عاطف خليفة
06-12-2008, 07:32 PM
The Mechanism of Reduction Reactions
Two fundamentally different reducing agents have been used to add hydrogen across a double bond. A metal can be used to catalyze the reaction between hydrogen gas and the C=C double bond in an alkene.
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A source of the hydride (H-) ion, on the other hand, is used to reduce C=O double bonds.
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The difference between these reactions is easy to understand. The first reaction uses a nonpolar reagent to reduce a nonpolar double bond. The atoms on the surface of a metal are different from those buried in the body of the solid because they cannot satisfy their tendency to form strong metal-metal bonds. Some metals can satisfy a portion of their combining power by binding hydrogen atoms and/or alkenes to the surface.
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Adding one of the hydrogen atoms to the alkene forms an alkyl group, which can bond to the metal until the second hydrogen atom can be added to form the alkene.
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Although the hydrogen atoms are transferred one at a time, this reaction is fast enough that both of these atoms usually end up on the same side of the C=C double bond. This can't be seen in most alkanes produced by this reaction because of the free rotation around Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifC bonds. Reduction of a cycloalkene, however, gives a stereoselective product.
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Reduction of an alkyne with hydrogen on a metal catalyst gives the corresponding alkane. By selectively "poisoning" the catalyst it is possible to reduce an alkyne to an alkene. Once again, the reaction is stereoselective, adding both hydrogen atoms from the same side of the Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/eqeqeq.gifC bond to form the cis-alkene.
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Because it is a polar reagent, LiAlH4 won't react with a C=C double bond. It acts as a source of the H- ion, however, which is a strong Brønsted base and a strong nucleophile. The H- ion can therefore attack the + end of a polar C=O double bond.
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The neutral AlH3 molecule formed when an AlH4- ion acts as a hydride donor is a Lewis acid that coordinates to the negatively charged oxygen atom in the product of this reaction. When, in a second step, a protic solvent is added to the reaction, an alcohol is formed.
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Nucleophilic Attack By Water
In the 1930s and 1940s, Dashiell Hammett (1894-1961) created the genre of the "hard-boiled" detective in books such as The Maltese Falcon and The Thin Man. A common occurrence in this literature was a character who "slipped someone a Mickey Finn" http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gif a dose of the sedative known as chloral hydrate dissolved in a drink that contains alcohol.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/22.gifChloral hydrate
Chloral hydrate is a white solid formed by adding a molecule of water across the C=O double bond in the corresponding aldehyde.
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The equilibrium constant for this reaction is sensitive to the substituents on the C=O double bond. Electron-withdrawing substituents, such as the Cl3C group in chloral, drive the reaction toward the dialcohol, or diol (Ka >> 1). Electron-donating substituents, such as the pair of CH3 groups in acetone, pull the equilibrium back toward the aldehyde (Ka = 2 x 10-3).
The rate of this reaction can be studied by following the incorporation of isotopically labeled water. The vast majority (99.76%) of water molecules contain 16O, but some contain 17O (0.04%) or 18O (0.2%). When acetone is dissolved in a sample of water that has been enriched in 18O, it gradually picks up the 18O isotope.
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The rate of this reaction is infinitesimally slow in a neutral solution (pH 7). But, in the presence of a trace of acid (or base), the reaction occurs very rapidly.
Acid and Base Catalyzed Hydration
The role of the acid catalyst is easy to understand. Protonation of the oxygen atom increases the polarity of the carbonyl bond.
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This increases the rate at which a water molecule can act as a nucleophile toward the positive end of the C=O double bond.
Acid-catalyzed hydration: Step 1
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The product of this reaction then loses an H+ ion to form the diol.
Acid-catalyzed hydration: Step 2
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The role of the base catalyst is equally easy to understand. The OH- ion is a much stronger nucleophile than water; strong enough to attack the carbonyl by itself.
Base-catalyzed hydration: Step 1
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The product of this reaction then picks up a proton from a water molecule to form the diol and regenerate the OH- ion.
Base-catalyzed hydration: Step 2
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There is a fundamental relationship between the mechanisms of the reactions at the carbonyl group introduced so far. In each case, a nucleophile or Lewis base attacks the positive end of the carbonyl group. And, in each case, the rate of reaction can be increased by coordinating a Lewis acid or electrophile at the other end of the carbonyl.
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There is a subtle difference between these reactions, however. Very strong nucleophiles, such as Grignard reagents or the hydride ion, add to the carbonyl in an irreversible reaction.
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Attack by a weaker nucleophile, such as water, is a reversible reaction that can occur in either direction.
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Nucleophilic Attack by an Alcohol
What would happen if we dissolved an aldehyde or ketone in an alcohol, instead of water? We would get a similar reaction, but now an ROH molecule is added across the C=O double bond.
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Once again, the reaction is relatively slow in the absence of an acid or base catalyst. If we bubble HCl gas through the solution, or add a small quantity of concentrated H2SO4, we get an acid-catalyzed reaction that occurs by a mechanism analogous to that described in the previous section.
Acid-catalyzed reaction of an alcohol with a carbonyl
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The product of this reaction is known as a hemiacetal (literally, "half of an acetal"). If an anhydrous acid is added to a solution of the aldehyde in a large excess of alcohol, the reaction continues to form an acetal.
