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In most general chemistry textbooks, the pKa of water at 25 ºC is listed as 14.0. In many organic chemistry textbooks and some biochemistry texts, however, the pKa of water at 25ºC is listed as 15.7. This module describes the derivation of these two values and describes why the value of 15.7 should not be used. Much of the information in this module comes from two review papers with more comprehensive coverage of the topic.1,2
Two Values for the Same Equilibrium Constant?
The value of 14.0 for the pKw of water at 25ºC has been experimentally measured.3,4 This value can also be obtained by the same process used to calculate the pKa of all other water-soluble compounds that can act as acids in aqueous solution: from the analysis of thermodynamic or electrochemical data for these aqueous solutions.
The origin of the value of 15.7 for the pKa of water can be tracked back at least as far as 1928, when Johannes Brønsted postulated that there could be two different values for the pKa of water.5 One value was the “conventional”, thermodynamically-sound equilibrium constant, in which the activity of the water is used, and assigned its correct value of unity (1) because it is the solvent in a dilute solution. Brønsted called the second constant the “rational” acidity constant (which we distinguish as K*a,H2O), in which he included a molar concentration of water. But Brønsted stated that this molar concentration “does not possess an explicit value” and thus it is not possible to calculate a value for K*a,H2O or pK*a,H2O. By 1937, Brønsted, seemed to have given this problem further thought and gave the “conventional” values precedence over the untenable “rational” values. Specifically, in his 1937 textbook Physical Chemistry,6 Brønsted defined an “autoprotolysis” constant in water, K(H2O) = [H+][OH-] which is not only equal to our modern autoionization constant, Kw, but also, by analogy to his discussion of the ionization of dilute acids, equal to Ka,H2O. Mathematically, Brønsted stated here that pKw = pKa,H2O. In addition, Brønsted included only the conventional values in a table in this textbook that lists the pKa values of several well-known acids.
Thus, nine years after presenting both the “rational” and “conventional” values, Brønsted seems to have set aside the “rational” values in favor of the activity-based, thermodynamically correct “conventional” values. Nevertheless, the erroneous 15.7 value, as well as the erroneous -1.7 value for the pKa of the aqueous proton (H3O+ or H+), was used by many organic chemists who employed the "Brønsted equation" in their research into the relationship between acid-base strength and catalytic power, even though these erroneous values led to poor correlations when plotted along with other aqueous acids.7,8,9
A second common source of the incorrect 15.7 value is the erroneous explanation of why the concentration of water, as the solvent, does not appear in the law of mass action for a reaction. This misunderstanding, and the corresponding miscalculation, will be fully described and remedied later in this module.
A third, deliberately developed, source of the value of 15.7 for the pKa of water was a theoretical proposal made by physical organic chemists who were under the false impression that solvents would not appreciably affect the relative strength of acids. These chemists experimentally determined the pKa of methanol dissolved in water at 25°C to be 15.5.10 They knew that previous research had shown that when 2-propanol was used as the solvent, methanol was roughly 3 times more acidic than water.11 They also knew that the commonly accepted pKa of pure water was 14.0. Thus, as it stood, the relative acidity of methanol and water dissolved in 2-propanol (methanol > water) differed from the relative acidity of water and methanol dissolved in water (water > methanol). In an attempt to make their experimental data conform to the incorrect assumption that the solvent identity does not affect relative acid strength, the chemists who measured the pKa of methanol mathematically manipulated the pKa of water so that the relative acidities in the two different solvents would agree. It will be shown below that this mathematical manipulation is incorrect and unnecessary because solvent is known to affect relative acid strength.12,13
Thermodynamically Correct Derivation of $K_a$ and $pK_a$
Thermodynamic equilibrium constants ($K_{eq}$) are defined based on the activity of each of the substances involved in the chemical reaction.14 The activity of a species is a measure of the “effective concentration” of that species that results from the attractive and repulsive interactions of particles in non-ideal mixtures. Thus, for the reaction:
$\ce{bB + cC <=> dD + eE} \label{1}$
The law of mass action is written as
$K_{eq} = \dfrac{a_D^d · a_E^e}{a_B^b · a_C^c} \label{1eq}$
where $a_\ce{D}^d$ indicates activity of species $\ce{D}$ raised to the $d$ power.
In an ideal dilute solution, the solutes are treated with Henry’s law so that activities can be approximated with concentrations. It is most appropriate to approximate the activity of a solute in a dilute solution with the molality of that solute (m = moles of solute per kilogram of solvent). However, for an ideal dilute aqueous solution at 25 ºC, the molality of a solute is often approximated with the molarity of that solute (M = moles of solute per Liter of solution) because the density of water at 25 ºC is 0.997 kilograms per Liter:
$a_{\ce{D}} \approx \text{molality}_\ce{D} \approx {[\ce{D}]} \nonumber$
where ${[\ce{D}]}$ indicates the concentration of the species $D$ in units of molarity.
