13.5 PREPARING ALCOHOLS VIA REDUCTION

13.5 PREPARING ALCOHOLS VIA REDUCTION

The stereochemical outcome is irrelevant because no stereocenters were formed.

Two of the same coin are represented by these two count ing methods.

The carbon atom appears to have four electrons of its own when we treat bonds as covalent.

When we compare the number of electrons carbon has with the number it is supposed to have, we see that everything is right.

It is supposed to have four electrons, but it is actually using four.

There is no formal charge.

Both electrons are given to the more negative atom for each bond.

Chlorine gets both electrons when it is in the C--Cl bond.
One electron is placed on each carbon atom for the C--C bond.
The carbon atom is using five electrons of its own when we count them.
We know that carbon only has four electrons.
The oxidation state of this carbon atom is -1 because it is using one extra electron.

Neither method is perfect.

Each method assumes an extreme is true.
In this book, we focused our attention on formal charges.

For purposes of this section, we will focus on oxidation states.

The identity of the atoms that are attached to the carbon atom will affect the oxidation state of the carbon atom.

Both electrons are given to the more negative atom for each bond.

Oxygen gets both electrons when it bonds to the C--O bond.
One electron on one carbon atom and the other electron on the other carbon atom are treated equally for each of the C--C bonds.

The carbon atom is using four electrons of its own when we count the electrons that carbon is using.

We compare that number to the number of valence electrons that carbon is supposed to have.
The oxidation state of this carbon atom is zero because it is using the right number of electrons.

The middle three compounds are alcohols, aldehydes, and carboxylic acids.

Alcohols are at a lower oxidation state than aldehydes, which in turn is at a higher oxidation state.

Imagine if we were to run a reaction that converts alcohol to aldehyde or carboxylic acid.

An increase in oxidation state is constituted by this reaction.

The oxidizer is the compound that causes the desired oxidation and must be reduced.

The product of reducing a bond is alcohol.

Reducing ketones or aldehydes can be used to make alcohols.
We will need a reducing agent to accomplish this reduction.
There are two reagents that can be used to do this.
Let's review something from the periodic table to understand the structures of these reagents.

The elements have three valence electrons.

In the compounds shown above, both aluminum and boron use their electrons to form bonds, but neither has an octet.

Each element has the ability to form a fourth bond in order to get an octet, but it will be saddled with a formal charge of -1.

The central atom has four bonds and a negative charge in both compounds.

The counter-ion is used in the first compound.

The choice of counter-ion isn't relevant for our discussion and we will ignore it.

The mechanism of this first step is the same whether we use LiAlH or NaBH.

In the second step, a source of protons is used to make alcohol.

The way that LiAlH and NaBH work has been seen.

The size of the atom is what determines nucleophilicity.

Large atoms are excellent nucleophiles because they are very polarizable.
Small atoms are not polarizable.

H- is not a great nucleophile because it is as small as they come.

H- is an excellent base, but it is not a good nucleophile.

The differences between LiAlH and NaBH are striking.

NaBH is more reactive than LiAlH.

The reduction notice states that two separate steps are required with LiAlH.

We have seen two sources of H-.

We can control the reactivity of the hydride reagent by carefully choosing the R groups above.

The two most used hydride reagents are LiAlH and NaBH.

In the previous section, we learned that H- can be used to attack ketones and aldehydes.

A suitable source of R- can be used to attack ketones and aldehydes.

Both H- and R- can give an alcohol.

The effect on the carbon skeleton is the main difference.
The carbon skeleton does not change with H-.
The carbon skeleton gets bigger with R-.
We are forming a bond.
This is important for synthesis problems.
We can make R- in the first place.
It is not trivial to make a negative charge on a carbon atom.

There are many ways to get a negative charge on a carbon atom.

You will learn a lot about special C- compounds in this course.

A magnesium atom is connected to a carbon atom in a Grignard reagent.

If we compare the values of C and Mg, we will see that C is more negative than positive.
Carbon will develop a negative charge if it pulls more strongly on the electron density.

We won't go into it because the mechanism for this is beyond the scope of this course.

We should know that we can put Mg into a C--X bond.

The C--X bond has a magnesium atom inserted between it, which makes it ionic in character.

Neither drawing is correct.

The reality is closer to being ionic than it is to being two extremes.

Grignard reagents attack other compounds that have a C+O bond.

Let's focus on Grignard reagents attacking aldehydes and ketones.

There are two perfectly correct answers to this problem.

From now on, we will not see synthesis problems that only have one solution.

Sometimes we find synthesis problems that have more than one answer.

In this section, we learned that a Grignard reagent can give an alcohol if it attacks a ketone or aldehyde.

We have only seen one type of C--C bond-forming reaction.

There is a second approach to creating a C--C bond.
When we have a synthesis problem, we always ask two questions: Is there a change in the carbon skeleton, and is there a change in the functional group?
We will have to create a C--C bond if we have a synthesis problem where the carbon skeleton is getting larger.
In this book, we have seen two ways to do that.
We can either alkylate a terminal alkyne or use a Grignard reagent.

It is important to be familiar with Grignard reactions because they will appear many times throughout the course.

Let's practice.
We will start with a few problems that are just one step in the process.