Surviving Alkene Reactions A logical approach Surviving Reactions • Seems like “surviving” is always the word that best describes what has to happen in this class… but I think with a little practice you’ll find this isn’t all that difficult… • Yes, its time for that memorization part so many people dread. Organic chemistry has this awful reputation and I think THIS is where it all sprung from. • Yes, you’ll need to remember some facts… but even worse, you’ll need to APPLY these in order to draw your product. Where to begin…? • There are several pieces that have to come together when drawing the product of a reaction. • You have to maintain a specific carbon skeleton for the molecule that contains the alkene. • You have to know what the reagents will transform the alkene into. • You have to know the details of the pieces that need to be attached. • Is the reaction “Markovnikov” or “NonMarkovnikov”? • Is the reaction going to proceed “Syn” or “Anti”? • If regiochemistry is the concern, you’ll need to determine which end of the alkene is more substituted, versus less substituted. • If stereochemistry is the concern, you’ll have to use wedges and dashes to communicate if the pieces are attaching on the same side or opposite sides. • And, of course, when all is said and done, you have to draw the actual product… • Hopefully this will help walk you through the trials and tribulations of your first reaction chapter… • Let’s begin… Understand your alkene • Alkenes come in all shapes and sizes. • When you want to do a reaction on an alkene, you first have to identify with the two carbons in the skeleton that constitute the alkene. • Depending on the reaction, you may need to determine which end is the more substituted end of the alkene. • Let’s start here… • Let’s look at an alkene so you can see what needs to be seen. • Identify how many carbons are attached to each sp2 carbon of the alkene. • First, just identify the alkene carbons (shown in blue). C C • Then count carbons attached to each. C C C C C TWO carbons attached C C ONE carbon attached C C C • The right-hand alkene carbon has two carbons (in red) attached. • The left-hand alkene carbon has only one carbon (in green) attached. TWO carbons attached C C ONE carbon attached C C C • With this in mind, you should be able to see that the right carbon is “more substituted” with more alkyl groups. • The left carbon is “less substituted” with only one alkyl group attached. • What about the next alkene? end is more substituted? Which C C C • The right side is more substituted with two carbon groups. The left side has only one carbon group. • Try another… • What about this alkene? Which end is more substituted? C C • The left side is more substituted with two carbon groups. The right side doesn’t have any carbon groups. • Got the idea? • Remember that any reaction that has regiochemistry (i.e. that reacts Markovnikov or Non-Markovnikov) will need you to determine which end is more (or less) substituted. Reagents… • For each reaction that is covered, there are many with mechanistic information that you need to remember (the “why is it this or that?”). • For each reaction, there are also some quick and easy pieces to remember (which will take some time to get embedded in yours brains)… Reagents… • The Halogenation Reaction uses HX to add “H” and “X” to the alkene in a Markovnikov manner. HCl • HX can be HCl, HBr or H3PO4 and KI. • “X” will attach to the more substituted side. Reagents… • Answer: HCl H • Need to remember: H X Mark Cl Reagents… • Addition of X2 will add “X” and “X” to the alkene in an ANTI manner. Cl2 • X2 can be Br2 or Cl2. Reagents… • Answer: Cl Cl2 Cl • Need to remember: X X ANTI Reagents… • Addition of X2 and H2O will add “X” and “OH” to the alkene in a Markovnikov, ANTI manner. Cl2 H2O • X2 can be Br2 or Cl2. Reagents… • Answer: Cl2 OH Cl H2O • Need to remember: X OH Mark ANTI Reagents… • Variation: Addition of X2 and ROH will add “X” and “OR” to the alkene in a Markovnikov, ANTI manner. Br2 CH3OH • X2 is Cl2 or Br2 and ROH can have any alkyl group (R). Reagents… • Answer: Br2 CH3O Br CH3OH • Need to remember: X OR Mark ANTI Reagents… • There are three reactions to make alcohols from alkenes. HO H Reagents… • Addition of H2O and H+ will add “H” and “OH” to the alkene in a Markovnikov manner. H2O H2SO4 • H+ can come from any acid like H2SO4 or H3PO4. Reagents… • Answer: H2O HO H2SO4 • Need to remember: H OH Mark H Reagents… • Addition of 1. Hg(OAc)2, H2O and 2. NaBH4 will add “H” and “OH” to the alkene in a Markovnikov manner. 