Question

What is the structure of a hydrocarbon that has \(\mathrm{M}^{+}=120\) in its mass spectrum and has the following 1 H NMR spectrum? \(7.25 \delta(5 \mathrm{H},\) broad singlet \() ; 2.90 \delta(1 \mathrm{H},\) septet \(, J=7 \mathrm{~Hz}) ; 1.22 \delta(6 \mathrm{H}\) doublet, \(J=7 \mathrm{~Hz}\) )

Short answer

The hydrocarbon is cumene (isopropylbenzene).

Step-by-step solution

Determine the Molecular Formula

The mass spectrum shows a molecular ion peak at 120. For common hydrocarbons, we assume a formula of the type \(C_xH_y\). Calculate the possible values by trial and error, checking if the mass matches when considering carbon (12 amu) and hydrogen (1 amu). A likely candidate is \(C_9H_{12}\) (9*12 + 12*1 = 120).

Analyze the NMR Spectrum

The NMR spectrum gives details about the hydrogen environment: 1. A broad singlet at 7.25 \(\delta\) for 5 hydrogens suggests an aromatic ring. 2. A septet at 2.90 \(\delta\) for 1 hydrogen indicates a hydrogen next to a group of 6 equivalent hydrogens, likely indicating an ethyl group. 3. A doublet at 1.22 \(\delta\) for 6 hydrogens indicates hydrogens adjacent to a single hydrogen, also supporting an ethyl group.

Construct the Molecular Structure

Combine the information: an aromatic ring explains the presence of 5 hydrogens as indicated by the broad singlet. To satisfy the ethyl group inference, we can place an isopropyl group attached to the benzene ring, leading to the structure of isopropyl benzene (cumene).

Verify the Proposed Structure

The proposed structure, cumene (isopropylbenzene), consists of a benzene ring with an isopropyl group. The benzene ring accounts for the 5 hydrogens at 7.25 \(\delta\). The isopropyl group contributes the septet (1 hydrogen adjacent to two methyl groups) and the doublet (6 hydrogens of two methyl groups). The overall molecular weight is 120, matching the given \(\mathrm{M}^{+}\) peak, confirming correctness.

Key concepts

Mass Spectrometry

Mass spectrometry is a powerful technique used to determine the molecular weight of a compound. It provides crucial information about the structure of a molecule by producing a spectrum of ions. Each ion displays a peak at its mass-to-charge ratio (m/z). In our exercise, the mass spectrum shows a peak at **120 m/z**, indicating the molecular weight of the unknown hydrocarbon.

In analyzing hydrocarbons, one commonly aims to find a formula of the type **CₓHᵧ** that matches this mass. This process typically involves checking different combinations of carbon and hydrogen atoms. Each carbon atom contributes **12 amu**, and each hydrogen atom contributes **1 amu**.

From the given peak, the likely combination is **C₉H₁₂**. When calculated, this results in **9 carbons (9 x 12)** plus **12 hydrogens (12 x 1)**, which adds up to a total mass of **120 amu**. This precise matching of the molecular formula to the mass spectrum peak is key in determining the structure of the compound.

1H NMR Spectroscopy

1H NMR spectroscopy is another essential technique in organic chemistry used to deduce the structure of a compound. It provides insights into the hydrogen environments in a molecule, which can be associated with different functional groups within the structure.

In the given 1H NMR spectrum, we observe signals due to different hydrogen environments. These signals appear as:

• A broad singlet at 7.25 ppm involving 5 hydrogens, indicative of an aromatic ring.
• A septet at 2.90 ppm involving 1 hydrogen, suggesting an environment where one hydrogen is adjacent to a group of 6 equivalent hydrogens.
• A doublet at 1.22 ppm involving 6 hydrogens, suggesting these are part of equivalent environments contributing to the same signal.

By interpreting these signals, a picture of the molecular structure begins to form.

Molecular Formula Determination

Determining the molecular formula is a fundamental step in hydrocarbon structure analysis. It sets the stage for understanding how atoms are arranged within the molecule.

The given mass spectrum hints at a **120 amu** molecular ion peak, pointing to a formula. Matching this with typical combinations of carbon and hydrogen, the **C₉H₁₂** formula emerges as a logical candidate. This formula provides:

• 9 carbon atoms contributing **108 amu** (9 x 12 amu).
• 12 hydrogen atoms adding **12 amu** (12 x 1 amu).

This adds up to the necessary **120 amu**, validating the proposed molecular formula as representative of the compound being analyzed. This formula plays a significant role in further 1H NMR interpretation.

Aromatic Compounds

Aromatic compounds are a specific class of structures characterized by their stability and planar ring of carbon atoms, commonly exemplified by benzene. The presence of an aromatic ring is usually indicated by NMR signals around **7.0 - 8.0 ppm**.

In this exercise, the **5-hydrogen broad singlet at 7.25 ppm** strongly suggests the presence of an aromatic ring, likely a benzene ring. These hydrogens resonate at similar shifts due to the uniform environment created by the aromatic system.

Aromatic compounds like benzene contribute significant insights into the molecular structure. In this case, linking it with additional segments helps determine the placement of other functional groups, such as **isopropyl** in isopropylbenzene (cumene). This link exploits the properties of aromaticity to complete the hydrocarbon's structural identification.

NMR Signal Interpretation

Interpreting NMR signals involves understanding the peak types and splitting patterns that reveal hydrogen locations and interactions. Each NMR signal corresponds to a specific hydrogen environment and is split based on nearby hydrogen atoms.

For example:

• Broad singlets, such as the one at **7.25 ppm**, suggest aromatic hydrogens that do not pair with adjacent H's significantly.
• The septet at **2.90 ppm** indicates a hydrogen neighbor poised between multiple chemically similar H's, typical for **isopropyl groups**.
• Doublets, like the one at **1.22 ppm**, arise from hydrogen pairs next to a distinctive adjacent H, supporting another segment of an **isopropyl** moiety.

This interpretation guides us from individual signal analysis to deducing a complete structure, finger-pointing the arrangement of an isopropyl group onto an aromatic ring, thus confirming the molecular identity of isopropylbenzene.