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:

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…

– from anion to cation… – from anion to partial positive charge…

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

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…

Dehydration of Alcohols Steps Involved: – Convert –OH to H 2 O (use that acid!) – Loss of H 2 O and carbocation formation – Removal of H +, resulting in formation of pi bond to complete the conversion to alkene – E1 mechanism – think Zaitsev and Trans!

Dehydration of Alcohols Step 1: Convert –OH to H 2 O

Dehydration of Alcohols Step 2: Loss of H 2 O (“spontaneous dissociation”) to form carbocation

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

Dehydration of Alcohols - Again Try another example: Alcohol to alkene (using acid) = dehydration (make water, lose water, form alkene)

Dehydration of Alcohols - Again Step 1: Convert –OH to H 2 O Remember the soul purpose for the acid is to turn the –OH into a water molecule. Now it wants to leave…

Dehydration of Alcohols - Again Step 2: Make the leaving group leave – “spontaneous dissociation” occurs:

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!

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…

Dehydration of Alcohols – And Again Step 1: Convert –OH to H 2 O Always draw the arrow from electron rich (lone pairs!) to electron poor (positive charge!)

Dehydration of Alcohols – And Again Step 2: Spontaneous Dissociation… Next step? Form the pi bond…

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!

Acid-Catalyzed Hydration Identify this mechanism – Starts with alkene, ends with alcohol…

Acid-Catalyzed Hydration Steps Involved: – Reaction of pi bond with H + (acid catalyst!) resulting in Markovnikov carbocation formation – Addition of H 2 O (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

Acid-Catalyzed Hydration Step 2: Addition of H 2 O

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

Acid-Catalyzed Hydration - Again Try Again… Identify the mechanism – alkene to alcohol, using acid and water…

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!

Acid-Catalyzed Hydration - Again Step 2: Add the H 2 O (the green H is still there, just didn’t show it in the second structure) Now - Finish it off… pull the extra proton…

Acid-Catalyzed Hydration - Again Step 3: Removal of extra proton (H + ) to finish formation of –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…

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!

Acid-Catalyzed Hydration – And Again Step 2: Add the H 2 O Finish it off… pull the extra proton…

Acid-Catalyzed Hydration – And Again Step 3: Removal the extra proton (H + ) to finish formation of –OH.

Addition of H-X Identify this mechanism – Starts with alkene, ends with single halide…

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.

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

Addition of H-X - Again Try it again… Identify the mechanism… Adding an Br and an H to an alkene…

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.

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

Addition of HX – Once Again Try it again… Identify the mechanism… Adding a chloride (and an H, not drawn in) to an alkene…

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.

Addition of HX – Once Again Step 2: Attack of X - to finish formation of product.

Addition of X 2 Identify this mechanism – Starts with alkene, ends with two halides…

Addition of X 2 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 X 2 Step 1: Attack by pi bond on polarized X- X with Halonium Ion formation (and loss of X - )

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

Addition of X 2 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…

Addition of X 2 - Again Identify this mechanism: Starts with alkene, ends with two halides… Yes, this reaction occurs “Anti”

Addition of X 2 - 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…

Addition of X 2 - 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…

Addition of X 2 – And Again Draw the mechanism steps for the example below and then check to see if you drew the arrows “flowing” correctly:

Addition of X 2 – 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…

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

Addition of X 2 and H 2 O Identify this mechanism – Starts with alkene, ends with one alcohol and one halide…

Addition of X 2 and H 2 O Steps Involved: – Attack by pi bond on polarized X-X with Halonium Ion formation – Attack by H 2 O 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 X 2 and H 2 O Step 1: Attack by pi bond on polarized X- X with Halonium formation

Addition of X 2 and H 2 O Step 2: Attack by H 2 O to pop open three- membered ring. Markovnikov addition means the water is attacking the more substituted side

Addition of X 2 and H 2 O Step 3: Removal of extra proton (H + ) by X - to complete the formation of –OH. Again, electron flow towards the positive charge…

Addition of X 2 and H 2 O - Again Identify the mechanism – Starts with alkene, adding an OH and an X… Markovnikov…

Addition of X 2 and H 2 O - Again Step 1: Attack by pi bond on polarized X- X with Halonium formation Halonium ion forms on one face of the alkene…

Addition of X 2 and H 2 O - Again Step 2: Attack by H 2 O to pop open three- membered ring. Note the inversion on the methyl group from the backside attack of the water…

Addition of X 2 and H 2 O - Again Step 3: Removal of extra proton (H + ) by X - to complete the formation of –OH. Again, electron flow towards the positive charge…

Formation of 3º halides from alcohols Identify this mechanism – Starts with alcohol, ends with a halide…

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 H 2 O and Carbocation formation – Attack of halide to form tertiary halide

Formation of 3º halides: Step 1: Convert –OH to H 2 O Good leaving group can now leave…

Formation of 3º halides Step 2: Loss of H 2 O (“spontaneous dissociation”) to form carbocation

Formation of 3º halides Step 3: Halide attacks the carbocation

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.

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

Formation of 3º alcohols Step 2: Addition of H 2 O

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…

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 S N 1 we just looked at…

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

E1 Elimination of 3º halides Step 2: Addition of H 2 O to remove beta- H and form pi bond

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