Mechanisms Some Basic Orgo I Reactions. Understanding the basics… Mechanisms are the most...

Mechanisms

Some Basic Orgo I Reactions

Understanding the basics…

• Mechanisms are the most mind-boggling part of organic chemistry.

• Students, generally speaking, have spent their time memorizing their way through science courses.

• Mechanisms require a student to UNDERSTAND the fundamentals of electron flow…

• Everyone knows that electrons are negatively charged.

• Everyone knows that electrons are attracted to things with positive charges.

• Yet, the understanding of a “mechanism” remains elusive to many students…

• Let’s review the basics…

• Electron flow is always from the electron-rich to the electron-poor species.

• The electron-rich species is a Lewis Base (must have a lone pair) and is called the “nucleophile”.

• The electron-poor species is a Lewis Acid (must have empty orbital) and is called the “electrophile”.

• Examples of Nucleophiles:

• Examples of Electrophiles:

X OH H

OH

X-XCl H+O

H H

• SO – electron flow is always from nucleophile to electrophile, electron-rich to electron-poor…

• Watch the direction of your arrows– from lone pairs to carbocation…

OH

H

– from anion to cation…

– from anion to partial positive charge…

X

+ BrBr

– from alkene pi bond to cation or partial positive charge…

Br-Br

H+

• When working through a mechanism, the goal is NOT to memorize the steps of a mechanism of a SPECIFIC molecule. When you do that, typically you become too focused on the structures provided in a single example.

• If that happens, you will get confused when the next mechanism problem has a DIFFERENT structure.

• What you want to do is make a game plan - break down the steps of the mechanism, into little parts or steps.

• The basic little steps are easier to remember.

• By knowing the steps, you know how the mechanism progresses, regardless of the structure you are given to work with.

• SO – break them down…

Dehydration of Alcohols

Identify this mechanism – Starts with alcohol, ends with alkene… losing water…

Dehydration, Acid-catalyzed… H+ from sulfuric or phosphoric acid…

OHH+

+ H2O


Steps Involved:– Convert –OH to H2O (use that acid!)

– Loss of H2O and carbocation formation

– Removal of H+, resulting in formation of pi bond to complete the conversion to alkene

– E1 mechanism – think Zaitsev and Trans!


Step 1: Convert –OH to H2O

OHH+

OH H


Step 2: Loss of H2O (“spontaneous dissociation”) to form carbocation

+ H-O-H

OH H


Step 3: Removal of H+, resulting in formation of pi bond to complete conversion to alkene

+ H-O-H+ H-O-H

H H

Dehydration of Alcohols - Again

Try another example:

Alcohol to alkene (using acid) = dehydration (make water, lose water, form alkene)

OHH2SO4

heat+ H3O

+



Remember the soul purpose for the acid is to turn the –OH into a water molecule. Now it wants to leave…

OH

H+ OH H


Step 2: Make the leaving group leave – “spontaneous dissociation” occurs:

OH H

OH H

+


Step 3: Form the pi bond, make an alkene

Find the beta-H on the more substituted side (Zaitsev!)… use your water as a Lewis Base and pull that H+ off, forming a pi bond… Done!

OH H

+

H

OH H

H

Dehydration of Alcohols – And Again

Try one more example:

Do the steps… and check that your arrows and intermediates look like those you’re about to see…

OHH3PO4

heat+ H3O

+



Always draw the arrow from electron rich (lone pairs!) to electron poor (positive charge!)

O H+H

OH H


Step 2: Spontaneous Dissociation…

Next step? Form the pi bond…

OH H

OH H+


Step 3: Form the pi bond, make an alkene

Find the beta-H on the more substituted side that has an H… use your water and pull it off, forming a pi bond… Done!

OH H

+

H

OH H

H

Acid-Catalyzed Hydration

Identify this mechanism – Starts with alkene, ends with alcohol…

H+, H2O

OH


Steps Involved:– Reaction of pi bond with H+ (acid

catalyst!) resulting in Markovnikov carbocation formation

– Addition of H2O (this is where the OH is coming from)

– Removal of extra proton (H+) to finish formation of –OH.


