Oxidations

Oxidations:

G. Chem. =

Loss of electron(s)

O. Chem. =

Lose bond to less electronegative atom +

Gain bond to more electronegative atom

H

[O]

Focus on the C

with the OH.

This C has 1

bond to a H.

Bond to H lost, another bond to

more Electronegative O gained.

[O] = Shorthand for Oxidation

In the General Chem. definition of Oxidation → e^-’s are fully lost from an atom.

In the Organic Chem. definition of Oxidation → e^- density is lost from an atom (almost always C).

Fe

Fe³⁺

+ 3 e^-

𝛅+

𝛅++

The Carbonyl C has a Partial + Ch. because of the single bond to the more Electronegative O.

The Carbonyl C now has more Partial + Ch. because of the double bond to the more Electronegative O.

It gained a second bond to O and lost a bond to the less Electronegative H.

Fe

Fe³⁺

+ 3 e^-

H

[O]

[O] = Shorthand for Oxidation

𝛅+

𝛅++

Click “Slideshow” to Animate

Oxidations

Oxidations:

G. Chem. =

Loss of electron(s)

O. Chem. =

Lose bond to less electronegative atom +

Gain bond to more electronegative atom

h𝝂

Oxidations don’t even need to involve the Oxygen atom.

A Radical Bromination of an Alkane is an Oxidation because Br is more Electronegative than H…

C has 2 H’s

C has 1 H’s

The C above lost a bond to a H and gained one to Br.

Br is more Electronegative than H so this is an Oxidation.

Fe

Fe³⁺

+ 3 e^-

𝛅+

𝛅-

Carbon 2 of the Product 2-Bromopropane has less e^- density than that of the Reactant Propane.

Oxidations

Oxidations:

G. Chem. =

Loss of electron(s)

O. Chem. =

Lose bond to less electronegative atom +

Gain bond to more electronegative atom

Now we’ll go through the full Oxidation of Alcohols starting with Methanol…

Fe

Fe³⁺

+ 3 e^-

[O]

Focus on the C with the Alcohol.

This Alcohol C has 3 H’s to start.

Methanol

Formaldehyde

Now the C has only 2 H’s.

The Alcohol O also loses a H.

[O]

Formic Acid

Now the C has only 1 H.

[O]

Carbonic Acid

Now the C has no H’s.

The C is fully Oxidized.

Carbonic Acid is what is made when we quench Acids with Sodium Bicarbonate (Baking Soda).

Sodium Bicarbonate (Baking Soda)

Carbonic Acid

Carbonic Acid is unstable and self decomposes to Carbon Dioxide and Water.

The Carbon Dioxide gas is the bubbles we see.

This step is not an Oxidation, the C started with 4 bonds to O and ended with 4.

Baking Soda is used when making Cookies and Cake!

Oxidations

Oxidations:

G. Chem. =

Loss of electron(s)

O. Chem. =

Lose bond to less electronegative atom +

Gain bond to more electronegative atom

Now we’ll go through the full Oxidation of the Primary Alcohol Ethanol…

Fe

Fe³⁺

+ 3 e^-

[O]

Focus on the C with the Alcohol.

This Alcohol C has 2 H’s to start.

Ethanol

An Aldehyde

Now the C has only 1 H.

The Alcohol O also loses a H.

[O]

Acetic Acid

Now the C has

no H.

The C is fully Oxidized.

Oxidations

Oxidations:

G. Chem. =

Loss of electron(s)

O. Chem. =

Lose bond to less electronegative atom +

Gain bond to more electronegative atom

Now we’ll go through the full Oxidation of the Secondary Alcohol Isopropyl Alcohol…

Fe

Fe³⁺

+ 3 e^-

[O]

Focus on the C with the Alcohol.

This Alcohol C has 1 H to start.

Isopropyl Alcohol

A ketone

Now the C has

no H’s.

The C is fully Oxidized.

