The seeds of the Costa Rican plant Ateleia herbert smithii are avoided by all seed eaters (except a weevil that adapts them for its defence) because they contain two toxic amino acids (IR spectra like other amino acids). Neither compound is chiral. What is the structure of these compounds? They can easily be separated because one $(\mathrm{A})$ is soluble in aqueous base but the other $(\mathrm{B})$ is not.
A is $\mathrm{C}_6 \mathrm{H}_9 \mathrm{NO}_4$ (mass spectrum) and has $\delta_{\mathrm{C}} 34.0(\mathrm{~d}), 40.0(\mathrm{t}), 56.2 (\mathrm{s}), 184.8(\mathrm{~s})$, and $186.0(\mathrm{~s})$. Its proton NMR has three exchanging protons on nitrogen and one on oxygen and two complex signals at $\delta_{\mathrm{H}} 2.68\left(4 \mathrm{H}, \mathrm{A}_2 \mathrm{~B}_2\right.$ part of $\mathrm{A}_2 \mathrm{~B}_2 \mathrm{X}$ system ) and $3.37(\mathrm{X}$ part of $\mathrm{A}_2 \mathrm{~B}_2 \mathrm{X}$ system) with $J_{\mathrm{AB}} 9.5, J_{\mathrm{AX}} 9.1$, and $J_{\mathrm{BX}}$ small.
B is $\mathrm{C}_6 \mathrm{H}_9 \mathrm{NO}_2$ (mass spectrum) and has $\delta_{\mathrm{C}} 38.0(\mathrm{~d}), 41.3(\mathrm{t}), 50.4 (\mathrm{t}), 75.2(\mathrm{~s})$, and $173.0(\mathrm{~s})$. Its proton NMR spectrum contains two exchanging protons on nitrogen and $\delta_{\mathrm{H}} 1.17(2 \mathrm{H}, \mathrm{ddd}, J 2.3$, $6.2,9.5), 2.31(2 \mathrm{H}$, broad m), $2.90(1 \mathrm{H}$, broad t, $J 3.2)$, and 3.40 ( 2 H , broad s).
Because the coupling pattern did not show up clearly as many of the coupling constants are small, decoupling experiments were used. Irradiation at $\delta_{\mathrm{H}} 3.4$ simplifies the $\delta_{\mathrm{H}} 2.3$ signal to $(2 \mathrm{H}$, $\mathrm{ddd}, J 5.8,3.2,2.3)$, sharpens each line of the ddd at 1.17 , and sharpens the triplet at 2.9 .
Irradiation at 2.9 sharpens the signals at 1.17 and 2.9 and makes the signal at 2.31 into a broad doublet, $J$ about 6 . Irradiation at 2.31 sharpens the signal at 3.4 slightly and reduces the signals at 2.9 and 1.17 to broad singlets. Irradiation at 1.17 sharpens the signal at 3.4 slightly so that it is a broad doublet, $J$ about 1.0 , sharpens the signal at 2.9 to a triplet, and sharpens up the signal at 2.31 but irradiation here had the least effect.
This is quite a difficult problem but the compounds are so small ( $\mathrm{C}_6$ only), have no methyl groups, and have some symmetry so you should try drawing structures at an early stage.13 The seeds of the Costa Rican plant Ateleia herbert smithii are avoided by all seed eaters (except a weevil that adapts them for its defence) because they contain two toxic amino acids (IR spectra like other amino acids). Neither compound is chiral. What is the structure of these compounds? They can easily be separated because one $(\mathrm{A})$ is soluble in aqueous base but the other $(\mathrm{B})$ is not.
A is $\mathrm{C}_6 \mathrm{H}_9 \mathrm{NO}_4$ (mass spectrum) and has $\delta_{\mathrm{C}} 34.0(\mathrm{~d}), 40.0(\mathrm{t}), 56.2 (\mathrm{s}), 184.8(\mathrm{~s})$, and $186.0(\mathrm{~s})$. Its proton NMR has three exchanging protons on nitrogen and one on oxygen and two complex signals at $\delta_{\mathrm{H}} 2.68\left(4 \mathrm{H}, \mathrm{A}_2 \mathrm{~B}_2\right.$ part of $\mathrm{A}_2 \mathrm{~B}_2 \mathrm{X}$ system ) and $3.37(\mathrm{X}$ part of $\mathrm{A}_2 \mathrm{~B}_2 \mathrm{X}$ system) with $J_{\mathrm{AB}} 9.5, J_{\mathrm{AX}} 9.1$, and $J_{\mathrm{BX}}$ small.
B is $\mathrm{C}_6 \mathrm{H}_9 \mathrm{NO}_2$ (mass spectrum) and has $\delta_{\mathrm{C}} 38.0(\mathrm{~d}), 41.3(\mathrm{t}), 50.4 (\mathrm{t}), 75.2(\mathrm{~s})$, and $173.0(\mathrm{~s})$. Its proton NMR spectrum contains two exchanging protons on nitrogen and $\delta_{\mathrm{H}} 1.17(2 \mathrm{H}, \mathrm{ddd}, J 2.3$, $6.2,9.5), 2.31(2 \mathrm{H}$, broad m), $2.90(1 \mathrm{H}$, broad t, $J 3.2)$, and 3.40 ( 2 H , broad s).
Because the coupling pattern did not show up clearly as many of the coupling constants are small, decoupling experiments were used. Irradiation at $\delta_{\mathrm{H}} 3.4$ simplifies the $\delta_{\mathrm{H}} 2.3$ signal to $(2 \mathrm{H}$, $\mathrm{ddd}, J 5.8,3.2,2.3)$, sharpens each line of the ddd at 1.17 , and sharpens the triplet at 2.9 .
Irradiation at 2.9 sharpens the signals at 1.17 and 2.9 and makes the signal at 2.31 into a broad doublet, $J$ about 6 . Irradiation at 2.31 sharpens the signal at 3.4 slightly and reduces the signals at 2.9 and 1.17 to broad singlets. Irradiation at 1.17 sharpens the signal at 3.4 slightly so that it is a broad doublet, $J$ about 1.0 , sharpens the signal at 2.9 to a triplet, and sharpens up the signal at 2.31 but irradiation here had the least effect.
This is quite a difficult problem but the compounds are so small ( $\mathrm{C}_6$ only), have no methyl groups, and have some symmetry so you should try drawing structures at an early stage.