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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.
Answer
Mechanism to form alkenes: Elimination (E2) mechanism, which allows the formation of alkenes by the removal of a halogen and hydrogen from adjacent carbons.
Mechanism to form alcohols: Nucleophilic substitution (SN2) mechanism, where the hydroxide ion substitutes for the bromide ion.
Step 2
Step 3
3. Deduce the name of isomer A. Explain why A does not exhibit stereoisomerism.
Answer
Name: 2-butene.
Explanation: Isomer A does not exhibit stereoisomerism because it has no double bond between carbons that can create geometric isomers. Additionally, it does not have chiral centers to form enantiomers.
Step 4
4. Outline the mechanism for the reaction of 2-bromobutane with sodium hydroxide to form alkene A.
Answer
The mechanism involves a base-induced elimination reaction.
- Sodium hydroxide ( ext{NaOH}) acts as a base, abstracting a proton (H) from the β-carbon.
- The bromine (Br) leaves from the α-carbon, resulting in the formation of a double bond between the α and β-carbons, yielding alkene A.
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.
Answer
Names: Isomer B is 2-butene and isomer C is cis-2-butene.
Explanation: The stereoisomerism in B and C arises from the presence of a double bond between the carbons, which restricts rotation and leads to different spatial arrangements of the substituents around the double bond.
Step 6
Step 7
7. State and explain the order in which precipitates appear.
Answer
Order in which precipitates appear: 1-iodobutane → 1-bromobutane → 1-chlorobutane.
Explanation: This order is due to the reactivity of the halides. Iodine is a better leaving group than bromine, which is better than chlorine, resulting in faster hydrolysis rates. Thus, the precipitate will form fastest with 1-iodobutane.