Question

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 \(\mathrm{GHB}\) ?

Short answer

The structures are:- A: 1,4-butanediol- B: succinic acid- C: γ-butyrolactoneGHB is gamma-hydroxybutyric acid.

Step-by-step solution

Determining the Molar Mass of Compound A

From the given conditions, use the Ideal Gas Law to find the molar mass (M) of compound A.Given:- Volume: 1.00 L- Mass: 2.48 g- Temperature: 160°C (converted to Kelvin is 433 K)- Pressure: Assuming 1 atmIdeal Gas Law: \(PV = nRT\)Where:\(P\) is pressure (in atm), \(V\) is volume (in L), \(n\) is number of moles, \(R\) is gas constant (0.0821 L atm K⁻¹ mol⁻¹), \(T\) is temperature (in K)Calculate the moles (n) of A:\(n = \frac{PV}{RT} = \frac{1 \, \text{atm} \times 1 \, \text{L}}{0.0821 \, \text{L atm K⁻¹ mol⁻¹} \times 433 \, \text{K}} = 0.0281 \, \text{mol}\)The molar mass (M) of A:\(M = \frac{\text{mass}}{\text{moles}} = \frac{2.48 \, \text{g}}{0.0281 \, \text{mol}} = 88.26 \, \text{g/mol}\)

Determining the Empirical Formula of Compound A

Calculate the composition of elements in compound A by its combustion products.Given:- 0.500 g of A produces 0.409 g of H₂O and 1.00 g of CO₂.Calculate the moles of C and H present in the products:- Moles of C in CO₂: \( \frac{1.00 \, \text{g}}{44.01 \, \text{g/mol}} = 0.0227 \, \text{mol of C} \)- Moles of H in H₂O: \( \frac{0.409 \, \text{g}}{18.02 \, \text{g/mol}} = 0.0227 \, \text{mol of H₂O} \). Since there are 2 H atoms in each H₂O molecule, \(0.0227 \, \text{mol} \times 2 = 0.0454 \, \text{mol of H} \)Since the total mass of compound A is 0.500 g, after accounting for mass of C and H, the remainder must be O:- Mass of C: \(0.0227 \, \text{mol} \times 12.01 \, \text{g/mol} = 0.2726 \, \text{g} \)- Mass of H: \(0.0454 \, \text{mol} \times 1.008 \, \text{g/mol} = 0.0457 \, \text{g} \)- Mass of O: \(0.500 \, \text{g} - 0.2726 \, \text{g} - 0.0457 \, \text{g} = 0.1817 \, \text{g} \)Moles of O: \( \frac{0.1817 \, \text{g}}{16.00 \, \text{g/mol}} = 0.0114 \, \text{mol} \)Empirical formula ratio: \( \text{C : H : O} = 0.0227 : 0.0454 : 0.0114 \)Simplify the ratio:\( \text{C}_2 \text{H}_4 \text{O} \)

Determine the Structure of Compounds A, B, and C

Given the molar mass and empirical formula, determine the molecular formula of A:The empirical formula mass of C₂H₄O = 44.05 g/mol, and the molar mass of A ≈ 88 g/mol, the molecular formula of A is C₄H₈O₂.(a) Since glycolic acid is synthesized from simple acetals or aldehydes and fits these criteria, the structure of:- Compound A: C₄H₈O₂ (likely 1,4-butanediol).- Compound B: C₄H₆O₃ (likely succinic acid formed through oxidation).- Compound C: C₄H₈O (gamma-butyrolactone formed through heating).The NMR spectrum of γ-butyrolactone (Compound C) shows one peak, consistent with its structure.

Identify the Structure and Name of GHB

Compound A is metabolized to GHB (\text{C₄H₈O₃}), gamma-hydroxybutyric acid, a controlled substance.The molecular structure and name of GHB are:\(\text{HO-CH₂-CH₂-CH₂-COOH} \) (4-hydroxybutanoic acid).