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Hemiacetals can be recognized by looking for a carbon atom that has both anhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOH and an http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOR group.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/34.gifA hemiacetal
Acetals, on the other hand, contain a carbon atom that has two http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOR groups.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/35.gifAn acetal
Hemiacetals and acetals play an important role in the chemistry of carbohydrates. Consider what would happen, for example, if the http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOH group on the fifth carbon atom in a glucose molecule attacked the aldehyde at other end of this molecule.
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The product of this reaction is a hemiacetal that contains a six-membered ring known as a pyranose. Two isomers of glucopyranose can be formed, depending on whether the http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOH group attacks from above or below the C=O group.
a-D-Glucopyranose http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/37.giforhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/39.gifb-D-Glucopyranose http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/38.giforhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/40.gif
An analogous intramolecular reaction can occur within a fructose molecule.
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In this case, a hemiacetal is formed that contains a five-membered furanose ring. Once again, there are two isomers, depending on how the http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOH group attacks the C=O group.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/42.gifhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/43.gifa-D-Fructofuranoseb-D-Fructofuranose
Sugars, such as glucose and fructose, can be linked to form complex carbohydrates by forming an acetal linkage between the http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOH group on one sugar and the hemiacetal on the other. Sucrose, or cane sugar, for example, is an acetal formed by linking -D-gluco-pyranose and -D-fructofuranose residues.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/44.gifSucrose
Addition/Elimination Reactions of Carboxylic Acid Derivatives
The following reaction can be used to illustrate the synthesis of an ester from a carboxylic acid
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These reactions occur very slowly in the absence of a strong acid. When gaseous HCl is bubbled through the solution, or a small quantity of concentrated H2SO4 is added, these reactions reach equilibrium within a few hours. Once again, the acid protonates the oxygen of the C=O double bond, thereby increasing the polarity of the carbonyl group, which makes it more susceptible to attack by a nucleophile.
As might be expected, the first step in this reaction involves attack by a nucleophile at the positively charged end of the C=O double bond. A pair of nonbonding electrons on the oxygen atom of the alcohol is donated to the carbon atom of the carbonyl to form a Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifO bond. As this bond forms, the electrons in the bond of the carbonyl are displaced onto the oxygen atom. A proton is then transferred back to the solvent to give a tetrahedral addition intermediate.
Nucleophilic addition
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One of the http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOH groups in this intermediate picks up a proton, loses a molecule of water, and then transfers a proton back to the solvent to give the ester.
Nucleophilic elimination
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The combination of addition and elimination reactions has the overall effect of substituting one nucleophile for another http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gif in this case, substituting an alcohol for water. The rate of these nucleophilic substitution reactions is determined by the ease with which the elimination step occurs. As a rule, the best leaving groups in nucleophilic substitutions reactions are weak bases. The most reactive of the carboxylic acid derivatives are the acyl chlorides because the leaving group is a chloride ion, which is a very weak base (Kb http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/congrnt.gif10-20).
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Esters are less reactive because the leaving group is an alcohol, which is a slightly better base (Kb http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/congrnt.gif 10-14).
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Amides are even less reactive because the leaving group is ammonia or an amine, which are significantly more basic (Kb http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/congrnt.gif 10-5).
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Free Radical Reactions
The starting point for reactions at a carbonyl involves attack by a nucleophile on the carbon atom of the C=O double bond.
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Or it involves the heterolytic splitting of a bond to form a nucleophile that can attack the carbonyl group.
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In either case, the reaction is carried by a reagent that donates a pair of electrons to a carbon atom to form a new covalent bond.
Free-radical halogenation of an alkane occurs by a very different mechanism. The first step in these reactions is the homolytic splitting of a bond to give a pair of free radicals.
Chain initiation
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A series of reactions then occurs that involves a chain-reaction. Consider the chlorination of propane, for example. A Cl· atom can attack the CH3 group at one end of the molecule.
Chain propagation
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Or it can attack the CH2 group in the center of the molecule.
Chain propagation
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The free radicals generated in these reactions then react with chlorine to form either 1-chloro-propane or 2-chloropropane and regenerate a Cl· radical.
Chain propagation
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There are six hydrogen atoms in the two CH3 groups and two hydrogens in the CH2 group in propane. If attack occurred randomly, six-eighths (or three-quarters) of the product of this reaction would be 1-chloropropane. The distribution of products of this reaction, however, suggests that 1-chloropropane is formed slightly less often than 2-chloropropane.
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This can be explained by noting that the 2 radical formed by removing a hydrogen atom from the CH2 group in the center of the molecule is slightly more stable than the 1 radical produced when a hydrogen atom is removed from one of the CH3 groups at either end of the molecule.
The difference between these radicals can be appreciated by considering the energy it takes to break the Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifH bond in the following compounds.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/59.gifhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/delta.gifH = -381 kJ/molrxn http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/60.gifhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/delta.gifH = -395 kJ/molrxn http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/61.gifhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/delta.gifH = -410 kJ/molrxn
These data suggest that it takes less energy to break a Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifH bond as the number of alkyl groups on the carbon atom that contains this bond increases. This can be explained by assuming that the products of the bond-breaking reaction become more stable as the number of alkyl groups increases. Or, in other words, 3º radicals are more stable than 2º radicals, which are more stable than 1º radicals.
