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Propose a structure for compound \(\mathbf{E}, \mathrm{C}_{7} \mathrm{H}_{12} \mathrm{O}_{2},\) which has the following 13 C NMR spectral data: Compound \(\mathbf{E}\) Broadband-decoupled \({ }^{13} \mathrm{C} \mathrm{NMR}: 19.1,28.0,70.5,129.0,129.8,165.8 \delta\) DEPT-90: \(28.0,129.8 \delta\) DEPT-135: positive peaks at \(19.1,28.0,129.8 \delta\); negative peaks at 70.5 , \(129.0 \delta\)

Short Answer

Expert verified
Ethyl cyclopent-2-enyl carboxylate is a likely structure for compound \( \mathbf{E} \).

Step by step solution

01

Analyze the Molecular Formula

The molecular formula given is \( \mathrm{C}_7 \mathrm{H}_{12} \mathrm{O}_2 \). This suggests a structure with some degree of unsaturation. We calculate the degree of unsaturation as follows: \[ \text{Unsaturation} = \frac{2+2 \times \text{C} + \text{N} - \text{H} - \text{X}}{2} = \frac{2+2\times 7 - 12}{2} = 3 \] The presence of 3 degrees of unsaturation likely indicates a combination of double bonds, rings, or an ester/carboxyl group.
02

Interpret the DEPT-90 and DEPT-135 Spectra

The DEPT-90 peaks at \(28.0\) and \(129.8 \delta\) indicate the presence of \(\mathrm{CH}\) carbons. The DEPT-135 with positive peaks at \(19.1, 28.0, 129.8 \delta\) and negative peaks at \(70.5, 129.0 \delta\) shows that the positive peaks are for \(\mathrm{CH}_2\) groups and the negative ones for \(\mathrm{CH}_3\) or \(\mathrm{C}\).
03

Assign the Carbon Environments

We see that \(129.0 \delta\) and \(129.8 \delta\) are consistent with alkene carbons or aromatic systems. The DEPT suggests \(129.8\) is a \(\mathrm{CH}\) and \(129.0\) is a \(\mathrm{CH}_2\). Furthermore, the peak at \(165.8 \delta\) is likely indicative of an ester since it's in the range of ester or carbonyl groups. The \(70.5 \delta\) is likely an oxygenated \(\mathrm{CH}\), probably from a \(\mathrm{CH}_2\) group attached to a more electronegative element or group.
04

Propose Possible Structure

Considering the provided data, the compound likely possesses an ester functionality suggested by the \(165.8 \delta\) carbon, an unsaturation, and identifiable \(\mathrm{CH}, \mathrm{CH}_2\), and \(\mathrm{CH}_3\) groups. A likely structure could be a cyclohexene derivative with an ester group, such as ethyl cyclopent-2-enyl carboxylate, gratifying the chemical shifts and DEPT data.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carbon-13 NMR Spectroscopy
Carbon-13 NMR spectroscopy is a vital technique in organic chemistry for elucidating molecular structures. Unlike proton NMR, carbon-13 NMR focuses on the carbon atoms in a molecule. In the given compound with the formula \( \text{C}_7 \text{H}_{12} \text{O}_2 \), the carbon environments provide insightful clues into the molecule's structure.

When using carbon-13 NMR spectroscopy, we consider chemical shifts, which are specific frequencies where carbons resonate. Each peak in the spectrum corresponds to different carbon environments based on their electronic surroundings. For example, carbon atoms bonded to oxygen resonate differently than those bonded only to hydrogen or other carbon atoms. Hence, these chemical shifts allow chemists to identify different functional groups within a molecule, like carbonyls, alkenes, or alcohols.

The DEPT (Distortionless Enhancement by Polarization Transfer) techniques such as DEPT-90 and DEPT-135 are used in carbon-13 NMR to determine the type of protons directly attached to the carbon atoms. A DEPT-90 spectrum highlights \( \text{CH} \) groups, while DEPT-135 reveals positive peaks for \( \text{CH}_2 \) groups and negative peaks for \( \text{CH}_3 \) or quaternary carbons. This data helps in precisely assigning each carbon environment within the molecule.
Degree of Unsaturation
The degree of unsaturation in a molecule informs us about the number of rings and/or multiple bonds present. It is calculated using the formula:\[\text{Degree of Unsaturation} = \frac{2+2 \times C + N - H - X}{2}\] where \( C \) is the number of carbons, \( N \) is the number of nitrogens, \( H \) is the number of hydrogens, and \( X \) is the number of monovalent atoms like halogens.