Oxidations

Oxidations:

G. Chem. =

Loss of electron(s)

O. Chem. =

Lose bond to less electronegative atom +

Gain bond to more electronegative atom

Now we’ll go through the full Oxidation of the Tertiary Alcohol t-Butanol…

Fe

Fe³⁺

+ 3 e^-

[O]

Focus on the C with the Alcohol.

This Alcohol C

no H’s to start.

t-Butanol

No Reaction

Oxidations

We’ll start with two types of Oxidations:

“Take just one or

Jones Oxidation:

Jones Oxidation Mixture =

Chromate (CrO₃), H₂SO_4(aq), Acetone

CrO₃

H₂SO_4(aq)

Acetone

Oxidations:

G. Chem. =

Loss of electron(s)

O. Chem. =

Lose bond to less electronegative atom +

Gain bond to more electronegative atom

Methyl

Alcohol

(Methanol)

1^o Alcohol

(Ethanol)

CrO₃

H₂SO_4(aq)

Acetone

Carboxylic Acid

(Acetic Acid)

2^o Alcohol

(Isopropyl Alcohol)

CrO₃

H₂SO_4(aq)

Acetone

Ketone

(Acetone)

H

3^o Alcohol

(t-Butanol)

CrO₃

H₂SO_4(aq)

Acetone

No

Rxn

No H’s

“TAKE ‘EM All!”

“Om, Nom, Nom, Nom…”

Jones Oxidation eats all the H’s on the C with the Alcohol, like the Cookie Monster eats as many Cookies as he can!

Jones ate all three H’s! Om, Nom, Nom, Nom…!

Ewart Ray Herbert Jones

Jones ate both the H’s!

He’s the Cookie Monster of H’s on Alcohol C’s! “Om nom nom nom!”

Jones ate the only H!

“Om!”

TAKE ’EM All!”

PCC:

PCC =

CH₂Cl₂

1^o Alcohol

(Ethanol)

Aldehyde

(Acetaldehyde)

2^o Alcohol

(Isopropyl Alcohol)

Ketone

(Acetone)

H

3^o Alcohol

(t-Butanol)

No

Rxn

No H’s

Pyridinium Chlorochromate, CH₂Cl₂

CH₂Cl₂

PCC

CH₂Cl₂

PCC

CH₂Cl₂

PCC

CH₂Cl₂

PCC

Aldehyde

(Formaldehyde)

Methyl

Alcohol

(Methanol)

“Just One Please.”

Oxidations

Oxidations

“Take just one or

TAKE ’EM All!”

“And just one Slice of Cake…”

Just one of the three H’s taken.

Rxn Stops at the Aldehyde.

Just one of the two H’s taken.

Rxn Stops at the Aldehyde.

Only one H to take.

Rxn Stops at the Ketone.

No H, No Rxn.

Same as Jones Oxidation.

Same as Jones Oxidation.

Pyridinium

Chlorochromate

Solvent

Oxidations

Why does Jones Oxidation [CrO₃, Acetone, H₂SO_4(aq)] “TAKE ‘EM ALL!” and PCC just takes one?

The answer is hidden in this little subscript!

Water is the Key! That’s why I’ve been emphasizing it with blue font.

The presence of Water allows a Hydrate to form that can continue to be Oxidized further.

CrO₃

H₂SO_4(aq)

Acetone

First H taken.

H₂O

Hydrate

CrO₃

H₂SO_4(aq)

Acetone

The Hydrate must form before more H’s can be taken.

The Hydrate will not form without Water.

PCC uses CH₂Cl₂ as its Solvent and no Water is added.

First H taken.

H₂O

Hydrate

PCC

CH₂Cl₂

No Water, No Hydrate, No more H’s taken!

No more H’s to take from Carbonyl

Hydrate

R = C or H

Reductions

Most Reductions:

(We’ll see)

“Just add 1 H to Aldehyde or Ketone Carbonyl C.”

