Chemistry 307 Chapter 7

Transcription

1 Organic hemistry Nucleophilic Substitution. D. Roth hemistry 307 hapter 7 When we discussed the S N 2 reaction, we had considered two possible mechanisms, involving either a single step (nucleophilic substitution with second order kinetics, S N 2) or two steps, i.e., 1. R LG R LG 2. R Nu R Nu The reactions you learned about so far followed the onestep mechanism; you learned about the kinetic order and about the role of nucleophile, leaving group, the structure of the substrate, and of the solvent. These reactions proceeds via transition states. The second mechanism we considered was a two-step reaction: in the first step, the haloalkane undergoes ionic 1

2 Organic hemistry Nucleophilic Substitution. D. Roth dissociation forming a carbocation; in the second step reaction with a nucleophile completes the substitution. This mechanism would have first-order kinetics: the relative slow formation of the carbocation, R (step 1), would determine the rate (k α[r LG]); once the cation is generated, it would capture the nucleophile rapidly, forming the product. Does this alternative type of nucleophilic substitution, involving two steps, actually occur? We can answer this question by comparing the reactions of two nucleophiles with two different substrates: a) the reaction of iodide ion (Na I ) with bromoalkanes in acetone (Table 6.8); bromomethane reacts much faster with Na I than 2-bromo-2-methylpropane; this result is exactly what you learned to expect from an S N 2 reaction; b) the reaction of water, 2 O, with the same bromoalkanes (Table 7.1); this reaction proceeds much faster for 2- bromo-2-methylpropane than for bromomethane; the reversed order of reactivity is incompatible with an S N 2 reaction and requires a different mechanism. A more detailed study shows that the reaction with water follows a first order rate, as expected for the two-step reaction. The reaction also has different stereochemical consequences: 2

3 Organic hemistry Nucleophilic Substitution. D. Roth the stereochemistry at the reacting carbon is randomized, i.e., iodide ion converts enantiomerically pure 3-bromo-3- methylhexane to racemic 3-iodo-3-methylhexane. This result requires an achiral intermediate, identified as the planar carbocation (cf., Figure 7.6). The two-step mechanism of these reactions and the nature of the intermediate causes a range of important features: 1. the reaction proceeds fastest for tertiary substrates; 2. the reaction requires a good leaving group, LG; 3. the solvent influences the rate significantly; 4. the nucleophile does not play a significant role [because it is not involved in the rate-determining step]. 1. The fast reaction of tertiary substrates is related to carbocation stability; carbocations with alkyl substituents are 3

4 Organic hemistry Nucleophilic Substitution. D. Roth stabilized by hyperconjugation (remember?), that is, donation of electron density from a σ-bond to the empty p orbital (Figure 7.5). The relative stability of carbocations follows the order: ( 3 ) 3 > ( 3 ) 2 > 3 2 > 3 The formation of the most stable (lowest-energy) carbocation intermediate has the lowest-lying transition state and, thus, the fastest rate. This is opposite to the S N 2 reaction, where the least hindered substrate has the lowest transition state. Salient features of S N 2 and S N 1 reactions are compared below. 2. The S N 1 reaction works best with good leaving groups, e.g., OSO 2 > I > > l 4

5 Organic hemistry Nucleophilic Substitution. D. Roth 3. Polar solvents accelerate the reaction; polar protic solvents are optimal. For example, the reaction of water with 2-bromo-2-methylpropane is 400,000 times as fast in pure (100%) water than in acetone containing 10% water. The role of the solvent can be understood by comparing the rate determining transition states for the S N 2 and S N 1 reactions (Figure 7.4): in the S N 2 reaction the negative charge is distributed between Nu and LG; in the S N 1 reaction a polar bond is broken with separation of two opposite charges, for the carbocation, for the LG. 4. The nucleophile does not play a significant role in the S N 1 reaction. Even weak nucleophiles, such as water or alcohols, react with carbocations. Reactions in which water or alcohols serve as solvents and nucleophiles are called hydrolysis and solvolysis, respectively. These reactions proceed in three steps (Figure 7.2): a. dissociation (rate determining); b. addition (capture) of an oxygen lone pair unto the empty pi orbital, forming an oxonium ion; c. deprotonation of the oxonium ion forming an alcohol (when water is the nucleophile) or ether (when an alcohol serves as nucleophile). 5

