Question

The reagents which on reaction with CC1CCC(=O)OC1 gives a product which give positive Tollen's test is (A) \(\mathrm{LiAlH}_{4}\) followed by \(\mathrm{H}_{2} \mathrm{O}\) (B) \(\mathrm{NaBH}_{4}\) followed by \(\mathrm{H}_{2} \mathrm{O}\) (C) Na-EtOH followed by \(\mathrm{H}_{2} \mathrm{O}\) (D) DIBAL-H followed by \(\mathrm{H}_{2} \mathrm{O}\)

Short answer

The reagents which will produce a product that gives a positive Tollen's test when reacted with the given molecule are (D) DIBAL-H followed by \(\mathrm{H}_{2} \mathrm{O}\).

Step-by-step solution

Convert the SMILES to its chemical structure

The SMILES representation of the molecule is CC1CCC(=O)OC1. Converting it to its chemical structure, we find that the molecule is a cyclic ester: ``` O || C // \ C C | \ C O | | C -- C // ```

Understand the Tollen's test

Tollen's test is used to detect the presence of aldehydes in a compound. The test relies on the reduction of silver ions to metallic silver in the presence of an aldehyde. A positive Tollen's test is indicated by the formation of a silver "mirror" on the inner surface of the test tube. Thus, we are looking for reactions that would convert the given molecule into a compound with an aldehyde functional group, which is -C(=O)H.

Analyze the reaction of the given reagents with the molecule

We will consider each given reagent and see if its reaction with the molecule would produce an aldehyde. (A) \(\mathrm{LiAlH}_{4}\) followed by \(\mathrm{H}_{2} \mathrm{O}\) \(\mathrm{LiAlH}_{4}\) is a strong reducing agent that can reduce esters to primary alcohols. The reaction of our molecule with \(\mathrm{LiAlH}_{4}\) followed by \(\mathrm{H}_{2} \mathrm{O}\) would produce a primary alcohol, not an aldehyde. (B) \(\mathrm{NaBH}_{4}\) followed by \(\mathrm{H}_{2} \mathrm{O}\) \(\mathrm{NaBH}_{4}\) is a milder reducing agent, typically used to reduce ketones and aldehydes to alcohols. Since it is milder, it will not be strong enough to reduce the ester to an aldehyde in this case. (C) Na-EtOH followed by \(\mathrm{H}_{2} \mathrm{O}\) Na-EtOH is a strong base causing elimination reactions on alkyl halides or esters. In this case, it would produce an alkene, not an aldehyde. (D) DIBAL-H followed by \(\mathrm{H}_{2} \mathrm{O}\) DIBAL-H (Diisobutylaluminum hydride) is a selective reducing agent that can reduce esters to aldehydes without continuing the reduction to an alcohol. In this case, reacting the given cyclic ester with DIBAL-H followed by \(\mathrm{H}_{2} \mathrm{O}\) will convert the ester to an aldehyde, making it capable of giving a positive Tollen's test. As a result, the correct choice is (D) DIBAL-H followed by \(\mathrm{H}_{2} \mathrm{O}\).