Extension: organic spectroscopic analysis

A pure sample of a straight-chain saturated aliphatic hydrocarbon is analysed in an electron-impact mass spectrometer. The spectrum shows the molecular ion peak at a mass/charge ratio $m/z = 58$, together with main fragments at $m/z = 43$ and $m/z = 29$.

1. a) State the molecular mass of the compound and determine its general molecular formula, given that it is an alkane ($\mathrm{C}_n\mathrm{H}_{2n+2}$).
2. b) Deduce the semi-structural formula and systematic name of the alkane, identifying which cationic fragment corresponds to the signal at $m/z = 43$. (Relative atomic mass data: $\mathrm{C}=12.0$; $\mathrm{H}=1.0$).

Mass spectrometry analysis of a saturated monohalogenated derivative reveals two molecular signals separated by two mass units: a peak $M$ at $m/z = 64$ and a peak $M+2$ at $m/z = 66$. The intensity ratio of these two peaks is approximately $3:1$.

The approximate natural isotopic composition is given:

• Bromine: $^{79}\mathrm{Br}\,(50.7\%)$ and $^{81}\mathrm{Br}\,(49.3\%)$, ratio $\approx 1:1$.

• Chlorine: $^{35}\mathrm{Cl}\,(75.8\%)$ and $^{37}\mathrm{Cl}\,(24.2\%)$, ratio $\approx 3:1$.

(Atomic masses of the main nuclides: $^{12}\mathrm{C}=12$, $^{1}\mathrm{H}=1$, $^{35}\mathrm{Cl}=35$, $^{37}\mathrm{Cl}=37$).

1. a) Determine, with justification, which of the two halogens is present in the molecule.
2. b) Deduce the molecular formula and systematic name of the haloalkane, justifying the origin of both molecular ion peaks.

A laboratory has two unlabelled bottles; one contains propan-1-ol ($\mathrm{CH_3-CH_2-CH_2-OH}$) and the other propanone ($\mathrm{CH_3-CO-CH_3}$). To identify them, the infrared (IR) spectrum of both liquids is recorded.

Consider the typical absorption ranges: $\mathrm{O-H}$ bond in alcohols at $3200-3600\,\text{cm}^{-1}$ (broad, intense band), $\mathrm{C=O}$ bond at $1700-1750\,\text{cm}^{-1}$ (sharp, intense band) and $\mathrm{C-H}\,(\text{sp}^3)$ bonds at $2850-2960\,\text{cm}^{-1}$.

1. a) Explain which main diagnostic absorption will appear exclusively in the spectrum of propan-1-ol.
2. b) State which characteristic band will allow propanone to be distinguished from the alcohol between these two candidates.

The simplified infrared spectrum of a pure organic liquid is available, whose molecular formula corresponds to one of the following three candidates: acetic acid ($\mathrm{CH_3-COOH}$), acetone ($\mathrm{CH_3-CO-CH_3}$) or ethanol ($\mathrm{CH_3-CH_2-OH}$). Two main absorption troughs can be seen on the graph: a very broad, diffuse band between $2500$ and $3300\,\text{cm}^{-1}$, and a narrow, deep, symmetrical band around $1710\,\text{cm}^{-1}$. Observe the recorded spectrum:

1. a) Identify, with justification, which of the three proposed compounds this spectrum belongs to, arguing carefully for the exclusion of the other two.
2. b) Justify why the infrared spectrum alone cannot unambiguously demonstrate the complete unique structure of a molecule and which additional technique would be required. IR references: O–H of carboxylic acids, broad band between $2500$ and $3300\,\mathrm{cm}^{-1}$; O–H of alcohols, between $3200$ and $3600\,\mathrm{cm}^{-1}$; C=O, approximately between $1700$ and $1750\,\mathrm{cm}^{-1}$.

An organic compound $X$ consisting exclusively of carbon, hydrogen and chlorine gives, in its mass spectrum, a molecular ion split into two signals: $m/z = 78$ ($M$, abundance $100\%$) and $m/z = 80$ ($M+2$, abundance $32.5\%$).

Its infrared spectrum shows no bands above $3000\,\text{cm}^{-1}$ and no absorption at all in the region of $1600-1850\,\text{cm}^{-1}$, but it shows bands between $2850$ and $2970\,\text{cm}^{-1}$ and a moderate absorption at $650\,\text{cm}^{-1}$ ($\mathrm{C-Cl}$ stretch).

Its $\mathrm{^{1}H}$-NMR spectrum shows only two signals:

• A doublet at $\delta = 1.50\,\text{ppm}$ integrating for $6$ protons.

• A septet at $\delta = 4.10\,\text{ppm}$ integrating for $1$ proton.

(Atomic masses: $^{12}\mathrm{C}=12$, $^{1}\mathrm{H}=1$, $^{35}\mathrm{Cl}=35$, $^{37}\mathrm{Cl}=37$).

1. a) Calculate the exact molecular formula of the compound, justifying the deduction from the mass spectrometry and infrared data.
2. b) Deduce the semi-structural formula and systematic name of the compound, justifying in detail the multiplicity observed in the NMR. For first-order coupling, a set of n equivalent neighbouring protons splits a signal into n + 1 lines; non-equivalent neighbours can produce successive splittings. Assume molecular ions with a +1 charge. A single chlorine atom produces an M/M+2 isotopic doublet with intensities of approximately 3:1; the C–H bonds of saturated carbons absorb between 2850 and 3000 cm⁻¹.

