Organic synthesis

WCH15Topic 5.53 questions

Multi-step synthetic routes, Grignard reagents, practical techniques and spectroscopic analysis.

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Original practice questions written in the style of the exam.

Question 14 marks
Compound Y, $\text{C}_3\text{H}_6\text{O}$ , has a strong infrared absorption at $1715\,\text{cm}^{-1}$ and no absorption between $3200$ and $3600\,\text{cm}^{-1}$ . Its proton NMR spectrum shows a single peak.

Use this evidence to identify Y, explaining each deduction.
Mark scheme
- The absorption at $1715\,\text{cm}^{-1}$ shows a $\text{C{=}O}$ group (1)
- No absorption at $3200$ – $3600\,\text{cm}^{-1}$ means there is no group (1)
Question 24 marks
Outline a two-step synthesis of butanoic acid, $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ , from 1-bromopropane. Give the reagents and conditions for each step and name the intermediate.

3600

cm−1

\text{O{-}H}

A single NMR peak means all six hydrogen atoms are in the same environment (1)

Y is propanone,

\text{CH}_3\text{COCH}_3

(1)

Mark scheme

Step 1 reagent: potassium cyanide (or sodium cyanide) in ethanol (1)
Step 1 conditions: heat under reflux (1)
Intermediate: butanenitrile, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CN}$ (1)
Step 2: heat under reflux with dilute hydrochloric (or sulfuric) acid to hydrolyse the nitrile (1)