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CHEMISTRY, M7 2025 HSC 15 MC

Consider the following sequence of reactions.

  • Prop-2-en-1-ol was reacted with hydrogen gas to form liquid \(X\).
  • \(X\) was oxidised, producing liquid \(Y\) that formed bubbles of a gas when reacted with aqueous sodium carbonate.
  • \(Y\) was heated under reflux with methanol and a drop of concentrated sulfuric acid, producing an organic liquid, \(Z\).

This process has been presented in the flow chart below.
 

  

Which option correctly identifies the structures for \(X\), \(Y\) and \(Z\)?
 

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\(D\)

Show Worked Solution
  • The first reaction that occurs is a hydrogenation addition reaction in which prop-2-en-1-ol reacts under a Pd catalyst to produce propan-1-ol.
  • The primary alcohol propan-1-ol then undergoes oxidation to produce the carboxylic acid propanoic acid. This is confirmed as when it is reacted with sodium carbonate, it undergoes an acid-carbonate reaction to produce carbon dixoide which is observed through the bubbles.
  • The third reaction is an estification reaction in which propanic acid reacts with methanol under relfex to produce methyl-propanoate.
  • All three of the correctly drawn compounds can be observed in \(D\)

\(\Rightarrow D\)

CHEMISTRY, M7 2025 HSC 2 MC

Consider this reaction.
 

Which reaction type is shown?

  1. Addition
  2. Oxidation
  3. Reduction
  4. Substitution
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\(B\)

Show Worked Solution
  • The reaction shows butan-2-ol being converted to butanone.
  • This occurs via oxidation where a secondary alcohol is converted to a ketone.

\(\Rightarrow B\)

CHEMISTRY, M7 2024 HSC 5 MC

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

  1. Bromine water
  2. Potassium nitrate solution
  3. Sodium carbonate solution
  4. Acidified potassium permanganate solution
Show Answers Only

\(D\)

Show Worked Solution
  • Can use an oxidation reaction to determine the unknown substance.
  • 2-methylpropan-1-ol is a primary alcohol whereas 2-methylpropan-2-ol is a tertiary alcohol.
  • Hence when undergoing oxidation with acidified potassium permanganate solution, the solution containing the primary alcohol will oxidise and experience a colour change from purple to colourless. The tertiary alcohol will not undergo oxidation.

\(\Rightarrow D\)

CHEMISTRY, M8 2022 HSC 27

A bottle labelled 'propanol' contains one of two isomers of propanol.

  1. Draw the TWO isomers of propanol.   (2 marks)
     

  1. Describe how \( \ce{^13C NMR}\) spectroscopy might be used to identify which isomer is in the bottle.   (2 marks)
  1. Each isomer produces a different product when oxidised.
  2. Write equations to represent the oxidation reactions of the two isomers. Include reaction conditions.   (3 marks)
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a.    Isomer 1:

Isomer 2:

b.   Identifying isomers with \( \ce{^13C NMR} \) spectroscopy:

  • this can be used to identify the isomers in the bottle because they show a different number of signals which helps deduce the carbon environment.
  • Propan-1-ol contains 3 \( \ce{C}\) environments so it would have 3 peaks on a \( \ce{^13C NMR}\) spectrum whereas propan-2-ol only contains 2 \( \ce{C}\) environments (due to symmetry), so it would only have 2 signals on a \( \ce{^13C NMR}\) spectrum.

c.  


  

Show Worked Solution

a.    Isomer 1:

   

Isomer 2:

   

b.   Identifying isomers with \( \ce{^13C NMR} \) spectroscopy:

  • this can be used to identify the isomers in the bottle because they show a different number of signals which helps deduce the carbon environment.
  • Propan-1-ol contains 3 \( \ce{C}\) environments so it would have 3 peaks on a \( \ce{^13C NMR}\) spectrum whereas propan-2-ol only contains 2 \( \ce{C}\) environments (due to symmetry), so it would only have 2 signals on a \( \ce{^13C NMR}\) spectrum.

c.  


  


♦ Mean mark (c) 52%.

CHEMISTRY, M6 2021 HSC 35

A manufacturer requires that its product contains at least 85% v/v ethanol.

The concentration of ethanol in water can be determined by a back titration. Ethanol is first oxidised to ethanoic acid using an excess of acidified potassium dichromate solution.

