Chapter 15, Problem 15.110P

(a)

Interpretation Introduction

Interpretation:

Compound A containing $\text{C, H and O}$ in branched structure is optically active.  A $\text{0}\text{.500 g}$ of sample burns in excess ${\text{O}}_{\text{2}}$ to yield $\text{1}{\text{.25 g of CO}}_{\text{2}}$ and $\text{0}{\text{.613 g of H}}_{\text{2}}\text{O}$.  The empirical formula of the compound has to be determined.

(b)

Interpretation Introduction

Interpretation:

When $\text{0}\text{.225 g}$ of compound A vaporizes at $\text{755 torr}$ and ${\text{97}}^{\text{0}}\text{C}$, the vapour occupies $\text{78}\text{.0 mL}$.  The molecular formula of compound A has to be determined.

Concept Introduction:

The equation used to find the molecular weight of the compound is

  $\begin{array}{l}\text{PV = }\frac{\text{mRT}}{\text{mass}}\\ \\ \text{mass = }\frac{\text{mRT}}{\text{PV}}\end{array}$

Where, $\begin{array}{l}\text{P = pressure}\\ \\ \text{V = volume}\\ \\ \text{R = gas constant}\\ \\ \text{T = temperature}\\ \\ \text{m = mass of compound}\end{array}$

(c)

Interpretation Introduction

Interpretation:

On careful oxidation, Compound A yields a ketone.  The compound A has to be drawn and named.  The chiral centre has to be circled.

Concept Introduction:

Carbonyl compounds are formed by the oxidation of alcohols.  Ketone is formed on oxidation of secondary alcohol.