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CHEMISTRY, M8 2025 HSC 37

Compound A has the molecular formula \(\ce{C5H10}\). The information below shows some chemical reactions beginning with this compound.

  • Compound A reacts with \(\ce{H+/H2O}\) to produce compounds B and C .
  • Compound B does not react with \(\ce{H+/Cr2O7^{2-}}\).
  • Compound C reacts with \(\ce{H+/Cr2O7^{2-}}\) to produce compound D .
  • Compound D does not react with \(\ce{Na2CO3(aq)}\).

Determine the structure of compound A. Justify your answer using all the data given.   (4 marks)

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  • The molecular formula \(\ce{C5H10}\) corresponds to a 5-carbon alkene.
  • Compound A reacts with \(\ce{H+/H2O}\) (hydration) to produce compounds B and C, which must be alcohols.
  • Compound B does not react with acidified dichromate, so it cannot be oxidised. Therefore, B is a tertiary alcohol.
  • Compound C reacts with acidified dichromate to form compound D. This means C can be oxidised and is either a primary or secondary alcohol.
  • Compound D does not react with \(\ce{Na2CO3(aq)}\), so D is not a carboxylic acid. Therefore, D must be a ketone.
  • Since primary alcohols oxidise to carboxylic acids, compound C must be a secondary alcohol (which oxidises to a ketone).
  • For compound A to produce both a tertiary and secondary alcohol, the \(\ce{C=C}\) double bond must be positioned so that one carbon is bonded to three other carbons and the other to two carbons.

\(\Rightarrow\) Compound A is 2-methylbut-2-ene.

Show Worked Solution
  • The molecular formula \(\ce{C5H10}\) corresponds to a 5-carbon alkene.
  • Compound A reacts with \(\ce{H+/H2O}\) (hydration) to produce compounds B and C, which must be alcohols.
  • Compound B does not react with acidified dichromate, so it cannot be oxidised. Therefore, B is a tertiary alcohol.
  • Compound C reacts with acidified dichromate to form compound D. This means C can be oxidised and is either a primary or secondary alcohol.
  • Compound D does not react with \(\ce{Na2CO3(aq)}\), so D is not a carboxylic acid. Therefore, D must be a ketone.
  • Since primary alcohols oxidise to carboxylic acids, compound C must be a secondary alcohol (which oxidises to a ketone).
  • For compound A to produce both a tertiary and secondary alcohol, the \(\ce{C=C}\) double bond must be positioned so that one carbon is bonded to three other carbons and the other to two carbons.

\(\Rightarrow\) Compound A is 2-methylbut-2-ene.

CHEMISTRY, M8 2022 VCE 28 MC

The \({ }^{13}\text{C NMR}\) spectrum of an organic compound is shown below.

The organic compound could be

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\(D\)

Show Worked Solution
  • The \({ }^{13}\text{C NMR}\) has five peaks indicating 5 different carbon environments within the molecule.
  • The peak at 140 indicates the presence of the \(\ce{C=C}\).

\(\Rightarrow D\)

♦ Mean mark 42%.
COMMENT: Solving by elimination is an effective strategy here.

CHEMISTRY, M8 2023 HSC 36

An organic reaction pathway involving compounds \(\text{A, B,}\) and \(\text{C}\) is shown in the flow chart.
 

The molar mass of \(\text{A}\) is 84.156 g mol\(^{-1}\).

A chemist obtained some spectral data for the compounds as shown.
 

