Upload
others
View
6
Download
0
Embed Size (px)
Citation preview
5/24/11
1
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+ OH 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
5/24/11
2
– 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
Dehydration of Alcohols
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!
5/24/11
3
Dehydration of Alcohols
Step 1: Convert –OH to H2O
OHH+
OH H
Dehydration of Alcohols
Step 2: Loss of H2O (“spontaneous dissociation”) to form carbocation
+ H-O-H
OH H
Dehydration of Alcohols
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+
Dehydration of Alcohols - Again
Step 1: Convert –OH to H2O
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
Dehydration of Alcohols - Again
Step 2: Make the leaving group leave – “spontaneous dissociation” occurs:
OH H
OH H+
5/24/11
4
Dehydration of Alcohols - Again
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
+
HO
H 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+
Dehydration of Alcohols – And Again
Step 1: Convert –OH to H2O
Always draw the arrow from electron rich (lone pairs!) to electron poor (positive charge!)
O H+H
OH H
Dehydration of Alcohols – And Again
Step 2: Spontaneous Dissociation…
Next step? Form the pi bond…
OH H
OH H+
Dehydration of Alcohols – And Again
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
5/24/11
5
Acid-Catalyzed Hydration
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.
Acid-Catalyzed Hydration
Step 1: Reaction of pi bond with H+ (acid cat.) resulting in Carbocation formation
H+
H
Acid-Catalyzed Hydration
Step 2: Addition of H2O
OH H
+ H-O-H
H H
Acid-Catalyzed Hydration
Step 3: Removal of extra proton (H+) to finish formation of –OH.
+ H-O-H
OHOH H + H-O-H
H
H
Acid-Catalyzed Hydration - Again
Try Again… Identify the mechanism – alkene to alcohol, using acid and water…
H2O, H2SO4
OH
Acid-Catalyzed Hydration - Again
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
5/24/11
6
Acid-Catalyzed Hydration - Again
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
Acid-Catalyzed Hydration - Again
Step 3: Removal of extra proton (H+) to finish formation of –OH.
OH
HH
OH
OH
+
HO
H
H
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
Acid-Catalyzed Hydration – And Again
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
Acid-Catalyzed Hydration – And Again
Step 2: Add the H2O
Finish it off… pull the extra proton…
OH
HH H
OH
H
Acid-Catalyzed Hydration – And Again
Step 3: Removal the extra proton (H+) to finish formation of –OH.
OH
HOH
HOH
OH
HH+
5/24/11
7
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
Addition of H-X - Again
Step 1: Reaction of pi bond with H+ (of H-X), concurrent separation of X-, and formation of carbocation intermediate.
H + BrH
Br
5/24/11
8
Addition of H-X - Again
Step 2: Attack of X- to finish formation of product.
BrH 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
Addition of HX – Once Again
Step 1: React the pi bond with H+ (of H-X), separate off the X-, form the more substituted carbocation intermediate.
H ClH
+ Cl
Addition of HX – Once Again
Step 2: Attack of X- to finish formation of product.
ClH H
Cl
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.
5/24/11
9
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-Cl Cl
Cl
Addition of X2 - Again
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
Addition of X2 - Again
Step 2: Attack of X- to pop open three-membered ring and finish formation of product.
As with the last example, remember the flow of electrons is towards the positive charge…
Cl + ClCl
Cl
5/24/11
10
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
Addition of X2 – And Again
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-BrBr
Addition of X2 – And Again
Step 2: Attack of X- to pop open three-membered ring and finish formation of product.
+ Br
BrBrBr
Addition of X2 and H2O
Identify this mechanism – Starts with alkene, ends with one alcohol and one halide…
+ HXX2
OH
XH2O
Addition of X2 and H2O
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.
Addition of X2 and H2O
Step 1: Attack by pi bond on polarized X-X with Halonium formation
+ XX X
X
5/24/11
11
Addition of X2 and H2O
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
Addition of X2 and H2O
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
Addition of X2 and H2O - Again
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
Addition of X2 and H2O - Again
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
Addition of X2 and H2O - Again
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
5/24/11
12
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:
Step 1: Convert –OH to H2O
Good leaving group can now leave…
H-Br+ X
OH
OH
H
Formation of 3º halides
Step 2: Loss of H2O (“spontaneous dissociation”) to form carbocation
OH HO
HH +
Formation of 3º halides
Step 3: Halide attacks the carbocation
+ BrBr
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
5/24/11
13
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…
Br
H2O
E1 Elimination of 3º halide
Steps Involved: – Loss of Leaving group and carbocation
formation – Attack by water on beta-H to form alkene
5/24/11
14
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…
Dr.Discordia [email protected]