Organic Chemistry: Week One Notes
Organic Chemistry: Week One Notes CH 2240
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This 4 page Class Notes was uploaded by Amanda Biddlecome on Sunday August 28, 2016. The Class Notes belongs to CH 2240 at Clemson University taught by Dr Schroeder in Fall 2016. Since its upload, it has received 38 views. For similar materials see Organic Chemistry in Chemistry at Clemson University.
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Date Created: 08/28/16
Organic Chemistry 2240 August 17, 2016 Amanda Biddlecome Chapter 14: Mass Spectrometry and Infrared Spectrometry 1) Addition Reactions -‐more substituted halide is the major product because the cation forms there *regiospecific -‐add hydrogen peroxide to get the minor as prominent product -‐3 questions *What forms? *Where does it form? *How does it form? 2) Radical Reactions -‐What happened to go in the anti direction? -‐hydrogen peroxide breaks and you have single hook arrows indicating that one electron goes to each side -‐compounds with unpaired electrons are radicals/oxidents and are very reactive -‐HX breaks apart like how the O-‐O bond breaks *products are hydrogen peroxide and X radical and is still reactive o *break the double bond between the carbons and electron goes to 3 carbon that forms a bond with H and X goes to the terminal carbon -‐anytime you see hν 3) Addition Reactions -‐tertiary carbocation is the most stable *Markovnikov Rule -‐Radical Halogenation Reactions *forms alkyl halides *forms on the most substituted carbon *stereochemistry is racemic 4) Identification of Compounds -‐mass spectrometer forms radicals and cations *measures molecular mass and finds the formula *how?-‐sample in solution goes in and goes through electron beam and goes through a tube with magnets and fragments are analyzed by the cations and radicals that are produced (we see the cations and use math to get the radicals) *tells the mass and abundance of particles formed ^abundance corresponds to stability *molecular ion peak is the molecular mass and is at the first peak *two peaks are important: one corresponding to molecular mass and the base peak (most abundant) *to find the radical, subtract base peak from molecular weight to get the mass of the radical; find the formula through trial until the mass of the formula is equal to the masses that you calculated *break the original bond with the highest priority so you can have the highest priority cation *first, focus on the cation stability and if they’re the same, look at the stability of the radical when figuring out why base peaks form -‐Hydrocarbons *saturated has general formula C H n 2n+2 *if you have carbon, hydrogen, and oxygen, you always have an even molecular mass -‐Nitrogen *molecular peak with an odd number means we have nitrogen *molecular peak with an even number means no nitrogen -‐one CH =one oxygen 4 -‐degrees of unsaturation 5) Definitions to keep in mind -‐Molecular Ion Peak=molecular mass of compound -‐base peak=most stable fragment that can form 6) Alkyl Halides: Identifying Compounds -‐Chlorides *2 naturally occurring isotopes: 35 (most abundant) and 37 *get two molecular mass values *3:1 ratio of Cl: Cl in terms of abundance *M=molecular ion peak *M+2 peak is the second peak that is 1/3 the size of M *if you have two peaks at 78 and 80, then that indicates that you have chlorine in your compound -‐Bromides *2 naturally occurring isotopes that are more even in value *79 and 81 * Br: Br=1:1 ratio *this is how you can tell if a compound has bromine instead of chlorine -‐When you put CH CHCH Cl into the 3 machine, you 3 lose electrons *the Cl-‐C bond is broken because it’s the weakest; break away both electrons *get CH CHCH w3th a mass 3of 43, which accounts for the peak at 43 and a Cl radical *also break a C-‐C bond which explains the peak at 63 *this all happens because of resonance structures -‐heterolytic cleavage=breaking bond between two things that are different *Pathway one *Alpha *double hooks -‐homolytic cleavage=breaking bond between two things that are the same *Pathway two *Beta *single hook -‐halides=heterolytic *halide is lost as radical most often because of bond strength and that’s what’s responsible for base peak 7) Ethers: Identifying Compounds -‐break C-‐O bond with a double hooked arrow *heterolytic cleavage *the oxygen becomes the radical *get another carbo-‐cation as well -‐can break on right or left of the oxygen 8) Alcohols: Identifying Compounds -‐put in the mass spectrometer, an electron comes off of the oxygen, and is replaced with a cation -‐alcohol doesn’t fall off easily because they’re very electronegative -‐homolytic cleavage happens because alcohol is a poor leaving group -‐fragment by the beta carbon and one electron goes to each side *get a radical on both carbons 9) Electromagnetic Spectrum -‐mass spectrometer doesn’t help figuring out what functional groups are present 10) Wave Characteristics: Identifying Compounds -‐wavelength, frequency -‐relationships *wavelength+frequency=speed of light *E=hv ^h=Planck’s Constnat *E=hc/wavelength 11) Infrared Spectroscopy -‐smaller -‐bonds aren’t static -‐tells us what functional groups we have in our sample 12) Functional Group Identification -‐match things to the table -‐team this information with the mass spectrometer -‐alcohol has a big peak -‐amine has a similar peak to alcohol, just slightly smaller 3 2 -‐find sp , sp , and sp to know if there’s a double bond -‐fingerprint region=advanced, so we ignore it -‐IR doesn’t tell you how many alcohols you have or structures *not as much functionalism
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