The solvent in an ideal dilute solution is treated with Raoult’s law so that the solvent can be approximated as the pure liquid. When the solvent is approximated as a pure liquid, its activity is approximated as unity:
$a_{solvent} \approx 1 \nonumber$
The concentration of the solvent does not explicitly appear in the equilibrium expression (Equation \ref{1eq}). The approximation being made is that intermolecular forces are negligible. If the solution is non-ideal, then the effects of the intermolecular forces must be accounted for by employing activity coefficients as correction factors.
The use of activity coefficients is the thermodynamically valid method to describe nonideal solutions, and is unrelated to the method of involving the solvent concentration in the calculations described below.
Thus, for a reaction in which water is a participant and also the solvent, the activity of water is not “left out” of the law of mass action. Rather, it remains in the law of mass action, but it is assumed to have a value of unity (the number 1), and so its value has no effect on the value of $K_{eq}$. This process is demonstrated in Example $1$.
Example $1$: Dissociation of Hypochlorous Acid
Consider the acid dissociation of hypochlorous acid
$\ce{HClO(aq) + H2O(l) <=> ClO^{-}(aq) + H3O^{+}(aq)} \label{2a}$
with the following equilibrium constant
$K_a = \dfrac{a_{\ce{ClO^{-}}} · a_{\ce{H3O^{+}}}}{a_{\ce{HClO}} · a_{\ce{H_2O}}} \approx \dfrac{[\ce{ClO^{-}}][\ce{H_3O^{+}}]}{[\ce{HClO}]\{1\}} = \dfrac{[\ce{ClO^{-}}][\ce{H_3O^{+}}]}{[\ce{HClO}]} \nonumber$
Using values of $\Delta G_f^o$ at 25ºC15 for the species involved, and the Equation $ΔGº = -RT\ln K$, it can be shown that for Equation $\ref{2a}$:
\begin{align*} \Delta G_{reaction}^o &= {\{\Delta G_f^o\, \ce{ClO^{-}(aq)} + \Delta G_f^o\, \ce{H3O^{+}(aq)}}\} - {\{\Delta G_f^o \,\ce{HClO(aq)} + \Delta G_f^o\, \ce{H2O(l)}}\} \[4pt] &= {\{-36.8\, \text{kJ} + -237.1\, \text{kJ}}\} - {\{-79.9\, \text{kJ} + -237.1\, \text{kJ}}\} \[4pt] &= 43.1\, \text{kJ} \end{align*}
thus $K_a = 2.8 \times 10^{-8} \nonumber$ or $pK_a=7.55 \nonumber$.
Some chemists prefer to use the short hand notation of $\ce{H^{+}}$ rather than $\ce{H_3O^{+}}$ to represent the proton exchanged during the acid/base reaction, resulting in the reaction equation:
$\ce{HClO(aq) <=> ClO^{-}(aq) + H^{+}(aq)} \label{2b}$
for which
$K'_a = \dfrac{a_{\ce{ClO^{-}}} · a_{\ce{H^{+}}}}{a_{\ce{HClO}}} \approx \dfrac{[\ce{ClO^{-}}][\ce{H^{+}}]}{[\ce{HClO}]} \nonumber$
Using values of $\Delta G_f^o$ at 25ºC15 for the species involved, and the Equation $\Delta G^o = -RT\ln K$, one can show that for Equation $\ref{2b}$:
\begin{align*} \Delta G_{reaction}^o &= {\{\Delta G_f^o \,\ce{ClO(aq)^{-}} + \Delta G_f^o\, \ce{H(aq)^{+}}}\} - {\{\Delta G_f^o\, \ce{HClO(aq)}}\} \[4pt] &= {\{-36.8\, \text{kJ} + 0\, \text{kJ}}\} - {\{-79.9\, \text{kJ}}\} \[4pt] &= 43.1\, \text{kJ} \end{align*}
thus $K'_a = K_a = 2.8 \times 10^{-8} \nonumber$ or $pK_a=7.55 \nonumber$.
As expected, the value of $K_a$ equals the value of $K'_a$ and thus does not depend on the manner in which the exchanged proton is represented.
