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How Sn2 Reactions Occur

Bonding is weak between carbon and bromine and carbon and oxygen in the transition state [Pg.309]

FIGURE 8.2 Hybrid orbital description of the bonding changes that take place at carbon during nucleophilic substitution by the Sn2 mechanism. [Pg.309]

Once past the transition state, the leaving group is expelled and carbon becomes tetracoordinate, its hybridization returning to sp.  [Pg.309]

During the passage of starting materials to products, three interdependent and synchronous changes take place  [Pg.309]

Stretching, then breaking, of the bond to the leaving group [Pg.309]

Formation of a bond to the nucleophile from the opposite side of the bond that is broken [Pg.309]


Now that we have a good picture of how SN2 reactions occur, we need to see how they can be used and what variables affect them. Some SN-2 reactions are fast, and some are slow some take place in high yield and others, in low yield. Understanding the factors involved can be of tremendous value. Let s begin by recalling a few things about reaction rates In general. [Pg.365]

The tosylate leaving group was used in one of the classic experiments that was used to determine the stereochemistry of the SN2 reaction. Now that we know about this leaving group, let s look at the experiment and see how it helped establish that SN2 reactions occur with inversion of configuration. [Pg.282]

The cleavage reaction occurs in three steps O protonation of the epoxide, Sn2 nucleophilic attack on the protonated epoxide, and deprotonation of the ring-opened product. Draw the complete mechanism. How many intermediates are there Which step determines diol stereochemistry ... [Pg.129]

Exercise 8-7 Equations 8-3 through 8-5 show how Kenyon and Phillips established that inversion of configuration accompanies what we now recognize to be SN2 substitutions. For each reaction, we indicate whether R—0 or O—H is broken by an appropriately placed vertical line. Explain how the sequence of steps shows that inversion occurs in the SN2 reaction of Equation 8-5. The symbols (+) or (—) designate for each compound the sign of the rotation a of the plane of polarized light that it produces. [Pg.221]

The fact that SN2 reactions always occur with inversion of configuration enables us to form a better picture of the transition state. The nucleophile must approach the carbon from the side opposite the leaving group (back-side attack). The structure of the transition state, with partial bonds to the entering hydroxide and the leaving chloride, is shown in the following structure. Figure 8.3 uses orbitals to show how this process occurs. [Pg.262]

Problem 7.12 (a) Give an orbital representation for an SN2 reaction with (S)-RCHDX and Nu, if in the transition state the C on which displacement occurs uses sp2 hybrid orbitals, (b) How does this representation explain (i) inversion, (ii) the order of reactivity 3- > 2 l > l M... [Pg.123]

Table 7.7 summarizes the factors that determine whether a reaction occurs by the S l or Sn2 mechanism. Sample Problems 7.4 and 7.5 illustrate how these factors are used to determine the mechanism of a given reaction. [Pg.267]

A vast number of SN2 reactions at carbon centres has been studied. In every case, it is found that inversion of configuration occurs. This certainty now allows us to use inversion of configuration as a diagnostic test if inversion does not take place, the reaction cannot have a simple Sn2 mechanism. SnI reactions show partial or complete racemization, depending on how free the intermediate carbocation becomes. [Pg.65]

Stereoselectivity comes from a stereospecific syn or anti addition to an alkene of fixed and known geometry. These last reactions, applied to cyclohexene, lead to anti bromohydrin 14 while epoxidation occurs stereospecifically and syn. It doesn t matter which end of the epoxide 12 or bromonium ion 13 is attacked by the nucleophile anti addition occurs in both cases since inversion is demanded by the mechanism of the SN2 reaction. Cyclohexene must be Z but in open chain compounds syn addition to the -isomer would lead to the same diastereoisomer of the product as anti addition to the Z-isomer. In this chapter we explore more advanced versions of these reactions in which usually several types of selectivity will be combined and show how they are used in synthesis. [Pg.278]

This discussion suggests that an alkane will react with a dihalogen, when heated or irradiated with light, to form an alkyl halide. This reaction is illustrated by the reaction of 2-methylbutane (141), an alkane with no polarized bonds, and chlorine gas under photochemical conditions ( hv means irradiation with light). When this experiment was done in the 1930s, four different products were isolated 142 (34%), 143 (22%), 144 (28%), and 145 (16%). This is clearly a substitution reaction (Cl for H), but substitution occurs for every hydrogen atom in 141, at every different carbon atom. A different carbon is defined as one that gives a different product when H is replaced with Cl. Experiments have determined that the mechanism of this reaction is not SnI or Sn2. How does it occur How can H be replaced in a substitution reaction ... [Pg.548]

Is this correct It depends on how long the reaction is allowed to proceed. If the reaction is stopped after 1 or 2 hours, the answer is, indeed, no reaction. If the reaction is heated for several days or several weeks, two different products may be observed 2-iodo-2-methylbutane and 2-ethoxy-2-methyl-butane. Both products arise by an SnI process. Given sufficient time, slow ionization to a tertiary carbocation is followed by trapping with iodide ion or with ethanol. This process is known as solvolysis (see Chapter 11, Section 11.6). This contradicts the assumption that ionization can only occur in water because it is an assumption, as pointed out in the earlier section and thus not always correct. Indeed, ionization occurs in protic solvents such as ethanol, but it is slow relative to ionization in water. If 2-bromobutane is heated with KI in ethanol, the Sn2 product (Chapter 11, Section 11.2) is formed, indicating that the Sn2 reaction is much faster than the very slow ionization reaction in ethanol. [Pg.632]

Mechanistic studies are an important tool to understand how the biosynthesis of natural products occurs in nature and how to mimic their synthesis in the laboratory. Progress in this field includes the biomimetric synthesis of flavonolignan diastereomers in milk thistle (13JOC7594) and the chemoenzymatic synthesis of tetrahydropyran-containing polyketides (13AGE13215). In the synthetic domain, DFT calculations and experimental assays have been carried out to explain the multiple mechanisms involved in the palladium(II)-catalyzed Sn2 reactions of aUylic alcohols to prepare chiral tetrahydropyran derivatives (13JOC7664). [Pg.465]

Given a particular starting material and nucleophile, how do we know whether a reaction occurs by the SnI or Sn2 mechanism Four factors are examined ... [Pg.262]

Waxes are naturally occurring esters (alkyl alkanoates) containing long, straight aUcyl chains. Whale oil contains the wax 1-hexadecyl hexadecanoate, as shown below. How would you synthesize this wax, using an Sn2 reaction ... [Pg.322]

Now that we have seen all four factors individually, we need to see how to put them all together. When analyzing a reaction, we need to look at all four factors and make a determination of which mechanism, SnI or Sn2, is predominating. It may not be just one mechanism in every case. Sometimes both mechanisms occur and it is difficult to predict which one predominates. Nevertheless, it is a lot more common to see situations that are obviously leaning toward one mechanism over the other. For example, it is clear that a reaction will be Sn2 if we have a primary substrate with a strong nucleophile in a polar aprotic solvent. On the flipside, a reaction will clearly be SnI if we have a tertiary substrate with a weak nucleophile and an excellent leaving group. [Pg.223]

SnI, Sn2, El, E2 How can you keep it all straight How can you predict what will happen in any given case Will substitution or elimination occur Will the reaction be bimolecular or unimolecular There are no rigid answers to these questions, but it s possible to recognize some trends and make some generalizations (Table 11.4). [Pg.423]


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