Under suitable conditions, 2-bromobutane reacts with sodium hydroxide to produce a mixture of five products, A, B, C, D and E - AQA - A-Level Chemistry - Question 3 - 2022 - Paper 3
Question 3
Under suitable conditions, 2-bromobutane reacts with sodium hydroxide to produce a mixture of five products, A, B, C, D and E.
Products A, B and C are alkenes.
A i... show full transcript
Worked Solution & Example Answer:Under suitable conditions, 2-bromobutane reacts with sodium hydroxide to produce a mixture of five products, A, B, C, D and E - AQA - A-Level Chemistry - Question 3 - 2022 - Paper 3
Step 1
1. Give the names of the two concurrent mechanisms responsible for the formation of the alkenes and the alcohols.
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Answer
Mechanism to form alkenes: Elimination mechanism (dehydrohalogenation).
Mechanism to form alcohols: Nucleophilic substitution mechanism.
Step 2
2. Define the term stereoisomers.
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Stereoisomers are compounds that have the same molecular formula and structural formula but differ in the spatial arrangement of atoms in space.
Step 3
3. Deduce the name of isomer A. Explain why A does not exhibit stereoisomerism.
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Answer
Name: 2-butene (A can be 1-butene).
Explanation: A does not exhibit stereoisomerism because it lacks a double bond or any element of chirality; hence, it cannot exist in different spatial arrangements.
Step 4
4. Outline the mechanism for the reaction of 2-bromobutane with sodium hydroxide to form alkene A.
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The mechanism involves a base-induced elimination process, where sodium hydroxide acts as a base to facilitate the removal of the halide (Br), leading to the formation of an alkene through dehydrohalogenation.
Step 5
5. Deduce the name of isomer B and the name of isomer C. Explain the origin of the stereoisomerism in B and C.
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Answer
Names: Isomer B is 2-butene and isomer C is cis-2-butene or trans-2-butene (if specified).
Explanation: The stereoisomerism in B and C arises from the presence of a double bond between carbons, where different spatial arrangements (cis and trans) around the double bond lead to variations in properties.
Step 6
6. Draw 3D representations of enantiomers D and E to show how their structures are related.
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Answer
Enantiomers D and E can be represented in 3D using wedge-and-dash notation to show their non-superimposable mirror image structure.
Step 7
7. A student compares the rates of hydrolysis of 1-chlorobutane, 1-bromobutane and 1-iodobutane. State and explain the order in which precipitates appear.
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Answer
Order in which precipitates appear: 1-iodobutane > 1-bromobutane > 1-chlorobutane.
Explanation: The order is primarily due to the bond strengths of the C-X bonds, where C-I bonds are weaker than C-Br and C-Cl. Hence, 1-iodobutane will hydrolyze faster, followed by 1-bromobutane and then 1-chlorobutane.