Sodium Borohydride (NaBH₄):

Reagent and Solvent =

NaBH₄, Alcohol (Methanol or Ethanol)

Reductions:

G. Chem. =

Gain of electron(s)

O. Chem. =

Gain bond to less electronegative atom +

Lose bond to more electronegative atom

H

[H]

Ketone

(Acetone)

Aldehyde

(Propanal)

1^o Alcohol

(1-Propanol)

“Mild, Whinny”

NaBH₄

CH₃OH

2^o Alcohol

(Isopropyl Alcohol)

NaBH₄

H

H

H

Ketone

(3,3-Dimethylcyclopentanone)

2^o Alcohol

(3,3-Dimethylcyclopentanol)

NaBH₄

H

“Focus on the C

with the O”

“This C has 0

bonds to a H”

“Bond to O lost,

another bond to

less electronegative

H gained”

[H] = Shorthand for Reduction

Carbonyl C gained one H.

Carbonyl had one H and gained another.

Carbonyl C gained one H.

Key 1st step of the long Mechanism

Key 1st step of the long Mechanism

Key 1st step of the long Mechanism

Mg²⁺ + 2e^- → Mg_(s)

𝛅+

𝛅++

We’ll learn two types of Reductions, here’s the first…

Most Reductions

“Just add 1 H to Aldehyde or Ketone Carbonyl C.”

Lithium Aluminum Hydride:

Reagent and Solvent =

1a. LiAlH₄, an Ether Solvent (Diethyl Ether or THF), b. H₃O⁺_(aq)

Ketone

(Acetone)

Aldehyde

(Propanal)

1^o Alcohol

(1-Propanol)

“Strong, Angry!”

1a. LiAlH₄

THF

b. H₃O⁺_(aq)

2^o Alcohol

(Isopropyl Alcohol)

1a. LiAlH₄

H

H

H

Ketone

(3,3-Dimethylcyclopentanone)

2^o Alcohol

(3,3-Dimethylcyclopentanol)

1a. LiAlH₄

H

Ether

b. H₃O⁺_(aq)

THF

b. H₃O⁺_(aq)

Carbonyl C gained one H.

Carbonyl had one H and gained another.

Carbonyl C gained one H.

Reductions

Key 1st step of the long Mechanism

Key 1st step of the long Mechanism

Key 1st step of the long Mechanism

Oxidation Practice

Reduction Practice

No

Rxn

No

Rxn

No

Rxn

No

Rxn

Oxidations and Reductions

Oxidations and Reductions

Add ¹H and/or D with NaBH₄ (NaBD₄) and LiAlH₄ (LiAlD₄)

Hydrogen Isotope Review

~99.97%

Natural Abundance

Trace

Natural Abundance

¹H =

1 Proton

+ 1 Electron

²H =

1 Proton

+ 1 Electron

D =

³H =

1 Proton

+ 1 Electron

T =

+ 1 Neutron

+ 2 Neutron

+ 0 Neutron

~0.001%

Natural Abundance

We will mainly just work with Protium and Deuterium. They react nearly identically.

Oxidations and Reductions

Add ¹H and/or D with NaBH₄ (NaBD₄) and LiAlH₄ (LiAlD₄)

NaBH₄

A =

B =

1a. LiAlH₄

b. H₃O⁺_(aq)

Adds no D’s

Adds no D’s

1st H added to Carbonyl C in 1st step of a long mechanism

2nd H added to O

1st H added to Carbonyl C in 1st step of a long mechanism

​

2nd H added to O

More steps of the mechanism

More steps of the mechanism

Oxidations and Reductions

Add ¹H and/or D with NaBH₄ (NaBD₄) and LiAlH₄ (LiAlD₄)

NaBD₄

C =

Adds 2 D’s

D =

1a. LiAlD₄

b. D₃O⁺_(aq)

Adds 2 D’s

1st D added to Carbonyl C in 1st step of a long mechanism

​

2nd D added to O

1st D added to Carbonyl C in 1st step of a long mechanism

​

​

2nd D added to O

​

More steps of the mechanism

More steps of the mechanism

Oxidations and Reductions

Add ¹H and/or D with NaBH₄ (NaBD₄) and LiAlH₄ (LiAlD₄)

