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Mock Exam III
CH 235-‐OQ Summer 2015
CHEM 101 | 5:30-‐6:30 pm
Ques%on 1
Select the weaker nucleophile from each of the following pairs: (a) NH3 or NH2
-‐ (b) Cl-‐ or Br-‐
Anion is a beDer Nu-‐ than the neutral species
Nucleophilicity increases with halides as ion size increases
Ques%on 2 Which of the following would react slowest under SN2 reac%on condi%ons?
SN2 reac%vity: 1° > 2° > 3°
As size of halide increases leaving group ability increases
Ques%on 3 Explain why the alkyl halide shown below will not undergo either SN1 or SN2 reac%ons.
3° alkyl halide SN2 highly unlikely
What is the hybridiza%on and geometry of this carboca%on?
• sp3 and should be trigonal planar
This molecule cannot adopt a trigonal planar geometry due to the bulkiness of the bridgehead
Ques%on 4 For each of the following reac%on, indicate whether the reac%on will proceed through an SN1 or an SN2 mechanism and give the final product. Indicate stereochemistry where necessary.
3° alkyl halide + polar pro%c solvent + weak base
SN1
racemic mixture
2° alkyl halide + polar apro%c solvent + strong base
SN2
inversion of stereochemistry
Ques%on 5 Which of the following reac%ons will be faster (I) or (II). What is the type of elimina%on that will occur and product of that reac%on?
3° alkyl halide + polar pro%c solvent + weak base
E1
2° alkyl halide + polar pro%c solvent + weak base
E1
E1 reac%vity: 1° < 2° < 3°
more stable carboca%on -‐ FASTER
Ques%on 6 What are the subs%tu%on and elimina%on products of the following reac%on? Show the full mechanism for each product. (Hint: there are four.)
2° alkyl halide + polar pro%c solvent + weak base
E1/SN1
E1
SN1
Ques%on 7 Explain the difference in the elimina%on products produced from the two reac%ons with the same star%ng material.
3° alkyl halide + polar pro%c solvent + weak base
E1
3° alkyl halide + polar apro%c solvent + strong base
E2
The less subs%tuted double bond will be formed because the an%-‐coplanar orienta%on is needed
Carboca%on is formed so proton removed to form most stable double bond
Ques%on 8 The following reac%on displays second order kine%cs. What would be the major product formed?
Second order kine%cs = SN2/E2 2° alkyl halide + strong base + no heat
SN2
Ques%on 9 The labels of 4 boDles in the lab fell off. We don’t know which label goes with which chemical. Below are the IR spectra of the 4 compounds. Match the compounds with the correct chemical name. Please give a brief reason for the match. a) 3-‐heptanone b) heptanenitrile c) ethyl heptanoate d) 2-‐heptanol
Ques%on 9
C=O
sp3 C-‐H sp2 C-‐O
sp3 C-‐O
Ques%on 9
O-‐H sp3 C-‐H
Ques%on 9
CΞN
sp3 C-‐H
Ques%on 9
C=O sp3 C-‐H
Radicals
Radical Chlorina%on
• Exothermic reac%on • Proceeds very quickly • Early transi%on state à liDle to no regioselec%vity
57% 43%
Radicals
Radical Bromina%on
• Endothermic reac%on • Proceeds slowly • Late transi%on state à regioselec%vity
92% 8%
Radicals
NBS • Brominates the allylic carbon • Due to radical intermediates rearrangment possible
Radicals
Peroxides • Effects hydrohalogena%on with bromine only, not chlorine • Causes an%-‐Markovnikov addi%on of Nu-‐
Ques%ons?