A similar discussion applies to the autoprotolysis of water, in which one water molecule acts as an acid, and the other water molecule acts as a base:
$\ce{H2O(l) + H2O(l) <=> OH^{-}(aq) + H3O^{+}(aq)} \label{3a}$
or
$\ce{H2O(l) <=> OH^{-}(aq) + H^{+}(aq)} \label{3b}$
resulting in the following equilibrium constant:
$K_{eq} = \dfrac{a_{OH^-} · a_{\ce{H3O^{+}}}}{a_{\ce{H2O}}^2} \approx \dfrac{[\ce{OH^{-}}][\ce{H_3O^{+}}]}{1^2} = [\ce{OH^{-}}][\ce{H3O^{+}}] \nonumber$
or
$K'_{eq} = \dfrac{a_{\ce{OH^{-}}} · a_{\ce{H^{+}}}}{a_{\ce{H_2O}}} \approx \dfrac{[\ce{OH^{-}}][\ce{H^{+}}]}{1} = [\ce{OH^{-}}][\ce{H^{+}}] \nonumber$
Using values of $\Delta G_f^o$ at 25ºC15 for the species involved, and the Equation $\Delta G^o = -RT\ln K$, one can show that for reaction $\ref{3a}$:
\begin{align*} \Delta G_{reaction}^o &= {\{\Delta G_f^o\, \ce{OH^{-}(aq)} + \Delta G_f^o\, \ce{H3O^{+}(aq)}}\} - {\{2 \times \Delta G_f^o\, \ce{H2O(l)}}\} \[4pt] &= {\{-157.2\, \text{kJ} + -237.1\,\text{kJ}}\} - {\{2(-237.1\, \text{kJ})}\} \[4pt] &= 79.9\, \text{kJ} \end{align*}
thus $K_{eq} = 1 \times 10^{-14} = K_a \nonumber$ or $pK_a =14.0 \nonumber$.
Using values of $\Delta G_f^o$ at 25ºC15 for the species involved, and the Equation $\Delta G^o = -RT\ln K$, one can show that for Equation $\ref{3b}$:
\begin{align*} \Delta G_{reaction}^o &= {\{\Delta G_f^o\, \ce{OH^{-}(aq)} + \Delta G_f^o \,\ce{H^{+}(aq)}}\} - {\{\Delta G_f^o\, \ce{H2O(l)}}\} \[4pt] &= {\{-157.2\,\, kJ + 0\, \text{kJ} }\} - {\{-237.1\, \text{kJ} }\} \[4pt] &= 79.9\, \text{kJ} \end{align*}
thus $K'_{eq} = 1 \times 10^{-14} = K_{eq} = K_a \nonumber$ or $pK_a=14.0 \nonumber$.
In addition, it is possible to show that for both reactions $\ref{3a}$ and $\ref{3b}$,
$K_{eq} = K_a = 1 \times 10^{-14} \nonumber$
using values of $E^o$ at 25ºC and the Equation $nFE^o = RT\ln K$.1 It is a standard and accepted practice to calculate $K_a$ values in both of these ways, and experimentally measured values (or extrapolated estimates)9 confirm these calculated values.
Once again, the value of $K_{eq}$ (and thus $K_a$) does not depend on the manner in which the exchanged proton is represented. It will be shown below that this equivalency of equations is not possible if the $pK_a$ of water is set as 15.7.
Non-Thermodynamic Derivations
Some general chemistry textbooks list the thermodynamically correct values of the $K_a$ for species in aqueous solutions, but use an incorrect explanation as to why water does not appear in the law of mass action equation. These textbooks state that the concentration of water is constant and then claim that the constant value of the concentration has been included in the value of Ka. Other texts explicitly state that the concentration of water is 55.33 Molar at 25ºC (1.000L of water at a density of 0.9970 g/mL, and a molar mass of 18.02 grams/mole), and then explicitly multiply $K_{eq}$ by 55.33 to obtain what they claim is a different constant, Ka. As pointed out above, the concentration of water does not have a place in the determination of the value of Ka for water, or for the K of any reaction involving water that occurs in aqueous solution. Thus, although the correct Ka values are shown in these texts, the assumptions that led to the simplification of the law of mass action are lost, especially the assumption that the activity of the water, as solvent, is unity.