\(\ce{3C2H5OH($aq$) + 2Cr2O7^2-($aq$) + 16H^+($aq$) ->3CH3COOH($aq$) + 4Cr^3+($aq$) + 11H2O($l$)}\)   

The remaining dichromate ions are reacted with excess iodide ions to produce iodine \(\ce{(I2)}\)

\(\ce{Cr2O7^2-($aq$) + 14H^+($aq$) + 61^-($aq$) -> 2Cr^3+($aq$) + 7H2O($l$) + 3I2($aq$)}\)

The iodine produced is then titrated with sodium thiosulfate \(\ce{(Na2S2O3)}\).

\(\ce{I2($aq$) + 2S2O3^2-($aq$) -> 2I^-($aq$) + S4O6^2-($aq$)}\)   

A 25.0 mL sample of the manufacturer's product was diluted with distilled water to 1.00 L. A 25.0 mL aliquot of the diluted solution was added to 20.0 mL of 0.500 mol L¯1 acidified potassium dichromate solution in a conical flask. Potassium iodide (5.0 g) was added and the solution titrated with 0.900 mol L¯1 sodium thiosulfate. This was repeated three times.

The following results were obtained.
 


 

The density of ethanol is 0.789 g mL¯1.

Does the sample meet the manufacturer's requirements? Support your answer with calculations.   (7 marks)

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\( \text{V}_\text{avg}\  (\ce{Na2S2O3}) \) `= (28.7 + 28.4 + 28.6) / 3`
  `= 28.5666…  text{mL}`
  `= 0.0285666…  text{L}`
   
  •  The first titration is an outlier and so is excluded from the average.

`text{n(Na}_2 text{S}_2 text{O}_3text{)}=\ text{c} xx text{V}= 0.900 xx 0.0285666…= 0.02571  text{mol}`

\( \ce{n(S2O3^2-) = n(Na2S2O3) = 0.02571\ \text{mol}} \)
 

`text{I}_2  text{and S}_2 text{O}_3  ^(2-)\ text{are in a}\ 1 : 2\ text{ratio:}`

`text{n(I}_2text{)} = 1/2 xx  text{n(S}_2 text{O}_3^( 2-) text{)}= 1/2 xx 0.02571= 0.012855\ text{mol}`
 

`text{Excess Cr}_2 text{O}_7^(\ \ 2-)  text{and I}_2  text{are in}\ 1:3\ text{ratio:}`

`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) excess}= 1/3 xx  text{n(I}_2 text{)}= 1/3 xx 0.012855= 0.004285\ text{mol}`

`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) initial}= text{c} xx text{V}= 0.500 xx 20/1000= 0.01\ text{mol}`

`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) reacted with ethanol}`

`= text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) initial}-text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) excess}`

`= 0.01-0.004285`

`= 0.005715\ text{mol}`
 

`text{n(C}_2 text{H}_5 text{OH)}= 3/2 xx text{n(Cr}_2 text{O}_7^(\ \ 2-) text{)}= 3/2 xx 0.005715= 0.0085725\ text{mol}`

`text{m(C}_2 text{H}_5 text{OH)}= text{n} xx text{MM}=0.0085725 xx (2 xx 12.01 + 6 xx 1.008 + 16.00)= 0.3949\ text{g}`
 

  • Thus, 0.3949 g of ethanol is in a diluted 25 mL solution.

Find the mass of ethanol in the original solution:

`text{m(C}_2 text{H}_5 text{OH) original}= 0.3949… xx 1000/25= 15.796…\ text{g}`

`text{D}= text{m} / text{V}\ \ =>\ \ text{V}= text{m} / text{D}`

`text{V(C}_2 text{H}_5 text{OH)}= 15.796 / 0.789= 20.021\ text{mL}`

`text{% (C}_2text{H}_5text{OH)}= text{V(ethanol)} / text{V(sample)}= 20.021 / 25.0= 80.08… %\ text{v}//text{v}`
 

  • Therefore, the product doesn’t meet the manufacturer’s requirement as the concentration is less than 85%.
Show Worked Solution
\( \text{V}_\text{avg}\  (\ce{Na2S2O3}) \) `= (28.7 + 28.4 + 28.6) / 3`
  `= 28.5666…  text{mL}`
  `= 0.0285666…  text{L}`
   
  •  The first titration is an outlier and so is excluded from the average.

