c.(15) Rank the following compounds by anion stability. The most stable anion is 1 while the least stable anion is 5.
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1 1. Rank the following series by the trend requested. a.(15) Rank the following compounds by rate of S N 1 reactivity. Molecule that will react the fastest in an S N 1 reaction is 1 while the slowest is 5. I b.(15) Rank the following compounds by radical stability. The most stable radical is 1 while the least stable radical is c.(15) Rank the following compounds by anion stability. The most stable anion is 1 while the least stable anion is 5. C 3 O C 3 N C 3 C 2 (C 3 ) 3 C C 3 CC
2 2. Provide answers for the following halogenation questions. a.(7) Supply the correct monobromination product. 2, hν b.(7) Supply the correct monobromination product. 2, hν c.(7) Supply the correct monobromination product. 2, hν d.(9) Supply all monochlorination products. 2, hν e.(5) In part d would the rate be faster or slower if a monofluorination was attempted instead of the monochlorination? fluorination reaction is faster than a chlorination
3 3.(8 pts each) Supply the correct product for the following reactions. Only supply one structure representing the preferred pathway. Indicate any regio or stereochemistry as required. a. 3 C 2 C C 3 NaOC 3 3 C C 2 C 3 E2 b. 3 C NaOC 3 3 C 3 C 3 C OC 3 c. NaOC(C 3 ) 3 C 3 C 3 E2 d. KI E2 e. AgNO 3 C 3 O OC 3 SN1
4 f. I C 3 SNa S g. NaOC 3 any one of these E1 type h. NaI I i. C 3 N 2 NC 3 excess j. C 3 C 2 O OC 2 C 3 Δ SN1
5 4.(15) Indicate how the following molecule can be synthesized starting with an alkyl halide. Need to show what reagents will produce this product with the quickest rate. O I O 5.(15) The starting material shown, 1-propanol, can be converted to 1-bromopropane with. When 1-propanol is reacted with Na, however, no reaction occurs. O Na no reaction a. Draw the mechanism indicating the S N 2 pathway that creates 1-bromopropane with. In order to do this use the appropriate arrow conventions which are used in organic chemistry to show a mechanism, thus indicating what is the nucleophile, what is the electrophile and what is the leaving group. O O 2 the first step MUST protonate the hydroxy group in order to make a good leaving group (the hydroxide will NEVER be a leaving group in an S N 2 reaction) and then in the second step the nucleophilic bromide displaces the water to give the product b. Why does the reaction not work when Na is used instead? If no acid is used then the hydroxy cannot be protonated and hence will never leave as a leaving group (strong bases are horrible leaving groups)
6 6. When converting chlorocyclohexane to bromocyclohexane a student should consider whether the reaction occurs by an S N 1 or S N 2 pathway. Compare the RATE of each of these pathways if the following changes are made to the reaction. Therefore for each question (a-f) answer by stating if that change will make either an S N 1 or S N 2 pathway faster, slower, or no change compared to the reaction shown. Each question has two answers: what happens to the S N 1 and what happens to the S N 2 pathways with the indicated change. Na C 3 O a.(10) The chlorocyclohexane is changed to 1-chloro-1-methylcyclohexane. : the rate is slowed ( never occurs at tertiary carbon) SN1: the rate is faster (tertiary carbon reacts faster than secondary carbon) b.(10) The Chlorine is replaced with an Iodine. : rate is faster, better leaving group SN1: the rate is faster, better leaving group c.(10) Sodium bromide is replaced with sodium iodide. : rate is faster, more polarizable nucleophile SN1: no change d.(10) The temperature the reaction is run at is raised. : rate is faster SN1: rate is faster e.(10) Change chlorocyclohexane to the starting material shown. : small change, slightly faster SN1: rate is dramatically faster, formed carbocation is resonance stabilized f.(10) The solvent is changed from methanol to hexanes. : small change SN1: rate is dramatically slower, less polar solvent will not stabilize carbocation intermediate
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