The mathematical and chemical rationales presented by the chemists attempting to reconcile the relative acidity of substances in water and in non-aqueous solvents contain similar mistakes. The correct and commonly accepted ways of writing the equations for the autoprotolysis of water are Equations $\ref{3a}$ or $\ref{3b}$. Instead, Ballinger and Long claim that is possible to differentiate in solution between the water molecules that are acting as solvent molecules and the water molecules that are acting as solute molecules in the role of Brønsted acids.10 This claim suggests the following equations for the autoprotolysis of water:
$\ce{H2O(aq) + H2O(l) <=> OH^{-}(aq) + H3O^{+}(aq)} \label{4a}$
or
$\ce{H2O(aq) <=> OH^{-}(aq) + H^{+}(aq)} \label{4b}$
Note the difference in the phases of water between Equations $\ref{4a}$ or $\ref{4b}$ with Equations $\ref{3a}$ or $\ref{3b}$. Ballinger and Long make the commonly accepted assumption that the solvent water molecules be treated as the solvent part of an ideal dilute solution, with an activity of 1. They then assume that the solute water molecules be treated as the solute part of an ideal solution whose activity is approximated as the numerical value of concentration of water, 55.33.
With these assumptions, the law of mass action for Equations $\ref{4a}$ would be expressed as:
$K'_{a} = \dfrac{a_{\ce{OH^{-}}} · a_{\ce{H3O^{+}}}}{a_{\ce{H2O(aq)}} · a_{\ce{H2O(l)}}} \approx \dfrac{[\ce{OH^{-}}][\ce{H3O^{+}}]}{[\ce{H2O(aq)}] · 1} = \dfrac{[\ce{OH^{-}}][\ce{H3O^{+}}]}{[\ce{H2O(aq)}]} \nonumber$
and the law of mass action for Equation $\ref{4b}$ would be expressed as:
$K''_{a} = \dfrac{a_{OH^-} · a_{H_3O^+}}{a_{H_2O_{(aq)}}} \approx \dfrac{[OH^-][H_3O^+]}{[H_2O_{(aq)}]}$
At 25 ºC, the product of $[\ce{OH^{-}(aq)}][\ce{H3O^{+}(aq)}] = 1 \times 10^{-14}$ and $[\ce{H2O(aq)}] = 55.33$, thus
$K'_{a}=K''_{a }\approx\dfrac{[\ce{OH^{-}(aq)}][\ce{H3O^{+}(aq)}]}{[\ce{H2O(aq)}]}= \dfrac{K_{eq}}{55.33} = \dfrac{1 \times 10^{-14}}{55.33} = 2 \times 10^{-16} = K_a^* \label{5}$
Therefore
$pK^*_a=15.7 \nonumber$
The assumptions made in this non-thermodynamic derivation require the acceptance of the following conundrums:
1. The idea that the few water molecules acting as the solute have a concentration of 55.33 M and that the water molecules acting as the solvent are pure liquid implies that we have two sets of water molecules (solute ones and solvent ones) both with a concentration of 55.33 M. This is false. The sum of the two water concentrations would have to be 55.33 M: $[\ce{H2O(l)}] + [\ce{H2O(aq)}]=55.33\;M \nonumber$ Additionally, setting the activity of the “solute” water molecules as 55.33 and the activity of the “solvent” water molecules as 1 implies that one set of water molecules is 55.33 times more reactive than the other set. This is an absurd assertion. Activity is not merely a convention. It is a description of the potential for a chemical species to react.16
2. Regarding item 1, some chemists argue that the acid dissociation constant for a mixture of labeled "solute" water acting as an acid in unlabeled water solvent should be calculated as $\dfrac{(H_3O^+)(OH^-)(mole \ fraction \ labeled "solute" water)}{(molarity \ labeled "solute" water)} = \dfrac{(1.0 x 10^{-7})(1.0 x 10^{-7})(mole \ fraction \ labeled "solute" water)}{(molarity \ labeled "solute" water)}$If we are to entertain the premise of "solute" water molecules, then it is important to realize that both the "solute" and the solvent water molecules in the autoionization reaction are water molecules, making the solution an ideal solution. In an ideal solution in which the "solute" and the solvent particles experience identical intermolecular forces, both "solute" and solvent molecules obey Raoult's Law.17 Therefore, the activity coefficients of the "solute" and the solvent are both 1. In such cases, the mole fraction equals the activity. Thus, it is incorrect to use the concentration of the "solute" to replace the activity of the "solute" in the calculation of the equilibrium constant. Instead, the mole fraction of the "solute" molecules must be used. $\dfrac{(H_3O^+)(OH^-)(mole \ fraction \ labeled "solute" water)}{(mole \ fraction \ labeled "solute" water)} = \dfrac{(1.0 x 10^{-7})(1.0 x 10^{-7})(mole \ fraction \ labeled "solute" water)}{(mole \ fraction \ labeled "solute" water)} = 1.0 x 10^{-14}$
3. As stated above, it is common practice to represent the autoprotolysis of water as Equations $\ref{4a}$ or $\ref{4b}$. The fact that these two reactions are equivalent is absolutely dependent upon the fact that all of the water molecules are identical.