Key concepts

Ideal Gas Law

The Ideal Gas Law is essential for determining various properties of gases. It combines several gas laws into one equation: \[ PV = nRT \]Where:

• P is pressure in atmospheres (atm)
• V is volume in liters (L)
• n is the number of moles of gas
• R is the gas constant, 0.0821 L·atm·K⁻¹·mol⁻¹
• T is temperature in Kelvin (K)

To solve for one of the variables, you rearrange the equation. For example, in a problem where you need to find the number of moles ( n ), you would write: \[ n = \frac{PV}{RT} \]You'll use given values of P, V, and T to find n. This calculation helps us understand how much gas we have under certain conditions. It is particularly useful in experiments where gases are heated or cooled and their behavior needs to be tracked.

Empirical Formula Calculation

The empirical formula represents the simplest whole-number ratio of elements in a compound. To find it, follow these steps: 1. Convert the masses of each element to moles by dividing by their atomic masses. 2. Find the simplest whole-number ratio of moles. Take the given amounts of carbon (C), hydrogen (H), and oxygen (O) and convert them to moles. For example, if 1.00 g of CO₂ gives 0.0227 mol of C, you follow a similar path for H₂O to find the moles of H. 3. Use the remainder of the mass to find the moles of oxygen. 4. The ratio of the elements in moles is your empirical formula. Simplify the ratio by dividing by the smallest number of moles to get whole numbers. If the ratio of C, H, and O is 0.0227 : 0.0454 : 0.0114 , it simplifies to C₂H₄O. This empirical formula is crucial for deducing the molecular formula and understanding the compound's basic structure.

Molar Mass Determination

Determining the molar mass of a compound is essential for identifying it. You often get this information by using the Ideal Gas Law. Here’s how to do it:

• Calculate the moles of the gas using PV = nRT.
• Use the mass of the gas and the calculated moles to find the molar mass (M) by using the formula: \[ M = \frac{\text{mass}}{\text{moles}} \]

For example, if a gas sample has a mass of 2.48 g and occupies 1.00 L at a certain temperature and pressure, the calculated moles might be 0.0281 mol. Dividing the mass by the moles, you get the molar mass. This value helps in deriving the molecular formula by comparing it with the empirical formula mass. If the empirical formula mass is 44 g/mol and the molar mass is approximately 88 g/mol, the molecular formula would be C₄H₈O₂ because the molar mass is twice the empirical formula mass.

NMR Spectroscopy

Nuclear Magnetic Resonance (NMR) spectroscopy is a powerful tool for determining the structure of organic compounds. It looks at the magnetic properties of atomic nuclei. Here’s what you need to know:

• It provides information on the number of hydrogen atoms (protons) in different environments.
• Peaks in an ¹H NMR spectrum show different proton environments in a molecule.
• The number of peaks indicates how many distinct environments are present.

In this exercise, compound C showed one peak in its NMR spectrum. This suggests all the hydrogen atoms are in a similar environment, fitting the structure of γ-butyrolactone, which has symmetrical hydrogen atoms. Understanding NMR spectra helps in the detailed structural analysis of compounds.

Chemical Combustion Analysis

Chemical combustion analysis helps determine the empirical formula of a compound. Here's the process:

• Burn the compound completely in oxygen.
• Measure the masses of carbon dioxide (CO₂) formed from carbon and water (H₂O) formed from hydrogen.
• Convert these masses to moles of carbon and hydrogen.

Example:
For 0.500 g of compound A, if combustion produces 0.409 g of H₂O and 1.00 g of CO₂:

• Convert masses to moles: 0.409 g of H₂O is 0.0227 mol of H₂O and 1.00 g of CO₂ is 0.0227 mol of C.


• Since every H₂O contains 2 H atoms, multiply by 2 to get 0.0454 mol of H.



• Subtract the mass of C and H from the total mass to find the mass of O remaining.

The empirical formula is derived from the mole ratio of elements. This method ensures a precise calculation of the compound's basic elemental composition.