Chapter 8 Reactions of Alkenes

Transcription

1 Chapter 8 Reactions of Alkenes Electrophilic Additions o Regio vs stereoselectivity Regio where do the pieces add? Markovnikov s rule hydrogen will go to the side of the double bond with most hydrogens. Some reactions give anti-markovnikov products Remember that if both sides of the double bond are equally substituted, then you will get a mixture of products Stereo Are there any specific stereochemical results? o Addition of X Anti addition the two pieces come from opposite sides of the double bond Syn addition the two pieces come from the same side of the double bond Markovnikov with rearrangement No stereoselectivity because intermediate is trigonal planar Electrons of pi bond come out to grab proton Generates carbocation and halide ion Proton goes to less substituted side of double bond because it creates a more stable carbocation intermediate. Carbocations can rearrange! X - attacks carbocation Cl - -Cl rearrangement Cl

2 Different halogens F is very slow, so you won t often see it in this reaction Rate increases as you go down the group, so Cl<<I Think about it: o Addition of with peroxide o The first step is breaking the -X bond, generating X - In terms of stability, I - > - >Cl - >F - Anti-Markovnikov, no rearrangement No stereoselectivity because intermediate is trigonal planar Initiation The oxygen-oxygen single bond is relatively weak and can break apart to yield two radicals An oxygen radical encounters - to generate an alcohol and a bromine radical Initiation 1) R R heat 2 R* 2) R* R * Propagation The bromine radical goes to the less substituted side of the double bond to generate a more stable carbon radical. * The carbon radical encounters - and generates an alkyl halide and another bromine radical.

3 * * Termination Any two radicals meet o Acid-Catalyzed ydration of Alkenes The reverse of dehydration of alcohols Markovnikov addition of water with rearrangements possible No stereoselectivity because intermediate is trigonal planar The electrons of the pi bond grab a proton from the acid, generating a carbocation on the other side of the double bond 2 Remember that carbocations can rearrange! Water attacks the carbocation, and then deprotonates 2

4 What would you get if you used an alcohol as the solvent instead of water? o xymercuration-demercuration Markovnikov addition of water without rearrangement No stereoselectivity because intermediate is free to spin around its single bond Generates an alcohol. Add g(ac)2 (aq) followed by NaB4 in subsequent step Form mercurinium ion as intermediate can t rearrange The mechanism below may make it look like you get anti addition, but there is no stereoselectivity for this reaction You won t be drawing out the mechanism of this on any test, but you need to understand it. Understanding the intermediates is essential to understanding why you get the products. First, the electrons of the double bond attack the mercury creating a three-membered ring. g(ac)2, 20 g Ac This ring is strained because of its size. It is trying to pull apart to alleviate this and as a result is creating partial positive charge at the two carbons. The water comes in and attacks the molecule at the carbon with the stronger partial positive (the more substituted side) and breaks the ring open.

5 2 g Ac gac gac Adding NaB4 (sodium borohydride) removes the mercury and replaces it with a hydrogen. gac NaB4 o Alkoxymercuration-Demercuration Markovnikov addition of an alcohol without rearrangement Generates an ether, not an alcohol 1) g(ac)2, C3 2) NaB4 o ydroboration-xidation C3 Anti-Markovnikov addition of water without rearrangement Syn addition you will get a maximum of two stereoisomers of the product ne important point here is to make sure you add the and the to the same side. It is very common for students to remember that this is syn addition but to accidentally mess this up.

6 1) B3 2) 22, - You think "syn" and want to put a "cis" product, but this product is wrong. 1) B3 2) 22, - If you draw in your hydrogen, you will be less likely to make this mistake. B3 acts as electrophile B3 is called borane. o Because pure borane does not exist alone in a liquid state, you will often see it as one of the following variants of borane. You will commonly see borane complexed with the solvent TF (listed as B3 TF). You might also see diborane listed, B26. Not responsible for drawing this mechanism on a test, but understanding it can help you understand why you get anti-markovnikov, syn addition You get syn addition because the hydrogen and boron add in one step from the same side. B B Looking at the above step of the mechanism, you see that the electrons of the pi bond are coming out to grab the boron, not the hydrogen. This step also defines the regiochemistry of the reaction (anti-markovnikov). The boron is then oxidized off, but the stereochemistry created in the first step is maintained.

7 22, - B hand waving Pretty much insignificant point You can add water to three alkenes per molecule of borane. o Catalytic ydrogenation Just adding hydrogen across the double bond ydrogen is conjugated onto metal catalyst Pt, Pd, or Ni If you see Raney Nickel, just think nickel The double bond just plops down onto the catalyst The hydrogens always add on the same side Syn addition metal catalyst This makes sense because the alkene just plonks down onto the two hydrogens, so there is no way for the two hydrogens to add anti. This matters because it reduces the number of possible stereoisomers eterogeneous catalysis

8 o Simmons-Smith Term used when the catalyst is in a different phase as the reagents In this case, the catalyst is a solid (the metal). C2I2 and Zn generate a carbenoid. The Zn inserts between the carbon and one iodine The carbon-zinc bond is very polar with a partial negative on the carbon If you imagine this taken to the extreme, you get a carbene. I C Zn I I - C Zn I The carbenoid forms a three-membered ring across the double bond ne step, concerted reaction Because there is no chance for the original single bond to rearrange, stereochemistry will be retained C trans double bond methyl and ethyl are trans on the ring o α-elimination Strong base turns C2X2 or CX3 into carbene Base abstracts proton why is this possible? o The electronegativity of the halogens makes that proton acidic enough for strong bases to abstract it.

