NESA Chemistry Analysis of Organic Substances

8 sample questions with marking guides and sample answers · Avg. score: 50%

Q6
2025
SCSA
1 mark
Q6
1 mark

Which of the following will decolourise a solution of bromine water?

A

1.0 mol L1^{-1} Fe(NO3)3Fe(NO_3)_3 solution

B

1.0 mol L1^{-1} KCl solution

C

CH3CH2CH2CH2CHOCH_3CH_2CH_2CH_2CHO

D

CH3CH2CHCHCH3CH_3CH_2CHCHCH_3

Reveal Answer
A

1.0 mol L1^{-1} Fe(NO3)3Fe(NO_3)_3 solution

This is a solution of Iron(III) nitrate. Neither Fe3+Fe^{3+} nor NO3NO_3^- ions react with bromine water, so the solution would not be decolourised.

B

1.0 mol L1^{-1} KCl solution

Chloride ions (ClCl^-) cannot reduce bromine (Br2Br_2) because chlorine is a stronger oxidising agent than bromine. Therefore, no reaction occurs to remove the bromine colour.

C

CH3CH2CH2CH2CHOCH_3CH_2CH_2CH_2CHO

Correct Answer

Pentanal (C₅H₁₀O) can cause a colour change with bromine because it is an aldehyde. Aldehydes are easily oxidised to carboxylic acids, and in this reaction, bromine (Br₂), which is orange-brown, is reduced to colourless bromide ions (Br⁻). As a result, the orange colour of bromine water disappears when it reacts with pentanal.

D

CH3CH2CHCHCH3CH_3CH_2CHCHCH_3

Correct Answer

The condensed formula represents 2-pentene, which contains a carbon-carbon double bond (C=CC=C). Alkenes undergo a rapid addition reaction with bromine water, consuming the Br2Br_2 and turning the orange solution colourless.

Q30
2021
VCAA
1 mark
Q30
1 mark

The 1H^1\text{H} NMR spectrum of an organic compound has three unique sets of peaks: a single peak, seven peaks (septet) and two peaks (doublet).

The compound is

A

3-methyl butanoic acid.

B

2-methyl propanoic acid.

C

2-chloro-2-methylpropane.

D

1,2-dichloro-2-methylpropane.

Reveal Answer
A

3-methyl butanoic acid.

3-methylbutanoic acid has four unique sets of protons, which would produce a singlet (OH), a doublet (CH3 groups), a multiplet/nonet (CH), and another doublet (CH2).

B

2-methyl propanoic acid.

Correct Answer

2-methylpropanoic acid has three unique sets of protons: the carboxylic acid proton (singlet), the CH proton split by six methyl protons (septet), and the six equivalent methyl protons split by the CH proton (doublet).

C

2-chloro-2-methylpropane.

2-chloro-2-methylpropane has only one unique set of protons (nine equivalent methyl protons), which would produce a single peak (singlet) in its 1H^1\text{H} NMR spectrum.

D

1,2-dichloro-2-methylpropane.

1,2-dichloro-2-methylpropane has two unique sets of protons that are not adjacent to any other protons, which would result in a spectrum with two singlets.

Q5
2024
NESA
1 mark
Q5
1 mark

Which would be the best reagent to use to determine whether an unknown substance was 2-methylpropan-1-ol or 2-methylpropan-2-ol?

A

Bromine water

B

Potassium nitrate solution

C

Sodium carbonate solution

D

Acidified potassium permanganate solution

Reveal Answer
A

Bromine water

Bromine water is used to test for the presence of carbon-carbon double bonds (unsaturation). It would not react with either of these saturated alcohols.

B

Potassium nitrate solution

Potassium nitrate is generally unreactive with alcohols and would not produce a visible change to help distinguish between the two substances.

C

Sodium carbonate solution

Sodium carbonate is typically used to test for acidic substances like carboxylic acids by observing effervescence (carbon dioxide gas), but it does not react with alcohols.

D

Acidified potassium permanganate solution

Correct Answer

Acidified potassium permanganate is a strong oxidizing agent. It will oxidize the primary alcohol (2-methylpropan-1-ol), changing color from purple to colorless, but will not react with the tertiary alcohol (2-methylpropan-2-ol).

Q16
2021
VCAA
1 mark
Q16
1 mark

Which one of the following statements about IR spectroscopy is correct?

A

IR radiation changes the spin state of electrons.

B

Bond wave number is influenced only by bond strength.

C

An IR spectrum can be used to determine the purity of a sample.

D

In an IR spectrum, high transmittance corresponds to high absorption.

Reveal Answer
A

IR radiation changes the spin state of electrons.

IR radiation causes changes in the vibrational states of molecules, not the spin states of electrons (which is associated with Electron Paramagnetic Resonance spectroscopy).

B

Bond wave number is influenced only by bond strength.

The wavenumber of a bond's vibration depends on both the bond strength (force constant) and the reduced mass of the atoms involved, as described by Hooke's Law.

C

An IR spectrum can be used to determine the purity of a sample.

Correct Answer

An IR spectrum can reveal the presence of impurities if unexpected absorption peaks appear that do not belong to the pure compound.

