Using Lithium Aluminum Hydride - Organic Chemistry

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.

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

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.

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.

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.

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

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.

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.

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.

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

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.

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.

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.

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

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.