Note: Regarding the Erroneous claim that Ka = [H2O]Keq
Many documents make the erroneous claim that the $K_a$ for any acid is equal to $[H_2O]K_{eq}$. Not only is this statement incorrect because of its improper use of $[H_2O]$, but it is also not the Equation that Ballinger and Long derived (incorrectly!) in their attempt to make the pKa of water equal 15.7.
Notice that in equation$\ref{5}$, $K_a^* = \dfrac{K_{eq}}{55.33}$. Rearranging this Equation gives $K_a^*[55.33] = K_{eq}$, not $K_a^*=[55.33]K_{eq}$.
The first instance of the use of $K_a=\ce[H_2O}]K_{eq}$ has not yet been tracked down, but it must have been made by someone who not only misunderstood thermodynamics, but also misread the Ballinger and Long paper.
Notation Issues
The following discussion shows that keeping track of the phase notation of $\ce{H2O(l)}$ versus $\ce{H2O(aq)}$ allows one to show that the $\ce{H2O(aq)}$ notation is untenable.
Using the Correct Notation
The process used to prove the equivalency of Equations $\ref{3a}$ and $\ref{3b}$ employs a basic rule of equilibrium chemistry; if two reactions are added together to make a new reaction, the equilibrium constant of the new reaction is the product of the equilibrium constants of the two original reactions. Applying that rule to convert from Equation $\ref{3a}$ to Equation $\ref{3b}$ requires a third equation:
$\ce{H3O^{+}(aq) <=> H2O(l) + H^{+}(aq)}\label{5a}$
Equation $\ref{5a}$ is developed from Equation $\ref{3a}$ as follows. One of the $\ce{H2O(l)}$ molecules in Equation $\ref{3a}$ acts as an acid, and its conjugate base is OH-. The other $\ce{H2O(l)}$ molecule must then act as a base, and its conjugate acid is $\ce{H3O^{+}}$. (There is no chemical difference between the two H2O(l) molecules in the equation. The likelihood of a particular water molecule’s acting as an acid or as a base is governed by the statistics of random events.) The value of K for Equation $\ref{5a}$ can be obtained from the equation
$K_a \times K_b = 1 \times 10^{-14}$
which holds for conjugate acid/base pairs in water at 25ºC. Because Kb = 1.0 x 10-14 for H2O, the Ka = 1 for H3O+ . It is also true that pKa + pKb = 14.0 for conjugate acid/base pairs in water. Thus, because the pKb of H2O is 14.0, the pKa of H3O+ is 0.0
The conversion of Equation $\ref{3a}$ to Equation $\ref{3b}$ is then:
$H_2O_{(l)} + H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H_3O^+_{(aq)}$
$+ H_3O^+_{(aq)} \rightleftharpoons H_2O_{(l)} + H^+_{(aq)}$
$K_{6} = 1 \times 10^{-14}$
$K_{14} = 1$
pK = 14.0
pK = 0.0
$H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H^+_{(aq)}$
$K_{7} = K_{6} · K_{14} = 1 \times 10^{-14}$ pK = 14.0
Equations $\ref{3a}$ and $\ref{3b}$ are balanced, result in equivalent products, and because $K_{6} = K_{7}$, the two equations have the same value of $K_{eq}$. These are equivalent equations.
Using the Incorrect Notation
Attempting this same conversion with Equations $\ref{4a}$ and $\ref{4b}$ leads to at least two impossible scenarios caused by the differentiation between $\ce{H2O(l)}$ and $\ce{H2O(aq)}$.
To obtain an Equation equivalent to Equation $\ref{5a}$, conjugate acid/base pairs must be assigned in Equation $\ref{4a}$:
$\underbrace{\ce{H2O(aq)}}_{\text{acid}} + \underbrace{\ce{H2O(l)}}_{\text{base}} \rightleftharpoons \underbrace{\ce{OH^{-}(aq)}}_{\text{conjugate base}} + \underbrace{\ce{H3O^{+}(aq)}}_{\text{conjugate acid}} \label{4aEx}$
For example, if $\ce{H2O(aq)}$ is the acid, then its conjugate base is $\ce{OH^{-}}$. This means that $\ce{H2O(l)}$ and $\ce{H3O^{+}}$ must be a conjugate pair. There is no conjugate acid related to $\ce{H2O(aq)}$ in this reaction. Without a conjugate acid for $\ce{H2O(aq)}$, the conversion from $\ref{4a}$ to $\ref{4b}$ cannot be carried out.
