Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Information gained from a 1H NMR SpectrumSlide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Slide 41Slide 42Slide 43Slide 44Slide 45Slide 46Slide 47Slide 48Slide 49Slide 50Slide 51Give the formula and structure of the compound with this IR and a molecular ion peak at 116.How many peaks in the 13C NMR spectra?OHOHFOHFF2 3 43ClFHOClFClClClCl642Which compound matches this 13C NMR spectrum?BrBrBrBrBrBrBrBr BrBrBrBrOOHABCOABCBrBCARank the labeled C’s of each compound in order of increasing chemical shift?B ~ 33ppmC ~ 35ppmA ~ 115ppmA ~ 8ppmB ~ 37ppmC ~ 210ppmC ~ 9ppmB ~ 30ppmA ~ 177ppmInformation gained from a 1H NMR Spectrum•Number of Signals•Position of Signals•Intensity of Signals•Spin-Spin Splitting of SignalsHow many different types of protons are in each compound?H3COCH3HAHA1 typeH3CH2CClHAHB2 typesH3CH2COCH3HAHBHC3 typesH3CCHH3CCH2H2C CH3HAHAHBHCHDHE5 typesDetermining the number of different protons in compounds with  bonds.ClClHHcis to Clcis to Cl1 typeClBrHHcis to Cltrans to Cl2 typesClHHHcis to Clcis to H3 typesDetermining the number of different protons in compounds with rings..HHHHHHAll H’s are equivalent, 1 typeHHHHClHcis to Clcis to H3 types HHHHCH3CH3Protons on methyls are equivalentEach of these is cis to a methyl2 typesOOOOHOOHOOHSpin-Spin CouplingWhen determining the spin-spin coupling, look at the number of protons on the adjacent carbon. For the methyl group, look at the methylene group. There are 2 protons, so using the N+1 rule tells us that the peak should be a triplet in a 1:2:1 ratio.For the methylene group, look at the methyl group. There are 3 protons, so using the N+1 rule tells us that the peak should be a quartet in a 1:3:3:1 ratio.Protons attached via a double bond show a unique splitting pattern., a doublet of doublets.OHOHcHaHbOOHOOHAn unknown molecule A has 4 signals in the 1H NMR spectrum. Which of the following corresponds to molecule AHow many nonequivalent protons does the following structure have? 4Reading from left to right, what multiplicity would be found for the three nonequivalent sets of protons in the 1H NMR spectrum of the following compound? d, d, sIntroductionHomolytic bond cleavage leads to the formation of radicals(also called free radicals)Radicals are highly reactive,, short lived speciesSingle headed arrows are used to show the movement fo single electrons.Production of RadicalsHomolysis of relatively weak bonds such as O-O and X-X bonds can occur with the addition of energy in the form of heat or light.RCH2R2CH R3CHCarbon radicals are categorized as primary (1°), secondary (2°) and tertiary (3°) based on the number of attached R groups.1°2° 3°A carbon radical is sp2 hybridized with a trigonal planar geometry with the unpaired electron in the unhybridized p orbital.Bond dissociation energy is used as a measure of radical stability.Two different radicals can be formed with the cleavage of a C-H bond.Basically, the more alkyl groups attached to the radical carbon the more stable it is. Also the more stable the radical, the less energy it takes to break the C-H bond.What type of radical are each of the following?H3CH3CCH2CH3•H3CCHH3CCHCH3•H3CCHH2CCH2CH3•1°2°3°Of these three radicals, which is the most stable?H3CH3CCH2CH3••H+X+H X•Radical Reactions of AlkanesAbstraction of a H from a C-H  bond in which one electron is sued to form H-X while the other is left on the new alkyl radical. XX••A radical can also add to a alkene by adding onto a double bond and leaving the other carbon that was part fo the double bond as a radical.X+XX XRadicals are highly reactive and unstable and usually react quickly with a sigma or pi bond. However sometimes they can react with another radical.O O+XO O XWhen oxygen, a diradical, is present it acts as a radical inhibitor or scavenger. Meaning it prevents the radical from attacking any alkanes or alkenes.••• • • •In the presence of heat and light, alkanes and halogens will react to form alkyl halides.H3C H+ Cl2H3C Cl+HClH+Br2Br+HBrH3CH2CCH3+2Cl2H3CH2CCH2Cl+H3CHCCH3ClPredict the products from the monobromination of the foloowing compound?CH CHH3CH3CCH3CH3Br2hv/ heatCH CHH3CH3CCH2ClCH3CH CClH3CH3CCH3CH3CH CHH3CH3CCH3CH2Cl CH CHClH2CH3CCH3CH3Step 1 - InitiationCl Clhv/heat2Cl•Step 2 – Propagation H3CH2CH+ClH3C CH2+HClH3C CH2+Cl ClH3CH2CCl+Cl••• •Step 3 - TerminationClCl+Cl ClH3C CH2CH3H2C+H3CH2CH2CCH3H3C CH2+ClH3C CH2ClIn each step of the propagation a bond is broken and formed. And because the overall step has a -H it is exothermic. Step 1 is called the rate determini g step because it is higher in energy.Transition StatesCl----H----CH2CH3 Cl---Cl---- CH2CH3H3CH2CCH3+Cl ClH3CH2CCH2Cl+H3CHClC CH3There are 6 Methyl H’s and 2 Methylene H s. Based on this, the ratio fo the two products should be 3:1(primary to secondary).However, the ratio is 1:1.The more stable the radical being formed is, the easier it is to cleave the C-H bond.Which C-H bond in each compound is most reactive?HHHHChlorination Vs. Bromination1:1H3CH2CCH3+Cl2H3CH2CCH2Cl+H3CHClC CH3H3CH2CCH3+Br2H3CH2CCH2Br+H3CHBrCCH399%Chlorination is faster and nonselective. This is due to it’s rate determining step being exothermic.Bromination is slower and chooses the most stablew radical. This due to it’s rate determining step being endothermic.H3CH2C CHCH3CH3+BR2H3CH2C CClCH3CH3+Br2BrHalogenation is useful in the formation of alkenes.+Cl2Cl+K+-OHAn elimination in the presence of a strong base is responsible for the formation of the alkene.Cl-OHH+H2O+KBrH+++-HSO4HOHOHH-HSO4OHaq. H2SO4OH+Conversion of the alkene to an alcohol via nucleophillic substitution is an extension of the utility of radical halogenation.+HgOAc+HgOAcMeOHHCH3HgOAcOCH3HgOAcNaBH4OCH3Hg(OAc)2NaBH4OCH3Oxymercuration-demercuration of an alkene results in the formation of an ether.Radical halogenations give a racemic mixture of pRodcuts when possible. This halogenation of an achiral compound results in 3 products. A primary and secondary alkyl halide. The secondary halide exist as a pair of enantiomers due to the creation of a stereogenic center upon