Question

Molecular formula: C10H15N; 1H NMR (CDCl3)  6.9 and 6.6 (AA’BB’ pattern, 4H, J = 8.0 Hz), 3.5 (broad singlet, 2H, D2O exchangeable), 2.5 (t, 2H, J = 7.5 Hz), 1.5 (quintet, 2H, J = 7.5 Hz), 1.3 (sextet, 2H, J = 7.5 Hz), 0.9 (t, 3H, J = 7.5 Hz)

Answer 1

Answer:

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Explanation:

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