One possible solution is to obtain a conjugate acid for $\ce{H2O(aq)}$ by proposing that the very active $\ce{H2O(aq)}$ molecules are also able to act as Brønsted bases. These $\ce{H2O(aq)}$ molecules would thus have $\ce{H3O^{+}}$ as their conjugate acid. The equation for this reaction would be:
$\ce{H3O^{+}(aq) <=> H2O(aq) + H^{+}(aq)}\label{5b}$
Although this proposal is not unreasonable, it is still a part of an impossible scenario because the conversion of Equation $\ref{4a}$ to Equation $\ref{4b}$ would be:
$\ce{H2O(aq) + H2O(l) <=> OH^{-}(aq) + H3O^{+}(aq)}$
$\ce{ + H3O^{+}(aq) <=> H2O(aq) + H^{+}(aq)} \nonumber$
$\ce{H2O(l) <=> OH^{-}(aq) + H^{+}(aq)} \nonumber$
The resulting Equation is not $\ref{4b}$, but $\ref{3b}$. This is a glaring error that clearly and simply shows that Ballinger and Long’s proposal is unsupportable.
A second method for obtaining a conjugate acid for $\ce{H2O(aq)}$ is commonly found in most of the textbooks espousing Ballinger and Long’s value of pKa. In this method, Equation $\ref{5a}$ is accepted as the Equation describing $\ce{H3O^{+}}$ as the conjugate acid of $\ce{H2O(aq)}$, with complete disregard for the phase designation of the $\ce{H2O}$. Nevertheless, using the fact that
$pK_a + pK_b = 14.0 \nonumber$
in water at room temperature, and the claim that because the pKa of water is 15.7, the pKb of water must also be 15.7, the pKa of H3O+ is calculated to be -1.7. To clarify, the Equation $\ref{5a}$ is (incorrectly!) assigned a pKa = -1.7, with a corresponding value of 50 for the $K_a$.
The resulting combination of $\ref{4a}$ and $\ref{5a}$, with their (incorrectly!) assigned $K$ values givess:
$\; H_2O_{(aq)} + H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H_3O^+_{(aq)}$
$+\; H_3O^+_{(aq)} \rightleftharpoons H_2O_{(l)} + H^+_{(aq)}$
$K_{8} = 2 \times 10^{-16}$
$K_{14} = 50$
pK = 15.7
pK = -1.7
$\; H_2O_{(aq)} \rightleftharpoons OH^-_{(aq)} + H^+_{(aq)}$
$K_{9} = K_{8} · K_{14} = 1 \times 10^{-14}$ pK = 14.0
In this case, the two equations are balanced and they result in the equivalent products, but because $K_{8}$ does not equal $K_{9}$, the two equations do not have the same value of $K_{eq}$. These are not equivalent equations. This is a second error that clearly and simply shows that Ballinger and Long’s proposal is unsupportable.
For Further Consideration: Dissociation Constant of Methanol
The great irony behind all of this discussion is that the Ballinger and Long10 paper not only presents data that clearly show that methanol is a weaker acid than water in aqueous solution, but also employs the value of 14.0 for the pKa of water to calculate the pKa of methanol.
The authors determined that for the reaction
$\ce{CH3OH(aq) + OH^{-}(aq) <=> CH3O^{-}(aq) + H2O(l)} \label{41}$
the equilibrium constant, K, has a value of 0.029 at 25.0oC. This value clearly shows that water is the stronger acid than methanol, because the methanol is unlikely to protonate the hydroxide ion (the conjugate base of water). Instead, the opposite is true; water is likely to protonate the methoxide ion (the conjugate base of the methanol.)
To obtain the direct reaction between $\ce{CH3OH}$ and $\ce{H2O}$,
$\ce{CH3OH(aq) + H2O(l) <=> CH3O^{-}(aq) + H3O^{+}(aq)} \label{42}$
along with the value of the equilibrium constant for this reaction, Ballinger and Long add reactions $\ref{41}$ and $\ref{3a}$ to obtain $\ref{42}$:
$\; CH_3OH_{(aq)} + OH^-_{(aq)} \rightleftharpoons CH_3O^-_{(aq)} + H_2O_{(l)}$
$+\; H_2O_{(l)} + H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H_3O^+_{(aq)}$
$K_{18} = 0.029$
$K_{6} = 1 \times 10^{-14}$
pK = 1.54
pK = 14.0
$\; CH_3OH_{(aq)} + H_2O_{(l)} \rightleftharpoons CH_3O^-_{(aq)} + H_3O^+_{(aq)}$
$K_{19} = K_{18} · K_{6} = 2.9 \times 10^{-16}$ pK = 15.54
Thus, there are two more pieces of evidence in the paper that seems to have started the discussion, that the entire discussion concerning the relative strength of methanol and water as acids in aqueous solution has been completely misunderstood.