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The activation energy for the chain-propagation steps in free-radical bromination reactions is significantly larger than the activation energy for these steps during chlorination. As a result, free-radical bromination reactions are more selective than chlorination reactions. Bromination reactions are far more likely to give the product predicted from the relative stability of the free-radical intermediate. Bromination of 2-methylpropane, for example, gives almost exclusively 2-bromo-2-methylpropane, not the statistically more likely 1-bromo-2-methylpropane.
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Bimolecular Nucleophilic Substitution or SN2 Reactions
Most of our knowledge of the mechanisms of chemical reactions has come from the study of the factors that influence the rate of these reactions. The type of reaction that has been studied more than any other involves attack by a nucleophile on a saturated carbon atom. Consider the following reaction, for example, which converts an alkyl bromide into an alcohol.
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In the course of this reaction, one nucleophile (the OH- ion) is substituted for another (the Br- ion). This is therefore a nucleophilic substitution reaction.
The rate of this reaction is first-order in both CH3Br and the OH- ion, and second-order overall.
Rate = k(CH3Br)(OH-)
In the 1930s, Sir Christopher Ingold proposed a mechanism for this reaction in which both the alkyl halide and the hydroxyl ion are involved in the rate-limiting or slowest step of the reaction. The OH- ion attacks the "backside" of the CH3Br molecule. (It attacks the carbon atom at a point directly opposite to the Br substituent or leaving group.) When this happens, a pair of nonbonding electrons on the OH- ion are used to form a covalent bond to the carbon atom at the same time that the carbon-bromine is broken, as shown in the figure below.
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Because the rate-limiting step in this reaction involves both the CH3Br and OH- molecules, it is called a bimolecular nucleophilic substitution, or SN2, reaction.
The most important point to remember about the mechanism of SN2 reactions is that they occur in a single step. The species in the middle of Figure O3.2 is known as a transition state. If you envision this reaction as an endless series of snapshots that capture the infinitesimally small changes which occur as one bond forms and the other bond breaks, the transition state is the snapshot in this series that has the highest energy http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifand is therefore the least stable. The transition state has an infinitesimally short lifetime, on the order of 10-12 seconds.
In the course of an SN2 reaction, the other three substituents on the carbon atom are "flipped" from one side of the atom to the other. This inevitably leads to inversion of the configuration at a stereocenter. Consider the following reaction, for example, in which cis-1-bromo-3-methylcyclopentane is converted into trans-3-methylcyclopentanol.
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Or the reaction in which the 2-butanol. R isomer of 2-bromobutane is transformed into the S isomer of
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Unimolecular Nucleophilic Substitution or SN1 Reactions
The kinetics of nucleophilic substitution reactions have been studied in greater detail than any other type of reaction because they don't always proceed through the same mechanism. Consider the reaction between the OH- ion and t-butyl bromide, for example.
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The rate of this reaction depends only on the concentration of the alkyl bromide. (Adding more OH- ion to the solution has no effect on the rate of reaction.)
Rate = k((CH3)3CBr)
Ingold and coworkers argued that this rate law is consistent with a mechanism in which the rate-limiting or slowest step involves the breaking of the carbon-bromine bond to form a pair of ions. As one might expect, the pair of electrons in the Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifBr bond end up on the more electronegative bromine atom.
Rate-limiting step:
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Because the bromine atom has formally gained an electron from the carbon atom, it is now a negatively charged Br- ion. Because the carbon atom has formally lost an electron, it is now a "carbocation."
The first step in this mechanism is a relatively slow reaction. (The activation energy for this step is roughly 80 kJ/mol.) If this reaction is done in water, the next step is extremely fast. The (CH3)3C+ ion is a Lewis acid because it has an empty orbital that can be used to accept a pair of electrons. Water, on the other hand, is a reasonably good Lewis base. A Lewis acid-base reaction therefore rapidly occurs in which a pair of nonbonding electrons on a water molecule are donated to the carbocation to form a covalent Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifO bond.
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The product of this reaction is a stronger acid than water. As a result, it transfers a proton to water.
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Because the slowest step of this reaction only involves t-butyl bromide, the overall rate of reaction only depends on the concentration of this species. This is therefore a unimolecular nucleophilic substitution, or SN1, reaction.
The central carbon atom in the t-butyl carbocation formed in the first step of this reaction is planar, as shown in the figure below.
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This means that water can attack this carbocation in the second step with equal probability from either side of the carbon atom. This has no effect on the products of this reaction, because the starting material is not optically active. But what would happen if we started with an optically active halide, such as 2-bromobutane?
Regardless of whether we start with the R or S isomer, we get the same intermediate when the Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifBr bond breaks.
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The intermediate formed in the first step in the SN1 mechanism is therefore achiral.
Mixtures of equal quantities of the +/- or R/S stereoisomers of a compound are said to be racemic. This term traces back to the Latin racemus, which means "a cluster of grapes." Just as there is an equal probability of finding grapes on either side of the stem in a cluster of grapes, there is an equal probability of finding the R and S enantiomers in a racemic mixture. SN1 reactions are therefore said to proceed with racemization. If we start with a pure sample of (R)-2-bromobutane, for example, we expect the product of the SN1 reaction with the OH- ion to be a racemic mixture of the two enantiomers of 2-butanol.
