Week 14 Notes
Week 14 Notes CHEM 2321
Popular in Organic Chemistry I
verified elite notetaker
Popular in Chemistry
This 10 page Class Notes was uploaded by Hayley Lecker on Monday November 30, 2015. The Class Notes belongs to CHEM 2321 at University of Texas at El Paso taught by Dr. James Salvador in Fall 2015. Since its upload, it has received 71 views. For similar materials see Organic Chemistry I in Chemistry at University of Texas at El Paso.
Reviews for Week 14 Notes
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 11/30/15
OrganicChemistry Week 14 Important Information: Professor’s Email: email@example.com Class Website: organic.utep.edu/courses/2324 Class Code (E-book): utep232Xfall2015 *Sorry For No Notes on Week 13, it was just repeated notes from last week ALL IMAGES/CHARTS AT THE BOTTOM1* IR SPECTRA So I am going to go over examples from the the homework to explain the IR Spectrum. 3300(br,s), 3024 (m), 2877(m), 1659(w), 1021, (s) So first you need to know br means broad, s is strong, m is medium, and w is weak. If you have the lab book there is a nice chart that will help you (I will scan it and insert it later). So if you have the book, look at 241 (lab NOTEBOOK)!, first thing it asked is Strong band at 1820-1650? Yes or No, our answer is so, so now it asks medium to weak band at 1680-1620? Which is Yes. We now know we have a carbon-carbon double bond and only 1 isomer of C4H8O2 has a double bond that was shown to us. *Hydrogens not shown Let’s look at a few more: 2983(w),1743(s),1243(s),1048(m) We know there is a strong band from 1820-1650, so we have C-O double bonding. And because of the 1743(s) we know we have an ester. This narrows the choices to two esters. The first choice And 2 NDchoice: For this problem they are very similar, so its hard to know which one is. However, it is the first one. So let’s do one more: 3500-2500(br) 1712(s). We know we have a C-O double bond because of 1820-1650. And beause of the broad brand at 3500- 2500 we know we have a carboxylic acid. 1 NMR H For this I am going to examples from the homework: Draw the C6H14O isomer with 1H NMR (ppm) 3.4 (4H, t), 1.6 (4H, sex), 0.9 (6H, t). 1 step: We see there is NO O-H Bonding that would show up as (1H, broad s) This lets us know we have an ether. So we know we have a O-C-O now its best to work out. We know we have 6H in a triple bond, these number can be broke up so its best to do a methyl to a methylene x2. So as you can see there is two 3H to 2H which equal the 6H triple bond. Now let’s do another example: Draw the C6H14O isomer with 1H NMR (ppm) 1.53 (1H, broad s), 1.49 (4H, q), 1.1 (3H, s), 0.9 (6H, t). We know there is is an alcohol, but look closely there is NOTHING about 1.53 meaning there is no hydrogens on the carbon that the oxygen is attached to. We see that were is 3H, s. So lets put it on the carbon that will have NO hydrogen, that will make it show up in a singlet. Now as above with the 6H, t we can break that up to get: So let’s do one more so show if there is a signal above the O-H.: Draw the (R)-C6H14O isomer with 1H NMR (ppm) 3.5 (1H, q), 1.6 (1H, broad s), 1.1 (3H, d), 0.9 (9H, s). We know this is an alcohol, and the carbon with the oxygen has only 1 hydrogen. There is also a 3H in a doublet which means it has one hydrogen neighbor, so let’s add it to the Carbon with 1. Now there is a 9H in a s this mean they are attached to ONE carbon with no hydrogens. Once you have all the parts together, it is best to look at the question again, it asked for (R), so the chiral center is carbon with the oxygen bonding. Oxygen is 1, the carbon with 3 methyls is 3, and the single methyl is 3 since the hydrogen is in the back this is R. 13 NMR C – FIRST DRAW OUT ALL ISOMERS Isomers of C4H10 Isomers of C6H14 Isomers of C7H16 Let’s first look at C4H10 for how many NMR signals it would have: In methyl propane there will be two signals, the three methyls will appear as one signal, and the middle C will appear as one in 13C NMR. In 135-DEPT the middle carbon will not show up. The straight chain butane would have 2 signals as well, but also have 2 negative signals. Signals will only be negative if you have methylenes (CH2). Which makes since since butane has two methylenes. Let’s first look at C5H12 for how many NMR signals it would have: Quiz: 1. Pentane 2. 2-methylbutane 3. 2,2-dimethylpropane A. 4 B. 3 C. 2 Match which would have the number of signals! Answer: Pentane – 3 signals, the ends would make one, the end in the middle would make one signals, and the center carbon will make one. Trick question: How many negative signals? 3 As well because there is 3 methylenes. 2-methylbutane – 4 signals, the two methyls make one signals, and the other 3 carbons make the additional 3 carbons. Now how many negative? 1 because there is one methylene. 2,2-dimethylpropane- 2 Signals. The 4 methyls and 1 Carbon make 2 signals. In 135 Dept there will only be one signal because a carbon with no hydrogens don’t show up.
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'