Using Lithium Aluminum Hydride

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Organic Chemistry › Using Lithium Aluminum Hydride

Questions 1 - 4

What is the product of the given reaction?

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III

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

What reagents are needed to satisfy the given reaction?

$1. LiAlH_{4}$

$2. H_{3}O^{+}$

$3. SOCl_{2}$

$1. H_{3}CMgBr$

$2. H_{3}O^{+}$

$3. SOCl_{2}$

$1.CCl_{4}$

$2. H_{3}O^{+}$

Explanation

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 $LiAlH_{4}$ to form an alcoxide. The alcoxide undergoes workup (the process whereby a negatively charged oxygen gains a proton) via $H_{3}O^{+}$ , depicted above as simply " $H^{+}$ ". We now have a secondary alcohol. From here, we can simply use the reagent $SOCl_{2}$ to convert the alcohol into the desired chlorine.

Which of the following can be reduced when mixed with $LiAlH_{4}$ ?

$CH_{3}CH_{2}CONH_{2}$

$CH_{3}CH_{2}OCH_{2}CH_{3}$

$CH_{3}CH_{2}CH_{2}OH$

$CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}CH_{3}$

Explanation

$LiAlH_{4}$ is a very powerful reducing agent that works to reduce almost any carbonyl compound. $CH_{3}CH_{2}CONH_{2}$ is an amide and the only carbonyl compound given of the answer choices.

What is the result of the following reaction?

All of the above

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.

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