Interpretation:
A structure for the compound A, C8H10O2, with the following spectral data is to be proposed.
IR: 1750 cm-1; 13CNMR: 219 δ (20), two signals between 35 δ to 45 δ.
Concept introduction:
In 13CNMR, carbons in double bond which are sp2 hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15δ, a secondary alkyl radical in the range 16-25δ while a tertiary alkyl in the range 25-35 δ.
To propose:
A structure for the compound A, C8H10O2, with the following spectral data.
IR: 1750 cm-1; 13CNMR: 219 δ (20), two signals between 35 δ to 45 δ.
Trending nowThis is a popular solution!
Chapter 19 Solutions
Organic Chemistry
- Treatment of benzoic acid (C6H5CO2H) with NaOH followed by 1-iodo-3methylbutane forms H. H has a molecular ion at 192 and IR absorptions at 3064, 3035, 2960−2872, and 1721 cm−1. Propose a structure for H.arrow_forwardPropose a structure for compound X (molecular formula C6H12O2), which gives a strong peak in its IR spectrum at 1740 cm−1. The 1H NMR spectrum of X shows only two singlets, including one at 3.5 ppm. The 13C NMR spectrum is given below. Propose a structure for X.arrow_forwardCompound Y (molecular formula C6H10) gives four lines in its 13C NMR spectrum (27, 30, 67, and 93 ppm) and the IR spectrum given here. Propose a structure for Y.arrow_forward
- Propose a structural formula for compound A, C5H13N, given its IR and calculated 1H-NMR spectra.arrow_forwardCompound X (molecular formula C10H12O) was treated with NH2NH2,−OH to yield compound Y (molecular formula C10H14). Based on the 1H NMR spectra of X and Y given below, what are the structures of X and Y?arrow_forwardCompound A has molecular formula C5H8Br4 but shows only one singlet in the 1H-NMR spectrum. Suggest a structure for A and explain your reasoning.arrow_forward
- Following are 1H-NMR spectra for compounds B (C6H12O2) and C (C6H10O). Upon warming in dilute acid, compound B is converted to compound C. Deduce the structural formulas for compounds B and C.arrow_forwardCompound A of molecular formula C3H6O shows a noteworthy infrared absorption at 1716 cm-1. Its 1H-NMR spectrum shows one singlet – δ 2.2 (6H) ppm. Its 13C-NMR spectrum has two signals – δ 30, 207 ppm. Suggest a structure for this compound.arrow_forwardTreatment of ketone A with ethynyllithium (HC≡CLi) followed by D3O+ afforded a compound B of molecular formula C12H13DO3, which gave an IR absorption at approximately 1715 cm−1. What is the structure of B and how is it formed?arrow_forward
- Deduce the structures of compounds A and B, two of the major components of jasmine oil, from the given data. Compound A: C9H10O2; IR absorptions at 3091–2895 and 1743 cm-1; 1H NMR signals at 2.06 (singlet, 3 H), 5.08 (singlet, 2 H), and 7.33 (broad singlet, 5 H) ppm. Compound B: C14H12O2; IR absorptions at 3091–2953 and 1718 cm-1; 1H NMR signals at 5.35 (singlet, 2 H) and 7.26–8.15 (multiplets, 10 H) ppm.arrow_forwardCompound A, MW 72, shows an IR absorption at 1711 cm-1 and a 1H NMR spectrum with peaks at 2.4 8 (2 H, quartet), 2.1 8 (3H, singlet) and 1.1 8 (3 H, triplet). Propose a structure for A. Drawingarrow_forwardCompounds Y and Z are isomers with the molecular formula C10H12O. The IR spectrum of each compound shows a strong absorption band near 1710 cm−1 . The 1H NMR spectra of Y and Z are given below. Propose structures for Y and Z.arrow_forward