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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
Step 1
Step 2
Step 3
Answer
Name: But-2-ene (or 2-butene depending on context)
Explanation: A does not exhibit stereoisomerism because it does not have a C=C double bond capable of existing in cis/trans forms due to the lack of different substituents on the double-bonded carbons.
Step 4
Answer
The hydroxide ion acts as a base and removes a proton from a hydrogen atom adjacent to the bromine-bearing carbon (deprotonation).
This results in the formation of a double bond between the carbon atoms, while the bromide ion is expelled as a leaving group.
Step 5
Answer
Names: B is cis-2-butene, C is trans-2-butene.
Explanation: The origin of stereoisomerism in B and C arises from the restricted rotation around the C=C double bond, which leads to different spatial arrangements of the substituents around the double bond.
Step 6
Step 7
Answer
Order: 1-iodobutane > 1-bromobutane > 1-chlorobutane.
Explanation: The order of hydrolysis rates is based on the reactivity of the halogen; iodine is the best leaving group, followed by bromine and then chlorine. Therefore, 1-iodobutane hydrolyzes fastest, leading to the quickest formation of the precipitate.
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