CH3(CH2)5CHBrCH3 exists as a pair of stereoisomers.

1. (a)Draw the three-dimensional structures of the two stereoisomers of CH3(CH2)5CHBrCH3. R can be used to represent CH3(CH2)5[2]
2. (b)A sample of CH3(CH2)5CHBrCH3 reacts with NaOH to make CH3(CH2)5CH(OH)CH3 in an SN1 mechanism. Complete Fig. 4.1 to show the mechanism for the reaction of CH3(CH2)5CHBrCH3 and NaOH. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate. H CH3(CH2)5-C-Br CH3 H CH3(CH2)5-C-OH CH3 Fig. 4.1[3]
3. (c)Separate samples of CH3(CH2)5CHBrCH3, CH3(CH2)5CH(OH)CH3 and CH3(CH2)5CHCH2 are tested with different reagents. Complete Table 4.1. If no reaction occurs, write x in the relevant box. Table 4.1 reagent added observation with CH3(CH2)5CHBrCH3 observation with CH3(CH2)5CH(OH)CH3 observation with CH3(CH2)5CHCH2 Br2(l) in the dark PCl5(s) AgNO3(aq)[3]
4. (d)(i)CH3(CH2)5CHBrCH3 is heated with D to produce three different molecules, E, F and G. [Figure 4.2 showing structures E, F, G] Name the type of reaction.[1]
5. (d)(ii)Identify D and the conditions used.[1]
6. (e)(i)Both σ and π bonds are present in a molecule of E as a result of different types of hybridisation in the carbon atoms. Complete Table 4.2 to show the number of carbon atoms with each type of hybridisation in a molecule of E. Table 4.2 number of carbon atoms sp hybridised sp2 hybridised sp3 hybridised E[2]
7. (e)(ii)Describe the essential feature of an unbranched hydrocarbon that causes its molecules to show stereoisomerism. Explain how this feature leads to stereoisomerism.[3]

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