\( \text{Data from} \ ^{1} \text{H NMR spectrum of compound C} \)
 \( Chemical \ Shift \ \text{(ppm)} \) \( Relative \ peak \ area \) \( Splitting \ pattern \)
\(1.01\) \(3\) \(\text{Triplet}\)
\(1.05\) \(3\) \(\text{Triplet}\)
\(1.65\) \(2\) \(\text{Multiplet}\)
\(2.42\) \(2\) \(\text{Triplet}\)
\(2.46\) \(2\) \(\text{Quartet}\)
\( ^{1} \text{H NMR chemical shift data}\)
\( Type \ of \ proton \)  \(  \text{δ/ppm} \)
\( \ce{R - C\textbf{H}3,R - C\textbf{H}2 - R}\) \(0.7-1.7\)
\( \left.\begin{array}{l}\ce{\textbf{H}3C - CO - \\-C\textbf{H}2 - CO -}\end{array}\right\} \begin{aligned} & \text { (aldehydes, ketones,} \\ &\text{carboxylic acids or esters) }\end{aligned}\) \(2.0-2.6\) 
\( \ce{R - C\textbf{H}O} \) \(9.4-10.00\)
\( \ce{R - COO\textbf{H}} \) \(9.0-13.0\)
 

Identify the functional group present in each of compounds \(\text{A}\) to \(\text{C}\) and draw the structure of each compound. Justify your answer with reference to the information provided.  (9 marks) 

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Compound \(\text{A}\): Alkene
 

   

Compound \(\text{B}\): Secondary alcohol

Compound \(\text{C}\): Ketone
 

   

Reasoning as follows:

  • Compound \(\text{A}\) is able to undergo an addition reaction to add water across a \(\ce{C=C}\) bond  \(\Rightarrow \)  Alkene
  • Compound \(\text{B}\) is the product of the above hydration reaction and is therefore an alcohol.
  • The \(\ce{^{13}C\ NMR}\) spectrum of Compound \(\text{A}\) confirms it is an alkene (132 ppm peak corresponding to the \(\ce{C=C}\) atoms). 3 spectrum peaks indicate 3 carbon environments. The molar mass of compound \(\text{A}\) is 84.156 g mol\(^{-1}\) which suggests symmetry within the molecule.
  • The Infrared Spectrum of Compound \(\text{B}\) has a broad peak at approximately 3400 cm\(^{-1}\). This indicates the presence of an hydroxyl group and confirms \(\text{B}\) is an alcohol.
  • Compound \(\text{C}\) is produced by the oxidation of Compound \(\text{B}\) with acidified potassium permanganate.
  • Compound \(\text{C}\) is a carboxylic acid if \(\text{B}\) is a primary alcohol or a ketone if \(\text{B}\) is a secondary alcohol.
  • Since the \(\ce{^{1}H NMR}\) spectrum of \(\text{C}\) does not show any peaks between 9.0 − 13.0 ppm, it cannot be a carboxylic acid. Compound \(\text{C}\) is therefore a ketone and Compound \(\text{B}\) is a secondary alcohol.
  • The \(\ce{^{1}H NMR}\) spectrum shows 5 peaks \(\Rightarrow \) 5 hydrogen environments.
  • Chemical shift and splitting patterns information indicate:
  • 1.01 ppm – 1.05 ppm: \(\ce{CH3}\) (next to a \(\ce{CH2}\))
  •   1.65 ppm: \(\ce{CH2}\) (with multiple neighbouring hydrogens)
  •   2.42 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH2}\))
  •   2.46 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH3}\))
Show Worked Solution

Compound \(\text{A}\): Alkene
 

   

Compound \(\text{B}\): Secondary alcohol

Compound \(\text{C}\): Ketone
 

   

Reasoning as follows:

  • Compound \(\text{A}\) is able to undergo an addition reaction to add water across a \(\ce{C=C}\) bond  \(\Rightarrow \)  Alkene
  • Compound \(\text{B}\) is the product of the above hydration reaction and is therefore an alcohol.
  • The \(\ce{^{13}C\ NMR}\) spectrum of Compound \(\text{A}\) confirms it is an alkene (132 ppm peak corresponding to the \(\ce{C=C}\) atoms). 3 spectrum peaks indicate 3 carbon environments. The molar mass of compound \(\text{A}\) is 84.156 g mol\(^{-1}\) which suggests symmetry within the molecule.
  • The Infrared Spectrum of Compound \(\text{B}\) has a broad peak at approximately 3400 cm\(^{-1}\). This indicates the presence of an hydroxyl group and confirms \(\text{B}\) is an alcohol.
  • Compound \(\text{C}\) is produced by the oxidation of Compound \(\text{B}\) with acidified potassium permanganate.
  • Compound \(\text{C}\) is a carboxylic acid if \(\text{B}\) is a primary alcohol or a ketone if \(\text{B}\) is a secondary alcohol.
  • Since the \(\ce{^{1}H NMR}\) spectrum of \(\text{C}\) does not show any peaks between 9.0 − 13.0 ppm, it cannot be a carboxylic acid. Compound \(\text{C}\) is therefore a ketone and Compound \(\text{B}\) is a secondary alcohol.
  • The \(\ce{^{1}H NMR}\) spectrum shows 5 peaks \(\Rightarrow \) 5 hydrogen environments.
  • Chemical shift and splitting patterns information indicate:
  • 1.01 ppm – 1.05 ppm: \(\ce{CH3}\) (next to a \(\ce{CH2}\))
  •   1.65 ppm: \(\ce{CH2}\) (with multiple neighbouring hydrogens)
  •   2.42 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH2}\))
  •   2.46 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH3}\))

CHEMISTRY, M8 2021 HSC 21

Four organic liquids are used in an experiment. The four liquids are

    • hexane
    • hex-1-ene
    • propan-1-ol
    • propanoic acid
  1. State ONE safety concern associated with organic liquids and suggest ONE way to address this safety concern.   (2 marks)

  2. The organic liquids are held separately in four flasks but the flasks are not labelled. Tests were conducted to identify these liquids. The outcomes of the tests are summarised below.   (2 marks)
     
           
    Identify the FOUR liquids.
     
           
  3. What chemical test, other than those used in part (b), could be used to confirm the identification of ONE of the liquids? Include the expected observation in your answer.   (2 marks)

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a.    A safety concern is that the organic liquids are flammable.

To address this, keep substance away from open flames and keep away from ignition sources.
 

b.    Flask 1: propanoic acid  (carboxylic acids can’t be oxidised and are polar)

Flask 2: hex-1-ene (alkenes can be oxidised and are non-polar)

Flask 3: propan-1-ol (primary alcohols can be oxidised and are polar)

Flask 4: hexane (alkanes don’t react with acidified oxidants and are non-polar)
 

c.   Hex-1-ene

  • Could be identified using the bromine water test.
  • The addition of brown bromine water to an alkene causes an addition reaction where the solution changes colours from brown to colourless.

Propanoic acid

  • Could be identified through a neutralisation reaction using `text{Na}_2text{CO}_3`.
  • Effervescent reaction will result.

Propan-1-ol

  • Could be identified through an oxidation reaction using acidified dichromate.
  • The reaction would cause the solution to change from green to orange.
Show Worked Solution

a.    A safety concern is that the organic liquids are flammable.

To address this, keep substance away from open flames and keep away from ignition sources.
 

b.    Flask 1: propanoic acid  (carboxylic acids can’t be oxidised and are polar)

Flask 2: hex-1-ene (alkenes can be oxidised and are non-polar)

Flask 3: propan-1-ol (primary alcohols can be oxidised and are polar)

Flask 4: hexane (alkanes don’t react with acidified oxidants and are non-polar)
 

c.   Hex-1-ene

  • Could be identified using the bromine water test.
  • The addition of brown bromine water to an alkene causes an addition reaction where the solution changes colours from brown to colourless.

Propanoic acid

  • Could be identified through a neutralisation reaction using `text{Na}_2text{CO}_3`.
  • Effervescent reaction will result.

Propan-1-ol

  • Could be identified through an oxidation reaction using acidified dichromate.
  • The reaction would cause the solution to change from green to orange.