We are now ready to address a pair of important questions. First, why does CH3Br react with the OH- ion by the SN2 mechanism if (CH3)3CBr does not? The SN2 mechanism requires direct attack by the OH- ion on the carbon atom that carries the Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifBr bond. It is much easier for the OH- ion to get past the small hydrogen atoms in CH3Br than it is for this ion to get past the bulkier CH3 groups in (CH3)3CBr.
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Thus, SN2 reactions at the 1º carbon atom in CH3Br are much faster than the analogous reaction at the 3º carbon atom in (CH3)3CBr.
Why, then, does (CH3)3CBr react with the OH- ion by the SN1 mechanism if CH3Br does not? The SN1 reaction proceeds through a carbocation intermediate, and the stability of these ions decreases in the following order.
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Organic chemists explain this by noting that alkyl groups are slightly "electron releasing."
They can donate electron density to a neighboring group. This tends to delocalize the charge over a larger volume of the molecule, which stabilizes the carbocation.
When we encountered a similar phenomenon in the chemistry of free radicals (http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/organic.html#free) we noted that 3º radicals are roughly 30 kJ/mol more stable than 1 radicals. In this case, the difference is much larger. A 3º carbocation is 340 kJ/mol more stable than a 1º carbocation! As a result, it is much easier for (CH3)3CBr to form a carbocation intermediate than it is for CH3Br to undergo a similar reaction.
In theory, both starting materials could undergo both reaction mechanisms. But the rate of SN2 reactions for CH3Br are much faster than the corresponding SN1 reactions, whereas the rate of SN1 reactions for (CH3)3CBr are very much faster than SN2 reactions
Elimination Reactions
Why do we need to worry about whether a nucleophilic substitution reaction occurs by an SN1 or SN2 mechanism? At first glance, it would appear that the same product is obtained regardless of the mechanism of the reaction. Consider the following substitution reaction, for example.
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The only apparent difference between the two mechanisms is the stereochemistry of the product. If the reaction proceeds through an SN2 mechanism, it gives inversion of configurationhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gif conversion of an R starting material into an S product, or vice versa. If the reaction proceeds through a carbocation intermediate via an SN1 mechanism, we get a racemic mixture.
The importance of understanding the mechanism of nucleophilic substitution reactions can best be appreciated by studying the distribution of products of the example given above. When 2-bromopropane is allowed to react with the methoxide ion in methanol, less than half of the starting material is converted into methyl isopropyl ether; the rest is transformed into 2-propene.
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The reaction that produces the alkene involves the loss of an HBr molecule to form a C=C double bond. It is therefore an example of an elimination reaction.
Starting materials that are likely to undergo an bimolecular SN2 reaction undergo elimination reactions by a bimolecular E2 mechanism. This is a one-step reaction in which the nucleophile attacks a Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifH bond on the carbon atom adjacent to the site of SN2 reaction.
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Starting materials that are likely to undergo a unimolecular SN1 reaction undergo elimination reactions by a unimolecular E1 mechanism. As might be expected, the rate-limiting step is the formation of the carbocation.
Rate-limiting Step:
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The solvent then acts as a base, removing an H+ ion from one of the alkyl groups adjacent to the carbocation. The electrons in the Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifH bond that is broken are donated to the empty orbital on the carbocation to form a double bond.
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Substitution Versus Elimination Reactions
There are three ways of pushing the reaction between an alkyl halide and a nucleophile toward elimination instead of substitution.
Start with a highly substituted substrate, which is more likely to undergo elimination. Only 10% of a primary alkyl bromide undergoes elimination to form an alkene, for example, when it reacts with an alkoxide ion dissolved in alcohol. The vast majority of the starting material goes on to form the product expected for an SN2 reaction.
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More than half of a secondary alkyl bromide undergoes elimination under the same conditions, as we have already seen.
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When the starting material is a tertiary alkyl halide, more than 90% of the product is formed by an E1 elimination reaction.
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Use a very strong base as the nucleophile. When we use a relatively weak base, such as ethyl alcohol, only about 20% of t-butyl bromide undergoes elimination.
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In the presence of the ethoxide ion, which is a much stronger base, the product of the reaction is predominantly the alkene.
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Increase the temperature at which the reaction is run. Because both E1 and E2 reactions lead to an increase in the number of particles in the system, they are associated with a positive entropy term. Thus, increasing the temperature of the reaction makes the overall free energy of reaction more negative, and the reaction more favorable.
Summary of Substitution/Elimination Reactions
Methyl halides and primary alkyl halides http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gif such as CH3CH2Brhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gif undergo nucleophilic substitution reactions.
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Secondary alkyl halides undergo SN2 reactions when handled gently http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gif at low temperatures and with moderate strength nucleophiles.
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At high temperatures, or in the presence of a strong base, secondary halides undergo E2 elimination reactions.
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Tertiary halides undergo a combination of SN1 and E1 reactions. If the reaction is kept cool, and the nucleophile is a relatively weak base, it is possible to get nucleophilic substitution. At high temperatures, or with strong bases, elimination reactions predominate.
