difficult to control; they can easily be reduced beyond the aldehyde stage to the alcohol stage. 3. Ketones (III) are typically reduced less selectively by DIBAL-H, leading to over-reduction to the corresponding alcohol.

Compare the reactivity of each compound and select the correct answer

Based on the reactivity of each functional group, we can predict the selectivity (rate of reduction) of each compound from most reactive to least reactive: III (ketone) > I (ester) > II (acyl chloride). This corresponds to option (A): III > I > II. The correct answer is (A).