Homework problems Chapters 6 and Give the curved-arrow formalism for the following reaction: CH 3 OH + H 2 C CH +

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1 omework problems hapters 6 and 7 1. Give the curved-arrow formalism for the following reaction: : 3 - : 2 : In each of the following sets, arrange the compounds in order of decreasing pka and explain your reasoning. As pka decreases, acidity increases. ence, Relative acidity: 3 2 < l 2 2 < l 2 2 S Relative acidity: 3 < 3 2 < 3 2 ( 3 ) (c) Relative acidity: 3 2 < ( 3 ) 2 N 2 2 < ( 3 ) 3 N (d) Relative acidity: < l < 3 2 < 3 2 l l In each case, the order of acidity is determined by inductive effects. 3. Rank the following compounds in order of increasing basicity: N - 3 N < < The donor atoms are from the same period (row) of the periodic table, hence electronegativity wins, and the less electronegative atom is more basic. For bases with identical donating atoms, positive charge on the molecule decreases basicity. - 1

2 4. In each pair, choose the species that reacts more rapidly with methyl bromide in ethanol solvent, and explain your choices. 2 5 or 2 5 When the attacking atom is the same, nucleophilicity roughly parallels onsted basicity. Since ethoxide ( 2 5 ) is a much stronger base than ethanol, it will react more rapidly with methyl bromide via S N 2 mechanism. F - or I - In protic solvents, nucleophilicity increases with increasing atomic number within a group of the periodic table. Thus, I - is the better nucleophile and will react more rapidly with methyl bromide. 5. In each pair, choose the combination of nucleophile and solvent that would give the faster S N 2 reaction with ethyl iodide Na Ac in formamide, or Na Ac in DMF Formamide [()N 2 ] is a protic polar solvent while DMF [()NMe 2 ] is an aprotic polar solvent.a protic solvent consists of molecules than can act as hydrogen-bond donors. Polar protic solvents slow down S N 2 reaction by stabilizing the nucleophile via hydrogen bonds to the nucleophile and hence decreasing its reactivity towards electrophiles. Since the nucleophilicity of an anion is much greater in polar aprotic solvents, Na Ac in DMF would give the faster S N 2 reaction with ethyl iodide. Na Ac in acetic acid, or Na Ac in DMF. Acetic acid is a protic solvent. Thus, Na Ac in DMF would give the faster S N 2 reaction with ethyl iodide for the same reason as above. 6. hoose the alkyl halide(s) from the following list of 6 13 isomers that meet each criterion below: (1) 1-bromohexane (2) 3-bromo-3-methylpentane (3) 1-bromo-2,2-dimethylbutane (4) 3-bromo-2-methylpentane (5) 2-bromo-3methylpentane 2

3 (1) (2) (3) (4) (5) (c) the compound that gives the fastest S N 2 reaction with sodium methoxide 1-omohexane gives the fastest S N 2 reaction with sodium methoxide because primary alkyl halides undergo S N 2 substitution with good nucleophiles (such as methoxide), whether they are good or poor bases. 1-omohexane reacts faster than 1-bromo-2,2-dimethylbutane (which is also a primary alkyl halide) due to decreased steric crowding of the transition state. the compound that is least reactive to sodium methoxide in methanol As a primary alkyl halide that contains no - bonds adjacent to the electrophilic carbon center, 1-bromo-2,2- dimethylbutane can only undergo S N 2 substitution and cannot undergo E2 elimination. Similarly, it cannot undergo S N 1 or E1 reactions since primary carbocations are too high in energy whereas secondary and tertiary alkyl halides readily dissociate in protic solvents due to stabilization of the resulting carbocation by solvation (via hydrogen bonding). Since S N 2 reactions are slowed down in protic solvents due to solvation of the nucleophile and the transition state for S N 2 reaction is more crowded for 1- bromo-2,2-dimethylbutane than for 1-bromobutane, it is least reactive to sodium methoxide in methanol. the compound(s) that give only one alkene in the E2 reaction Anti periplanar geometry, meaning that all four reacting atoms- the hydrogen, the two carbons, and the leaving group- lie in the same plane, is required for E2 elimination. nly 1-bromobutane will give only one alkene because it contains only primary - bonds adjacent to the electrophilic carbon. The remaining alkenes either have 3

