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Chapter 15: Problem 312

Excess methylmagnesium iodide and \(0.1776 \mathrm{~g}\) of compound A of formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}_{3}\) Sreact to give \(84.1 \mathrm{cc}\) of methane collected over mercury at \(740 \mathrm{~mm}\) and \(25^{\circ} \mathrm{C}\). How many active hydrogens does compound A possess? Suggest a possible structure for the compound given that the infrared spectrum shows no carbonyl absorption and the n.m.r. spectrum shows only three types of hydrogen with areas in the ratio of \(1: 2: 2\).

Short Answer

Expert verified
In conclusion, compound A appears to be 1,2,3-propanetriol (glycerol) with the structural formula CH2OH-CHOH-CH2OH and possesses one active hydrogen. The moles of methane produced (0.00332 mol) correspond to the moles of active hydrogen in compound A, with two molecules of compound A reacting for every mole of methane produced. The infrared spectrum shows no carbonyl absorption, and the NMR spectrum displays three types of hydrogen in a ratio of 1:2:2, in coherence with the compound's structure.

Step by step solution

01

Calculate moles of methane formed

Determine the moles of methane formed using the Ideal Gas Law equation, PV = nRT. The volume (84.1 cc) and pressure (740 mm) have been given along with the temperature (25°C). First, let's convert the given volume, pressure, and temperature to standard units. Volume (V): 84.1 cc = 0.0841 L Pressure (P): 740 mmHg = (740 / 760) atm = 0.973684 atm Temperature (T): 25°C = 298.15 K Now, we can use the Ideal Gas Law formula to determine the moles of methane: PV = nRT n = PV / RT = (0.973684 atm * 0.0841 L) / (0.0821 L atm/mol K * 298.15 K) ≈ 0.00332 mol We obtained 0.00332 mol of methane.
02

Connect moles of methane to active hydrogen atoms

It is important to know that the moles of methane produced are equal to the moles of active hydrogen in compound A (since each active hydrogen atom in compound A reacts with an equivalent of methylmagnesium iodide to produce methane). Therefore, compound A has 0.00332 mol of active hydrogens.
03

Calculate active hydrogen atoms in one molecule of compound A

Now, we can determine the moles of compound A used in the reaction by using its mass and molar mass: Moles of compound A = mass / molar_mass = 0.1776 g / (4*12.01 + 10*1.008 + 3*16.00) g/mol ≈ 0.0060 mol Now, divide the moles of active hydrogens by the moles of compound A to find active hydrogens per molecule: Active hydrogens per molecule = (0.00332 mol) / (0.0060 mol) ≈ 0.55 Since we cannot have a fractional number of active hydrogens, it should be interpreted that there are 1 active hydrogen and 2 molecules of compound A reacting for every 1 mole of methane produced.
04

Suggest a possible structure for compound A

Considering that the infrared spectrum shows no carbonyl absorption, we can rule out the presence of a carbonyl group. The NMR spectrum displays only three types of hydrogen and a ratio of 1:2:2. A possible structure that meets these criteria is 1,2,3-propanetriol (glycerol). The structural formula is CH2OH-CHOH-CH2OH. We can observe three types of hydrogens: one from the central carbon in the ratio of 1 and the other two types from both terminal carbons in the ratio of 2:2. The active hydrogen is the one bonded to the first carbon atom in glycerol. In conclusion, compound A appears to be 1,2,3-propanetriol and possesses one active hydrogen.

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Most popular questions from this chapter

The following projected synthesis for \(\mathrm{n}\) -butane is not very efficient. Why? \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{CH}_{3} \mathrm{Br}+2 \mathrm{Na} \rightarrow \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3}+2 \mathrm{NaBr}\)

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