`text{n(Na}_2 text{S}_2 text{O}_3text{)}=\ text{c} xx text{V}= 0.900 xx 0.0285666…= 0.02571  text{mol}`

\( \ce{n(S2O3^2-) = n(Na2S2O3) = 0.02571\ \text{mol}} \)
 

`text{I}_2  text{and S}_2 text{O}_3  ^(2-)\ text{are in a}\ 1 : 2\ text{ratio:}`

`text{n(I}_2text{)} = 1/2 xx  text{n(S}_2 text{O}_3^( 2-) text{)}= 1/2 xx 0.02571= 0.012855\ text{mol}`
 

`text{Excess Cr}_2 text{O}_7^(\ \ 2-)  text{and I}_2  text{are in}\ 1:3\ text{ratio:}`

`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) excess}= 1/3 xx  text{n(I}_2 text{)}= 1/3 xx 0.012855= 0.004285\ text{mol}`

`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) initial}= text{c} xx text{V}= 0.500 xx 20/1000= 0.01\ text{mol}`

`text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) reacted with ethanol}`

`= text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) initial}-text{n(Cr}_2 text{O}_7^(\ \ 2-) text{) excess}`

`= 0.01-0.004285`

`= 0.005715\ text{mol}`
 

`text{n(C}_2 text{H}_5 text{OH)}= 3/2 xx text{n(Cr}_2 text{O}_7^(\ \ 2-) text{)}= 3/2 xx 0.005715= 0.0085725\ text{mol}`

`text{m(C}_2 text{H}_5 text{OH)}= text{n} xx text{MM}=0.0085725 xx (2 xx 12.01 + 6 xx 1.008 + 16.00)= 0.3949\ text{g}`
 

  • Thus, 0.3949 g of ethanol is in a diluted 25 mL solution.

Find the mass of ethanol in the original solution:

`text{m(C}_2 text{H}_5 text{OH) original}= 0.3949… xx 1000/25= 15.796…\ text{g}`

`text{D}= text{m} / text{V}\ \ =>\ \ text{V}= text{m} / text{D}`

`text{V(C}_2 text{H}_5 text{OH)}= 15.796 / 0.789= 20.021\ text{mL}`

`text{% (C}_2text{H}_5text{OH)}= text{V(ethanol)} / text{V(sample)}= 20.021 / 25.0= 80.08… %\ text{v}//text{v}`
 

  • Therefore, the product doesn’t meet the manufacturer’s requirement as the concentration is less than 85%.

♦♦♦ Mean mark 39%.

CHEMISTRY, M7 2021 HSC 26

A sequence of chemical reactions, starting with 2-methylprop-1-ene, is shown in the flow chart.

  1. Complete the flow chart by drawing structural formulae for compounds `text{A}`, `text{B}`, `text{C}`, and `text{D}`.   (4 marks)
     
     

     
  2. Reflux is used in the synthesis of methyl 2-methylpropanoate.
  3. Provide TWO reasons for using this technique.   (2 marks)

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a.   Compound A:

Compound B:

Compound C:

Compound D:


 

b.  Reasons for reflux technique:

  • Reflux heats the reaction mixture which increases the average kinetic energy, and thus increases the reaction rate.
  • Heating causes the volatile substances to form vapour molecules. Refluxing uses a condenser to cool the vapour molecules into liquids, and thus retains the substances.
Show Worked Solution

a.   Compound A:

Compound B:

Compound C:

Compound D:


 

b.  Reasons for reflux technique:

  • Reflux heats the reaction mixture which increases the average kinetic energy, and thus increases the reaction rate.
  • Heating causes the volatile substances to form vapour molecules. Refluxing uses a condenser to cool the vapour molecules into liquids, and thus retains the substances.

♦ Mean mark (b) 46%.

CHEMISTRY, M7 2021 HSC 13 MC

A chemist synthesises a substance using the following pathway.

\[\ce{ X ->[{hydration}] {Y} ->[{oxidation}] Z}\]

What are compounds `text{X, Y, Z}` ?

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`C`

Show Worked Solution

By elimination:

  • Hydration reaction is an addition reaction that can only occur on alkenes, thus  `X` = prop-1-ene  (eliminate A and B)
  • `Y` = propan-2-ol
  • The oxidation of secondary alcohol creates a ketone, thus `Z` = propanone

`=> C`