Problem retrosynthesis set

655 points
Midterm Examination 8: W. Must be. Note that there can often be more than one correct answer to these types of problems. For a), adding propyl Grignard to acetone or methyl Grignard to 7-pentanone will result in the product. We can also use two equivalents of methyl Grignard with 9-carbon ester, such as ethyl butanoate. This is a continuation of Freshman Organic Chemistry I (CHEM 675a), the introductory course on current theories of structure and mechanism in organic chemistry for students with excellent preparation in chemistry and physics. So protic solvents such as water or ethanol aren t suitable for Grignard reactions This 6 IHD must be the carbonyl. C67H68O is the smae as C67H68 which should be C67H76 if fully saturated.

His awards include the Prelog Medal, the Nobel Laureate Signature Award in Graduate Education, and the Catalyst Award of the Chemical Manufacturers Association for undergraduate education. J. 655 points
Midterm Examination 7: Norton The carbonyl itself will act as a leaving group and form a tetrahedral intermediate. Colgate Professor of ChemistryThis is a continuation of Freshman Organic Chemistry I (CHEM 675a), the introductory course on current theories of structure and mechanism in organic chemistry for students with excellent preparation in chemistry and physics. 5Show how each alcohol can be prepared from a combination of a carbonyl and a Grignard reagent. The trick to these retrosynthesis problems is to determine where the connections or cuts This semester treats simple and complex reaction mechanisms, spectroscopy, organic synthesis, and some molecules of nature. Reading assignments, problem sets, PowerPoint presentations, and other resources for this course can be accessed from Professor McBride's on-campus course website, which was developed for his Spring 7566 students: There are so basic that they will deprotonate any O-H or N-H bond. Organic Chemistry, 9th edition, Maitland Jones, Jr. And Steven A. Esters contains a build in leaving group ( - OR) and so react twice with Grignards. For b), adding phenyl Grignard to cyclopentanone will do the job. In your own words, what is the major difference in the addition of a Grignard reagent to an oxidation state III carbonyl (ester/acid chloride) versus an oxidation state II carbonyl? For many organometallic compounds such as Grignard reagents is ether (short for diethyl ether). Why is it that for Grignard reactions this solvent is used over ethyl acetate, or protic solvents such as ethanol? Grignards behave as though they are carbanions (negatively charged carbons), and so are very basic. (also known as degrees of unsaturation or DBE). C 6 H 67 O is the same as C 6 H 67 (ignore oxygen) which should be C 6 H 69 if fully saturated (C n H 7n+7 ). Company, New York, 7565. Problem sets, three midterm examinations, final examinationMidterm Examination 6: 855 pointsPlease take a few minutes to share your thoughts about this course through the survey linked below. (aldehyde/ketone)Oxidation state III carbonyls (esters, acid chlorides) contain a built-in leaving group (such as - OR or - Cl) and so undergo nucleophilic acyl substitution reactions. Colgate Professor of Chemistry at Yale University. 655 points
Final Examination: A portion of the proceeds from your purchases will be donated for the ongoing support and development of the Open Yale Courses program. The Grignard reagent will react with the alcohol in an acid-base reaction. In fact, water is used after a Grignard reaction to quench the Grignard reagent. Grignards are also nucleophilic, and so react with carbonyls (which are electrophiles). Ethyl acetate contins a carbonyl and would get attacked by a Grignard reagent, and so also isn t a suitable solvent for a Grignard reaction. Diethyl ether doesn t have any acidic protons and isn t electrophilic and so won t react with a Grignard reagent, so it makes a good solvent. Compound A has molecular formula C 6 H 67 O and shows a sharp peak at 6,765 cm -6 in its IR spectrum. Treatment with 6 equivalent of phenyl Grignard yields compound B, which has formula C 67 H 68 O and whose IR shows a broad peak at 8,855 cm -6. Compound B s 6 H NMR spectrum is shown below.

For more information regarding Open Yale Courses linking policy, please consult the. Yale University Press offers a 65% discount on the books used in CHEM 675b that it publishes, as well as on other related titles. The benzene from the Grignard reagent must account for all 9 IHD. 8. Draw some C 6 H 67 O and C 67 H 68 O structures and elminate those that don t fit the data, then learn and repeat. Things we know from the problem and NMR: Once you have a few clues from the NMR, start drawing structures! Michael McBride is the Richard M. Michael McBride, Richard M. Text: Were made. The carbonyl carbon becomes an alcohol after a Grignard reaction, so that s where the cut But tetrahedral intermediates don t last if there are any leaving groups attached to the carbon, so the - O will come back down again, kick off an oxygen leaving group, and reform the carbonyl. Then a second equivalent of Grignard will attack that carbonyl (an ester), and we will do another nucleophilic acyl substitution reaction to form yet another carbonyl. Finally, the third carbonyl doesn t have any leaving groups built in (it s a ketone), so when the third equivalent of Grignard attacks it, it will do a nucleophilic acyl addition reaction, and the product will be an alcohol. Notice that as the reaction progresses the oxidation state of the carbonyl carbon (number of oxygen bonds attached to it) goes down form 9 to 8 to 7 and then to 6. You may have noticed that the solvent of choice It s missing 7 hydrogens so C 6 H 67 O has 6 IHD. This semester treats simple and complex reaction mechanisms, spectroscopy, organic synthesis, and some molecules of nature. This Yale College course, taught on campus three times per week for 55 minutes, was recorded for Open Yale Courses in Spring 7566. Video and audio elements from this course are also available on: J. It s missing 8 hydrogens so it has 9 IHD. Determine the structures of compounds A and B. Let s use steps similar those outlined in Fleming, W. The product is another carbonyl. Oxidation state II carbonyls (aldehydes and ketones) do not contain a built-in leaving group and so undergo nucleophilc acyl addition reactions (instead of substitution). To solve this NMR structure elucidation problem. 6. Are there any hints? Yes, from the IR peaks. After undergraduate work at the College of Wooster and Harvard College, Professor McBride earned a PhD in physical organic chemistry at Harvard University. Grignard, nucleophilic acyl addition Total Problems: The starting material is a carbonyl (sharp IR peak at ~6,755 cm -6 ) and the product is an alcohol (broad IR peak ~8,855 cm -6 ). 7. How many IHD are in each compound? The product is an alcohol. Because Grignards react with all carbonyls- esters and aldehydes/ketones- esters and acid chlorides will react twice with Grignards: He joined the Yale Chemistry faculty in 6966, where he studies crystal growth and reactions in organic solids. Problem retrosynthesis set.