E =

1a. LiAlH₄

b. D₃O⁺_(aq)

Adds 1 D to O

F =

b. H₃O⁺_(aq)

Adds 1 D to

the Carbonyl C

1a. LiAlD₄

H added to Carbonyl C in 1st step of a long mechanism

D added to O

D added to Carbonyl C in 1st step of a long mechanism

​

H added to O

More steps of the mechanism

More steps of the mechanism

Oxidations and Reductions

Add ¹H and/or D with NaBH₄ (NaBD₄) and LiAlH₄ (LiAlD₄)

NaBH₄

A =

B =

1a. LiAlH₄

b. H₃O⁺_(aq)

NaBD₄

C =

D =

1a. LiAlD₄

b. D₃O⁺_(aq)

E =

1a. LiAlH₄

b. D₃O⁺_(aq)

F =

b. H₃O⁺_(aq)

1a. LiAlD₄

A or B

C or D

F

E

2 H’s Added

1 D added to O

1 D added to Carbonyl C

2 D’s added

Oxidations and Reductions

Oxidation and Reduction practice on next slide.

PCC

CH₂Cl₂

PCC

CH₂Cl₂

PCC

CH₂Cl₂

CrO₃

H₂SO_4(aq)

Acetone

CrO₃

H₂SO_4(aq)

Acetone

CrO₃

H₂SO_4(aq)

Acetone

NaBH₄

CH₃OH

NaBD₄

CH₃OD

NaBD₄

CH₃OD

1a. LiAlD₄

Ether

b. H₃O⁺_(aq)

NaOH

CrO₃

H₂SO_4(aq)

Acetone

CrO₃

H₂SO_4(aq)

Acetone

PCC

CH₂Cl₂

PCC

CH₂Cl₂

D₃O⁺_(aq)

Acetone =

PCC =

Pyridinium

ChloroChromate

1

2

3

4

Differences between LiAlH₄ and NaBH₄

LiAlH₄ is much stronger!

Oxidations and Reductions

+ A Strong S_N2 Nuc

It’s a Strong Carbonyl Nuc

+ A Strong Base

A Strong Carbonyl Nuc

A Strong Base

A Strong S_N2 Nuc

Ether

The weaker NaBH₄ is not as strong of a Base and is not an S_N2 Nuc.

LiAlH₄ Deprotonate Alcohols, Amines, Carboxylic Acids (Protic H’s)

LiAlH₄ needs to be in an Ether Solvent while Reducing Ketones & Aldehydes

Differences between LiAlH₄ and NaBH₄

Oxidations and Reductions

If a Protic solvent is used for a LiAlH₄ Carbonyl Reduction the solvent quenches the LiAlH₄…

1a.

b. H₃O⁺_(aq)

Acid/Base Rxn is Faster

Ether is not Protic (Acidic)

No Carbonyl Reduction

Carbonyl is Reduced

Hydronium added after Carbonyl Reduction

NaBH₄ is much weaker!

Differences between LiAlH₄ and NaBH₄

Oxidations and Reductions

A Strong Carbonyl Nuc

+ A Weak Base

+ Not an S_N2 Nuc

A Strong Carbonyl Nuc

A Weak Base

Not an S_N2 Nuc

No Rxn

No Rxn

Protic Solvent is fine during Carbonyl Reductions

Differences between LiAlH₄ and NaBH₄

Oxidations and Reductions

Why is NaBH₄ weaker than LiAlH₄?

We mostly see H attached to atoms that it’s about as Electronegative as, like C

or ones it’s less Electronegative than like O and N.

Electronegative

Values →

~Same Electronegativity

Less Electronegativity

𝛅+

𝛅-

𝛅+

𝛅-

𝛅+

𝛅+

But when H is bonded to B and Al, it’s more Electronegative!

And the difference in Electronegativity is greater between H and Al than H and B.

H (2.20)

H (2.20)

H’s of LiAlH₄ are more Negative, more Nucleophilic and more Basic.