9 ne halogen falls off C - C - C Carbene adds across double bond to form ring C The ring will have 1 fewer halogen on it than the starting material. Stereochemistry is retained o Addition of halogens (X2) Need to have inert solvent Usually a chlorinated methane (C3Cl, C2Cl2, or CCl3) Double bond attacks X2, forming halonium ion inert solvent alide opens halonium ring inert solvent - alide goes to more substituted side, because the partial positive is stronger there.

10 Anti addition nce the reaction has occurred, you have a single bond. o Remember singles bonds can rotate! In multiple-choice questions, be careful to look and see if the answer choices have the carbon chain in the same place. o If you rotate around the single bond after the product is formed, your product can look like it underwent syn addition. = o alohydrin formation Use water instead of inert solvent First step is the same as with addition of halogens 2 Double bond attacks X2, forming halonium ion Water attacks more substituted side of halonium ring - 2 The more substituted side has a greater partial positive charge

11 Deprotonate Anti addition o Epoxidation Addition of peroxy acid (RC) to alkene A peroxy acid is just a carboxylic acid with an extra oxygen R MCPBA is a common peroxy acid o You don t need to know the structure of this specific peroxy acid. Form a little ring with an oxygen in it Stereochemistry is retained Not responsible for drawing out this mechanism R o Epoxidation with ring opening Can do this in the same reaction vessel as just the epoxidation, or can do this to an existing epoxide Protonate oxygen of epoxide

12 Water attacks more substituted side of ring and deprotonates Anti addition xidation You don t need to know these mechanisms. o smium tetroxide Expensive reagent Syn addition of two s Product is called a vicinal diol s s 22 o KMn4 (mild) What makes it mild? Another syn addition of two s Far less expensive, but also far less efficient o zonolysis Cleaves the double bond, giving ketones and/or aldehydes

13 1) 3 2) DMS Mr. Baker often gives you the products and asks for the starting material Just put the pieces together like a puzzle! Ex. If you subject a certain compound to ozonolysis, you get the following products: 2 molecules of acetone What was the original compound? The only way to put these together to make one molecule is this: o KMn4 (vigorous) Similar to ozonolysis, but aldehydes are further oxidized to carboxylic acids KMn4 (vig.)

14 Summary Name Additional Reagents What Adds? Regioselectivity Stereoselectivity Rearrangement Possible? Addition of X X -, -X Markovnikov N/A Yes Addition of with BR, -, - Anti-Markovnikov N/A No 22 Peroxide Acid-Catalyzed 2, -, - Markovnikov N/A Yes ydration xymercuration- Demercuration 1)g(Ac)2, 2 -, - Markovnikov N/A No Alkoxymercuration- Demercuration ydroboration- xidation ydrogenation Simmons-Smith α-elimination 2) NaB4 1)g(Ac)2, R 2) NaB4 1) B3 2) 22, - 2 and Pt, Pd, or Ni C2I2, Zn(Cu) C2X2 or CX3 with strong base -, -R Markovnikov N/A No -, Anti-Markovnikov Syn No - -, - N/A Syn No -C2- N/A Retention of stereochemistry No -CX- N/A Retention of No or stereochemistry -CX2- -X, -X N/A Anti No Addition of halogens X2, inert solvent alohydrin X2, 2 -X, - - to more Anti No formation substituted side Epoxidation R-C3 -- N/A Retention of No stereochemistry Epoxidation Ring R-C3, -, - N/A Anti No pening (aq) smium tetroxide s4 -, - N/A Syn No KMn4 (mild) KMn4 -, - N/A Syn No (mild) zonolysis 1)3 N/A Splits the bond N/A N/A 2)DMS (or and turns into a something carbonyl else) KMn4 (vig) KMn4 (vig) N/A Splits the bond Further oxidizes aldehydes To C

15 *** If the name of the reaction is in bold in the above chart, then you need to be able to draw out the step-by-step mechanism. You may still be asked about the mechanisms of some of the others (intermediates, why you get certain stereochemistry, etc.) but you won t have to draw them. Simply memorizing the above chart will not get you an A on this test. You need to know how to apply it. If you don t know what Markovnikov vs. anti-markovnikov means, or if you can t identify the more- or less-substituted side of the double bond then you won t apply this chart correctly. That said, you definitely need to know everything on that chart! If you don t know what the different reagents do, you ll have a rough time with test 3. For example, I couldn t remember whether g(ac)2 led to Markovnikov addition with no rearrangement or anti-markovnikov, syn addition (which is the result of a different reaction). As a result, I missed a lot of points. Don t be lazy like me!