Chapter 15, Problem 15.107P

(a)

Interpretation Introduction

Interpretation:

The structures of A, B and C has to be given.

Concept Introduction:

The ideal gas equation can be used to calculate the volume using pressure and temperature.  The ideal gas equation is,

  $\begin{array}{l}\text{PV = nRT}\\ \\ \text{n =}\frac{\text{PV}}{\text{RT}}\end{array}$

Where, $\begin{array}{l}\text{P}\text{ = pressure}\\ \\ \text{V}\text{ = volume}\\ \\ \text{R}\text{ = gas constant}\\ \\ \text{T}\text{ = temperature}\\ \\ \text{n}\text{ = number of moles}\text{.}\end{array}$

Explanation of Solution

The structures of compounds A, B and C can be determined by calculating the moles of the compounds from the given mass of the compounds.

The moles of A can be calculated by using the ideal gas formula.

Given,

  $\begin{array}{l}\text{pressure = 1}\text{.00atm}\\ \\ \text{volume = 1}\text{.00L}\\ \\ {\text{temperature = 160}}^{\text{0}}\text{C}\end{array}$

  $\begin{array}{l}\text{moles of A (n) = }\frac{\text{PV}}{\text{RT}}\\ \\ \text{ (n) = }\frac{\text{(1}\text{.00 atm)(1}\text{.00L)}}{\left(\frac{\text{0}\text{.0821L}\text{.atm}}{\text{mol}\text{.K}}\right)\left(\text{273+160}\right)\text{K}}\\ \\ \text{ (n) = 0}\text{.02812995 mol}\end{array}$

Molar mass of A can be calculated using the number of moles obtained.

Given: $\text{2}\text{.48 g}$ of A.

  $\begin{array}{l}\text{molar mass of A = }\frac{\text{2}\text{.48gA}}{\text{0}\text{.02812995mol}}\\ \\ \text{ = 88}\text{.16226 g/mol}\end{array}$

The molecular formula of A can be determined by the combustion analysis.

Given,

• $\text{0}\text{.409 g}$ of ${\text{H}}_{\text{2}}\text{O}$
• $\text{1}{\text{.00 g of CO}}_{\text{2}}$

Moles of $\text{H}$ can be calculated from the mass of ${\text{H}}_{\text{2}}\text{O}$.

  $\begin{array}{l}\text{moles of H = (0}\text{.409 g H2O)}\left(\frac{\text{2 mol H}}{\text{18}{\text{.02 g H}}_{\text{2}}\text{O}}\right)\\ \\ \text{ = 0}\text{.0454 mol H}\\ \end{array}$

By using the moles of $\text{H}$ in ${\text{H}}_{\text{2}}\text{O}$, the mass of $\text{H}$ is calculated.

  $\begin{array}{l}\text{mass of H }\text{= }\left(\text{0}\text{.0454 mol H}\right)\left(\frac{\text{1}\text{.008 g H}}{\text{1 mol H}}\right)\\ \\ \text{= 0}\text{.04576 g H}\end{array}$

Moles of $\text{C}$ can be calculated from the mass of ${\text{CO}}_{\text{2}}$.

  $\begin{array}{l}\text{moles of C}\text{= }\left(\text{1}{\text{.00 g CO}}_{\text{2}}\right)\left(\frac{\text{1 mol C}}{\text{44}{\text{.01 g CO}}_{\text{2}}}\right)\\ \\ \text{= 0}\text{.0227 mol C}\end{array}$

By using the moles of $\text{C}$ in ${\text{CO}}_{\text{2}}$, the mass of $\text{C}$ is calculated.

  $\begin{array}{l}\text{mass of C}\text{= }\left(\text{0}\text{.0227 mol C}\right)\left(\frac{\text{12}\text{.01 g C}}{\text{1 mol C}}\right)\\ \\ \text{= 0}\text{.2726 g C}\end{array}$

Moles of $\text{O}$ can be calculated by subtracting the mass of $\text{C}$ and $\text{H}$ from the total mass.

