QUESTION 146 (5 marks)
2-methylpropan-2-ol reacts with HBr to form a haloalkane and water. The haloalkane is then heated with ethanolic NaOH to form an alkene.
(a) Write the condensed structure and IUPAC name of the haloalkane. [2 marks]
(b) Name the alkene and classify its formation. [2 marks]
(c) Predict the major organic product of acid-catalysed hydration of this alkene. [1 marks]
Practice marking scheme
Answer
2-bromo-2-methylpropane, ; 2-methylpropene by elimination; hydration gives 2-methylpropan-2-ol.
Working
The first reaction replaces OH with Br: . Elimination from the haloalkane gives . All adjacent methyl groups are equivalent. Markovnikov hydration places OH on the substituted carbon, restoring the tertiary alcohol.
Marking criteria
- Gives correct haloalkane structure (CH₃)₃CBr. [1 mark]
- Names it 2-bromo-2-methylpropane. [1 mark]
- Names the alkene 2-methylpropene. [1 mark]
- Classifies alkene formation as elimination. [1 mark]
- Identifies the major hydration product as 2-methylpropan-2-ol. [1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabusCompare your working with the guide above.