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Two fundamentally different reducing agents have been used to add hydrogen across a double bond. A metal can be used to catalyze the reaction between hydrogen gas and the C=C double bond in an alkene.
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A source of the hydride (H-) ion, on the other hand, is used to reduce C=O double bonds.
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The difference between these reactions is easy to understand. The first reaction uses a nonpolar reagent to reduce a nonpolar double bond. The atoms on the surface of a metal are different from those buried in the body of the solid because they cannot satisfy their tendency to form strong metal-metal bonds. Some metals can satisfy a portion of their combining power by binding hydrogen atoms and/or alkenes to the surface.
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Adding one of the hydrogen atoms to the alkene forms an alkyl group, which can bond to the metal until the second hydrogen atom can be added to form the alkene.
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Although the hydrogen atoms are transferred one at a time, this reaction is fast enough that both of these atoms usually end up on the same side of the C=C double bond. This can't be seen in most alkanes produced by this reaction because of the free rotation around Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifC bonds. Reduction of a cycloalkene, however, gives a stereoselective product.
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Reduction of an alkyne with hydrogen on a metal catalyst gives the corresponding alkane. By selectively "poisoning" the catalyst it is possible to reduce an alkyne to an alkene. Once again, the reaction is stereoselective, adding both hydrogen atoms from the same side of the Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/eqeqeq.gifC bond to form the cis-alkene.
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Because it is a polar reagent, LiAlH4 won't react with a C=C double bond. It acts as a source of the H- ion, however, which is a strong Brønsted base and a strong nucleophile. The H- ion can therefore attack the + end of a polar C=O double bond.
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The neutral AlH3 molecule formed when an AlH4- ion acts as a hydride donor is a Lewis acid that coordinates to the negatively charged oxygen atom in the product of this reaction. When, in a second step, a protic solvent is added to the reaction, an alcohol is formed.
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Nucleophilic Attack By Water
In the 1930s and 1940s, Dashiell Hammett (1894-1961) created the genre of the "hard-boiled" detective in books such as The Maltese Falcon and The Thin Man. A common occurrence in this literature was a character who "slipped someone a Mickey Finn" http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gif a dose of the sedative known as chloral hydrate dissolved in a drink that contains alcohol.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/22.gifChloral hydrate
Chloral hydrate is a white solid formed by adding a molecule of water across the C=O double bond in the corresponding aldehyde.
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The equilibrium constant for this reaction is sensitive to the substituents on the C=O double bond. Electron-withdrawing substituents, such as the Cl3C group in chloral, drive the reaction toward the dialcohol, or diol (Ka >> 1). Electron-donating substituents, such as the pair of CH3 groups in acetone, pull the equilibrium back toward the aldehyde (Ka = 2 x 10-3).
The rate of this reaction can be studied by following the incorporation of isotopically labeled water. The vast majority (99.76%) of water molecules contain 16O, but some contain 17O (0.04%) or 18O (0.2%). When acetone is dissolved in a sample of water that has been enriched in 18O, it gradually picks up the 18O isotope.
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The rate of this reaction is infinitesimally slow in a neutral solution (pH 7). But, in the presence of a trace of acid (or base), the reaction occurs very rapidly.
Acid and Base Catalyzed Hydration
The role of the acid catalyst is easy to understand. Protonation of the oxygen atom increases the polarity of the carbonyl bond.
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This increases the rate at which a water molecule can act as a nucleophile toward the positive end of the C=O double bond.
Acid-catalyzed hydration: Step 1
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The product of this reaction then loses an H+ ion to form the diol.
Acid-catalyzed hydration: Step 2
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The role of the base catalyst is equally easy to understand. The OH- ion is a much stronger nucleophile than water; strong enough to attack the carbonyl by itself.
Base-catalyzed hydration: Step 1
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The product of this reaction then picks up a proton from a water molecule to form the diol and regenerate the OH- ion.
Base-catalyzed hydration: Step 2
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There is a fundamental relationship between the mechanisms of the reactions at the carbonyl group introduced so far. In each case, a nucleophile or Lewis base attacks the positive end of the carbonyl group. And, in each case, the rate of reaction can be increased by coordinating a Lewis acid or electrophile at the other end of the carbonyl.
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There is a subtle difference between these reactions, however. Very strong nucleophiles, such as Grignard reagents or the hydride ion, add to the carbonyl in an irreversible reaction.
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Attack by a weaker nucleophile, such as water, is a reversible reaction that can occur in either direction.
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Nucleophilic Attack by an Alcohol
What would happen if we dissolved an aldehyde or ketone in an alcohol, instead of water? We would get a similar reaction, but now an ROH molecule is added across the C=O double bond.
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Once again, the reaction is relatively slow in the absence of an acid or base catalyst. If we bubble HCl gas through the solution, or add a small quantity of concentrated H2SO4, we get an acid-catalyzed reaction that occurs by a mechanism analogous to that described in the previous section.
Acid-catalyzed reaction of an alcohol with a carbonyl
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The product of this reaction is known as a hemiacetal (literally, "half of an acetal"). If an anhydrous acid is added to a solution of the aldehyde in a large excess of alcohol, the reaction continues to form an acetal.