4 no - bonds adjacent to the electrophilic carbon or more than one type of - bond (primary, secondary, or tertiary). (d) the compound(s) that give an E2 but no S N 2 reaction with sodium methoxide in methanol Tertiary alkyl halides are so hindered that they never undergo S N 2 substitution. E2 elimination occurs when a good base, such as methoxide, is used. Thus, only 3- bromo-3-methylpentane undergoes no S N 2 reaction. Remember, secondary alkyl halides undergo both S N 2 and E2 elimination, and a mixture of products often results. (e) the compound(s) that undergo an S N 1 reaction to give rearranged products When 3-bromo-2-methylpentane and 2-bromo-3- methylpentane dissociate bromide, they form secondary carbocations. The latter can undergo a hydride shift to give a more stable tertiary carbocation, which then reacts with a nucleophile to produce a rearranged product Nu - 3 hydride shift Nu Nu Nu (f) the compound that gives the fastest S N 1 reaction 3-bromo-3-methylpentane gives the fastest S N 1 reaction because it forms the most stable carbocation, and carbocation formation is rate determining in S N 1 reactions. 7. Give the products expected when 1-bromo-3-methylbutane reacts with the following reagents: KI in aqueous acetone 4

5 KI 2 /acetone I Under the above conditions, which are essentially neutral, primary alkyl halides undergo S N 2 substitution with good nucleophiles. K in aqueous ethanol K 2 /ethanol Under basic conditions, primary alkyl halides undergo S N 2 substitution with good nucleophiles, whether they are good or poor bases. (c) K t-bu - on t-bu K( 3 ) 3 t-butanol Under basic conditions, primary alkyl halides undergo E2 elimination with good bases/poor nucleophiles. 8. Give the products expected when 2-bromo-2-methylhexane reacts with the following reagents: KI in aqueous acetone I KI 2 /acetone A B This is a tertiary halide in a polar protic solvent mixture (water/acetone) containing a good nucleophile that is a weak base (I - ). The polar protic solvent mixture promotes carbocation 5

6 formation, and hence S N 1 and E1 reactions are observed. The S N 1 products are the alkyl iodide (A) and the alcohol (B). A will predominate, the more iodide is present since it is a much better nucleophile than water. Some alkenes will also be formed since E1 reaction always accompanies the S N 1 reaction of an alkyl halide with β-hydrogens. No rearrangement products are predicted because the carbocation intermediate is tertiary. Sodium ethoxide in ethanol This tertiary alkyl halide is subjected to a strong base in a protic solvent. Under these conditions, tertiary alkyl centers are so hindered that they never undergo S N 2 substitution but the E2 elimination can occur. While protic solvents promote S N 1 and E1 reactions, high concentration of base will favor E2 elimination since rates of S N 1 and E1 reactions are independent of the concentration of base. The alkene products are shown below. Note that the more substituted alkene D will predominate according to Zaitsev s rule. Na 2 5 ethanol D If the concentration of ethoxide is low, then S N 1 and E1 reactions will compete more effectively with E2 elimination. E1 elimination will give the same alkene products as above while S N 1 result in the ether product. (c) 1:1 ethanol-water Neither S N 2 nor E2 reactions can occur for tertiary alkyl halides in polar protic solvents in the presence of a poor nucleophile and a weak base. owever, polar protic solvents promote S N 1 and E1 reactions, and hence these reactions are observed. The substitution products would predominate because of high concentration of nucleophile, and the alcohol will be the predominant substitution product since water is the better nucleophile. The more substituted alkene will predominate according to Zaitsev s rule. 6

