{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

C13NMR Notes - Nuclear Magnetic Resonance Spectroscopy 13C...

Info icon This preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Nuclear Magnetic Resonance Spectroscopy 13C NMR — Determining Number ofSignals Nuclear Magnetic Resonance (NMR) is a spectroscopic technique that allows us to determine the molecular connectivity of organic molecules. There are many different nuclei we can use, but the most common are the hydrogen (Ill or proton) nucleus and the carbon ~ 13 (BC) nucleus. This worksheet deals with I3C spectra and the information we can obtain from them. Information obtained from a ”(3 NMR s ectrum. ' the number ol‘carbons in the molecule ' the chemical environment of the carbon (referred to as chemical shift) Determining the number of carbon signals in the "‘C NMR spectrum , t . . . z t . . lhe number 0t carbons obtained in the "C NMR spectrum is determined by whether the carbons are similar either by rotation around a (7 bond or via a plane ol‘symmetry. Examples Determine the number of signals for the molecules given below lixample #1 : 'l‘his molecule has a plane of symmetry and therefore the carbons that reflect into each other are similar and will only appear as a single line on the I5C. NMR spectrum. The molecule will have 4 signals in the ”(3 NMR spectrum. lixample #2: 0 CH3 0 CH3 Br CH3 This molecule does not have a plane of symmetry. so we might expect all carbons to be different and would expect 8 signals in the ]’C NMR, However. the three methyls of the tert—butyl are similar through rotation. 0 CH3 *— all three » oi~ these eurbons 0 CH3 ‘— urc similar Br CH3 via rotation Problem: For each molecule below, determine the number 01‘ signals that will appear in the 13C NMR spectrum. Nuclear Magnetic Resonance Spectroscopy “c NMR — Chemical Shift orsrgnais Nuclear Magnetic Resonance (NMR) is a spectroscopic technique that allows us to determine the molecular connectivity of organic molecules. There are many different nuclei we can use, but the most common are the hydrogen ('11 or proton) nucleus and the carbon .- 13 (BC) nucleus. 'l‘his worksheet deals with I3C spectra and the information we can obtain from them. Information obtained from a ”(T NMR s ectrum. ' the number ofcarbons in the molecule ° the chemical cm ironment of the carbon (referred to as chemical shift) Determining th_c__chemical shift ofthe carbon siunal in the I3C_NMR spectrum Where are the peaks on the 13 C spectrum? This has to do with what‘s connected to the carbon in question. Table #1 shows typical ranges for different types of carbons. 'l‘able #1: Chemical Shift oi‘CTarbons in I3C NMR Spectrum ____-_. .-..-fLrBt‘0fC‘arb-op _____ - l» Chemital-Shift-Rangmpm) Alkanes Primary 0—40 Secondary ‘ 10-50 'l'ertiar) 15-50 __.- -.__-$Et_l?£nal;\;_s..._m_-_-F__._ _ 30-40 a C—X l Alkyl halide or amine 10-65 Alcohol or ether _“_w _ u _ _‘ 50—90 Alkz'nes __7 _ 60-90 ._ Alkenes _ ‘__- 100—170 Aromatic -_- . - m __ N ‘_ _100-170__ _ (.‘arboxylic Acids & [Derivatives Carboxvlic Acids & l’isters 160-185 Amide-s 150-180 _-- Nitrite-S". __ - -.. - .- _ ___120—1.3-<2_ olden-9951.22251-Kcmnss- _ - 1340-215 a- Examples Example it] : The first thing we do is to determine the number of signals we would expect to see. The molecule above has a plane of symmetry so this will reduce the number of signals in the 13C NMR spectrum. Based on the numbering above. we would expect to see 6 signals in the 13C spectrum. We don‘t need to consider the individual aromatic carbons ~— we can simply lump them together. This means that we have to arrange the order of C1—C4. C5 and C6 in the correct order. Aromatic carbons are further downfield than one might expect. They usually show up between 100 and 170 ppm. Both CS and (.‘6 are connected to an electronegative atom and this will give it a higher chemical shift (larger number). A closer look- at C5 and we Pall/e that it is connected to an aromatic ring aromatic rings will also shift the location of a carbon signal to a higher chemical shil‘t. 'l'hese cll‘eets are additive in NMR spectroscopy. Since (‘5 has two groups that shill i1 downlield. we expect the relative order of the signals to be as shown below. Ix) Ix) U1 (3 (ppm) 0 l3 xamplflefi .23; \ In the molecule above. there is a plane 01‘ symmetry that make the two methyl carbons identical. ‘l‘herelbre. we would expect 5 signals in the I’C NMR spectrum. 5 Alkene carbons between 100—170 ppm Alkane carbons primary 0-40 secondary 10-50 tertiary 15—50 ’l'here is some ambiguity in this molecule about the relative order of the carbon signals since the allsane hydrogens appear in overlapping regions of the NMR spectrum. We would not be able to definitively identify the order of carbon signals without performing i a different NMR experiment. 1 The most important part is that the number of signals matches the number of carbons expected. Alkene carbons r—A—fi Alkane carbons r—‘A‘fi [\J Example #3: O HN The molecule above does not have a plane of symmetry and none 01‘ the carbons are identical via rotation. 'l'heret‘ore. each carbon is different and will give rise to its‘ own . - 13‘ . Signal in the (. NMR spectrum. This molecule will have 5 signals in the UC NMR spectrum. Alkenc carbons 100—170 ppm (these tend to be on the lower end 01' the range) Amide carbon 150—180 ppm Alkane carbons 'l'hcre are two alkane carbons: one connected to an sp2 hybridized carbon and one connected to nitrogen. Nitrogen is clectronegative and therefore will shilt it to a higher chemical shift. 0 HN12 s 3 4 . 225 6 (ppm) 0 Problem: 101' cach molecule bclow. assign each carbon 21 number and arrange the chemical shifts ol‘ the carbons from lowest chcmi ‘al to highest chemical shift on the ”C NMR spectrum. ll‘a molcculc has an aromatic ring. _\ou can lump all aromatic carbons into thc same rcgion. O NH2 fi ‘0“ wr‘v'ald S LLbJAcA ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern