What is the structure of a hydrocarbon that has M^+=120 in its mass spectrum and has the followi

step 1 Solving problem 40 of chapter 15. Let's write down the information we have. So we know that the

What is the structure of a hydrocarbon that has M^+=120 in...

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, $J=7 \mathrm{~Hz}$ )

Key Concepts

Structure Elucidation

Structure elucidation involves the integration of data from mass spectrometry and NMR spectroscopy to build a coherent molecular structure. By considering molecular weight, degrees of unsaturation, chemical shift patterns, integration, and splitting information, one can reconstruct the arrangement of atoms within the molecule.

Mass Spectrometry

Mass spectrometry provides the molecular ion peak, which is key to determining the molecular weight of a compound. This information helps narrow down possible molecular formulas and, when combined with other spectral data, guides the overall structure determination process.

NMR Spectroscopy

NMR spectroscopy is a pivotal analytical technique that reveals the different chemical environments of hydrogen atoms in a molecule. Analysis of chemical shifts, integration values, and splitting patterns enables the identification of various substructures and the connectivity between atoms.

Chemical Shift and Integration

Chemical shifts offer insights into the electronic environments around protons, while integration quantitatively reflects the number of protons contributing to each signal. This dual information is essential for assigning groups such as aromatic rings, alkyl chains, or other specific proton environments.

Spin-Spin Splitting and Coupling Constants

Spin-spin splitting patterns occur due to interactions between neighboring protons, resulting in multiplets like doublets and triplets. The coupling constants (J values) further help elucidate the proximity and arrangement of adjacent proton groups, which is critical in confirming structural details.

Transcript

00:01 Solving problem 40 of chapter 15.

00:05 Let's write down the information we have.

00:09 So we know that the structure of the hydrocarbon we need to find as a mass plus equal to 120.

00:20 Then we know that the proton and nmr we have a signal at 7 .25 corresponding to 5 protons.

00:32 And this is a broad singlet.

00:37 Then we have a signal 2 .90, 2 .90 corresponding to one proton, one proton, and this is a septet.

00:58 And then finally we have a signal 1 .22, 6 proton doubled.

01:09 Okay, was the molecule that correspond to this information.

01:17 So first of all, starting from the protein mr signal, we recognize immediately that a chemical shift of 7 .25 correspond for sure to an aromatic, an aromatic benzene ring.

01:35 So let's try, let's start drawing this.

01:43 Then so this one it means we have an aromatic ring of course we need to attach something first of all because we need to reach this mass second because we know that there are other groups attached to that also because it's saying five protons and not six so it means we have only one substitution on the on the benzene ring so that we will then have one, two, three, four and five protons, five protons.

02:17 So we have something and what do we have? these other part is going to give us the information.

02:23 So we know that we have one proton that appear as accepted...

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