Using Lithium Aluminum Hydride

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Organic Chemistry › Using Lithium Aluminum Hydride

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1

What is the product of the given reaction?

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IV

CORRECT

I

0

II

0

III

0

V

0

Explanation

First step: esterification

Second step: lithium aluminum hydride reduction

Third step: neutralization to form primary alcohol

Fourth step: SN2 reaction to form final chlorinated product

2

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What reagents are needed to satisfy the given reaction?

CORRECT

0

0

0

Explanation

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This problem requires that we convert our ketone group into a chlorine. However, this cannot be done directly, and requires multiple steps.

We begin by reducing the ketone with to form an alcoxide. The alcoxide undergoes workup (the process whereby a negatively charged oxygen gains a proton) via , depicted above as simply "". We now have a secondary alcohol. From here, we can simply use the reagent to convert the alcohol into the desired chlorine.

3

Which of the following can be reduced when mixed with ?

CORRECT

0

0

0

Explanation

is a very powerful reducing agent that works to reduce almost any carbonyl compound. is an amide and the only carbonyl compound given of the answer choices.

4

What is the result of the following reaction?

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CORRECT

0

0

All of the above

0

Explanation

Lithium Aluminum Hydride is a potent reducing agent; it has the ability to turn esters and aldehydes into primary alcohols, and ketones into secondary alcohols. The starting material is an aldehyde, so the correct answer is thus a primary alcohol ONLY.