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Chapter 18: Problem 26

The reagent diisobutylaluminum hydride (DIBALH) reduces esters to aldehydes. When nitriles are treated with DIBAI.H, followed by mild acid hydrolysis, the product is also an aldehyde. Propose a mechanism for this reduction.

Short Answer

Expert verified
Answer: DIBALH acts as a reducing agent in the reduction of nitriles to aldehydes. The process involves: 1) Reduction of nitrile by DIBALH, forming a nucleophilic imine anion, 2) Protonation of the imine anion by an aluminum alkoxide, forming a neutral imine compound, 3) Mild acid hydrolysis of the imine compound, forming a hydroxylamine compound, and 4) Loss of an ammonia molecule and formation of the aldehyde as the final product.

Step by step solution

01

Identify the structure of the reagents and substrates

To start, we need to know the structure of the diisobutylaluminum hydride (DIBALH) and nitrile substrates. DIBALH is a reducing agent that has the chemical formula \(\text{AlH}(\text{C}_4\text{H}_9)_2\). A nitrile is an organic compound with a carbon-nitrogen triple bond, represented as RC≡N, where R is an organic group.
02

Reduction of nitrile by DIBALH

The first part of the mechanism involves the reduction of the nitrile by DIBALH. The hydride ion (H-) from DIBALH attacks the electrophilic carbon atom in the carbon-nitrogen triple bond, forming a new carbon-hydrogen bond and breaking one carbon-nitrogen bond. This results in the formation of a Nucleophilic imine anion species.
03

Protonation of the imine anion

In the next step, the nucleophilic imine anion is protonated by an aluminum alkoxide species that is formed upon hydride transfer. As a result, the nucleophilic imine anion is converted into a neutral imine compound.
04

Hydrolysis of the imine compound

After the formation of the imine compound, it undergoes mild acid hydrolysis. Water molecules attack the electrophilic carbon of the carbon-nitrogen double bond, resulting in the addition of a hydroxy group (OH) to the carbon and breaking the double bond between C and N. Consequently, a hydroxylamine compound is formed.
05

Formation of the aldehyde

Lastly, a proton transfer within the hydroxylamine compound takes place, and another water molecule acts as a nucleophile, which results in the loss of an ammonia molecule (\(\text{NH}_3\)). This step effectively removes the nitrogen atom from the compound, leaving behind an aldehyde as the final product.

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Most popular questions from this chapter

Draw a structural formula of the principal product formed when ethyl benzoate is treated with each reagent. (a) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{NaOH}\), heat (b) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{SO}_{4}\), heat (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (d) DIBALH \(\left(-78^{\circ} \mathrm{C}\right)\), then \(\mathrm{H}_{2} \mathrm{O}\) (e) \(\mathrm{LiAlH}_{4}\), then \(\mathrm{H}_{2} \mathrm{O}\) (f) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{MgBr}\) (two equivalents), then \(\mathrm{HCl} / \mathrm{H}_{2} \mathrm{O}\)

Predict the distribution of oxygen-18 in the products obtained from hydrolysis of ethyl benzoate labeled in the ethoxy oxygen under the following conditions. (a) In aqueous NaOH (b) In aqueous \(\mathrm{HCl}\) (c) What distribution would you predict if the reaction were done with the tert-butyl ester in \(\mathrm{HCl}\) ?

Using your roadmaps as a guide, show how to convert 1-bromopropane and carbon dioxide into 4-heptanone. You must use 1-bromopropane and carbon dioxide as the source of all carbon atoms in the target molecule. Show all reagents and all molecules synthesized along the way.

Both the melting point and boiling point of acetamide are higher than those of its \(N, N\)-dimethyl derivative. How do you account for these differences?

Will phthalimide dissolve in aqueous sodium bicarbonate?
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