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Hemiacetals can be recognized by looking for a carbon atom that has both anhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOH and an http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOR group.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/34.gifA hemiacetal
Acetals, on the other hand, contain a carbon atom that has two http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOR groups.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/35.gifAn acetal
Hemiacetals and acetals play an important role in the chemistry of carbohydrates. Consider what would happen, for example, if the http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOH group on the fifth carbon atom in a glucose molecule attacked the aldehyde at other end of this molecule.
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The product of this reaction is a hemiacetal that contains a six-membered ring known as a pyranose. Two isomers of glucopyranose can be formed, depending on whether the http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOH group attacks from above or below the C=O group.
a-D-Glucopyranose http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/37.giforhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/39.gifb-D-Glucopyranose http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/38.giforhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/40.gif
An analogous intramolecular reaction can occur within a fructose molecule.
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In this case, a hemiacetal is formed that contains a five-membered furanose ring. Once again, there are two isomers, depending on how the http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOH group attacks the C=O group.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/42.gifhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/43.gifa-D-Fructofuranoseb-D-Fructofuranose
Sugars, such as glucose and fructose, can be linked to form complex carbohydrates by forming an acetal linkage between the http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOH group on one sugar and the hemiacetal on the other. Sucrose, or cane sugar, for example, is an acetal formed by linking -D-gluco-pyranose and -D-fructofuranose residues.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/44.gifSucrose
Addition/Elimination Reactions of Carboxylic Acid Derivatives
The following reaction can be used to illustrate the synthesis of an ester from a carboxylic acid
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These reactions occur very slowly in the absence of a strong acid. When gaseous HCl is bubbled through the solution, or a small quantity of concentrated H2SO4 is added, these reactions reach equilibrium within a few hours. Once again, the acid protonates the oxygen of the C=O double bond, thereby increasing the polarity of the carbonyl group, which makes it more susceptible to attack by a nucleophile.
As might be expected, the first step in this reaction involves attack by a nucleophile at the positively charged end of the C=O double bond. A pair of nonbonding electrons on the oxygen atom of the alcohol is donated to the carbon atom of the carbonyl to form a Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifO bond. As this bond forms, the electrons in the bond of the carbonyl are displaced onto the oxygen atom. A proton is then transferred back to the solvent to give a tetrahedral addition intermediate.
Nucleophilic addition
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One of the http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifOH groups in this intermediate picks up a proton, loses a molecule of water, and then transfers a proton back to the solvent to give the ester.
Nucleophilic elimination
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The combination of addition and elimination reactions has the overall effect of substituting one nucleophile for another http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gif in this case, substituting an alcohol for water. The rate of these nucleophilic substitution reactions is determined by the ease with which the elimination step occurs. As a rule, the best leaving groups in nucleophilic substitutions reactions are weak bases. The most reactive of the carboxylic acid derivatives are the acyl chlorides because the leaving group is a chloride ion, which is a very weak base (Kb http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/congrnt.gif10-20).
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/48.gif
Esters are less reactive because the leaving group is an alcohol, which is a slightly better base (Kb http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/congrnt.gif 10-14).
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Amides are even less reactive because the leaving group is ammonia or an amine, which are significantly more basic (Kb http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/congrnt.gif 10-5).
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Free Radical Reactions
The starting point for reactions at a carbonyl involves attack by a nucleophile on the carbon atom of the C=O double bond.
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Or it involves the heterolytic splitting of a bond to form a nucleophile that can attack the carbonyl group.
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In either case, the reaction is carried by a reagent that donates a pair of electrons to a carbon atom to form a new covalent bond.
Free-radical halogenation of an alkane occurs by a very different mechanism. The first step in these reactions is the homolytic splitting of a bond to give a pair of free radicals.
Chain initiation
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A series of reactions then occurs that involves a chain-reaction. Consider the chlorination of propane, for example. A Cl· atom can attack the CH3 group at one end of the molecule.
Chain propagation
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Or it can attack the CH2 group in the center of the molecule.
Chain propagation
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The free radicals generated in these reactions then react with chlorine to form either 1-chloro-propane or 2-chloropropane and regenerate a Cl· radical.
Chain propagation
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There are six hydrogen atoms in the two CH3 groups and two hydrogens in the CH2 group in propane. If attack occurred randomly, six-eighths (or three-quarters) of the product of this reaction would be 1-chloropropane. The distribution of products of this reaction, however, suggests that 1-chloropropane is formed slightly less often than 2-chloropropane.
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This can be explained by noting that the 2 radical formed by removing a hydrogen atom from the CH2 group in the center of the molecule is slightly more stable than the 1 radical produced when a hydrogen atom is removed from one of the CH3 groups at either end of the molecule.
The difference between these radicals can be appreciated by considering the energy it takes to break the Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifH bond in the following compounds.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/59.gifhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/delta.gifH = -381 kJ/molrxn http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/60.gifhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/delta.gifH = -395 kJ/molrxn http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/61.gifhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/delta.gifH = -410 kJ/molrxn
These data suggest that it takes less energy to break a Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifH bond as the number of alkyl groups on the carbon atom that contains this bond increases. This can be explained by assuming that the products of the bond-breaking reaction become more stable as the number of alkyl groups increases. Or, in other words, 3º radicals are more stable than 2º radicals, which are more stable than 1º radicals.
