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A-Level Chemistry – Organic Synthesis and Analysis

Synthesis routes, mass spectrometry, IR and NMR.

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12 questions  ·  46 marks  ·  answers included  ·  A-Level Chemistry (Edexcel/AQA/OCR, Year 2)
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This worksheet covers:

  • This worksheet covers multi-step organic synthesis routes and structure determination using mass spectrometry
  • Infrared (IR) spectroscopy
  • $^1\text{H}$ NMR spectroscopy

Use $A_r$ values C = 12, H = 1, O = 16, N = 14. Allow about 45 minutes.

Questions

Section A – Synthesis routes (skills)

1

State the reagent(s) and condition(s) needed to convert ethanol into ethene.

Show answer
Concentrated $\text{H}_2\text{SO}_4$ (or $\text{Al}_2\text{O}_3$ as a solid catalyst), heat – acid-catalysed dehydration (elimination of water).
[2]
2

State the reagent(s) and condition(s) needed to convert propene into propan-2-ol, and name the type of reaction taking place.

Show answer
Steam ($\text{H}_2\text{O}(g)$) with a phosphoric acid ($\text{H}_3\text{PO}_4$) catalyst, high temperature and high pressure. This is an electrophilic addition (hydration) reaction; Markovnikov addition gives the secondary alcohol, propan-2-ol, as the major product.
[3]
3

State the reagent(s) and condition(s) needed to convert ethanoic acid into ethyl ethanoate, and name this type of reaction.

Show answer
Ethanol, with a small amount of concentrated $\text{H}_2\text{SO}_4$ catalyst, under reflux. This is esterification (condensation reaction): $\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$
[3]
4

Bromoethane can be converted into ethanoic acid in two steps.Give the reagents and conditions for each step, and name the type of reaction occurring in each.

Show answer
Step 1: warm with aqueous sodium hydroxide, under reflux – nucleophilic substitution, giving ethanol ($\text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2\text{OH}$).Step 2: heat under reflux with excess acidified potassium dichromate(VI) ($\text{K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4$) – oxidation, giving ethanoic acid ($\text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{COOH}$).
[4]

Section B – Spectroscopic analysis (exam-style)

5

A hydrocarbon has molecular formula $\text{C}_4\text{H}_6$.(a) Calculate its degree(s) of unsaturation (index of hydrogen deficiency) using $\text{DoU} = \dfrac{2C+2-H}{2}$.(b) Suggest, giving a reason, one type of structural feature (or combination of features) consistent with this value.

Show answer
(a) $\text{DoU} = \dfrac{2(4)+2-6}{2} = 2$(b) Two degrees of unsaturation could be one ring plus one double bond, two double bonds (a diene), or one triple bond (an alkyne) – e.g. but-1-yne or buta-1,3-diene, since each ring or $\pi$ bond accounts for one degree of unsaturation, and a triple bond counts as two.
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6

In the mass spectrum of butanone ($\text{CH}_3\text{COCH}_2\text{CH}_3$), the molecular ion peak is observed at $m/z = 72$, with fragment peaks at $m/z = 57$ and $m/z = 43$.(a) Confirm that $72$ is consistent with the $M_r$ of butanone.(b) Identify the species responsible for the peak at $m/z = 43$, and explain the loss that produces it from the molecular ion.(c) Identify the species responsible for the peak at $m/z = 57$.

Show answer
(a) $M_r(\text{C}_4\text{H}_8\text{O}) = 4(12)+8(1)+16 = 72$ – consistent.(b) $m/z = 43$ is the acylium ion $\text{CH}_3\text{CO}^{+}$, formed by loss of a $\cdot\text{C}_2\text{H}_5$ radical (mass 29) from the molecular ion: $72 – 29 = 43$.(c) $m/z = 57$ corresponds to loss of a $\cdot\text{CH}_3$ radical (mass 15) from the molecular ion, $72-15=57$, giving the ion $\text{CH}_3\text{CH}_2\text{CO}^{+}$.
[5]
7

An unknown compound has $M_r = 60$. Its infrared spectrum shows a very broad, strong absorption across approximately $2500$–$3300\,\text{cm}^{-1}$ and a strong sharp absorption at $1710\,\text{cm}^{-1}$.(a) Identify the functional group(s) indicated by each absorption.(b) Suggest the identity of the compound, showing that its $M_r$ is consistent with your answer.

Show answer
(a) The very broad absorption at $2500$–$3300\,\text{cm}^{-1}$ is the O–H stretch of a carboxylic acid (much broader than an alcohol O–H due to strong hydrogen bonding/dimerisation); the sharp absorption at $1710\,\text{cm}^{-1}$ is a C=O stretch.(b) Ethanoic acid, $\text{CH}_3\text{COOH}$: $M_r = 2(12)+4(1)+2(16) = 60$ – consistent, and it contains both a C=O and a carboxylic O–H group.
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8

Propan-1-ol, $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$, gives a $^1\text{H}$ NMR spectrum with four signals.Using the $n+1$ rule, state the splitting pattern (multiplicity) expected for each of the four proton environments, explaining your reasoning for the two $\text{CH}_2$/$\text{CH}_3$ environments with non-adjacent OH.