  $\begin{array}{l}\text{moles of O}\text{= }\left(\text{0}\text{.500 g A - }\left(\text{0}\text{.04576+0}\text{.2726}\right)\text{ g O}\right)\left(\frac{\text{1 mol O}}{\text{16}\text{.00 g O}}\right)\\ \\ \text{= 0}\text{.01135 mol O}\end{array}$

To get the moles of the $\text{C}$, $\text{H}$ and $\text{O}$, the moles calculated is divided by the smallest number.

  $\begin{array}{l}\text{C: }\frac{\text{0}\text{.0227 mol C}}{\text{0}\text{.01135 mol}}\text{= 2}\\ \\ \text{H: }\frac{\text{0}\text{.0454 mol H}}{\text{0}\text{.01135 mol}}\text{= 4}\\ \\ \text{O: }\frac{\text{0}\text{.01135 mol O}}{\text{0}\text{.01135 mol}}\text{= 1}\end{array}$

The empirical formula of compound A is obtained as ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}$, with a molar mass $\text{44}\text{.05 g/mol}$.

As the molar mass of compound A is calculated as $\text{88}\text{.16 g/mol}$, the molecular formula of compound A is ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$.

As compound B is acidic, it will be a carboxylic acid.  Moles of $\text{COOH}$ can be calculated using the given information.

Given,

$\text{1}\text{.000 g}$ Of B is neutralized with $\text{33}\text{.9 mL of 0}\text{.5 M}$ sodium hydroxide.

  $\begin{array}{l}\text{moles of COOH}\text{= }\left(\frac{\text{0}\text{.5 mol NaOH}}{\text{L}}\right)\left(\text{33}\text{.9 mL}\right)\left(\frac{{\text{10}}^{\text{-3}}\text{ L}}{\text{1 mL}}\right)\left(\frac{\text{1 mol COOH}}{\text{1 mol NaOH}}\right)\\ \\ \text{= 0}\text{.01695 mol COOH}\end{array}$

The 1.00 g sample of compound B contains 0.01695 moles of $\text{COOH}$.

  $\begin{array}{l}\text{mass of COOH}\text{= }\frac{\text{1}\text{.00 g}}{\text{0}\text{.01695 mol COOH}}\\ \\ \text{= 59 g/mol COOH}\end{array}$

As the molar mass of compound A is $\text{88}\text{.16 g/mol}$, the molar mass of compound B will be $\text{118 g/mol}$, it will be a dicarboxylic acid.  The molecular formula can be given as ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}$.

Compound B on loss of water molecule by heating forms compound C.  Hence, it will be an anhydride.  The molecular formula of compound C can be given by the loss of water molecule from molecular formula of B.

Molecular formula of compound C = $\left({\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{4}}{\text{-H}}_{\text{2}}\text{O}\right){\text{ = C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{O}}_{\text{3}}$.

As the NMR spectrum of compound C gives only one peak, the compound must be symmetrical with only one type of hydrogens.  The structures of compound B and compound C according to the NMR spectrum are given as

Compound A is not acidic.  It will have an alcohol and aldehyde groups.  The structure of compound A can be given as

(b)

Interpretation Introduction

Interpretation:

Compound A is a controlled substance because it is metabolized to the weakly acidic date rape drug GHB, ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}{\text{O}}_{\text{3}}$.  The structure and name of GHB has to be given.

Concept Introduction:

Oxidation:

Addition of oxygen atoms to the molecule or removal hydrogens from the molecule is said to be oxidation.

Aldehyde on oxidation forms carboxylic acid.

Explanation of Solution

Compound A is 4-hydroxybutanal with an alcohol and aldehyde groups.

GHB is an acidic compound, hence compound A is oxidized and the aldehyde group converts to carboxylic acid group.  The structure and name of GHB is