Organic preparation: separation, product choice and purity

What you’ll learn

• Explain the purposes of reflux, distillation, extraction and drying.
• Calculate yield and use physical evidence to evaluate purity.

Understand the idea

Core practicals 6 and 7 involve organic preparation and the interpretation of separation methods. The key written skill is to explain what each stage achieves. Reflux returns condensed vapour to the reaction mixture, allowing prolonged heating with reduced loss of volatile material. Distillation collects a volatile component separately. In primary-alcohol oxidation, removing an aldehyde as it forms limits its further oxidation; retaining material under suitable oxidising conditions supports formation of the carboxylic acid.

A separating funnel separates immiscible liquid layers. The organic layer is not always the upper layer: use the supplied density or identification evidence. A suitable anhydrous drying agent removes traces of water from an organic liquid, rather than removing every dissolved impurity. A narrow boiling range near a reference value supports purity, but a boiling point alone does not prove identity. Explain the role of each observation together with the chemical evidence.

The limiting reactant determines theoretical yield. Divide the isolated, dry product mass by that theoretical mass to obtain percentage yield. Transfer losses and incomplete conversion can lower yield; residual solvent or water can inflate the recorded mass. In these supervised preparation contexts, corrosive reagents and volatile flammable liquids require the prescribed ventilation, splash protection and suitable heating equipment. These lessons analyse supplied results and techniques rather than give a synthesis procedure.

Key formulae

$$\mathrm{CH_3CH_2CH_2OH}+[\mathrm O]\rightarrow\mathrm{CH_3CH_2CHO}+\mathrm{H_2O}$$

$$\mathrm{CH_3CH_2CHO}+[\mathrm O]\rightarrow\mathrm{CH_3CH_2COOH}$$

$$m_{\mathrm{theoretical}}=n_{\mathrm{limiting}}\times\frac{\nu_{\mathrm{product}}}{\nu_{\mathrm{limiting}}}\times M_{\mathrm{product}}$$

$$\mathrm{yield}\ (\%)=\frac{m_{\mathrm{dry\ product}}}{m_{\mathrm{theoretical}}}\times100$$

Worked example

A supplied preparation record gives $0.0800\ \mathrm{mol}$ of a limiting alcohol and a $1:1$ amount ratio to a chloroalkane of molar mass $92.5\ \mathrm{g\,mol^{-1}}$. The dry isolated product mass is $5.55\ \mathrm{g}$. Calculate the percentage yield.

1. The maximum product amount is $0.0800\ \mathrm{mol}$ because the ratio is $1:1$.
2. The theoretical product mass is $0.0800\times92.5=7.40\ \mathrm{g}$.
3. The percentage yield is $100(5.55/7.40)=75.0\%$.

Answer: $75.0\%$.