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The activation energy for the chain-propagation steps in free-radical bromination reactions is significantly larger than the activation energy for these steps during chlorination. As a result, free-radical bromination reactions are more selective than chlorination reactions. Bromination reactions are far more likely to give the product predicted from the relative stability of the free-radical intermediate. Bromination of 2-methylpropane, for example, gives almost exclusively 2-bromo-2-methylpropane, not the statistically more likely 1-bromo-2-methylpropane.
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Bimolecular Nucleophilic Substitution or SN2 Reactions
Most of our knowledge of the mechanisms of chemical reactions has come from the study of the factors that influence the rate of these reactions. The type of reaction that has been studied more than any other involves attack by a nucleophile on a saturated carbon atom. Consider the following reaction, for example, which converts an alkyl bromide into an alcohol.
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In the course of this reaction, one nucleophile (the OH- ion) is substituted for another (the Br- ion). This is therefore a nucleophilic substitution reaction.
The rate of this reaction is first-order in both CH3Br and the OH- ion, and second-order overall.
Rate = k(CH3Br)(OH-)
In the 1930s, Sir Christopher Ingold proposed a mechanism for this reaction in which both the alkyl halide and the hydroxyl ion are involved in the rate-limiting or slowest step of the reaction. The OH- ion attacks the "backside" of the CH3Br molecule. (It attacks the carbon atom at a point directly opposite to the Br substituent or leaving group.) When this happens, a pair of nonbonding electrons on the OH- ion are used to form a covalent bond to the carbon atom at the same time that the carbon-bromine is broken, as shown in the figure below.
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Because the rate-limiting step in this reaction involves both the CH3Br and OH- molecules, it is called a bimolecular nucleophilic substitution, or SN2, reaction.
The most important point to remember about the mechanism of SN2 reactions is that they occur in a single step. The species in the middle of Figure O3.2 is known as a transition state. If you envision this reaction as an endless series of snapshots that capture the infinitesimally small changes which occur as one bond forms and the other bond breaks, the transition state is the snapshot in this series that has the highest energy http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifand is therefore the least stable. The transition state has an infinitesimally short lifetime, on the order of 10-12 seconds.
In the course of an SN2 reaction, the other three substituents on the carbon atom are "flipped" from one side of the atom to the other. This inevitably leads to inversion of the configuration at a stereocenter. Consider the following reaction, for example, in which cis-1-bromo-3-methylcyclopentane is converted into trans-3-methylcyclopentanol.
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Or the reaction in which the 2-butanol. R isomer of 2-bromobutane is transformed into the S isomer of
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Unimolecular Nucleophilic Substitution or SN1 Reactions
The kinetics of nucleophilic substitution reactions have been studied in greater detail than any other type of reaction because they don't always proceed through the same mechanism. Consider the reaction between the OH- ion and t-butyl bromide, for example.
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The rate of this reaction depends only on the concentration of the alkyl bromide. (Adding more OH- ion to the solution has no effect on the rate of reaction.)
Rate = k((CH3)3CBr)
Ingold and coworkers argued that this rate law is consistent with a mechanism in which the rate-limiting or slowest step involves the breaking of the carbon-bromine bond to form a pair of ions. As one might expect, the pair of electrons in the Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifBr bond end up on the more electronegative bromine atom.
Rate-limiting step:
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Because the bromine atom has formally gained an electron from the carbon atom, it is now a negatively charged Br- ion. Because the carbon atom has formally lost an electron, it is now a "carbocation."
The first step in this mechanism is a relatively slow reaction. (The activation energy for this step is roughly 80 kJ/mol.) If this reaction is done in water, the next step is extremely fast. The (CH3)3C+ ion is a Lewis acid because it has an empty orbital that can be used to accept a pair of electrons. Water, on the other hand, is a reasonably good Lewis base. A Lewis acid-base reaction therefore rapidly occurs in which a pair of nonbonding electrons on a water molecule are donated to the carbocation to form a covalent Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifO bond.
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The product of this reaction is a stronger acid than water. As a result, it transfers a proton to water.
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Because the slowest step of this reaction only involves t-butyl bromide, the overall rate of reaction only depends on the concentration of this species. This is therefore a unimolecular nucleophilic substitution, or SN1, reaction.
The central carbon atom in the t-butyl carbocation formed in the first step of this reaction is planar, as shown in the figure below.
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This means that water can attack this carbocation in the second step with equal probability from either side of the carbon atom. This has no effect on the products of this reaction, because the starting material is not optically active. But what would happen if we started with an optically active halide, such as 2-bromobutane?
Regardless of whether we start with the R or S isomer, we get the same intermediate when the Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifBr bond breaks.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/25_1.gif
The intermediate formed in the first step in the SN1 mechanism is therefore achiral.
Mixtures of equal quantities of the +/- or R/S stereoisomers of a compound are said to be racemic. This term traces back to the Latin racemus, which means "a cluster of grapes." Just as there is an equal probability of finding grapes on either side of the stem in a cluster of grapes, there is an equal probability of finding the R and S enantiomers in a racemic mixture. SN1 reactions are therefore said to proceed with racemization. If we start with a pure sample of (R)-2-bromobutane, for example, we expect the product of the SN1 reaction with the OH- ion to be a racemic mixture of the two enantiomers of 2-butanol.