Proton nuclear magnetic resonance spectroscopy ($\mathrm{^{1}H}$-NMR) provides information about the different chemical environments of the hydrogen atoms in an organic molecule.

1. a) Determine the number of distinct signals that will be observed in the $\mathrm{^{1}H}$-NMR spectrum of propanone ($\mathrm{CH_3-CO-CH_3}$) and of dimethyl ether ($\mathrm{CH_3-O-CH_3}$).
2. b) Deduce the number of signals that the $\mathrm{^{1}H}$-NMR spectrum of propanal ($\mathrm{CH_3-CH_2-CHO}$) will show, briefly justifying the chemical equivalence or non-equivalence of its protons. Consider as equivalent the protons related by symmetry and by fast rotations in an achiral medium.

The $\mathrm{^{1}H}$-NMR spectrum of bromoethane ($\mathrm{CH_3-CH_2-Br}$) shows two separate groups of signals. To interpret the spin-spin coupling, the multiplicity rule $n+1$ is applied, where $n$ represents the number of equivalent neighbouring protons.

Consider that electronegative halogen atoms draw electron density towards themselves, deshielding the adjacent nuclei and shifting their signals towards higher chemical shift values ($\delta$, in ppm).

1. a) State the expected multiplicity (singlet, doublet, triplet, quartet...) for the signal of the $\mathrm{-CH_3}$ group and for that of the $\mathrm{-CH_2-}$ group.
2. b) Determine the theoretical ratio of the integration areas between the signal appearing at the higher chemical shift and the one at the lower chemical shift.

The $\mathrm{^{1}H}$-NMR spectrum of an ester of molecular formula $\mathrm{C_4H_8O_2}$, obtained by the esterification of an aliphatic carboxylic acid with an aliphatic alcohol, is recorded.

The spectrum shows three clear signals with the following parameters:

• Signal A: singlet at $\delta = 2.05\,\text{ppm}$, relative integration of $3$ protons.

• Signal B: quartet at $\delta = 4.12\,\text{ppm}$, relative integration of $2$ protons.

• Signal C: triplet at $\delta = 1.26\,\text{ppm}$, relative integration of $3$ protons.

Given that an alkyl group bonded directly to the oxygen of an ester ($\mathrm{-O-CH_2-}$) typically resonates at $\delta = 3.7-4.3\,\text{ppm}$, whereas a group bonded to the carbonyl carbon ($\mathrm{-CO-CH_3}$) does so at $\delta = 2.0-2.3\,\text{ppm}$:

1. a) Identify, with justification, the semi-structural formula and the systematic name of the compound, uniquely assigning each of the signals (A, B and C) to the corresponding protons.
2. b) Justify the explicit exclusion of its constitutional isomer methyl propanoate ($\mathrm{CH_3-CH_2-COO-CH_3}$) using the expected multiplicity and chemical shift. For first-order coupling, a set of n equivalent neighbouring protons splits a signal into n + 1 lines; non-equivalent neighbours can produce successive splittings. Assume molecular ions with a +1 charge.

A volatile liquid has a molar mass of $58.0\,\text{g/mol}$ (determined by mass spectrometry with the molecular ion at $m/z = 58$) and elemental analysis indicates $62.07\,\%$ of $\mathrm{C}$, $10.34\,\%$ of $\mathrm{H}$ and $27.59\,\%$ of $\mathrm{O}$.

(Atomic masses: $\mathrm{C}=12.0$; $\mathrm{H}=1.0$; $\mathrm{O}=16.0$).

Two isomers with this composition are proposed: propanal ($\mathrm{CH_3-CH_2-CHO}$) and propanone ($\mathrm{CH_3-CO-CH_3}$).

Consider the following tabulated spectroscopic data:

• IR: ketone $\mathrm{C=O}$ at $1715\,\text{cm}^{-1}$; aldehyde $\mathrm{C=O}$ at $1725\,\text{cm}^{-1}$ together with two weak aldehydic $\mathrm{C-H}$ stretching bands (Fermi bands) at $2720$ and $2820\,\text{cm}^{-1}$.

• $\mathrm{^{1}H}$-NMR: aldehyde proton ($\mathrm{-CHO}$) appears as a triplet at $\delta \approx 9.7\,\text{ppm}$.

1. a) Show by calculation the molecular formula of the liquid.
2. b) If the $\mathrm{^{1}H}$-NMR spectrum of the sample contains a single signal at $\delta = 2.15\,\text{ppm}$ and there are no signals in its IR between $2700$ and $2850\,\text{cm}^{-1}$, determine which compound it is and predict in detail the $\mathrm{^{1}H}$-NMR spectrum that the other isomer would show. For first-order coupling, a set of n equivalent neighbouring protons splits a signal into n + 1 lines; non-equivalent neighbours can produce successive splittings. Assume molecular ions with a +1 charge. For propanal, use as reference the methyl at around 1 ppm, the methylene adjacent to the carbonyl at around 2.4 ppm and the aldehydic proton at around 9.7 ppm. The methylene may couple differently with the methyl and with the aldehydic proton.