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(Rmgx) Problem Set You Will Be Using Grignard Reagents Throughout This Course to Make Carbon-Carbon Bonds

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(Rmgx) Problem Set You Will Be Using Grignard Reagents Throughout This Course to Make Carbon-Carbon Bonds CHEM 341 Problem Set 18 Problem Set 10/27/17 Grignard (RMgX) Problem Set You will be using Grignard reagents throughout this course to make carbon-carbon bonds. Making Grignard Reagents: Molecules containing a carbon halogen bond can be transformed into a Grignard reagent. Iodides react faster than bromides. Bromides react faster than chlorides. Fluorides react very slowly to form Grignards and are in general not used. Bromides are the most widely used because they balance good reactivity while not as high of cost as iodides. To make a Grignard reagent, simply add magnesium metal and diethyl ether to a molecule containing a carbon halogen bond. The magnesium will insert itself between the carbon halogen bond. (The actual mechanism probably involves unpaired electrons, but I will never ask for that mechanism) Mg R X R MgX O Grignards are highly reactive compounds. First, and foremost, they are REALLY, REALLY STRONG BASES. DO NOT FORGET THIS!!!!! Grignards are essentially carbanions. They will react with acidic protons such as OH’s, NH’s, SH’s, and terminal alkynes. So, the Grignard reagent cannot have an acidic hydrogen on it. Second, they will react with carbon nitrogen or carbon oxygen pi bonds. Grignards cannot have carbonyls or nitriles. They also cannot contain epoxides. Third, they cannot have a leaving group (even a poor one like OMe) attached to the same carbon next to the halogen. Such Grignards eliminate to give alkenes. Examples of Bad Grignard Reagents MgCl MgBr MeO HO MgBr 1.) Circle the following molecules that can be made into a Grignard reagent. Br Br O Cl O Br Cl HO Br Reactions of Grignard Reagents: A Grignard reaction consists of two steps. The first step is reacting the Grignard reagent with a carbonyl in an inert solvent such as diethyl ether. The carbon carbon bond forming reaction happens here forming + an alkoxide. The alkoxide is then protonated by a mild acid (H3O ) in + a second step. H3O is a general term form an aqueous acidic workup step. It is just assumed that it is a weak or dilute acid in water. + O H3O RMgBr R OH O R O racemic racemic The mechanism is straight forward. The electrons move from the carbon magnesium bond to the carbonyl carbon. The electrons from the pi bond go the oxygen which is protonated in the next step. H MgBr O O R R O H H R OH 2.) The following alcohols can be made by combining a Grignard reagent and a ketone/aldehyde followed by an aqueous acidic workup. Some can be made by multiple methods. Provide the Grignard and the appropriate electrophile. a.) 1.) Et2O + OH 2.) H3O racemic a.) 1.) Et2O + OH 2.) H3O racemic c.) 1.) Et2O + 2.) H3O OH d.) 1.) Et2O + OH 2.) H3O d.) 1.) Et2O + OH 2.) H3O 3.) Draw the following mechanism. O 1.) Et2O 2.) H O+ OH PhMgBr 3 Ph 4.) Provide a mechanism for the following reactions: a. O HBr Br excess Br b. O O H3PO4 O OH What other carbon containing product is formed? c. O trace H2SO4 O O OH d. NaOH O HO OH O racemic e. O NaOH HO OH O racemic 5.) Fill in the appropriate reaction arrow. The starting material is on the left, the product is on the right. Think about stability. Use or . O a. O O O b. O O O O O O c. O O 6.) There are three ways pf making the following alcohol from a Grignard reagent and a carbonyl. List them out. Number the carbons on the alcohol and number the corresponding carbons on the Grignard reagent. OH 1.) Et2O + + 2.) H3O Grignard reagent ketone racemic OH 1.) Et2O + + 2.) H3O Grignard reagent ketone racemic OH 1.) Et2O + + 2.) H3O Grignard reagent ketone racemic 7.) You should be able to do this one. This is a critical thinking problem. Using the base strengths of the anions, predict which reaction arrow you should use. You will need to know the pKa values of conjugate acids of the bases. Assume all of the products have approximately the same basicity. ( ) a. H O O O O H b. H O Cl Cl O H c. O O H O O H d. H O O H e. H O C N O N H One of the key problems you will be faced in organic chemistry is the idea of a synthesizing a molecule from simpler components. The key to doing these problems is to start at the end and work backwards. Which chemical bond was made in that last step? These first few synthesis problems have hints. Some have multiple answers. Beware, learning synthesis is the first step on the dark path that led to Walter White’s destruction.* 8.) Fill in the boxes: To deprotonate alcohols, you should use sodium hydride (NaH) as the base. Part a. has two possible starting alcohols. Part b. only has one possible starting alcohol. Why? a.) base alcohol alkoxide halide O b.) base alcohol alkoxide halide O optically active * I actually don’t know what happened to Walter White, because I have yet to see Breaking Bad. So don’t spoil it for me. 9.) To deprotonate phenols, thiols or carboxylic acids, you should use sodium hydroxide (NaOH) or potassium carbonate (K2CO3) as the base. Sodium hydride is more expensive, much more dangerous, and is well overkill. Why can you use sodium hydroxide to deprotonate phenols, thiols, and carboxylic acids but not use it to deprotonate alcohols? 10.) Fill in the boxes: a.) base thiol halide S optically active b.) base carboxylic acid halide O optically active O 11.) To deprotonate terminal alkynes, sodium hydroxide will not work. Instead, you should use sodium amide (NaNH2). Why won’t sodium hydroxide work? Fill in the boxes: a.) base terminal alkyne halide optically active b.) H H This problem set had some training wheels on it. Pretty soon, you will be given questions without the boxes. .
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