Which compounds can be reduced readily with sodium borohydride?

Sodium borohydride is a relatively selective reducing agent. Ethanolic solutions of sodium borohydride reduce aldehydes and ketones in the presence of epoxides, esters, lactones, acids, nitriles, or nitro groups. Reduction of aldehydes is straightforward.

Which is better NaBH4 or LiAlH4?

Both LiAlH4 and NaBH4 are reducing agents. But LiAlH4 is a very strong reducing agent than NaBH4 because the Al-H bond in the LiAlH4 is weaker than the B-H bond in NaBH4. This makes the Al-H bond less stable. As a result, LiAlH4 is a better hydride donor.

Why is LiAlH4 a stronger reducing agent than NaBH4?

The hydride anion is not present during this reaction; rather, these reagents serve as a source of hydride due to the presence of a polar metal-hydrogen bond. Because aluminium is less electronegative than boron, the Al-H bond in LiAlH4 is more polar, thereby, making LiAlH4 a stronger reducing agent.

Which type of reducing agent is NaBH4?

Sodium borohydride
What it’s used for: Sodium borohydride is a good reducing agent. Although not as powerful as lithium aluminum hydride (LiAlH4), it is very effective for the reduction of aldehydes and ketones to alcohols.

What Cannot be reduced by NaBH4?

NaBH4 is effective reducing agent for the reduction of aldehydes, ketones and acyl chloride to alcohol but it cannot reduce esters, carboxylic acids and amides.

What type of reducing agents are NaBH4 and LiAlH4?

Both NaBH4 and LiAlH4 are reducing agents. These are the most common sources of hydride nucleophiles we use in organic synthesis reactions. The name of NaBH4 is sodium borohydride while the name of LiAlH4 is lithium aluminum hydride.

Which compounds can be reduced by LiAlH4?

LiAlH4 is a strong, unselective reducing agent for polar double bonds, most easily thought of as a source of H-. It will reduce aldehydes, ketones, esters, carboxylic acid chlorides, carboxylic acids and even carboxylate salts to alcohols. Amides and nitriles are reduced to amines.

How do you make sodium borohydride?

Synthesis and handling In the Brown-Schlesinger process sodium borohydride is industrially prepared from sodium hydride (produced by reacting Na and H2) and trimethyl borate at 250–270 °C: B(OCH3)3 + 4 NaH → NaBH4 + 3 NaOCH.