For compound \( E \) with the molecular formula \( \text{C}_7 \text{H}_{12} \text{O}_2 \), the degree of unsaturation is calculated to be 3. This indicates that the compound may contain any combination of three rings or double bonds. In the context of the NMR data and the presence of an ester functional group, it’s likely that this unsaturation involves one or more double bonds or possibly aromatic groups alongside an ester bond.
Ester Functional Group
Esters are a type of organic functional group characterized by the formula \( \text{RCOOR'} \). In the context of carbon-13 NMR, esters typically show distinct chemical shifts due to their carbonyl carbon. This is apparent in Compound \( E \) where a peak at \( 165.8 \delta \) suggests an ester carbonyl.

In addition to the carbonyl carbon, ester groups also include other carbons which may be detected in an NMR spectrum, such as the carbon in the \(-OCH_2- \) group, usually displaying a unique shift due to the presence of electronegative oxygen. Such environments can be clearly distinguished from aliphatic or alkene carbons. Therefore, in constructing a plausible structure, recognizing the ester’s distinct NMR signature becomes valuable in confirming parts of the molecule's skeleton while correlating it with other unsaturation sources.
Carbocation Structures
Carbocations are positively charged carbon atoms that usually exist as intermediates in chemical reactions. While not directly present in stable molecular structures, understanding them helps clarify rearrangements and stability principles in organic chemistry.

However, in the context of interpreting carbon-13 NMR, carbocation structures, in particular, don't usually play a role unless one considers fragments in dynamic processes during reaction pathways. More relevant is the carbon environments that might resemble high-energy carbocation centers, often determined through shifts and chemical environments in more reactive compounds.

For Compound \( E \), the discussion around carbocations helps appreciate the structural possibilities, including rearrangement and resonance, which might affect stability and structure elucidation from NMR spectra. Although not applicable directly, awareness of this concept facilitates a deeper understanding when predicting molecular behavior and stability.

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Most popular questions from this chapter

Calculate the amount of energy required to spin-flip a proton in a spectrometer operating at \(300 \mathrm{MHz}\). Does increasing the spectrometer frequency from 200 to \(300 \mathrm{MHz}\) increase or decrease the amount of energy necessary for resonance?

We saw in Section 8 - 15 that addition of \(\mathrm{HBr}\) to a terminal alkyne leads to the Markovnikov addition product, with the Br bonding to the more highly substituted carbon. How could you use \({ }^{13} \mathrm{C}\) NMR to identify the product of the addition of 1 equivalent of HBr to hex-1-yne?

The following \({ }^{1} \mathrm{H}\) NMR absorptions were obtained on a spectrometer operating at \(200 \mathrm{MHz}\) and are given in hertz downfield from the TMS standard. Convert the absorptions to \(\delta\) units. (a) \(436 \mathrm{~Hz}\) (b) \(956 \mathrm{~Hz}\) (c) \(1504 \mathrm{~Hz}\)

The following \({ }^{1} \mathrm{H}\) NMR absorptions were obtained on a spectrometer operating at \(300 \mathrm{MHz}\). Convert the chemical shifts from \(\delta\) units to hertz downfield from TMS. (a) \(2.1 \delta\) (b) \(3.45 \delta\) (c) \(6.30 \delta\) (d) \(7.70 \delta\)

The amount of energy required to spin-flip a nucleus depends both on the strength of the external magnetic field and on the nucleus. At a field strength of \(4.7 \mathrm{~T}, \mathrm{rf}\) energy of \(200 \mathrm{MHz}\) is required to bring a \({ }^{1} \mathrm{H}\) nucleus into resonance but energy of only \(187 \mathrm{MHz}\) will bring a \(19 \mathrm{~F}\) nucleus into resonance. Calculate the amount of energy required to spin-flip a \({ }^{19} \mathrm{~F}\) nucleus. Is this amount greater or less than that required to spin-flip a \({ }^{1} \mathrm{H}\) nucleus?

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