6 Organic hemistry Nucleophilic Substitution. D. Roth While the nature of the nucleophile does not affect the rate of an S N 1 reaction, it determines the nature of the products; we call this product determining. In solutions containing more than one nucleophile, substitution products are formed with all nucleophiles present in competition with one another. Good nucleophiles react faster than poor nucleophiles (hardly surprising). In summary, we now have two reactions of haloalkanes with nucleophiles, depending on the nature of the substrate, a bimolecular, one-step (S N 2) and a unimolecular two-step substitution (S N 1). Please NOTE that for this general scheme R 1 and/or R 2 may be alkyl or (especially for the S N 2 reaction) 6

7 Organic hemistry Nucleophilic Substitution. D. Roth The simple features discussed above cannot be applied, without caution, to all reactions; some reactions may be more complicated. omplication 1: a large leaving group (I, ) may interfere with the approach of the nucleophile, causing the rate of substitution with retention (from the side left open by the leaving group, LG) to be slightly lower than the rate of substitution with inversion (from the backside). As a result the product may be slightly optically active, indicating some degree of inversion (NOT due to an S N 2 reaction). Sufficient separation of ions - unhindered access - racemization R 2 R 1 R R 3 3 R 2 R 1 O inversion O R 2 R 1 R 2 R 1 O retention Insufficient separation of ions - access for retention impeded incomplete racemization R R O 3 3 omplication 2: We have seen that the reaction of 3- bromo-3-methylhexane with iodide ion leads to racemic 3-7

8 Organic hemistry Nucleophilic Substitution. D. Roth iodo-3-methylhexane. What happens to substrates with two chiral centers, such as 2-bromo-3-methylpentane? The configuration at -2 is randomized, but -3 is not affected; it remains unchanged. As a result we obtain two diastereomers of 2-iodo-3-methylpentane Diastereomers omplication 3: We know that the strength of a Nu 3 3 nucleophile correlates with its base strength and that typical nucleophiles are also bases, that is, acceptors. For that reason, nucleophiles may abstract from the beta-carbon (β) of a carbocation, competing with "nucleophilic capture", that is, the second step of the S N 1 reaction. This is a new reaction, an elimination, which we call the E1 reaction (because it is unimolecular; Table 7.3, Figure 7.7). 1. R LG R LG 3. R Nu alkene Nu Nu 8

9 Organic hemistry Nucleophilic Substitution. D. Roth The E1 and the S N 1 reaction proceed via the same carbocation intermediate (formed in the rate determining step, the dissociation, 1). The distribution of the corresponding products is determined by the nature of the base/nucleophile(s) in solution. The nature of the leaving group affects the rate, but NOT the distribution of products. Rate E1 = k [R LG] Adding a base to the reaction increases the fraction of E1 over S N 1, but the overall rate of product formation {Rate α [R-al]} remains unchanged [because the different products are formed via the same intermediate. omplication 4: Addition of a weak base does not affect the rate law but changes the ratio of E1 to S N 1. In contrast, addition of a strong base may change the rate law as well as the ratio of substitution to elimination. This again is a new reaction, a bimolecular concerted reaction with second-order kinetics, which we call the E2 reaction. Rate E2 = k [R LG] [B ] Both the loss of the leaving group and the deprotonation occur in the rate-determining step; the mechanism is shown in Figure 7.8. Because of the bimolecular nature of this reaction, the base strength and the leaving group ability are important, 9

