FOURTH EXAMINATION. (1)..20 pts... (2)..24 pts... (3)..18 pts... (4)..12 pts... (5)..12 pts... (6).. 6 pts... (7).. 8 pts... Bonus 4 pts...

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1 Name: CHEM 331 FURTH EXAMINATIN All answers should be written on the exam in the spaces provided. Clearly indicate your answers in the spaces provided; if I have to guess as to what or where your answer is, it is wrong. Where applicable, outline the logic or mystical principle you used to arrive at your answer, as partial credit may be awarded for correct approaches. Work pages are provided at the end of the exam. Make sure your name is on each page of the examination; the management is not responsible for lost pages. You may assume that standard work-up conditions ( i.e. those required to obtain a stable, neutral product, like aqueous acid as example) follow each reaction on the exam. Clearly indicate stereochemistry where appropriate. Be careful not to show stereochemistry where none exists. Please take this test wisely for the time provided. You are strongly advised to read it through completely before you begin. (1)..20 pts... (2)..24 pts... (3)..18 pts... (4)..12 pts... (5)..12 pts... (6).. 6 pts... (7).. 8 pts... Bonus 4 pts... TTAL (100 pts)..... Percentage.

2 1. [20 pts] Answer the following questions about radical reactions: a. [6 pts] Label each of the following steps as Initiation, propagation or termination + termination CH 4 + Cl CH 3 + H-Cl propagation Cl 2 2 Cl initiation b. [6 pts] There are three constitutional isomers of molecular formula C 6 H 14 shown below. Assign the structures to their identities (A C) based on the information below: When treated with Cl 2 and light, isomer A gives a mixture of five monochlorinated products. Under the same conditions, isomer B gives a mixture of three monochlorinated products. Isomer C produces two different monochlorinated products when reacted under these conditions. Assign structural formulas to isomers A, B and C. Isomer A: Isomer B: Isomer C: c. [4 pts] Draw the most stable radical that could form upon hydrogen atom abstraction from the following molecule. d. [4 pts] Explain why bromine is regioselective for radical bromination, as shown in the reaction below: 2 light the most stable radical forms on the benzylic position

3 2. [24 pts] Predict whether substitution and/or elimination will predominate in the following reactions by: a. Identify what type of alkyl halide is present (1, 2, 3, allylic or benzylic if so) b. Identify whether you have a strong or weak base, strong or weak nucleophile, or what the solvent indicates for the reaction. c. Identify the most likely mechanism(s) (SN1, SN2, E1, E2) d. Draw the structures of the resulting product(s) and do not forget stereochemistry in your product if necessary. (i) H 2 H 2º benzylic halide weak nuc/weak base SN1 (E1 not possible) (ii) NaH H 1º halide strong nuc/strong base SN2 only on 1º systems (E2 only with SNNB) (iii) Ms NaCH 3 CH 3 2º halide strong nuc/strong base SN2 and E2 (iv) KtBu tbuh 3º halide Strong, Non-Nucleophilic Base Presence of tbuh solvent - Hofmann Product

4 3. [18 pts] Answer the following general questions about Substitutions and Eliminations (Circle the correct answer): a. Which of the following isomers reacts fastest in an S N 2-type reaction? i. 2-chloro-4,4-dimethylpentane ii. 1-chloro-3,3-dimethylpentane iii. 3-chloro-2,3-dimethylpentane b. What is the effect on the rate of the E2 reaction of tert-butyl bromide, (CH 3 ) 3 C, and NaH if the concentration of the base is halved? iv. Faster v. Slower vi. No Change c. Which would be the best alkyl halide to react with sodium methoxide, NaCH 3, to form the following ether: CH 3 i. 1-fluoro-4-methylpentane ii. 1-iodo-4-methylhexane iii. 1-bromo-4-methylpentane d. What is the effect on the S N 2 reaction of 1-bromobutane if the solvent is switched from dimethylsulfoxide (DMS) to methanol (CH 3 H)? i. Faster ii. Slower iii. No Change e. Which produces the most selective reaction and highest yielding product formation of 2-pentene from 2-bromopentane? 1. NaCH 3 in DMS ii. iii. KtBu (KC(CH 3 ) 3 ) in DMS KtBu (KC(CH 3 ) 3 ) in tbuh f. What is the effect on the S N 1 reaction of 3-bromo-3-methylpentane with water, H 2, if the concentration of the nucleophile is doubled? i. Faster ii. Slower iii. No Change

5 4. [12 pts] For each of the following reactions, draw the starting reagent or product(s) or provide the reagent necessary for each one-step conversion to occur. Assume standard conditions to produce stable, neutral product formations. a. HCl Cl b. H H c. H SCl 2, pyridine Cl d. (CH 3 ) 2 CuLi CH 3 5. [12 pts] Synthesis: Devise a short sequence (2 steps) of reactions for the construction of the following product. Draw the intermediate product that may form in your sequence to receive full credit. a. H (CH 3 CH 2 ) 2 CuLi b. Mg Mg H 2

6 6. [6 pts] Name the following compound according to IUPAC rules: 5-bromo-7-cyclopropyl-3-ethyl-6-isobutyl-1-decene 7. [16 pts] Short Answer Questions: Answer 2 of the following 3 questions: a. Explain why 1-bromo-3,4-dimethyloctane does not undergo S N 1 type reactions. Primary alkyl halides cannot form a carbocation intermediate and this compound is primary. b. Explain why S N 2 reactions proceed slower in a protic solvents like methanol, CH 3 H. Protic solvents surround the nucleophilic preventing it from being able to react with the electrophilic carbon of the alkyl halide. c. Front side attack is disfavored for an S N 2-type substitution reaction because the leaving group and the nucleophile both carry negative charges and thus repel each other. What is the other reason front side attack is disfavored? Sterics not enough physical space for a nucleophile to attack while the leaving group is leaving. Bonus: [4 pts] Explain why trans-1-bromo-2-methylcyclohexane yields the Hofmann product 3- methylcyclohexene on treatment with any strong base. H CH 3 For an E2 reaction on a cyclohexane ring, anitperiplanar means that the leaving halide must be trans diaxial to a H atom. The H atom that would lead to the Zaitsev product is cis, not trans to the bromide. Zaitsev cannot form. Hofmann is the only elimination product able to form.