Reactions Markovnikov Practice Problems and Keys Part 1

SN1 Reactions: By-Products - Whenever an SN1 reaction occurs there is always the possibility that an E1 by-product will form. This occurs if the nucleophile does not get to the carbocation soon enough. Therefore, if you see a reaction with SN1 conditions always assume that some E1 minor product will form. Racemic Mixtures - In SN1 reactions you also have the possibility of making a racemic mixture of products – that is, left and right hand versions of the same molecule. Always check to see if the leaving group is on a chiral carbon, if so, then a racemic mixture will be made. Cl H3C C OH C 2H5 KOH, H2O H3 C H C H C2H5 and H3 C H C C 2H5 OH Rearrangements – It is also possible for SN1 reactions to rearrange. Carbocations want to be on the most stable carbon, and this means 3 if it is available. Therefore carbocations will rearrange themselves to place the positive charge onto the most stable carbon ONLY IF the most stable carbon is right next door. CH3 H H3 C C C H Cl H3 C C CH3 H CH3 H CH3 H CH3 Hydride C CH3 Shift + H3 C C + C H H CH3 OH - H3 C C C OH H E2 Reactions: Big Bases – For an E2 reaction to take place you must use a big base. Classically this means using the t-butoxide ion (t-ButO-) but other large bases can be used (more on this later). You do not want to use a base that is so small that an SN2 reaction could occur. If you look carefully, you will see that the conditions of an E2 and an SN2 reaction are nearly identical except for the size of the base. So large bases must be used for E2 reactions. Hoffman vs. Saytzeff - E2 reactions make double and triple bonds by removing an HX from a molecule. But which HX? As can be seen below, you may have a couple of choices; H3C CH3 Cl H C C C H H H Saytzeff Product H Hoffman Product CH3 H3C CH3 Cl H C C C H H H CH3 H Base- H H3C C C CH3 CH3 H3 C Saytzeff the Hoffman "Inner" Product H C C H H C H Hoffman Of “outer”"Outer" product. Product The Saytzeff the “inner” product and is the the two, Saytzeff is Hoffman Saytzeff is the most stable because it produces a double bond with more carbons around it which can feed electrons to the double bond (by induction). It is possible to select between Saytzeff and Hoffman products by selecting the proper sized base. Hoffman products are always made when very large bases like t-ButO- are used. Large bases are simply to big to grab inner hydrogens and do an elimination. Big bases are force to attack on the outside of the molecule where there is less hinderance, thus the Hoffman product is formed. To get the Saytzeff product a smaller base must be used, but not one that is so small that you risk the possibility of SN2 product formation. For this purpose EtO- is often the base of choice. The EtO- ion sits between the region of large and small bases and will do either E2 or SN2 reactions depending on the substrate used. As long as the substrate is hindered enough, E2 reactions will predominate, but there is always the risk of SN2 byproducts. Proper Orientation - Perhaps the most important aspect of E2 elimination is the need forproper orientation of the molecule. E2 reactions occur anti which means that the hydrogen being attacked and the halogen that is leaving must be on opposite sides of the molecule. Syn - Interferes with the Cl- trying to leave (repulsion) H3C CH3 Cl H C C C H H H H Incoming Base- Anti - No interferance with the ClThe reason why this orientation is important is that if the base is on the same side as the leaving group (syn attack) then the negative charge of the base, and the negative charge of the leaving group will repulse one another and keep a reaction from occurring. In addition, since the hydrogen and the halogen must be on opposite sides of the molecule, you may have more than one hydrogen to choose from when doing the elimination. Not all hydrogens are equal. In order to choose the right hydrogen, you must draw the most stable Newman