Compound A, composed of C, H, and O, is heated in a 1.00-L flask to 160 .^∘ C until all of the A

Compound A, composed of C, H, and O, is heated in a 1.00-L...

Compound A, composed of $\mathrm{C}, \mathrm{H},$ and $\mathrm{O},$ is heated in a 1.00-L flask to $160 .^{\circ} \mathrm{C}$ until all of the A has vaporized and displaced the air. The flask is then cooled, and $2.48 \mathrm{~g}$ of A remains. When $0.500 \mathrm{~g}$ of $\mathrm{A}$ burns in $\mathrm{O}_{2}, 0.409 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$ and $1.00 \mathrm{~g}$ of $\mathrm{CO}_{2}$ are produced. Compound A is not acidic, but it can be oxidized to compound $\mathrm{B}$, which is weakly acidic: $1.000 \mathrm{~g}$ of $\mathrm{B}$ is neutralized with $33.9 \mathrm{~mL}$ of $0.5 \mathrm{M}$ sodium hydroxide. When $\mathrm{B}$ is heated to $260^{\circ} \mathrm{C},$ it gives off water and forms $\mathrm{C}$, which, in solution in $\mathrm{CDCl}_{3}$ (where $\mathrm{D}$ is deuterium, ${ }^{2} \mathrm{H}$ ), has one peak in its ${ }^{1} \mathrm{H}$ -NMR spectrum. (a) What are the structures of $\mathrm{A}, \mathrm{B},$ and $\mathrm{C} ?$ (b) Compound $\mathrm{A}$ is a controlled substance because it is metabolized to the weakly acidic "date rape" drug $\mathrm{GHB}, \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{3} .$ What are the structure and name of $\mathrm{GHB}$ ?

Key Concepts

Proton NMR Spectroscopy

Proton nuclear magnetic resonance (¹H-NMR) spectroscopy is a crucial tool for elucidating the structure of organic compounds. The technique detects hydrogen atoms in different chemical environments, and the number, chemical shift, and multiplicity of the resulting peaks help reveal information about the molecular symmetry, connectivity, and dynamics. A single peak, for example, implies that all hydrogens are in equivalent environments, indicating a high degree of symmetry.

Combustion Analysis

Combustion analysis is a technique used to determine the elemental composition of an organic compound. By burning a known mass of the substance and measuring the masses of the resulting CO2 and H2O, one can calculate the amounts of carbon and hydrogen present and, by difference with the original mass, infer the oxygen content. This method is foundational for deducing the empirical and often the molecular formula of organic compounds.

Acid–Base Titration

Acid–base titration is an analytical procedure utilized to quantify the acidic or basic character of a substance. In organic chemistry, titration can be used to determine the number of acidic protons a compound has, or to measure its overall acidity. The volume and concentration of titrant required to neutralize the compound provide key quantitative information about its acidic functional groups.

Oxidation Reactions

Oxidation reactions in organic chemistry involve the increase of the oxidation state of a molecule, often transforming one functional group into another (for example, oxidizing an alcohol to a carbonyl compound). These reactions are essential for understanding how compounds can be chemically modified, which provides insight into their structure and reactivity as well as their biological metabolism.

Dehydration Reactions

Dehydration reactions are processes in which a molecule loses a water molecule, typically leading to the formation of a double bond or cyclic structure. In organic chemistry, dehydration is often used to infer the presence of certain functional groups or the potential for structural rearrangement, and can produce compounds with highly symmetric structures that simplify spectroscopic analysis.

Transcript

00:02 Using the information provided, let's determine the chemical formulas of each of the compounds and then draw their leo structure.

00:10 So we have five compounds, which are listed a, b, c, d, and e.

00:22 So let's first solve for compound d because we're told that compound d is 40 ,000, 3 .7 % nitrogen and 50 % oxygen.

00:48 So for the mass of hydrogen, it'll be 100 % minus 50 % minus 43 .7%, which would be equal to 6 .3 % hydrogen.

01:07 So we're also told that the density, of compound d is equal to 2 .86 grams per liter at stp.

01:18 So now let's go ahead and find the formula for d.

01:25 We can find the moles of n which is going to be before we do that let's assume that we have 100 grams of d and so we have 43 .7 grams of n, 50 grams of o and 6 .3 grams of hydrogen.

01:57 Let's convert each of these to moles.

01:59 14 grams per moles, molar mass of nitrogen.

02:05 This would give us 3 .12 moles.

02:35 This would be 16 grams per mole.

02:38 So 50 divided by 16 gives me 3 .12 moles and times one gram per mole.

02:52 That's equal to 6 .3 moles.

02:55 We're going to divide by 3 .12.

03:01 This would give me 1 .1 and 2.

03:08 So my empirical formula for compound d would be equal to n.

03:20 Oh, h2, and we can solve for the empirical mass for d, which would be equal to n, sorry, h2 there, n, o, h2, so 32 .0 .0 .2.

03:43 So 32 .0.

03:46 We're told that the density is 2 .86 grams per liter.

03:53 So, let's solve for the molar mass of d.

04:02 We have 2 .86 grams per liter times 22 .4 liters per mole.

04:10 And this would work out two liters would cancel.

04:13 2 .86 times 22 .4.

04:18 64 .1 grams per mole.

04:23 Sulf for n, which is our whole number ratio, m .m over e.

04:27 64 .1 grounds per mole, divided by 32.

04:31 Grams per mole and so we kill the two so the molecular formula for d would be equal to two times the n o h2 which should be equal to n2 o2 h2 we rearrange this we can get a compound which is equal to nh4 n -o -2.

05:00 So there is compound d, h -4 -n -o -2.

05:10 So now there's d.

05:12 We can identify compound c.

05:16 We're told compound c.

05:19 It has one more oxygen than compound d.

05:31 So this is equal to n -h -4 -n -o -3 for the formula of c.

05:41 Now let's move on to compound a.

05:46 We're told that compound c and a have one ion in common.

05:55 So has one ion in common with compounds c.

06:08 And when a is acting as a strong electrolyte, the solution is strongly acidic.

06:14 So it's strongly acidic...