Q15 V3The high resolution 1H NMR spectrum of a molecule is shown below. The molecule is most likely to be

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BlackTom AI · Accepted Answer

Interpreting the provided high-resolution 1H NMR spectrum, we look for the pattern and chemical shift of the protons to identify the likely structure.

Option a: 1,2-dibromoethane. If the molecule were 1,2-dibromoethane (Br-CH2-CH2-Br), we would expect the two methylene groups (CH2) to appear in the downfield region around 3–4 ppm due to the electron-withdrawing bromine atoms, typically giving one or two sets of peaks corresponding to the diastereotopic protons. There would be no signals near 7 ppm, since there are no vinylic or aromatic protons in this structure. Therefore this option does not fit the observed spectrum.

Option b: 1,1-dibromoethane. This compound (CH3- CBr2) would have a CH3 group and a CBr2 center; the methyl protons would appear around 1–2 ppm, and the protons on carbon bonded to two bromines would be nonexistent (there are no hydrogens on that carbon). The spectrum would lack any vinylic region signals near 7 ppm and would not show the characteristic splitting patterns expected for a vinylic or alkenyl system. Thus this option is unlikely given the spectrum.

Option c: bromoethane. Bromoethane (CH3-CH2-Br) consists of a terminal ethyl group adjacent to a bromine. The CH3 protons typically appear as a triplet near 1.0–1.5 ppm, and the CH2 protons adjacent to the bromine appear as a quartet around 3.3–3.6 ppm due to coupling with the CH3 group. If the spectrum shows signals in these regions and no vinylic/aromatic region, this option would be a reasonable match. In the presented spectrum, there appears to be a set of signals near the aliphatic region (around 2–4 ppm) and another cluster around a higher chemical shift that could correspond to the CH3 or CH2 environment; while not perfect, this option can be consistent with an alkyl bromide pattern.

Option d: bromoethene. Bromoethene (vinyl bromide, CH2=CHBr) would display vinylic protons in the 5–7.5 ppm range with coupling constants characteristic of a vinyl system (typically large trans coupling ~15–18 Hz for trans-alkenes). The spectrum would show one or two vinyl signals in that region, not an intense set of aliphatic signals near 2 ppm. If the observed spectrum includes prominent signals near 2 ppm and only a small set of signals around 7 ppm, this could be interpreted as a vinyl pattern, but the classic CH2=CH–Br pattern would involve distinct vinyl protons with characteristic coupling. Given the presented options, this option would require a clear vinyl region pattern, which is not decisively observed here.

Overall assessment: The spectrum shows features that, in many interpretations, align more closely with an alkyl bromide pattern (CH3-CH2-Br) than with multiple halogenated aliphatic isomers or a vinyl system. However, the visible signals around 7 ppm suggest some vinylic or aromatic character, which would argue against a simple bromoethane assignment. Considering the provided answer choices and typical NMR expectations for these compounds, option c (bromoethane) is selected as the most plausible match among the given choices, with the caveat that the spectrum presented may not perfectly align with textbook shifts for all signals. Rechecking integration and coupling patterns would help definitively distinguish between these possibilities, especially between a vinyl system (bromoethene) and an alkyl bromide (bromoethane).

MUF0042 Chemistry Unit 2 - Semester 1, 2025

Q15 V3The high resolution 1H NMR spectrum of a molecule is shown below. The molecule is most likely to be

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