D

In an IR spectrum, high transmittance corresponds to high absorption.

Transmittance and absorbance are inversely related; high transmittance means that most of the light passed through the sample, indicating low absorption.

Q18
2022
SCSA
1 mark
Q18
1 mark

The presence of a carboxylic acid functional group in an unknown organic compound may be identified by observing a reaction with which of the following?

A

Cr2O72(aq)\mathrm{Cr_2O_7^{2-}(aq)}

B

Br2(aq)\mathrm{Br_2(aq)}

C

Na2SO4(s)\mathrm{Na_2SO_4(s)}

D

Na2CO3(s)\mathrm{Na_2CO_3(s)}

Reveal Answer
A

Cr2O72(aq)\mathrm{Cr_2O_7^{2-}(aq)}

Dichromate is a strong oxidizing agent used to test for primary and secondary alcohols or aldehydes, but carboxylic acids are already fully oxidized at the carbon atom and will not react.

B

Br2(aq)\mathrm{Br_2(aq)}

Bromine water is used to test for unsaturation (such as alkenes and alkynes) or highly activated aromatic rings, not carboxylic acids.

C

Na2SO4(s)\mathrm{Na_2SO_4(s)}

Sodium sulfate is a neutral salt and does not undergo an acid-base reaction with weak acids like carboxylic acids.

D

Na2CO3(s)\mathrm{Na_2CO_3(s)}

Correct Answer

Sodium carbonate is a weak base that reacts with carboxylic acids to produce carbon dioxide gas. The visible effervescence (bubbling) serves as a positive test for the presence of an acid.

Q6
2023
VCAA
1 mark
Q6
1 mark

Consider the following statements.

I. HPLC is a qualitative process.
II. HPLC is a quantitative process.
III. Triplets give information about molecule structure.

Which of the above statements apply to high-performance liquid chromatography (HPLC)?

A

I only

B

II only

C

I and II only

D

I, II and III

Reveal Answer
A

I only

While HPLC can be used qualitatively to identify substances based on retention time, this option is incomplete because HPLC is also a quantitative technique.

B

II only

While HPLC can be used quantitatively to determine concentration based on peak area, this option is incomplete because HPLC is also a qualitative technique.

C

I and II only

Correct Answer

HPLC is both a qualitative process (identifying components via retention time) and a quantitative process (determining concentration via peak area).

D

I, II and III

Statement III is incorrect because triplets refer to splitting patterns found in Nuclear Magnetic Resonance (NMR) spectroscopy, not HPLC.

Q8
2020
QCAA
Paper 1
1 mark
Q8
1 mark

An organic compound, X, reacts with sodium hydrogen carbonate to form carbon dioxide gas.
Compound X is

A

an amine.

B

a haloalkane.

C

a carboxylic acid.

D

a primary alcohol.

Reveal Answer
A

an amine.

Amines are basic compounds and do not react with sodium hydrogen carbonate to produce carbon dioxide.

B

a haloalkane.

Haloalkanes do not have acidic properties sufficient to react with sodium hydrogen carbonate to release carbon dioxide.

C

a carboxylic acid.

Correct Answer

Carboxylic acids are stronger acids than carbonic acid, so they react with sodium hydrogen carbonate (NaHCO3NaHCO_3) to release carbon dioxide gas (CO2CO_2).

D

a primary alcohol.

Alcohols are weaker acids than carbonic acid and do not react with sodium hydrogen carbonate to evolve carbon dioxide.

Q29
2023
VCAA
1 mark
Q29
1 mark

Which one of the following statements about mass spectrometry is always correct?

A

The relative molecular mass of a molecule is determined from the base peak.

B

The peaks in a mass spectrum are caused by the presence of different isotopes.

C

The base peak is formed when an uncharged species is removed from the molecule.

D

The height of each peak in the mass spectrum is measured relative to the height of the base peak.

Reveal Answer
A

The relative molecular mass of a molecule is determined from the base peak.

The relative molecular mass is determined from the molecular ion peak (M+M^+), whereas the base peak simply represents the most abundant ion fragment.

B

The peaks in a mass spectrum are caused by the presence of different isotopes.

While isotopes cause specific adjacent peaks (like M+1M+1), the vast majority of peaks in a mass spectrum are caused by the fragmentation of the molecule into smaller ions.

C

The base peak is formed when an uncharged species is removed from the molecule.

The base peak is simply the most intense peak representing the most stable ion; it can sometimes be the intact molecular ion itself, meaning no uncharged species was removed.

D

The height of each peak in the mass spectrum is measured relative to the height of the base peak.

Correct Answer

In mass spectrometry, the base peak is the most intense peak and is assigned a relative abundance of 100%, with all other peak heights measured relative to it.

Frequently Asked Questions

How many NESA Chemistry questions cover Analysis of Organic Substances?
AusGrader has 74 NESA Chemistry questions on Analysis of Organic Substances, all with instant AI grading and detailed marking feedback.

Ready to practise NESA Chemistry?

Get instant AI feedback on past exam questions, aligned to the syllabus

Start Practising Free