The boiling points for two series of compounds are listed. --- 0 WORK AREA LINES (style=lined) --- --- 12 WORK AREA LINES (style=lined) --- a. b. The alcohols have higher boiling points than amines of the same chain length. → Both alcohols and amines have polar hydrogen bonding as a result of the \(\ce{OH}\) and \(\ce{NH2}\) functional groups respectively. → Oxygen molecules have higher electronegativity than nitrogen molecules making the hydrogen bonding in the alcohols significantly stronger than the hydrogen bonding in amines. → Since the dispersion forces in amines and alcohols of the same chain length are very similar, the difference in the strength of the intermolecular bonding is dependent on the strength of the hydrogen bonding. → Therefore, a larger amount of thermal energy is required to separate the alcohol molecules resulting in higher boiling points than amines. As the chain length of the alcohols and amines increase, the boiling points also increase. → When the chain length increases, the number of electrons in the molecules also increase which corresponds to a larger number of dispersion forces between neighbouring molecules. → The stronger dispersion forces between the molecules increase the overall strength of the intermolecular forces in both the alcohols and amines therefore leading to higher boiling points as chain length increases. a. b. The alcohols have higher boiling points than amines of the same chain length. → Both alcohols and amines have polar hydrogen bonding as a result of the \(\ce{OH}\) and \(\ce{NH2}\) functional groups respectively. → Oxygen molecules have higher electronegativity than nitrogen molecules making the hydrogen bonding in the alcohols significantly stronger than the hydrogen bonding in amines. → Since the dispersion forces in amines and alcohols of the same chain length are very similar, the difference in the strength of the intermolecular bonding is dependent on the strength of the hydrogen bonding. → Therefore, a larger amount of thermal energy is required to separate the alcohol molecules resulting in higher boiling points than amines. As the chain length of the alcohols and amines increase, the boiling points also increase. → When the chain length increases, the number of electrons in the molecules also increase which corresponds to a larger number of dispersion forces between neighbouring molecules. → The stronger dispersion forces between the molecules increase the overall strength of the intermolecular forces in both the alcohols and amines therefore leading to higher boiling points as chain length increases.
CHEMISTRY, M7 2016 HSC 3 MC
What is the molecular formula of pentanoic acid?
- \(\ce{C_5H_9O}\)
- \(\ce{C_5H_10O}\)
- \(\ce{C_5H_10O_2}\)
- \(\ce{C_5H_11O_2}\)
CHEMISTRY, M6 2015 HSC 26
A sodium hydroxide solution was titrated against citric acid \(\ce{(C6H8O7)}\) which is triprotic.
- Draw the structural formula of citric acid (2-hydroxypropane-1,2,3-tricarboxylic acid). (1 mark)
- How could a computer-based technology be used to identify the equivalence point of this titration? (2 marks)
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- The sodium hydroxide solution was titrated against 25.0 mL samples of 0.100 mol L ¯1 citric acid. The average volume of sodium hydroxide used was 41.50 mL.
- Calculate the concentration of the sodium hydroxide solution. (4 marks)
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a.
b. Technology solution
→ A digital pH probe could be placed in the flask and used to collect data that plots the pH of the solution against the volume of sodium hydroxide added.
→ The equivalence point would be identified by a steep rise in the pH on the graph.
c. \(\ce{C6H8O7 + 3NaOH -> C6H5O7Na3 + 3H2O}\)
\(\ce{n(C6H8O7) = c \times V = 0.100 \times 0.0250 = 0.00250 mol}\)
\(\ce{n(NaOH) = 3 \times 0.00250 = 0.00750 mol}\)
\[\ce{[NaOH] = \frac{n}{V} = \frac{0.00750}{0.04150} = 0.181 mol L^{-1} (3 sig fig)}\]
Show Worked Solution
a.
♦ Mean mark (a) 46%.
b. Technology solution
→ A digital pH probe could be placed in the flask and used to collect data that plots the pH of the solution against the volume of sodium hydroxide added.
→ The equivalence point would be identified by a steep rise in the pH on the graph.
♦ Mean mark (b) 42%.
c. \(\ce{C6H8O7 + 3NaOH -> C6H5O7Na3 + 3H2O}\)
\(\ce{n(C6H8O7) = c \times V = 0.100 \times 0.0250 = 0.00250 mol}\)
\(\ce{n(NaOH) = 3 \times 0.00250 = 0.00750 mol}\)
\[\ce{[NaOH] = \frac{n}{V} = \frac{0.00750}{0.04150} = 0.181 mol L^{-1} (3 sig fig)}\]
Mean mark (c) 56%.
CHEMISTRY, M7 2019 HSC 21
- The structural formula for 2-methylpropan-2-ol is shown in the table.
- Draw one structural isomer of this alcohol and state its name. (2 marks)
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- The structural formulae for two compounds are shown below.
- Why are these two compounds classed as functional group isomers? (2 marks)
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- A chemical test is required to distinguish between the isomers in part (b).
- Identify a suitable test and explain the expected observations. (3 marks)
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a. Successful answers should have one of the following:
b. Functional Group isomers
→ Both isomers have the same number and type of atoms, but they have different arrangements of those atoms and therefore have different functional groups.
→ Isomer A has a ketone functional group, while isomer B has an aldehyde functional group.
c. Tollens’ Test:
→ The Tollens’ test can be used to differentiate between Isomer A (a ketone) and Isomer B (an aldehyde).
→ Isomer B will be readily oxidised to a carboxylic acid, whereas isomer A will not.
→ As a result, Isomer B will reduce the silver-ions in Tollens’ reagent to form a silver mirror inside the test tube, while Isomer A will not react.
Show Worked Solution
a. Successful answers should have one of the following:
b. Functional Group isomers
→ Both isomers have the same number and type of atoms, but they have different arrangements of those atoms and therefore have different functional groups.
→ Isomer A has a ketone functional group, while isomer B has an aldehyde functional group.
c. Tollens’ Test:
→ The Tollens’ test can be used to differentiate between Isomer A (a ketone) and Isomer B (an aldehyde).
→ Isomer B will be readily oxidised to a carboxylic acid, whereas isomer A will not.
→ As a result, Isomer B will reduce the silver-ions in Tollens’ reagent to form a silver mirror inside the test tube, while Isomer A will not react.
♦♦ Mean mark (c) 39%.
CHEMISTRY, M7 2019 HSC 10 MC
Which class of organic compound must contain at least three carbon atoms?
- Aldehydes
- Alkenes
- Carboxylic acids
- Ketones
CHEMISTRY, M7 2020 HSC 3 MC
Which of the following compounds is the most basic?
- Ethane
- Ethanol
- Ethanamine
- Ethyl ethanoate