Compound A is treated with a mixture of nitric and sulfuric acids to generate Compound B. The 1H-NMR spectrum of B shows two singlets, one at 2.52 pm and one at 8.13 ppm. The 13C-NMR spectrum of B shows five signals. The mass spectrum of B shows a peak at m/z = 260 and another peak at m/z = 262; the relative height of the two peaks is 1:1 respectively. - Identify compound B, explaining your reasoning

A graduate student was making some 4-hydroxybutanoic acid and she obtained an excellent yield of a different compound whose 13C NMR is shown here. Propose a structure for this copound.

The 'H NMR spectrum of compound A (C3H100) has four signals: a multiplet at 8 = 7.25-7.32 ppm (5 H), a singlet at d = 5.17 ppm (1 H), a quartet at d = 4.98 ppm (1 H), and a doublet at ô = 1.49 ppm (3 H). There are 6 signals in its 13C NMR spectrum. The IR spectrum has a broad absorption in the -3200 cm-1 region. Compound A reacts with KMNO4 in a basic solution followed by acidification to give compound B with the molecular formula C7H6O2. Draw structures for compounds A and B.

The 1H-NMR spectrum of Compound C shows five signals – δ 2.38 (1H, dt), 2.72 (1H, dt), 5.34 (1H, t), 5.49 (2H, ddd), 6.27 (2H, dd) ppm. Its 13C-NMR spectrum has four signals – δ 26, 58, 127, 129 ppm. In the compound’s mass spectrum, the M+1 peak appears at m/z = 115. An M+2 peak, whose intensity is roughly one-third that of the M+1 peak, also appears. Suggest a structure for this compound.