CHEM 345 Problem Set 4 Key Grignard (RMgX) Problem Set You will be using Grignard reagents throughout this course to make carbon-carbon bonds. To use them effectively, it will require some knowledge from Chem 343. Making Grignard Reagents: Molecules containing a carbon halogen bond can be transformed into a Grignard reagent. Iodides react faster than bromides. omides react faster than chlorides. Fluorides react very slowly to form Grignards and are in general not used. omides 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 radicals, but I will never ask for that mechanism) R X Mg R MgX Grignards are highly reactive compounds. First, and foremost, they are REALLY, REALLY STRNG BASES. D NT FRGET THIS!!!!! Grignards are essentially carbanions. They will react with acidic protons such as H 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 Me) attached to the same carbon or the carbon next to the halogen. Such Grignards eliminate to give alkenes or carbenes (that end up giving a mess) Examples of Bad Grignard Reagents H Me MgCl
1.) Circle the following molecules that can be made into a Grignard reagent. Cl contains an acidic proton Cl H contains an acidic proton contains a carbonyl Reactions of Grignard Reagents: In Chem 343, you learned that Grignards can react with epoxides to give alcohols. The Grignard typically attacks the less hindered side. This actually only works when a copper salt like CuCN is present and really is not the best use of a Grignard reagent. Grignards are primarily used to make carbon carbon bonds by reacting with carbonyl carbons to form alcohols. 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 (H 3 + ) in a second step. H 3 + is a general term form an aqueous acidic workup step. It is just assumed that it is a weak or dilute answer in water. H 3 + R R R H 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. R R H H H R H
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.) Et 2 2.) H 3 + H a.) 1.) Et 2 2.) H 3 + H c.) 1.) Et 2 2.) H 3 + H d.) 1.) Et 2 2.) H 3 + H d.) 1.) Et 2 2.) H 3 + H
3.) Draw the following mechanism. Ph 1.) Et 2 2.) H 3 + H Ph H Ph Ph H H Ph H Chem 343 Review: Formation of alkyl bromides: S N 2 reaction Change alkyl alcohols into alkyl bromides. Two common reagents: H and P 3. P 3 is a selective reagent that converts H s into bromides by S N 2. The alpha carbon (carbon attached to the H) must be sp 3 hybridized and unhindered. H is less selective and can react with alkenes, alcohols, alkynes, and ethers. It should be used to convert tertiary alcohols into alkyl bromides. 4.) Circle the alcohols below that can be converted into bromides using P 3. a. b. H H tertiary alpha carbon too hindered for S N 2 H could be used here. c. H d. e. H f. H H neopentyl alpha carbon too hindered for S N 2 alpha carbon is sp 2 and not sp 3. need sp 3 alpha carbon for S N 2 reaction
5.) Fill in the boxes: P 3 Mg H 1.) 2.) H 3 + Mg, Et2 P 3 H 1.) 2.) H 3 + H H
6.) Chem 343 Review: The difference between PCC/CH 2 Cl 2 and H 2 Cr 4. (PCC generates aldehydes, but does not react with them. H 2 Cr 4 converts aldehydes to carboxylic acids). Fill in the boxes: For no detectable reaction, write NDR. Chiral molecules must be labeled as or. a. H PCC, CH 2 Cl 2 b. H H 2 Cr 4, H 2 H c. H PCC, CH 2 Cl 2 + D A deuterium behaves just like a hydrogen. D d. H H 2 Cr 4, H 2 D H
6.) continued. e. H PCC, CH 2 Cl 2 f. H H 2 Cr 4, H 2 g. H PCC, CH 2 Cl 2 NDR h. H H 2 Cr 4, H 2 NDR i. H PCC, CH 2 Cl 2 or H j. or H H 2 Cr 4 H 2 H
9.) From the NMR provided, determine the structure of the following compound. Assign the peaks Ha, Hb, Hc etc. accordingly. The NMR spectra are from the Sigma-Aldrich Corporation. Integrations are in bold below each peak. C 13 H 10
10.) From the NMR provided, determine the structure of the following compound. Assign the peaks Ha, Hb, Hc etc. accordingly. The NMR spectra are from the Sigma-Aldrich Corporation. Integrations are in bold below each peak. C 5 H 10