10 Organic hemistry Nucleophilic Substitution. D. Roth R I > R > R-l The reaction works best when the α LG bond and the β bond are anti-parallel to each other; we call this an antiperiplanar arrangement and the reaction an anti elimination. The E2 elimination from cis- and trans-1-bromo-4-(1,1- dimethylethyl)cyclohexane is a good example: it occurs readily for the trans- but not for cis-isomer. The existence of two elimination reactions with different rate laws requires that you consider several reaction features before you can assign a reaction product with any degree of confidence. For example, sterically hindered nucleophile/bases favor elimination (E2) over substitution (S N 2). potassium tert-butoxide lithium diisopropylamide onsidering the alkoxide ions, methoxide is unhindered and, therefore, a good nucleophile (as well as a base). Ethoxide 10

11 Organic hemistry Nucleophilic Substitution. D. Roth and isopropoxide have increasing bulk and increasing steric hindrance. tert-butoxide is no longer a nucleophile. Bulk of Alkoxide Anions O O O O In summary we have learned that haloalkane can undergo four reactions of with nucleophiles (bases): S N 2, S N 1, E1, E2. Please note that for this general scheme R 1 and/or R 2 may be alkyl or (especially for the S N 2 reaction) okusai, 11

12 Organic hemistry Nucleophilic Substitution. D. Roth okusai, Miscellaneous S N 2, S N 1, E2, E1 12

13 Organic hemistry Nucleophilic Substitution. D. Roth Factors governing the regiochemistry of the E2 reaction Regiochemistry of the E2 reaction with a small base: K O 70% 30% Small bases preferentially abstract the more acidic from the more highly substituted carbon forming the more highly substituted (more stable) alkene. This was recognized by Saytzev (19 th century) who formulated a rule that the is abstracted from the carbon with fewer s (Saytzev s rule). Saytzev based his rule on a difference between the substrates that he noted. You understand today that the course of the reaction is determined by the energetics. Regiochemistry of the E2 reaction with a bulky base % O(3 ) % For bulky bases (t-butoxide) the approach to the more highly substituted carbon is sterically hindered; the less highly substituted alkene is favored, even though it is less stable. 13

14 Organic hemistry Nucleophilic Substitution. D. Roth Some cyclohexane derivatives show particularly interesting features (limited rotation around the key bond). For example two stereo-isomers, cis- and trans-1-bromo- 2-methylcyclohexane, generate different products by E2 elimination. O3 O3 an you think of a good reason? Drawing the chair forms of the isomers with the in the axial position (because of the stereochemistry required for the E2 reaction), you will notice that the trans-isomer (right) has only one antiperiplanar eligible for abstraction; therefore, it can form only the less highly substituted, less stable alkene (anti-saytzev). The stereochemistry required for an E2 reaction overrides the energetic considerations. The cis-isomer (left) has two antiperiplanar atoms and follows the favorable energetics

15 Organic hemistry Nucleophilic Substitution. D. Roth The E2 reactions of cis- and trans-1-bromo-4-tertbutylcyclohexane take place with vastly different rates. ( 3 ) 3 ( 3 ) 3 O ( 3 ) 3 O ( 3 ) 3 Why would one isomer react significantly faster than the other, and which one? an you think of a good reason? The trans-isomer has antiperiplanar s only in the conformer with an axial t-butyl group; any coformer with two bulky axial substituents is very unstable. Moreover, the approach of the base is further disfavored by the bulky t-butyl group. stable conformer 2 antiperiplanar s stable conformer 0 antiperiplanar s highly unstable conformer 2 antiperiplanar s 15

16 Organic hemistry Nucleophilic Substitution. D. Roth Stereochemistry of the E2 reaction the general case free rotation around the bond which of the eligible s is chosen and which alkene is formed? B R ' R 2 R 1 rotate around by 120 B ' R R 2 R 1 If the substrate contains bulky groups, the different conformers may have significantly different stabilities, e.g., B t-bu B t-bu t-bu 3 t-bu t-bu R R anti-elimination t-bu 3 t-bu B 3 B t-bu 3 t-bu t-bu 3 R 2 R 1 R 2 R 1 t-bu t-bu 3 16