All experimental data show that water has a pKw = pKa = 14.0 at 25.0oC, and all experimental data show that methanol is a weaker acid than water in aqueous solution. There is absolutely no experimental evidence to contradict either of these two statements.
Conclusions
There would be little or no confusion in this matter if it were made clear from the outset that solvents can affect the relative acidity of compounds.7,8 A straightforward method of clarifying the effect of solvent on relative acid strengths would be to present a separate table for each solvent. There would also be less confusion in this matter if it were made clear that the solvent does belong in the law of mass action in its complete form, but that the solvent does not usually appear in the final form of the law of mass action because the solvent in an ideal dilute solution has an activity of unity.
It is incorrect and misleading to present the value of 15.7 for the pKa of water, yet this value has entered the fields of organic chemistry and biochemistry. The proposed value of 1.8 x 10-16 for the Ka of water cannot be justified with thermodynamic data, nor are there any experimental data to support this value. In fact, 1.8 x 10-16 is a hypothetical value that was arrived at using specious arguments in order to justify an incorrect assumption that the relative strengths of acids were not affected by changes in solvent. There is no reason to use this value.
The Ka of water at 25 ºC is $1 \times 10^{-14}$ and the pKa is 14.0. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/What_is_the_pKa_of_water.txt |
The triesters of fatty acids with glycerol (1,2,3-trihydroxypropane) compose the class of lipids known as fats and oils. These triglycerides (or triacylglycerols) are found in both plants and animals, and compose one of the major food groups of our diet. Triglycerides that are solid or semisolid at room temperature are classified as fats, and occur predominantly in animals. Those triglycerides that are liquid are called oils and originate chiefly in plants, although triglycerides from fish are also largely oils. Some examples of the composition of triglycerides from various sources are given in the following table.
Saturated Acids (%)
Unsaturated Acids (%)
Source
C10
& less
C12
lauric
C14
myristic
C16
palmitic
C18
stearic
C18
oleic
C18
linoleic
C18
unsaturated
Animal Fats
butter
15
2
11
30
9
27
4
1
lard
-
-
1
27
15
48
6
2
human fat
-
1
3
25
8
46
10
3
herring oil
-
-
7
12
1
2
20
52
Plant Oils
coconut
-
50
18
8
2
6
1
-
corn
-
-
1
10
3
50
34
-
olive
-
-
-
7
2
85
5
-
palm
-
-
2
41
5
43
7
-
peanut
-
-
-
8
3
56
26
7
safflower
-
-
-
3
3
19
76
-
As might be expected from the properties of the fatty acids, fats have a predominance of saturated fatty acids, and oils are composed largely of unsaturated acids. Thus, the melting points of triglycerides reflect their composition, as shown by the following examples. Natural mixed triglycerides have somewhat lower melting points, the melting point of lard being near 30 º C, whereas olive oil melts near -6 º C. Since fats are valued over oils by some Northern European and North American populations, vegetable oils are extensively converted to solid triglycerides (e.g. Crisco) by partial hydrogenation of their unsaturated components. Some of the remaining double bonds are isomerized (to trans) in this operation. These saturated and trans-fatty acid glycerides in the diet have been linked to long-term health issues such as atherosclerosis.
Triglycerides having three identical acyl chains, such as tristearin and triolein (above), are called "simple", while those composed of different acyl chains are called "mixed". If the acyl chains at the end hydroxyl groups (1 & 3) of glycerol are different, the center carbon becomes a chiral center and enantiomeric configurations must be recognized.
The hydrogenation of vegetable oils to produce semisolid products has had unintended consequences. Although the hydrogenation imparts desirable features such as spreadability, texture, "mouth feel," and increased shelf life to naturally liquid vegetable oils, it introduces some serious health problems. These occur when the cis-double bonds in the fatty acid chains are not completely saturated in the hydrogenation process. The catalysts used to effect the addition of hydrogen isomerize the remaining double bonds to their trans configuration. These unnatural trans-fats appear to to be associated with increased heart disease, cancer, diabetes and obesity, as well as immune response and reproductive problems.
Fatty Acids
The common feature of these lipids is that they are all esters of moderate to long chain fatty acids. Acid or base-catalyzed hydrolysis yields the component fatty acid, some examples of which are given in the following table, together with the alcohol component of the lipid. These long-chain carboxylic acids are generally referred to by their common names, which in most cases reflect their sources. Natural fatty acids may be saturated or unsaturated, and as the following data indicate, the saturated acids have higher melting points than unsaturated acids of corresponding size. The double bonds in the unsaturated compounds listed on the right are all cis (or Z).