projection of the molecule. Consider the molecule below. Cl CH3 C C C H H C C C E2 Saytzeff C C H CH3 cis C C or C CH3 C C H C trans C Depending on the orientation of the molecule only one of the two possible products are formed, but which one? To know this, we must draw the Newman projection and then rotate the molecule so that the hydrogen is opposite the halogen. H3 C Cl CH3 C C H H C C-C CH3 Cl H3 C C Cl H3 C HH C-C H C-C CH3 H H3 C Put the halogen and hydrogen opposite one another CH3 H Eliminate the HCl and form a double bond C C C H E1 Reactions: E1 reactions are by the far the rarest reactions among this group. The reaction requires that there be a protic solvent and no nucleophile - a condition difficult, but not impossible, to satisfy. Most protic solvents are also weak nucleophiles. As we have seen, solvents like water and alcohol can are nucleophilic enough to give SN1 by-products even under the best of conditions. The trick is to use a protic solvent that is also such a poor nucleophile that the nucleophile does not want to react. This can be done in two ways, make it big, and make it a very weak base. As it turns out, strong acids, like sulfuric acid (H2SO4) and phosphoric acid (H3PO4), fit this profile. The sulfate and phosphate ions are very large and very poor nucleophiles as are the conjugate bases of most strong acids. Even relatively weak acids like acetic acid can be used, because the acetate ion is large and weakly basic. Most of the time, strong acids are used to do E1 eliminations. This is not exclusive of course – water and other protic solvents could be used, but you run the risk of making large amounts of SN1 by-product by using these solvents. As long as no strong nucleophile is present, solvents like water and alcohol could also be used. A typical E1 elimination is shown below. Note: The most stable product is always formed. For E1 reactions this always means trans. OH CH3 C C H H C C H2SO4, H3PO4 or both C CH3 C H + H2 O C C C CH3 The final product has a cis double bond By appropriate rotation and elimination you can see that the final product will be cis-3methyl-2-butene (or Z-3-methyl-2-butene). You cannot predict whether the product will be cis or trans (E or Z) unless you draw the Newman projection and then do the elimination. C C C About Solvents…. By now, I am sure that you are very confused about solvents and which solvent to use with which reaction. Some of it is common sense and some of it is experience, but there is much more to the common sense than the experience. Let me give you some examples. Alcohols and Alkoxides By far the most common solvent/nucleophile combination is the alcohol/alkoxide combination. Alkoxides (methoxide, ethoxide, t-butoxide, etc) are all made from their respective alcohols based on the following reaction; 2 R O + H2 2 R OH + 2 Na(s) The point is that the alkoxide is always made from the alcohol so both are present in solution – the alcohol being the solvent and the alkoxide being the nucleophile/base. Common pairs are given below. Solvent CH3OH C2H5OH t-ButOH Nucleophile/base CH3ONa C2H5ONa t-ButOK Also known as; (MeOH and MeO-) (EtOH and EtO-) (t-ButOH and t-ButO-) These solvent/base pairs are commonly used in SN2, E2, and even SN1 reactions. Strong Acids Strong acids are common solvents used in E1 reactions but they are also used in SN1, and even SN2 reactions (but never E2). Now why would a strong protic solvent like H2SO4 be needed in a reaction that prefers aprotic solvents (like SN2 reactions)? The answer is really very simple. Acids are commonly used to get rid of OH groups by turning them into good leaving groups (water!). So you frequently see acids used whenever the leaving group is an OH – even on SN2 reactions as shown below. H OH C C C H+ (H2SO4) C + O C H C Br- C C Br C + H2O SN1, SN2, E1, and E2 Reaction Conditions Reaction Type SN2 SN1 E2 E1 Substrate 1o, unhind 2o 3o, hind 2o 1o, 2o, or 3o 3o, hind 2o Nucleophile/Base Small