Show answer
$\text{CH}_3$ (adjacent to one $\text{CH}_2$, 2 neighbouring H): triplet.Middle $\text{CH}_2$ (adjacent to $\text{CH}_3$ and $\text{CH}_2\text{OH}$, 3+2 = 5 neighbouring H): sextet.$\text{CH}_2$ next to OH (adjacent to the middle $\text{CH}_2$, 2 neighbouring H): triplet.OH proton: singlet, because it exchanges rapidly with traces of water/other OH protons and does not show coupling to neighbouring C–H protons under normal conditions.
[4]

Section C – Structure elucidation and synthesis (extended)

9

An unknown compound X has the following data:• Mass spectrum: molecular ion peak at $m/z = 88$.• IR spectrum: strong sharp absorption at $1740\,\text{cm}^{-1}$; no broad absorption in the $2500$–$3300\,\text{cm}^{-1}$ region.• $^1\text{H}$ NMR spectrum: singlet, $3\text{H}$, $\delta \approx 2.05$; quartet, $2\text{H}$, $\delta \approx 4.12$; triplet, $3\text{H}$, $\delta \approx 1.26$.Deduce the structure of X, explaining how each piece of data supports your answer.

Show answer
X is ethyl ethanoate, $\text{CH}_3\text{COOC}_2\text{H}_5$.$M_r = 4(12)+8(1)+2(16) = 88$, matching the molecular ion.The IR absorption at $1740\,\text{cm}^{-1}$ is a C=O stretch (ester carbonyl); the absence of a broad O–H absorption rules out an alcohol or carboxylic acid, consistent with an ester.In the NMR: the $3\text{H}$ singlet at $\delta \approx 2.05$ is the $\text{CH}_3$ attached to the carbonyl carbon (no adjacent H, so no splitting); the $2\text{H}$ quartet at $\delta \approx 4.12$ is the $\text{O-CH}_2$ group (deshielded by the adjacent O, split into a quartet by the neighbouring $\text{CH}_3$’s 3 protons); the $3\text{H}$ triplet at $\delta \approx 1.26$ is the terminal $\text{CH}_3$ of the ethyl group, split into a triplet by the adjacent $\text{CH}_2$’s 2 protons. The 3:2:3 integration ratio matches $\text{CH}_3\text{COOCH}_2\text{CH}_3$ exactly.
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10

Propan-1-ol and propan-2-ol are structural isomers with the same molecular formula, $\text{C}_3\text{H}_8\text{O}$.Explain how $^1\text{H}$ NMR spectroscopy (splitting patterns and/or integration) could be used to distinguish between these two isomers.

Show answer
Propan-2-ol, $(\text{CH}_3)_2\text{CHOH}$, gives a doublet ($6\text{H}$, two equivalent $\text{CH}_3$ groups split by the single adjacent $\text{CH}$ proton) and a septet ($1\text{H}$, the $\text{CH}$ proton split by 6 equivalent neighbouring protons), plus an OH singlet – three signals with a 6:1:1 integration ratio. Propan-1-ol, $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$, gives four signals (triplet, sextet, triplet, OH singlet) with integration ratio 3:2:2:1, and no signal integrating for 6H. The very different number of signals, splitting patterns and integration ratios distinguish the two isomers.
[3]
11

Design a three-step synthetic route to convert propan-1-ol into propanoic acid via propanenitrile, $\text{CH}_3\text{CH}_2\text{CN}$, given the intermediate step of forming 1-bromopropane.For each step, state the reagent(s), condition(s) and the type of reaction.

Show answer
Step 1 (propan-1-ol $\rightarrow$ 1-bromopropane): reflux with sodium bromide and concentrated sulfuric acid (generates HBr in situ), or $\text{PBr}_3$ – substitution reaction.Step 2 (1-bromopropane $\rightarrow$ propanenitrile): heat under reflux with potassium cyanide (KCN) dissolved in ethanol – nucleophilic substitution, extending the carbon chain by one carbon.Step 3 (propanenitrile $\rightarrow$ propanoic acid): heat under reflux with dilute hydrochloric acid (or dilute $\text{NaOH}$ followed by acidification) – hydrolysis of the nitrile to the carboxylic acid.
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12

Explain why the mass spectrum of a compound containing one carbon atom shows a small peak at $M+1$ (one mass unit above the molecular ion), and state what information the relative height of this peak can, in principle, give about the compound.

Show answer
The $M+1$ peak arises because a small proportion (about 1.1%) of carbon atoms in nature are the $^{13}\text{C}$ isotope rather than $^{12}\text{C}$; a molecule containing one $^{13}\text{C}$ atom is one mass unit heavier than the main molecular ion. The relative height of the $M+1$ peak compared to the $M$ peak increases with the number of carbon atoms in the molecule, so it can be used to estimate how many carbon atoms the molecule contains.
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