The Negative Charge is not all on B (Al) as the Formal Charge suggests, it’s shared between all atoms.

𝛅-

𝛅--

𝛅-

𝛅----

- B (2.04)

= 0.16

- Al (1.61)

= 0.59

More Negative!

Differences between LiAlH₄ and NaBH₄

Oxidations and Reductions

Why is the weak NaBH₄ strong enough to be a Carbonyl Nuc but not an S_N2 Nuc?

Oxygen’s more Electronegative than Br…

Electronegative

Values →

This means the C attached to these atoms have different amounts of Partial Positive Charge.

𝛅++

𝛅+

𝛅--

And both are more Electronegative than C.

𝛅-

More Partial Positive Charge

Stronger Electrophile

Will react with weaker NaBH₄ Nuc.

Less Partial Positive Charge

Weaker Electrophile

Will Not react with weaker NaBH₄ Nuc.

Oxidations and Reductions

Now for some practice with it all mixed together.

Br₂

h𝝂

DMSO

DMSO = DimethylSulfoxide

(Polar Aprotic Solvent, great for S_N2)

CrO₃

Acetone, H₂SO_4(aq)

Jones oxidation

PCC

CH₂Cl₂

PCC =

Pyridinium

ChloroChromate

LiAlD₄

Ether

NaBD₄

No Rxn

1a. LiAlD₄

Ether

b. H₃O⁺_(aq)

NaBD₄

CH₃OD

Strong LiAlD₄ can S_N2

​

Weak NaBD₄ can not S_N2

​

Unhind. 1^o Elec. + Unhind. Strong Base

S_N2 Major

LiAlD₄ adds D to Carbonyl C

H₃O⁺ adds H to O.

NaBD₄ adds D to Carbonyl C

CH₃OD adds D to O.

Oxidations and Reductions

Oxidation Practice

Reduction Practice

Quiz Time!

(Key in Slides above).

Quiz continues on next slide. If you’d like, here is a Google Doc version of this Slide: Oxidation & Reduction Quiz

Oxidations and Reductions

Add ¹H and/or D with NaBH₄ (NaBD₄) and LiAlH₄ (LiAlD₄)

NaBH₄

A =

B =

1a. LiAlH₄

b. H₃O⁺_(aq)

NaBD₄

C =

D =

1a. LiAlD₄

b. D₃O⁺_(aq)

E =

1a. LiAlH₄

b. D₃O⁺_(aq)

F =

b. H₃O⁺_(aq)

1a. LiAlD₄

2 H’s Added

1 D added to O

1 D added to Carbonyl C

2 D’s added

Quiz continues on next slide.

Quiz Time!

(Key in Slides above).

PCC

CH₂Cl₂

PCC

CH₂Cl₂

PCC

CH₂Cl₂

CrO₃

H₂SO_4(aq)

Acetone

CrO₃

H₂SO_4(aq)

Acetone

CrO₃

H₂SO_4(aq)

Acetone

NaBH₄

CH₃OH

NaBD₄

CH₃OD

NaBD₄

CH₃OD

1a. LiAlD₄

Ether

b. H₃O⁺_(aq)

NaOH

CrO₃

H₂SO_4(aq)

Acetone

CrO₃

H₂SO_4(aq)

Acetone

PCC

CH₂Cl₂

PCC

CH₂Cl₂

D₃O⁺_(aq)

Acetone =

PCC =

Pyridinium

ChloroChromate

Quiz continues on next slide.

Quiz Time!

(Key in Slides above).

DMSO = DimethylSulfoxide

(Polar Aprotic Solvent, great for S_N2)

CrO₃

Acetone, H₂SO_4(aq)

Jones oxidation

PCC

CH₂Cl₂

PCC =

Pyridinium

ChloroChromate

LiAlD₄

Ether

NaBD₄

1a. LiAlD₄

Ether

b. H₃O⁺_(aq)

NaBD₄

CH₃OD

Oxidations and Reductions

Last slide of Quiz

Quiz Time!

(Key in Slides above).