17 Organic hemistry Nucleophilic Substitution. D. Roth Both conformers have and in an antiperiplanar arrangement; the top conformer is much less hindered (more stable) than the bottom one and it forms the more stable alkene. S N 2 at two differently substituted centers Na N l N N 2S,4S 2R,4R Na N l N N 2S,4R meso arbocation rearrangements Some reactions involving carbocations yield unexpected products. The normal (expected) product is obtained in low yield and an unexpected product in higher yield. 3 I 3 O 3 O 3 3 O expected product minor 3 unexpected product major These reactions occur, because 2 carbocations formed by dissociation of I from an iodoalkane can form more stable 17

18 Organic hemistry Nucleophilic Substitution. D. Roth carbocations by rearrangement. Migration (1,2-shifts) of or R yield the "unexpected substitution products. I 3O [the symbol signifies rearrangement] Both carbocations react with methanol but the intramolecular rearrangement is faster than the intermolecular capture by the nucleophile. In order to understand the mechanism by which these rearrangements occur, we recall the concept of hyperconjugation stabilizing a free radical or a carbocation (cf., Figure 9.2). yperconjugation involves distorting or shifting the electrons of a bond in the direction of the electron-poor carbon. For the rearrangement the hydrogen atom migrates along with the electron pair; this can be described as a hydride shift. The positive charge ends up at the carbon from where the hydride left. We will discuss three different types of rearrangement: (i) 1,2-hydride shift Rearrangement by 1,2-hydride shift is exemplified by the first reaction in this section, involving conversion of a 2 to a 3 carbocation. 18

19 Organic hemistry Nucleophilic Substitution. D. Roth Another example, also involving conversion of a 2 to a 3 carbocation, shows in addition that the 1,2-hydride shift can be followed either by capture (S N 1) or by deprotonation (E1); however, the cyclohexene derivative is a minor product. 3 3 O 3 3 o O O 3 3 O O The curly arrow indicates random stereochemistry, as you had expected for a reaction involving a carbocation. (ii) 1,2-alkyl shift In some cases, secondary carbocations undergo rearrangement by 1,2-methyl shift, followed by either nucleophilic capture (S N 1) or by deprotonation (E1) O O O O 3 3 O

20 Organic hemistry Nucleophilic Substitution. D. Roth (iii) ring expansion releasing angle strain omoethylcyclobutane generates a 2 carbocation with three neighboring groups that can migrate. The 1,2-hydride shift forms a 3 carbocation with increased ring strain; shift of one of the two equivalent alkyl groups forms a 2 carbocation with greatly reduced ring strain. Stay tuned for more in future chapters. 20

21 Organic hemistry Nucleophilic Substitution. D. Roth Summary (parting shot?) S N 2, S N 1, E2, E1 oncerning the competition between S N 2, S N 1, E2, and E1: Nature of the substrate: 1 substrates clearly favor S N 2 and E2 Solvent: 3 substrates clearly favor S N 1 and E1 Polar aprotic solvents favor S N 2 and E2 Polar protic solvents favor S N 1 and E1 Leaving Group: Only good leaving groups undergo S N 1 and E1 Poor leaving groups disfavor S N 1 and E1 The leaving group does not affect the ratio of S N 1 vs. E1 Nucleophile/Base: Strong nucleophiles favor S N 2 Strong bases favor E2 Strong unhindered alkoxides favor S N 2 Strong hindered alkoxides favor E2 S N 1 and E1 with rearrangement 2 carbocations formed may be stabilized by hydride shift, alkyl shift, ring enlargement; S N 1 is typically preferred, E1 becomes more pronounced if the rearranged carbocation has an adjacent 3 carbon. 21

22 Organic hemistry Nucleophilic Substitution. D. Roth Some further examples: Secondary carbocations whose rearrangements generate tertiary carbocations with an adjacent tertiary carbon produce greater yields of elimination products. 3 O o O 3 O 3 22

23 Organic hemistry Nucleophilic Substitution. D. Roth Overall we are dealing with the following palette of reactions ere is what your book says: 23

24 Organic hemistry Nucleophilic Substitution Kin-kaku-ji (Kyoto) 24. D. Roth