Operação Matemática

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Chapter 15 Suggested solutions for Chapter 15 109 Purpose of the problem of Chapter 11 and the first example of total structure determination. Suggested solution The starting material for 2A is C₇H₁₂O₃ and has apparently lost an oxygen atom. As the reagent is 1.02 (6H, s) 3.9 (2H, s) it is more likely to have gained two hydrogen atoms and lost a water molecule. The IR 0 shows a C=O group and suggests an ester or a strained ketone. The NMR shows two CH₂ groups, one (at 2.51 p.p.m.) next to a functional group but not oxygen, so presumably 1.66 (2H, t) 0 the intact CMe₂ group, and an isolated CH₂ group next to oxygen at 3.9 p.p.m.. There is only 2.51 (2H, t) structure. The mass spectrum of compound 2B shows that it has chlorine in it, the IR shows a CN group, CI 4.6 (2H, s) the proton NMR shows eight Hs. If we assume that no C atoms have been lost, the most formula is The compound has lost a water molecule. The proton NMR CN three linked CH₂ groups, a quintet in the middle , and two triplets at the ends. The shifts of 2.9 (2H, t) 2.5 (2H, t) CH₂s (2.5 and 2.9 p.p.m.) show that they are next to functional groups but not Cl is no oxygen in the molecule). This must mean that we have a unit: All 2.0 (2H, quintet) remains is an isolated CH₂ group with a large shift and the chlorine atom so they must be to S. The large shift (4.9 p.p.m.) comes from 1.5 + 1 (S) + 2 = 4.5 p.p.m. Again one emerges. Problem 3 Two alternative structures are shown for the possible products of the following reactions. Explain each case how you would decide which product is actually formed. Several pieces of evidence be required and estimated values are more convincing than general statements. 0 0 0 H Ho or Ho 0 0 OH 0 0 H H or O₂N O₂N O₂N of the problem you thinking the other way round - from structure to data. What are the important pieces of ence? solution are many acceptable ways in which you could answer this question ranging from choosing vital statistic for each pair to analysing all the data for each compound. We'll adopt a way and point out several important distinctions. In the first example, one main difference is size, seen mainly in the IR. Both are esters (about 1745 cm but we should add 30 cm⁻¹ five-membered ring. The functional group next to the OCH₂ group is different - an OH in one case and an ester in the other. There are other differences. 0 1775 cm⁻¹ in the IR 0 0 8H about 4.1 0 1745 in the IR HO about 3.6