strong Nuc- present Large base No base or nuc- Small Strong Bases (Nucleophiles): OHCH3O- (MeO-) C2H5O- (EtO-) CH3C2H5IHNH2CH3NH- Big Bulky Bases: t-ButOisoPrO- Protic Solvents H 2O Alcohols – MeOH, EtOH Organic Acids – HAC Inorganic Acids – H2SO4, H3PO4 Aprotic Solvents Acetone THF Diethyl ether DMSO Methylene Chloride Solvent Aprotic Protic Aprotic Protic Leaving Group Good LG Good LG Good LG Good LG Mechanisms: SN2 Mechanism: H H OH- H C HO Cl H C + Cl- C OH Note: Walden inversion H H SN1 Mechanism: Note: Racemic mixtures are possible H3 C SN1 H H H3 C H3 C C C H Cl H3 C H3 C E1 H3 C C+ CH3 CH3 H H C CH3 C + H+ CH3 E1 Mechanism: H H3 C E1 H3 C H3 C H3 C C Cl C+ H C C H CH3 H H + Cl- H CH3 H3 C C H3 C H SN1 C Sol+ Note: Racemic mixtures are possible C H CH3 E2 Mechanism: H H H C C Cl H H Base- H H C H + HBase + Cl- C H H H H3 C Sol C C CH3 H H + H+ Details…details… SN2 Reactions: Solvent – SN2 reactions prefer the use of aprotic solvents but that does not mean that protic solvents cannot be used – it simply means that the reaction will go slower if a protic solvent is used, but that should not hinder its use. Many reactions will require the use of a protic solvent because of the nature of the nucleophile used. A large number of nucleophiles are the conjugate bases of alcohols. These nucleophiles are made by adding pure sodium metal to the alcohol according to the following reaction; 2 Na(s) + 2 ROH  2 RO- Na+ + H2 The nucleophile (RO-) is produced in this reaction and then used to substitute for other poorer leaving groups. BUT because of the nature of the nucleophile, the solvent must be the alcohol from which it was made. Therefore you must use the corresponding alcohol for each of the following nucleophiles; NucAlcohol CH3O- & CH3OH C2H5O- & C2H5OH t-ButO- & t-ButOH isoPrO- & isoPrOH So, if you want to use a nucleophile that is made from an alcohol, you must use the alcohol as the solvent. The problem of course is that alcohols are protic, but this should not be cause for concern because they will work just fine even if they do slow down the reaction. For other nucleophiles like OH- (really NaOH), you can go into the stock room, get it, and throw it into any solvent you like (like THF or diethyl ether). This makes it easy. But most of the time this is not the case. Markovinikov Additions to Alkenes Cl HX Addition C C H+ C C C Cl- C + C C C H H Br Halogen Addition C C Br+ C C C Br- C C C C Br + Br H Hydration Addition of Water C C C H+ C C H 2O C + C H C C C C C H+ C C CH3OH C + OH C C + H C C O C C C H OH Halohydrin Reaction C C C Br+ C C C OH- Br + Oxymercuration/ Deoxymercuration C C HgOAc + C C C C H 2O C C C H + Br H C C C O Hg OAc+ Alkoxymercuration/ C Dealkoxymercuration C C C C C C C C H+ C C O C CH3OH C C C C C O C C C O- C H 2O + C C O H OH C C OH H 2O C C C H C OH C C OH OH OH C HgOAc H C OCH3 NaBH4 C O+ OH- C H H Epoxide Ring Opening - Base C H + O Hg OAc+ Epoxide Ring Opening - Acid OH NaBH4 HgOAc H3 C HgOAc+ + H+ OCH3 C H H C H H H H3C Alcohol Addition + O + OH- + H+ Anti-Markovinikov Additions Anti-Markovinikov HX Addition Chain Initiation O O C OH O C O Heat or Light + OH Cl Cl meta-chloroperoxybenzoic acid (MCPBA) OH + C C HBr H2 O + Br + Br Chain Propagation C Br C C C Br H C C HBr C C C C Br Br Chain Termination - Any two radicals (not shown) Hydroboration H C C BH3 C C H B C C H H H C C B H H H C C C C C C C B H C C C C C H C C H C C C H B C C C H C C 3x C C C B C C H OH C C H H H C + H3BO3 H C C C C C C H B C C C C Other Alkene Reactions Epoxide Formation C C H O O H O O O C H C O O C C O C O H O Epoxide C O C O C C C Cl Cl Cl meta-chloroperoxybenzoic acid Cl m-chlorobenzoic acid Carbene Addition C C C CH2 C C C C C C (other products possible) C H2 C H2 Catalytic Hydrogenation Alkane Produced H2 gas H Surface of Catalyst (Typically Platinum) H Surface of Catalyst (Typically Platinum) C C H H C Surface of Catalyst (Typically Platinum) C C H H C Surface of Catalyst (Typically Platinum)