Understanding the Reduction of 1,1-Diphenyl-2-propanone

The question asks about the reagents used to convert 1,1-Diphenyl-2-propanone into 1,1-Diphenyl-2-propanol. This transformation involves the reduction of a ketone (the -propanone part) to a secondary alcohol (the -propanol part).

Ketones can be effectively reduced to secondary alcohols using various reducing agents. One common and powerful reducing agent is Lithium Aluminium Hydride, represented as LiAlH$_4$.

The reduction using LiAlH$_4$ typically proceeds in two main steps:

1. The reduction itself, where the hydride ion from LiAlH$_4$ attacks the carbonyl carbon of the ketone. This step is usually carried out in an aprotic solvent, meaning a solvent that does not contain acidic protons. Common aprotic solvents for LiAlH$_4$ reductions include diethyl ether and Tetrahydrofuran (THF). The product of this step is an alkoxide complex.
2. The workup step, where water or a dilute acid/base solution is added to hydrolyze the alkoxide complex and any remaining LiAlH$_4$. This step liberates the desired alcohol.

In the given reaction, 1,1-Diphenyl-2-propanone is reacted with compound 'a' and then compound 'b' to form 1,1-Diphenyl-2-propanol.

Based on the standard procedure for LiAlH$_4$ reduction, compound 'a' would correspond to the reducing agent (LiAlH$_4$) added in a suitable aprotic solvent (like diethyl ether). Compound 'b' would then correspond to the substance used for the workup, which is typically water (H$_2$O).

Let's examine the options provided in this context:

• Option 1: a = LiAlH$_4$; b = diethyl ether. This suggests the reduction agent is added, followed by the solvent as a second step, which is not the standard procedure. The solvent is typically present during the addition of the reducing agent or the substrate.
• Option 2: a = LiAlH$_4$, H$_2$O; b = diethyl ether. Adding LiAlH$_4$ directly with water is dangerous as LiAlH$_4$ reacts violently with water. This sequence is incorrect.
• Option 3: a = LiAlH$_4$, diethyl ether; b = H$_2$O. This indicates that 'a' involves adding LiAlH$_4$ in the presence of diethyl ether (the reduction step with solvent) and 'b' is the addition of water (the workup step). This matches the typical procedure for LiAlH$_4$ reduction of ketones.
• Option 4: a = LiAlH$_4$; b = diethyl ether, H$_2$O. This suggests adding LiAlH$_4$ first, followed by adding both diethyl ether and water simultaneously as 'b'. While the workup involves water, adding the solvent again at this stage along with water is not the standard method.

Therefore, the sequence that correctly represents the reduction of 1,1-Diphenyl-2-propanone to 1,1-Diphenyl-2-propanol using LiAlH$_4$ is the addition of LiAlH$_4$ in diethyl ether (a), followed by the addition of water for workup (b).

The reaction can be summarized as:

Ketone + a (LiAlH$_4$ in diethyl ether) $\rightarrow$ Alkoxide Complex

Alkoxide Complex + b (H$_2$O) $\rightarrow$ Alcohol + Byproducts

Thus, 'a' corresponds to LiAlH$_4$ and diethyl ether, and 'b' corresponds to H$_2$O.