We are now ready to address a pair of important questions. First, why does CH3Br react with the OH- ion by the SN2 mechanism if (CH3)3CBr does not? The SN2 mechanism requires direct attack by the OH- ion on the carbon atom that carries the Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifBr bond. It is much easier for the OH- ion to get past the small hydrogen atoms in CH3Br than it is for this ion to get past the bulkier CH3 groups in (CH3)3CBr.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/23_1.gif
Thus, SN2 reactions at the 1º carbon atom in CH3Br are much faster than the analogous reaction at the 3º carbon atom in (CH3)3CBr.
Why, then, does (CH3)3CBr react with the OH- ion by the SN1 mechanism if CH3Br does not? The SN1 reaction proceeds through a carbocation intermediate, and the stability of these ions decreases in the following order.
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Organic chemists explain this by noting that alkyl groups are slightly "electron releasing."
They can donate electron density to a neighboring group. This tends to delocalize the charge over a larger volume of the molecule, which stabilizes the carbocation.
When we encountered a similar phenomenon in the chemistry of free radicals (http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/organic.html#free) we noted that 3º radicals are roughly 30 kJ/mol more stable than 1 radicals. In this case, the difference is much larger. A 3º carbocation is 340 kJ/mol more stable than a 1º carbocation! As a result, it is much easier for (CH3)3CBr to form a carbocation intermediate than it is for CH3Br to undergo a similar reaction.
In theory, both starting materials could undergo both reaction mechanisms. But the rate of SN2 reactions for CH3Br are much faster than the corresponding SN1 reactions, whereas the rate of SN1 reactions for (CH3)3CBr are very much faster than SN2 reactions
Elimination Reactions
Why do we need to worry about whether a nucleophilic substitution reaction occurs by an SN1 or SN2 mechanism? At first glance, it would appear that the same product is obtained regardless of the mechanism of the reaction. Consider the following substitution reaction, for example.
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The only apparent difference between the two mechanisms is the stereochemistry of the product. If the reaction proceeds through an SN2 mechanism, it gives inversion of configurationhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gif conversion of an R starting material into an S product, or vice versa. If the reaction proceeds through a carbocation intermediate via an SN1 mechanism, we get a racemic mixture.
The importance of understanding the mechanism of nucleophilic substitution reactions can best be appreciated by studying the distribution of products of the example given above. When 2-bromopropane is allowed to react with the methoxide ion in methanol, less than half of the starting material is converted into methyl isopropyl ether; the rest is transformed into 2-propene.
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The reaction that produces the alkene involves the loss of an HBr molecule to form a C=C double bond. It is therefore an example of an elimination reaction.
Starting materials that are likely to undergo an bimolecular SN2 reaction undergo elimination reactions by a bimolecular E2 mechanism. This is a one-step reaction in which the nucleophile attacks a Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifH bond on the carbon atom adjacent to the site of SN2 reaction.
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Starting materials that are likely to undergo a unimolecular SN1 reaction undergo elimination reactions by a unimolecular E1 mechanism. As might be expected, the rate-limiting step is the formation of the carbocation.
Rate-limiting Step:
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The solvent then acts as a base, removing an H+ ion from one of the alkyl groups adjacent to the carbocation. The electrons in the Chttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gifH bond that is broken are donated to the empty orbital on the carbocation to form a double bond.
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Substitution Versus Elimination Reactions
There are three ways of pushing the reaction between an alkyl halide and a nucleophile toward elimination instead of substitution.
Start with a highly substituted substrate, which is more likely to undergo elimination. Only 10% of a primary alkyl bromide undergoes elimination to form an alkene, for example, when it reacts with an alkoxide ion dissolved in alcohol. The vast majority of the starting material goes on to form the product expected for an SN2 reaction.
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More than half of a secondary alkyl bromide undergoes elimination under the same conditions, as we have already seen.
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When the starting material is a tertiary alkyl halide, more than 90% of the product is formed by an E1 elimination reaction.
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Use a very strong base as the nucleophile. When we use a relatively weak base, such as ethyl alcohol, only about 20% of t-butyl bromide undergoes elimination.
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In the presence of the ethoxide ion, which is a much stronger base, the product of the reaction is predominantly the alkene.
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Increase the temperature at which the reaction is run. Because both E1 and E2 reactions lead to an increase in the number of particles in the system, they are associated with a positive entropy term. Thus, increasing the temperature of the reaction makes the overall free energy of reaction more negative, and the reaction more favorable.
Summary of Substitution/Elimination Reactions
Methyl halides and primary alkyl halides http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gif such as CH3CH2Brhttp://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gif undergo nucleophilic substitution reactions.
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Secondary alkyl halides undergo SN2 reactions when handled gently http://chemed.chem.purdue.edu/genchem/topicreview/bp/3organic/graphics/em.gif at low temperatures and with moderate strength nucleophiles.
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At high temperatures, or in the presence of a strong base, secondary halides undergo E2 elimination reactions.
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Tertiary halides undergo a combination of SN1 and E1 reactions. If the reaction is kept cool, and the nucleophile is a relatively weak base, it is possible to get nucleophilic substitution. At high temperatures, or with strong bases, elimination reactions predominate.
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