1. Hg(OAc)2, H2O 2. NaBH4 Reagents… • Answer: 1. Hg(OAc)2, H2O 2. NaBH4 • Need to remember: H OH Mark HO H Reagents… • Addition of 1. BH3 and 2. NaOH, H2O2 will add “H” and “OH” to the alkene in a NON-Markovnikov and SYN manner. 1. BH3 2. NaOH, H2O2 Reagents… • Answer: 1. BH3 H OH 2. NaOH, H2O2 • Need to remember: H OH Non-Mark SYN Reagents… • Hydrogenation by the addition of H2, Pd/C will add “H” and “H” to the alkene in a SYN manner. H2 Pd/C Reagents… • Answer: H2 H Pd/C • Need to remember: H H SYN H Reagents… • Dihydroxylation will add “OH” and “OH” to the alkene in a SYN manner. 1. OsO4 2. NaHSO3, H2O • Can use either the addition of KMnO4, NaOH, H2O OR • 1. OsO4 and 2. NaHSO3, H2O Reagents… • Answer: OH 1. OsO4 2. NaHSO3, H2O OH • Need to remember: OH OH SYN Reagents… • There are reactions. three cyclopropanation Reagents… • Addition of CH2N2 will form a SYN cyclopropane ring, with two hydrogens on its apex. CH2N2 Reagents… • Answer: CH2N2 H H • SYN addition for ring formation. Reagents… • Addition of KOH, CHCl3 will form a SYN cyclopropane ring, with two chlorines on its apex. CHCl3 KOH Reagents… • Answer: CHCl3 Cl Cl KOH • SYN addition for ring formation Reagents… • Addition of CH2I2, Zn(Cu) will form a SYN cyclopropane ring, with two hydrogens on its apex. The hydrogens can be replaced by one or two alkyl groups. CH2I2 Zn(Cu) CH(CH3)I2 Zn(Cu) Reagents… • Answer: CH2I2 H H Zn(Cu) CH(CH3)I2 H CH3 Zn(Cu) • SYN addition for ring formation Reagents… • Addition of either of these last two reagents will do an oxidative cleavage reaction, splitting the molecule into pieces. Every double bond must be split. 1. O3 O 2. Zn, H3O+ H O • Ozonolysis, 1. O3 and 2. Zn, H3O+, will form aldehydes and ketones. Reagents… • In either of these last two reactions, an oxidative cleavage reaction will occur, splitting the molecule into pieces. Every double bond must be split. O + O Reagents… • Addition of KMnO4, H2O or H3O+ will form carboxylic acids, ketones or CO2 (H2CO3). KMnO4 O H2O or H3O+ KMnO4 H2O or H3O+ O OH HO O OH O Know your reagents… • Make a single list of all of your reagents and their important “must know” tidbits… • Writing this list out will help you remember all those little facts. • Writing the list multiple times, over and over again will REALLY ingrain it in your memory. • Start working on this early and the night before the exam won’t be so painful! Now, put it all together… • Identify what the reagent is doing to the alkene • Then identify the carbons in your alkene carbon skeleton • Find the two carbons of the alkene in your product skeleton • Draw in the appropriate parts, with correct regiochemistry and stereochemistry. • Let’s start with a simple example using the reagent, HCl. HCl, like all HX reactions, adds one H and one X (in this case, Cl) to the ends of your alkene in a Markovnikov fashion. HCl • Start by identifying the carbons in your alkene carbon skeleton. Attach all of the original carbons in a similar zig-zag format (unless of course, its an oxidative cleavage reaction). • Identify the two carbons that were in the original alkene. Your reaction will occur between these two carbons. • In the example here, the substituted end is on the right: ONE carbon attached TWO carbons attached more • For the addition of HCl, the reaction occurs Markovnikov-style with Cl attaching to more substituted end. HCl H Cl • This would give you the final product shown below: • Or simply: H Cl H Cl HCl H Cl Let’s try a few more… • The addition of X2 adds X and X in an ANTI fashion. • Try this problem: Br2 • Find the two carbons of the alkene: Br2 • Line up the pieces to be attached. Notice that if the same piece is being added to both ends, you don’t have to worry about regiochemistry. Br2 Br Br • Attach the pieces: Br2 Br Br • And finish with stereochemistry: your Br2 Br Br ANTI Let’s do another… • The addition of X2 and H2O adds X and OH, Markovnikov style and in an ANTI fashion. • Try this problem: Cl2, H2O • Find the two carbons of the alkene: Cl2, H2O • Line up the pieces to be attached. Markovnikov addition places the OH on the more substituted end, X on the less substituted end. Cl2, H2O Cl OH • Attach the pieces: Cl2, H2O OH Cl • And finish with stereochemistry: your Cl2, H2O OH Cl ANTI And another… • The addition of H2 with Pd/C adds H and H in a SYN fashion. • Try this problem: H2, Pd/C • Find the two carbons of the alkene: H2, Pd/C • Line up the pieces to be attached. No regiochemistry since you are adding the same thing to both