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Ethanol reacts with hydrogen cyanide, in the presence of a small amount of NaCN, as shown - CIE - A-Level Chemistry - Question 3 - 2015 - Paper 1
Question 3
Ethanol reacts with hydrogen cyanide, in the presence of a small amount of NaCN, as shown. CH₃CHO + HCN → CH₃(CH(OH)CN) (a) Use bond energies from the Data Booklet... show full transcript
Worked Solution & Example Answer:Ethanol reacts with hydrogen cyanide, in the presence of a small amount of NaCN, as shown - CIE - A-Level Chemistry - Question 3 - 2015 - Paper 1
Step 1
Use bond energies to calculate the enthalpy change
Answer
To calculate the enthalpy change for the reaction, we need to determine the total energy required for bond breaking and the total energy released from bond forming.
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Bond Breaking:
- C=O bond in ethanal: 740 kJ
- C-H bond in HCN: 410 kJ
- Total energy for breaking bonds:
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Bond Forming:
- C-C bond in the product: 350 kJ
- C-O bond in the product: 460 kJ
- O-H bond in the product: 1170 kJ
- Total energy from forming bonds:
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Enthalpy change (ΔH): Thus, the enthalpy change for the reaction is .
Step 2
Explain the meanings of the terms stereoisomerism and chiral centre
Answer
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Stereoisomerism: Stereoisomerism refers to molecules that have the same molecular formula and the same structural formula, but different spatial arrangements of atoms. This type of isomerism arises when the orientation of the atoms in space is different, which can lead to different chemical properties.
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Chiral Centre: A chiral centre is an atom, usually carbon, that has four different atoms or groups attached to it. This uniqueness allows for the existence of two non-superimposable mirror images, known as enantiomers.
Step 3
Suggest why the two optical isomers are produced in equal amounts by this reaction
Answer
The two optical isomers are produced in equal amounts because the reaction involves a planar carbonyl group, which can be attacked equally from either side by the nucleophile (HCN). As a result, there is an equal chance of forming each enantiomer.
Step 4
Complete the diagram showing the mechanism of this reaction
Answer
- Begin with the carbonyl functional group (C=O) in the reactant.
- Show a lone pair on the nitrogen atom of HCN attacking the carbonyl carbon, creating an intermediate with partial charges.
- Include curly arrows indicating the movement of electrons - from the lone pair to the carbonyl carbon and from the C=O bond to the oxygen.
- Show the formation of the intermediate: N-C(OH)(CN) with appropriate charges.
- Finally, indicate the conversion back into the product along with any regeneration of NaCN as a catalyst.
Step 5
Explain the role of NaCN in this reaction
Answer
NaCN acts as a catalyst in this reaction. It provides the cyanide ion (CN⁻), which is a good nucleophile. By facilitating the nucleophilic attack on the carbonyl carbon of ethanal, it enhances the reaction rate without being consumed in the overall process. Additionally, the regeneration of NaCN ensures that it continues to participate in subsequent reactions.