8 Identify the compound: practice bank

The workbook’s NCEA-type question and sample assessment tasks. Each gives you spectra and a set of six candidate structures. The format you will meet in the internal. Work them as written: evidence from each spectrum, then a justification of your chosen molecule. NCEA-type exam question Question One Use the following spectra (mass spectrum, infrared and carbon-13 NMR) to evaluate the identity of the molecule from the six possibilities given. Your answer should include: evidence of the molecule’s identity from each spectrum, and a justification of your chosen molecule. The six possibilities are: (a) 2-methylpropanamide · (b) 2-methylpropanoic acid · (c) ethyl ethanoate · (d) N-ethylethanamide · (e) N-ethyl-N-methylmethanamide · (f) 3-methylbutan-2-one. Work the mass spectrum first: what is the molecular ion, and is it odd or even? Then the IR: is there a C=O? Is there a broad 3000–3500 absorption, and if so, is it one band or two? Then the ¹³C: how many environments, and which region is the carbonyl in: 180–220 or 160–185? ✓ Use the “student NMR table” ranges above. The workbook directs you to that table for this question. Note that several of these six share a molecular formula, so the mass spectrum alone cannot separate them. The discrimination has to come from IR and ¹³C. Sample assessment · Task A Identify the compound (M = 88) A set of IR, ¹³C NMR and MS spectra for a single compound containing ONE functional group. Note: the highest significant peak is at m/z = 88, and it is not visible on the printed spectrum. The ¹³C shows three signals. 1. Discuss the information each of the spectra provides about the nature of the molecule. 2. Circle the structure that best fits: (i) N-isopropylmethanamide · (ii) isopropyl methanoate · (iii) ethyl ethanoate · (iv) butanoic acid · (v) 2-methylpropanoic acid · (vi) N,N-dimethylethanamide. 3. Justify the structure you have chosen, and why each of the other structures does not fit the spectra provided. ✓ Three ¹³C environments is the key constraint. Count the environments in each of the six candidates before doing anything else, and see how many survive. Then use the IR to decide between what is left. Remember: 88 is an even mass, so an even number of nitrogens. Sample assessment · Task B Identify the compound (M ≈ 100) A second set of IR, ¹³C NMR and MS spectra for a single compound containing one functional group. The ¹³C shows a signal in the 160–170 region plus three lower signals. 1. Discuss the information each of the spectra provides about the nature of the molecule. 2. Circle the structure that best fits: (i) N-methylbutanamide · (ii) 3-methylbutanoic acid · (iii) methyl butanoate · (iv) 3-methylbutanamide · (v) N,N-dimethylpropanamide · (vi) 2-methylbutanamide. 3. Justify the structure you have chosen, and why each of the other structures does not fit the spectra provided. ✓ A carbonyl in the 160–185 band immediately rules out any ketone or aldehyde. From there, the IR decides between acid (broad O–H), amide (one or two N–H bands) and ester (neither), and the number of ¹³C environments picks the specific isomer.

Practice for this standard covers the whole topic — flashcards and quiz.