CH2-solutions[1] - Answers for Student Exercise 2.1 2.9 1...

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Answers for Student Exercise 2.1 – 2.9 1 N a. benzonitrile H 2 N b. isobutylamine S O O c. diphenyl sulfone O O d. dioxane N + O O - e. 1-nitropropane O HO f. formic acid g. biphenyl O HO h. benzoic acid O H 2 N i. benzamide O O Butyl acetate O NH 2 Butyramide H 2 N Isobutylamine C 11 H 23 O HO Lauric acid Na + O O - Sodium propionate Spectrum E Spectrum H Spectrum I Spectrum F Spectrum G O Allyl phenyl ether O Benzaldehyde H HO o-Cresol O HO m-Toluic acid Spectrum L Spectrum M Spectrum K Spectrum J 2.1. Since the indicated carbon of phenylacetonitrile is sp 3 hybridized, it is reasonable for this compound to show C H stretching at less than 3000 cm -1 (2960-2940 cm -1 ). Where as benzonitrile has only aromatic C H stretching which is typically between 3100 - 3000 cm -1 (page 87). 2.2. 2.3. 2.4. 2.5. C CH 2 N phenylacetonitrile CN benzonitrile 1,3-Cyclohexadiene Diphenylacetylene 1-Octene 2-Pentene Spectrum B Spectrum A Spectrum D Spectrum C
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Answers for Student Exercise 2.1 – 2.9 2 H 2 N Aniline N N Azobenzene N HO Benzophenone oxime H 2 N Benzylamine NH 2 Cl Dimethylamine hydrochloride Spectrum Q Spectrum O Spectrum P Spectrum N
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CH2-solutions[1] - Answers for Student Exercise 2.1 2.9 1...

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