ends. H H2, Pd/C H • Attach the pieces: H H2, Pd/C H • And finish with stereochemistry: your H H2, Pd/C H SYN A different kind of reaction… • The formation of cyclopropane rings from alkenes occurs in a SYN fashion. • Try this problem: CH2N2 • Find the two carbons of the alkene: CH2N2 • Form the cyclopropane (with H’s at the apex) across the two carbons of the alkene: CH2N2 H H • And finish with stereochemistry: CH2N2 your SYN H H • Using CHCl3 and KOH forms a cyclopropane with Cl’s on the apex. • Using CH2I2 and Zn(Cu) forms a cyclopropane with H’s on the apex. Alternatively, you might have alkyl groups, instead of H’s. For example, you might use CH(CH3)I2 and form a cyclopropane that has one H and one CH3 group on the apex. More Flexibility! Those last reactions… • The oxidative cleavage reactions split an alkene into two pieces. • When using the ozonolysis reaction, the alkene splits to form ketones and/or aldehydes: R R H R R R O and O H R • Oxidatively cleave the following: 1. O3 2. Zn, H3O+ • It may help to draw in the H’s attached to the alkene. • Then STRETCH that alkene, and insert the oxygen atoms that will form the aldehydes or ketones. (This reaction doesn’t happen this way but it may help you draw the products.) O H H O • Finish the cleavage of both the pi and sigma bonds. • The final products should be: 1. O3 O 2. Zn, H3O+ H H O The final set of reactions… • When using the KMnO4 reaction, with either H2O or H3O+, the alkene splits to form ketones and/or carboxylic acids: R R OH R R R O and O OH R • Oxidatively cleave the following: KMnO4 H H2O or H3O+ • It may help to draw in the H’s attached to the alkene. • Then STRETCH that alkene, and insert the oxygen atoms that will form the carboxylic acids or ketones. Any aldehydes that typically form for ozonolysis are converted to carboxylic acids with KMnO4. O H OH O • The final products should be: KMnO4 H2O or H3O+ H O OH O Other things to consider… • As you get started, you will notice that you spend a lot of your time learning how to draw the products. • You should also be able to look at a reaction, given the starting material and product, and determine what reagent did the transformation shown. Br • Consider this reaction… • What reagent was used? What changed when the starting molecule turned into the product molecule? Br • View the alkene carbons – from both the perspective of the starting alkene and the product. • Identify those carbons • Then identify what changed on those carbons. Br • Notice that a Br is now attached to the left-hand side. An extra H is now attached to the right-hand carbon. • Ask yourself what reagent adds an H and a Br and does so in such a way that the Br is on the more substituted end? • HBr, of course… • Here’s another reagent is transformation? example causing – Cl OH what THIS • Identify the carbons of the alkenes – what pieces are adding to the alkene? Cl OH Cl OH • The more substituted carbon (bottom) now has an OH group, while the less substituted carbon (top) has a Cl now attached. • The reagent that was used was Cl2 and H2O. Two final reactions to think about… • Don’t forget that you need to know how to MAKE an alkene. • An acid like sulfuric acid or phosphoric acid is used to dehydrate an alcohol. • A strong base like sodium hydroxide or potassium tert-butoxide, KOtBu, is used to dehydrohalogenate an alkyl halide. Synthesis… • Synthesis problems require the stringing together of a series of reactions as one molecule is turned into another molecule. • Consider the following: Br Br Br Br Br Br • Synthesis problems need to be worked from both directions, forwards and reverse. • Identify what’s changing. • You started with a single Br. The product has two Br, one on the original position (note that I didn’t say the SAME Br) and one on the carbon next door. Br Br Br • You do NOT know how to randomly add Br’s to just any carbon. • You actually only know ONE reaction that alkyl halides do - dehydrohalogenate. • You only know ONE reaction that adds TWO halides (ANTI!) on two consecutive carbons. • Put the sequence together… • First add base (like KOtBu or NaOH) to form an alkene. • Then add Br2 to add two Br’s in an ANTI fashion. Br KOtBu Br2 Br Br • Alkene reactions will go easier for you if you start sooner not later for studying all those bits and pieces you need to know. This will also make practicing problems easier, so you can better learn to draw the parts and new molecules. • As always, if this has been helpful, drop me a line at email@example.com and let me know.