Compound 𝐘(C8 H4 O3) dissolves slowly when warmed with...

Compound $\mathbf{Y}\left(\mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{3}\right)$ dissolves slowly when warmed with aqueous sodium bicarbonate. The IR spectrum of $\mathbf{Y}$ has strong peaks at 1779 and at $1854 \mathrm{cm}^{-1}$. The broadband proton-decoupled ' $^{13}$ C spectrum of $Y$ exhibits signals at $\delta 125(\mathrm{CH}), 130(\mathrm{C}), 136(\mathrm{CH}),$ and 162 (C). Acidification of the bicarbonate solution of $\mathbf{Y}$ gave compound $\mathbf{Z}$. The proton-decoupled ' $^{13} \mathrm{C}$ NMR spectrum of $\mathbf{Z}$ showed four signals. When Y was warmed in ethanol, compound AA was produced. The '3C NMR spectrum of AA displayed 10 signals. Propose structures for $\mathbf{Y}, \mathbf{Z}$ and $\mathbf{A A}$

Key Concepts

Structure Elucidation Using Combined Spectroscopic Data

Determining the structure of an unknown compound involves integrating information from multiple spectroscopic techniques, such as IR and NMR. Each technique provides unique data regarding functional groups, symmetry, and chemical environment, and when combined, they allow for a comprehensive understanding of the molecular structure.

Esterification Reactions

Esterification is a fundamental reaction where a carboxylic acid derivative reacts with an alcohol to form an ester. This reaction often requires heating or a catalyst and is used to modify the functional groups present in a molecule, thereby changing its chemical properties and spectral characteristics.

Reactivity of Carboxylic Acid Derivatives with Bases

The reactivity of compounds containing carboxylic acid derivatives, such as anhydrides, toward bases is a key concept in organic chemistry. Reactions with bases like sodium bicarbonate typically lead to deprotonation or nucleophilic attack, resulting in ring opening or formation of carboxylate salts, which can be further manipulated by acidification.

Carbon-13 NMR Spectroscopy and Chemical Shift Analysis

Carbon-13 NMR spectroscopy provides detailed insight into the different carbon environments present in a molecule. By analyzing the number and position of the signals in a proton-decoupled spectrum, chemists can infer the symmetry and electronic characteristics of the carbons, which is essential for structure determination.

IR Spectroscopy of Carbonyl Functional Groups

IR spectroscopy is used to identify carbonyl functionalities based on their characteristic absorption bands. Carbonyl groups such as those found in anhydrides, acids, esters, or ketones absorb infrared light in specific regions, allowing chemists to determine the presence and environment of these functional groups in a molecule.

Transcript

00:02 This is the answer to chapter 20, problem number 73 from the smith organic chemistry textbook.

00:09 This problem says treatment of compound c with phenomagnosium bromide followed by water, for its compound d.

00:18 And we're given formulas for c and d.

00:21 We're given an ir for d and then proton nmr for c and d.

00:27 And we're asked to come up with structures for each of them.

00:32 Okay.

00:33 And so i'm just going to take this information a piece at a time and try to come up with these structures.

00:41 So the first thing i'm going to do is an hdi for c.

00:45 So it's going to be two times the number of carbons plus two minus the number of hydrogens, all of that over two.

00:58 So we have an hdi of one here.

01:00 Here.

01:00 So that tells us we have one ring or one double bond.

01:04 Okay.

01:07 So then looking at the nmr, we have a singlet at 1 .3 ppm with an integration of six.

01:16 And so i'm going to say that that's probably two methyl groups.

01:23 So two ch3s.

01:26 And then we have a singlet at 2 .4 p .p.

01:31 With an integration of 2.

01:37 Okay, and so this is probably a ch2.

01:42 Both of these have no neighbors.

01:45 So a four carbon molecule with a double bond or a ring and an oxygen in it.

01:54 And this spectral information, the only thing that really fits here is this.

02:03 So it's going to be an epoxide with two methyl groups.

02:10 And so our first signal, this signal, is going to be these two methyl groups.

02:20 So the protons on these two methyl groups.

02:23 And then our other signal, this signal, is going to be the two protons that are here on this carbon.

02:32 Okay.

02:32 Okay, so then for d, so if we treat this epoxide with phenomagnesium bromide, we would expect the grinier to act as a nucleophile there and open the epoxide at the less substituted side.

02:54 I guess, yeah, at the less substituted side.

02:57 And so we should bear that in mind, but we should also, you know, do the.

03:03 Exact same thing here.

03:04 So we'll calculate an hdi for this formula.

03:07 So it's going to be two times 10 plus 2 minus 14, all of that over 2.

03:16 So we have 22 minus 14 is 8 over 2 is 8 over 2 is 4...

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