Propose a structure consistent with each set of data. a. Compound A: Molecular formula: C8H10O I

Propose a structure consistent with each set of data. a....

Propose a structure consistent with each set of data. a. Compound $\textbf{A}$: Molecular formula: C$_8$H$_{10}$O IR absorption at 3150-2850 cm$^{-1}$ $^1$H NMR data: 1.4 (triplet, 3 H), 3.95 (quartet, 2 H), and 6.8-7.3 (multiplet, 5 H) ppm b. Compound $\textbf{B}$: Molecular formula: C$_9$H$_{10}$O$_2$ IR absorption at 1669 cm$^{-1}$ $^1$H NMR data: 2.5 (singlet, 3 H), 3.8 (singlet, 3 H), 6.9 (doublet, 2 H), and 7.9 (doublet, 2 H) ppm

Propose a structure consistent with each set of data.
a. Compound $\textbf{A}$:
Molecular formula: C$_8$H$_{10}$O
IR absorption at 3150-2850 cm$^{-1}$
$^1$H NMR data: 1.4 (triplet, 3 H), 3.95 (quartet, 2 H), and 6.8-7.3 (multiplet, 5 H) ppm
b. Compound $\textbf{B}$:
Molecular formula: C$_9$H$_{10}$O$_2$
IR absorption at 1669 cm$^{-1}$
$^1$H NMR data: 2.5 (singlet, 3 H), 3.8 (singlet, 3 H), 6.9 (doublet, 2 H), and 7.9 (doublet, 2 H) ppm

Key Concepts

Structure Elucidation Process

This process involves the integration of data from molecular formula, IR spectroscopy, and 1H NMR spectroscopy to propose a consistent molecular structure. Each piece of data helps narrow down possible structures and confirms the presence of specific functional groups and substitution patterns, making it a systematic approach to organic structure determination.

Proton Nuclear Magnetic Resonance (1H NMR) Spectroscopy

1H NMR spectroscopy provides detailed information about the hydrogen environments in a compound. Chemical shifts, splitting patterns, and integration values help to identify the nature of the hydrogen atoms (e.g., aromatic, alkyl, or near electronegative atoms) and their connectivity, which is crucial for deducing the overall molecular structure.

Molecular Formula Analysis

Analyzing the molecular formula helps determine the number and types of atoms present in a molecule as well as the degree of unsaturation. This key step allows one to infer the number of rings and multiple bonds, setting the stage for further structural deductions.

Infrared (IR) Spectroscopy

IR spectroscopy is used to identify functional groups within a molecule by detecting specific bond vibrational frequencies. For instance, characteristic absorptions can indicate the presence of carbonyl groups, hydroxyl groups, or specific types of C–H bonds, which are essential clues in building the structure.

Transcript

0:00 Hello everyone.

00:01 Today we're doing chapter 17 problem 56, and this problem asks us to propose a structure consistent with each set of data.

00:07 So compound a has this molecular formula, ir absorption, and this hnmr data.

00:12 So let's determine first the degrees of unsaturation for molecular formula a.

00:17 So we know that the degrees of unsaturation are a number of double bonds equals, no double bonds, equals a number of carbons, which is eight, minus the number of hydrogens plus halides over two.

00:31 So it's 10 over 2, which equals 5, minus the number of nitrogen's over 2, so 0 over 2 plus 1.

00:39 So if a minus 5 is 3 plus 1 equals 4.

00:43 So now we know here we have four double bonds or 4 degrees of unsaturation.

00:48 Well, what does the ir data tell us? the ir data tells us that we have an sb2 and an sb3 hybridized carbon hydrogen bond.

00:58 And what is the hnmr data tells us? well, we have a triplet which has two neighbors.

01:03 At 1 .4, a quartet representing three hydrogens, that means three neighbors, at 3 .95, and we have a multiplyt representing five hydrogens at 6 .8 to 7 .3.

01:18 So we know that these five hydrogens right away are going to be aromatic.

01:22 So right away, let's draw an aromatic benzene ring, and we know that these are aromatic because they fall in the range of 7 to 8 ppm.

01:33 And not only that, we know that there's going to be five hydrogens on this aerobatic.

01:38 Dramatic ring.

01:39 So now we know that one of these substituents is going to actually be the rest of our moiety to create this molecule.

01:49 Not only that, i just satisfied my four degrees of unsaturation.

01:53 Three of them are from the double bond.

01:55 The last, the fourth one is from the cyclical structure.

01:58 So we have one, two, three, four for the cyclical structure.

02:01 So now we know that we completed or we satisfied the degrees on saturation.

02:06 So now let's look at the stripping, the splitting pattern.

02:09 Well, we have five more protons that we're interested in.

02:12 We have a triplet that is neighbor to two and we have two hydrogens that neighbor to three.

02:16 So it's very obvious that if we have three hydrogens that are neighboring two, then these three hydrogens will act as a triplet just like this and these two hydrogens are neighboring three hydrogens so they would be a quartet.

02:34 So we know that this would be our structure but hold on we just forgot this oxygen.

02:38 So we now actually need to add an oxygen in between the benzene ring and these groups to satisfy our molecular formula.

02:49 Because we know that the benzene ring is not involved in any sort of splitting, so it's okay to add that oxygen here.

02:59 So now i actually need to draw an oxygen, and then i can bond this.

03:04 So now this becomes my overall structure...

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