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1) Molecular ion peak of 122 and 124 (3:1)

1) Molecular ion peak of 122 and 124 (3:1). 122-35 = 87/12 = 7 carbons. 87-84 = 3 hydrogens C 7 H 3 Cl. (2(7) + 2 – 3 – 1)/2 = 6 DOUS . C 6 H 15 Cl 2(6) + 2 – 15 – 1 = -2/2 = -1. C 5 H 11 OCl (2(5) + 2 – 11 – 1)/2 = 0 . C 4 H 7 O 2 Cl (2(4) + 2 – 7 – 1)/2 = 1.

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1) Molecular ion peak of 122 and 124 (3:1)

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  1. 1) Molecular ion peak of 122 and 124 (3:1) 122-35 = 87/12 = 7 carbons 87-84 = 3 hydrogens C7H3Cl (2(7) + 2 – 3 – 1)/2 = 6 DOUS C6H15Cl 2(6) + 2 – 15 – 1 = -2/2 = -1 C5H11OCl (2(5) + 2 – 11 – 1)/2 = 0 C4H7O2Cl (2(4) + 2 – 7 – 1)/2 = 1

  2. You have a carbonyl peak, so it’s the last formula, C4H7O2Cl And the peaks at 1000 and 1200 show a C-O so ester

  3. The singlet shows that you have 1 carbon attached to the ester and the 2H shows this is where the Cl is 1.310 triplet 3H 4.252 quartet 2H 4.062 singlet 2H

  4. And the triplet and the quartet represent the ethyl group attached to the carbonyl.

  5. The carbon between 160-180 is your carbonyl, so 3 other types of carbon.

  6. 2) Molecular ion peak at 150 and 152 (1:1) 150 – 79 = 71/12 = 5 carbons 71 – 60 = 11 hydrogens C5H11Br 2(5) + 2 – 11 – 1 = 0

  7. Nothing but C-H stretches so C5H11Br

  8. 3.417 triplet 2H 0.92 doublet 6H 1.86 sextet 2H 1.64 multiplet 1H

  9. 3.417 triplet 2H 0.92 doublet 6H 1.86 sextet 2H 1.64 multiplet 1H With 0 degrees of unsaturation we know this is a chain, so now what does it look like? We have a multiplet with an integration of 1H and a doublet with an integration of 6 so that tells us its an isopropyl group. So that is 3 carbons and 7 hydrogens, leaving 2 carbons, 4 hydrogens and a bromine.

  10. 3.417 triplet 2H 0.92 doublet 6H 1.86 sextet 2H 1.64 multiplet 1H So the sextet must be the carbon adjacent to the isopropyl group, because (1+1)(2+1)= 6 So that leaves the triplet that integrates to 2 hydrogens, the triplet tells you it is next to a CH2 group and the 2 H integration tells you the bromine is attached b/c it should have 3 H otherwise.

  11. The carbon matches this structure b/c it tell syou there are 4 types of carbon.

  12. 3) Molecular ion peak of 73. 73 – 14 = 59/12 = 4 carbons 59 – 48 = 11 hydrogens so C4H11N 2(4) + 2 – 11 + 1 = 0 C3H7NO 2(3) + 2 – 7 + 1 = 2/2 = 1

  13. Carbonyl peak tells you tha tit is C3H7NO also there are no N-H stretch so it’s a tertiary amide.

  14. 8.019 singlet 1H 2.970 singlet 3H 2.883 singlet 3H Singlet at 8.019 tells you it is attached to the carbonyl carbon.

  15. So we have the following structure thus far: 8.019 singlet 1H 2.970 singlet 3H 2.883 singlet 3H So two more singlets each representing 3 H so methyl groups.

  16. 4) Name the following compound: 2,2’-dimethylpropanoyl chloride 5) Name the following compound: N,N-dibenzylformamide

  17. 6) Name the following compound: Isopropyl formate Isopropyl methanoate 7) Name the following compound: Benzoic propanoic anhydride

  18. 8) Draw 3-methylhexanoyl chlordie. 9) Draw isopropyl 3-ethyl-4-methylpentanoate

  19. 10) Draw N-tert-butyl-N-ethyl-2-isopropy-3,4-dimethylpentanamide

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