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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 \(\mathrm{GHB}\) ?

Step by step solution

01

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掳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鈦宦� mol鈦宦�), \(T\) is temperature (in K)Calculate the moles (n) of A:\(n = \frac{PV}{RT} = \frac{1 \, \text{atm} \times 1 \, \text{L}}{0.0821 \, \text{L atm K鈦宦� mol鈦宦箎 \times 433 \, \text{K}} = 0.0281 \, \text{mol}\)The molar mass (M) of A:\(M = \frac{\text{mass}}{\text{moles}} = \frac{2.48 \, \text{g}}{0.0281 \, \text{mol}} = 88.26 \, \text{g/mol}\)

02

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 H鈧侽 and 1.00 g of CO鈧�.Calculate the moles of C and H present in the products:- Moles of C in CO鈧�: \( \frac{1.00 \, \text{g}}{44.01 \, \text{g/mol}} = 0.0227 \, \text{mol of C} \)- Moles of H in H鈧侽: \( \frac{0.409 \, \text{g}}{18.02 \, \text{g/mol}} = 0.0227 \, \text{mol of H鈧侽} \). Since there are 2 H atoms in each H鈧侽 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:\( \text{C}_2 \text{H}_4 \text{O} \)

03

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 C鈧侶鈧凮 = 44.05 g/mol, and the molar mass of A 鈮� 88 g/mol, the molecular formula of A is C鈧凥鈧圤鈧�.(a) Since glycolic acid is synthesized from simple acetals or aldehydes and fits these criteria, the structure of:- Compound A: C鈧凥鈧圤鈧� (likely 1,4-butanediol).- Compound B: C鈧凥鈧哋鈧� (likely succinic acid formed through oxidation).- Compound C: C鈧凥鈧圤 (gamma-butyrolactone formed through heating).The NMR spectrum of 纬-butyrolactone (Compound C) shows one peak, consistent with its structure.

04

Identify the Structure and Name of GHB

Compound A is metabolized to GHB (\text{C鈧凥鈧圤鈧儅), gamma-hydroxybutyric acid, a controlled substance.The molecular structure and name of GHB are:\(\text{HO-CH鈧�-CH鈧�-CH鈧�-COOH} \) (4-hydroxybutanoic acid).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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路atm路K鈦宦孤穖ol鈦宦�
• 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 CO鈧� gives 0.0227 mol of C, you follow a similar path for H鈧侽 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 C鈧侶鈧凮. 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鈥檚 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 C鈧凥鈧圤鈧� 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鈥檚 what you need to know:

• It provides information on the number of hydrogen atoms (protons) in different environments.
• Peaks in an 鹿H 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 纬-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 (CO鈧�) formed from carbon and water (H鈧侽) 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 H鈧侽 and 1.00 g of CO鈧�:

• Convert masses to moles: 0.409 g of H鈧侽 is 0.0227 mol of H鈧侽 and 1.00 g of CO鈧� is 0.0227 mol of C.


• Since every H鈧侽 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.