Draw an example using a diagram of Carbon, three "Y" substrates and an "L" indicating any leaving group the Sn1 nucleophilic substitution reaction. (3 marks) Which step is fastest (1 mark).
Y3CL -> [Y3C]+ + L-(1 mark)Ligand leaves in first step. Can still show ligand as product anion but no need. Also accept any halide as leaving group.
Step 2 a little harder as not signposted in question. Accept nucleophile, Nuc- or any equivalent.[Y3C]+ +Nuc--> Y3CNuc
The nucleophile attacks the carbocation intermediate. Attack on 2 planes of the compound leading to the same steroisomer produced, but can show the 2 different plane products leading to the same steroisomer product. (2 marks)
The first step of this reaction requires bond breaking (ie endothermic change) to lose the leavin group it is the slower step. By contract a nucleophile will rapidly attack the electrophilic carbon cation to form a new S bond. (1 mark)
JP
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