7 ethanol/ 2 (1:1) 2 5 major major minor minor 9. Rank the following compounds in order of increasing S N 2 reaction rate with KI in acetone. methyl bromide sec-butyl bromide (c) 3-(bromomethyl)-3-methylpentane (d) 1-bromopentane (e) 1-bromo-2-methylpentane (c) (d) (e) < (c) < (e) < (d) < S N 2 reactions can occur only at relatively unhindered sites. Thus, methyl halides are by far the most reactive in S N 2 reactions, followed by primary alkyl halides. anching one carbon removed from the leaving group greatly slows nucleophilic displacement, with steric hindrance increasing as branching increases. Thus, (c) reacts more slowly than (e). Alkyl branching next to the leaving slows the reaction even more, and hence only a few simple secondary alkyl halides are useful substrates in this reaction while tertiary alkyl halides do not undergo S N 2 reactions. 10. Give all of the product(s) expected, with pertinent stereochemistry, when each of the following compounds reacts with sodium ethoxide in ethanol. (R)-2-bromopentane 7

8 This secondary alkyl halide is subjected to a strong base/strong nucleophile in a protic solvent. Under these conditions, the S N 2 reaction is retarded by branching at the electrophilic carbon center and by solvation of the nucleophile but the E2 elimination can occur. Although protic solvents promote SN1 and E1 reactions, high concentration of base will favor E2 elimination since rates of S N 1 and E1 reactions are independent of the concentration of base. Thus, the products are the following two alkenes: 3 E and Z, with E predominanting major minor The ether product resulting from the minor S N 2 reaction pathway will show complete inversion of configuration at the electrophilic carbon stereocenter. Under basic conditions, primary alkyl centers undergo S N 2 substitution with good nucleophiles whether they are good or poor bases. Thus, ethoxide will be substituted for bromide with complete inversion of configuration at the electrophilic carbon stereocenter ethanol D D S-configuration R- configuration 11. Which of the following alkyl halides can give only one alkene, and which can give a mixture of alkene in the E2 reaction? 8

9 (c) B: B: - = Both compounds and can only one alkene product in the E2 reaction: because there is only one - bond adjacent to the - bond; because abstraction of either axial hydrogen by base will lead to the same product. In contrast, (c) contains primary and secondary - bonds adjacent to the - bond and can thus give more than one product. (c) B: R 3 B: - 9

10 12. Which alkyl halide and what conditions should be used to prepare the following alkene in good yield by an E2 elimination? 2 10

11 13. What products are formed, and by what mechanisms, in each of the following cases? methyl iodide and NaN in ethanol Because a methyl halide has no β-hydrogens (- bonds adjacent to the electrophilic center), it cannot undergo elimination reaction. The only possible reaction is S N 2 although polar protic solvents are not as effective for S N 2 reaction as polar aprotic ones. Thus, the product is 3 N because N is an excellent nucleophile. 2-bromo-3-methyl butane in ethanol This is a secondary halide under conditions involving no nucleophile or base other than solvent, and a polar aprotic solvent. Because ethanol is a poor nucleophile and a weak base, neither S N 2 nor E2 reactions can occur. Because polar protic solvents promote S N 1 and E1 reactions, these will be the only reactions observed: 2 5 ethanol S N 1 product E1 products 2 5 carbocation rearrangement products Notice the rearrangement products! Rearrangements should be considered anytime the S N 1 or E1 reaction is expected, especially if the initially formed carbocation is secondary. (c) 2-bromo-3-methylbutane in anhydrous acetone (d) 2-bromo-3-methylbutane in ethanol containing an excess of sodium ethoxide (e) 2-bromo-2-methylbutane in ethanol containing an excess of sodium iodide 11

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13 (f) neopentyl bromide in ethanol containing an excess of sodium ethoxide. Neopentyl bromide contains no β- hydrogens hence no elimination is possible. Although it is a primary halide, it is does not readily undergo S N 2 reaction because of great steric hindrance in the transition state of the reaction. And since primary alkyl halides do not form carbocations, neither S N 1 nor E1 reaction is possible. Thus, this alkyl halide is essentially inert and no reaction is predicted. (If the reaction mixture is strongly heated, some S N 2 reaction might occur after several days) 13