Step 1: Reaction of pi bond with H+

(acid cat.) resulting in Carbocation formation

H+

H


Step 2: Addition of H2O

OH H

+ H-O-H

H H


Step 3: Removal of extra proton (H+) to finish formation of –OH.

+ H-O-H

OHO

H H + H-O-H

H

H

Acid-Catalyzed Hydration - Again

Try Again… Identify the mechanism – alkene to alcohol, using acid and water…

H2SO4

H2O OH


Step 1: React the pi bond with H+ and form that carbocation:

We don’t have to show this new H but make sure you are drawing the Markovnikov carbocation!

H+ H


Step 2: Add the H2O

(the green H is still there, just didn’t show it in the second structure)

Now - Finish it off… pull the extra proton…

OH

HH

OH H


Step 3: Removal of extra proton (H+) to finish formation of –OH.

OH H

H+

OH H O

H H

OH

Acid-Catalyzed Hydration – And Again

‘Third time’s the charm”… Try one more example:

Do the steps… and check that your arrows and intermediates look like those you’re about to see…

H2SO4

H2O+ H3O+

OH


Step 1: React that nucleophilic pi bond with the proton H+:

We don’t have to show this new H but make sure you are drawing the Markovnikov carbocation!

H+

H


Step 2: Add the H2O

Finish it off… pull the extra proton…

OH

HH H

OH

H


Step 3: Removal the extra proton (H+) to finish formation of –OH.

OH

HO

H

HOH

OH

HH+

Addition of H-X

Identify this mechanism – Starts with alkene, ends with single halide…

H-X

X

Addition of H-X

Steps Involved:– Reaction of pi bond with H+ (of H-X),

concurrent separation of X- , and formation of Markovnikov carbocation intermediate.

– Attack on carbocation by X- to finish formation of product

Addition of H-X

Step 1: Reaction of pi bond with H+ (of H-X), concurrent separation of X-, and formation of carbocation intermediate.

H X+ XH

Addition of H-X

Step 2: Attack of X- to finish formation of product.

H X H

X

Addition of H-X - Again

Try it again… Identify the mechanism… Adding an Br and an H to an alkene…

H-Br Br


Step 1: Reaction of pi bond with H+ (of H-X), concurrent separation of X-, and formation of carbocation intermediate.

H + BrH

Br



Br

H H

Br

Addition of HX – Once Again

Try it again… Identify the mechanism… Adding a chloride (and an H, not drawn in) to an alkene…

H-ClCl


Step 1: React the pi bond with H+ (of H-X), separate off the X-, form the more substituted carbocation intermediate.

H ClH

+ Cl



Cl

H HCl

Addition of X2

Identify this mechanism – Starts with alkene, ends with two halides…

X2

X

X

Addition of X2

Steps Involved:– Attack by pi bond on polarized X-X with

Halonium Ion formation– Attack of X- to pop open three-

membered ring and finish formation of product.

Addition of X2

Step 1: Attack by pi bond on polarized X-X with Halonium Ion formation (and loss of X-)

+ XX X

X

Addition of X2

Step 2: Attack of X- to pop open three-membered ring and finish formation of product.

X

X

X

X

Addition of X2

Of course, you can “attack” the other end of the halonium and open in the other direction…

Regardless, electrons must always flow towards the positive charge…

X

XX

+ X

Addition of X2 - Again

Identify this mechanism: Starts with alkene, ends with two halides… Yes, this reaction occurs “Anti”

Cl-ClCl

Cl


Step 1: Attack by pi bond on polarized X-X with Halonium Ion formation

The halonium ion has to form on one face or the other… whichever you want…

Cl-Cl Cl + Cl



As with the last example, remember the flow of electrons is towards the positive charge…

Cl + ClCl

Cl

Addition of X2 – And Again

Draw the mechanism steps for the example below and then check to see if you drew the arrows “flowing” correctly:

Br-Br

BrBr


Step 1: Attack by pi bond on X-X followed by Halonium Ion formation

The halonium ion has to form on one face or the other… random choice for “up” on this problem…

Br-Br

Br



+ Br

BrBrBr

Addition of X2 and H2O

Identify this mechanism – Starts with alkene, ends with one alcohol and one halide…

+ HXX2

OH

XH2O


Steps Involved:– Attack by pi bond on polarized X-X with

Halonium Ion formation

– Attack by H2O on more substituted side to pop open three-membered ring.

