Final answer:
The correct reagent for a transformation involving HOCH2CH2OH and the donation of hydride ions is Lithium Aluminum Hydride (LiAlH4). LiAlH4 is a strong reducing agent that can provide the necessary H- for the reduction, followed by an acid to donate a proton to the oxygen to complete the alcohol formation.
Step-by-step explanation:
The question pertains to the selection of the appropriate reagent for a chemical transformation involving HOCH2CH2OH in the presence of hydrogen, designated as (HOCH2CH2OH / H). Given the potential reagents listed, the correct answer is d) Lithium Aluminum Hydride (LiAlH4). Lithium aluminum hydride is a strong reducing agent that is capable of donating hydride ions (H-) to perform reductions, such as turning a carbonyl compound into an alcohol. In this context, 'H' symbolizes a hydride ion, not elemental hydrogen. The donated H- would react in the first step, and then acid is added in a second step to donate a proton to the oxygen. This stepwise process is essential for proper reduction to occur.
Reagent a) Sodium Borohydride (NaBH4) is a milder reducing agent compared to lithium aluminum hydride and may not be strong enough for some reductions. Reagent b) Hydrochloric Acid (HCl) is not a reducing agent but rather an acid, which provides H+ ions. Reagent c) Diethyl Ether is typically used as a solvent and does not donate hydride ions. Therefore, the most appropriate reagent for the transformation would be Lithium Aluminum Hydride (LiAlH4), which donates hydride ions effectively and allows the subsequent protonation with an acid to yield the desired alcohol.