– Removal of extra proton (H+) by X- to complete the formation of –OH.


Step 1: Attack by pi bond on polarized X-X with Halonium formation

+ XX X

X


Step 2: Attack by H2O to pop open three-membered ring.

Markovnikov addition means the water is attacking the more substituted side

+ X+ H-O-H

X

O

X

H H


Step 3: Removal of extra proton (H+) by X- to complete the formation of –OH.

Again, electron flow towards the positive charge…

+ X

X

OH H

X

OH

+ H-X

Addition of X2 and H2O - Again

Identify the mechanism – Starts with alkene, adding an OH and an X… Markovnikov…

OH

Cl

Cl2, H2O


Step 1: Attack by pi bond on polarized X-X with Halonium formation

Halonium ion forms on one face of the alkene…

Cl-Cl Cl + Cl


Step 2: Attack by H2O to pop open three-membered ring.

Note the inversion on the methyl group from the backside attack of the water…

Cl

O

Cl

OH

H

H

H


Step 3: Removal of extra proton (H+) by X- to complete the formation of –OH.

Again, electron flow towards the positive charge…

O

Cl

+ ClO

Cl

H

H

H

+ H-Cl

Formation of 3º halides from alcohols

Identify this mechanism – Starts with alcohol, ends with a halide…

H-BrOH Br

Formation of 3º halides:

Steps Involved:– Tertiary OH must turn into a better

leaving group by picking up a H+ to form water

– Loss of H2O and Carbocation formation

– Attack of halide to form tertiary halide

Formation of 3º halides:


Good leaving group can now leave…

H-Br+ X

O

H

O

H

H

Formation of 3º halides

Step 2: Loss of H2O (“spontaneous dissociation”) to form carbocation

OH HO

H

H +

Formation of 3º halides

Step 3: Halide attacks the carbocation

+ Br

Br

Formation of 3º Alcohols

Identify the mechanism – Starts with 3º halide, ends with a 3º alcohol…

Of course, the E1 elimination happens simultaneously but we’ll deal with that separately.

H2O

Br OH

Formation of 3º alcohols:

Steps Involved:– Loss of Leaving group and carbocation

formation– Attack by water – Removal of extra proton to complete

formation of –OH group

Formation of 3º alcohols:

Step 1: Spontaneous dissociation of –Br to form carbocation

Br

+ Br

Formation of 3º alcohols

Step 2: Addition of H2O

OH H

OHH

Formation of 3º alcohols

Step 3: Loss of extra proton to finish formation of –OH group:

Again, flow of electrons towards positive charge, not a direct attack of it…

OHH

+ BrOH

+ HBr

E1 elimination of 3º alkyl halide

Identify the mechanism – Starts with 3º halide, ends with an alkene, weak base…

Happening at the same time as the SN1 we just looked at…

H2O

Br OH

E1 Elimination of 3º halide

Steps Involved:– Loss of Leaving group and carbocation

formation– Attack by water on beta-H to form

alkene

E1 Elimination of 3º halides

Step 1: Spontaneous dissociation of –Br to form carbocation

Br

+ Br

E1 Elimination of 3º halides

Step 2: Addition of H2O to remove beta-H and form pi bond

OH

H

H

AND REMEMBER…

If you UNDERSTAND the basic steps to these mechanisms, it won’t matter what the double bond in the molecule looks like...

Every alkene reacts the same way, every time, regardless of what’s attached…

Finally…

Track the pieces you need to add or subtract overall… See where you are starting and where you are ending…

Don’t memorize a specific molecule but go ahead and memorize the sequence of steps involved… get a feel for how to approach the problems… develop the instinct…

Thanks for walking through yet another really long power point… hope it helped… let me know what you think… if there’s something else that you think should be added in, I’ll be happy to try to fill in any other missing blanks…

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