The higher melting points of the saturated fatty acids reflect the uniform rod-like shape of their molecules. The cis-double bond(s) in the unsaturated fatty acids introduce a kink in their shape, which makes it more difficult to pack their molecules together in a stable repeating array or crystalline lattice. The trans-double bond isomer of oleic acid, known as elaidic acid, has a linear shape and a melting point of 45 ºC (32 ºC higher than its cis isomer). The shapes of stearic and oleic acids are displayed in the models below.
Stearic acid
Oleic aci
Two polyunsaturated fatty acids, linoleic and linolenic, are designated "essential" because their absence in the human diet has been associated with health problems, such as scaley skin, stunted growth and increased dehydration. These acids are also precursors to the prostaglandins, a family of physiologically potent lipids present in minute amounts in most body tissues.
Because of their enhanced acidity, carboxylic acids react with bases to form ionic salts, as shown in the following equations. In the case of alkali metal hydroxides and simple amines (or ammonia) the resulting salts have pronounced ionic character and are usually soluble in water. Heavy metals such as silver, mercury and lead form salts having more covalent character (3rd example), and the water solubility is reduced, especially for acids composed of four or more carbon atoms.
RCO2H
+
NaHCO3
RCO2(–) Na(+) + CO2 + H2O
RCO2H
+
(CH3)3N:
RCO2(–) (CH3)3NH(+)
RCO2H
+
AgOH
RCO2δ(–) Agδ(+) + H2O
Lipid Soluble Vitamins
The essential dietary substances called vitamins are commonly classified as "water soluble" or "fat soluble". Water soluble vitamins, such as vitamin C, are rapidly eliminated from the body and their dietary levels need to be relatively high. The recommended daily allotment (RDA) of vitamin C is 100 mg, and amounts as large as 2 to 3 g are taken by many people without adverse effects. The lipid soluble vitamins, shown in the diagram below, are not as easily eliminated and may accumulate to toxic levels if consumed in large quantity. The RDA for these vitamins are:
• Vitamin A 800 μg ( upper limit ca. 3000 μg)
• Vitamin D 5 to 10 μg ( upper limit ca. 2000 μg)
• Vitamin E 15 mg ( upper limit ca. 1 g)
• Vitamin K 110 μg ( upper limit not specified)
From this data it is clear that vitamins A and D, while essential to good health in proper amounts, can be very toxic. Vitamin D, for example, is used as a rat poison, and in equal weight is more than 100 times as poisonous as sodium cyanide.
From the structures shown here, it should be clear that these compounds have more than a solubility connection with lipids. Vitamins A is a terpene, and vitamins E and K have long terpene chains attached to an aromatic moiety. The structure of vitamin D can be described as a steroid in which ring B is cut open and the remaining three rings remain unchanged. The precursors of vitamins A and D have been identified as the tetraterpene beta-carotene and the steroid ergosterol, respectively.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Lipids/Properties_and_Classification_of_Lipids/Fats_and_Oils.txt |
Phospholipids are the main constituents of cell membranes. They resemble the triglycerides in being ester or amide derivatives of glycerol or sphingosine with fatty acids and phosphoric acid. The phosphate moiety of the resulting phosphatidic acid is further esterified with ethanolamine, choline or serine in the phospholipid itself. The following diagram shows the structures of some of these components. Clicking on the diagram will change it to display structures for two representative phospholipids. Note that the fatty acid components (R & R') may be saturated or unsaturated.
As ionic amphiphiles, phospholipids aggregate or self-assemble when mixed with water, but in a different manner than the soaps and detergents. Because of the two pendant alkyl chains present in phospholipids and the unusual mixed charges in their head groups, micelle formation is unfavorable relative to a bilayer structure. If a phospholipid is smeared over a small hole in a thin piece of plastic immersed in water, a stable planar bilayer of phospholipid molecules is created at the hole. As shown in the following diagram, the polar head groups on the faces of the bilayer contact water, and the hydrophobic alkyl chains form a nonpolar interior. The phospholipid molecules can move about in their half the bilayer, but there is a significant energy barrier preventing migration to the other side of the bilayer.
This bilayer membrane structure is also found in aggregate structures called liposomes. Liposomes are microscopic vesicles consisting of an aqueous core enclosed in one or more phospholipid layers. They are formed when phospholipids are vigorously mixed with water. Unlike micelles, liposomes have both aqueous interiors and exteriors.
A cell may be considered a very complex liposome. The bilayer membrane that separates the interior of a cell from the surrounding fluids is largely composed of phospholipids, but it incorporates many other components, such as cholesterol, that contribute to its structural integrity. Protein channels that permit the transport of various kinds of chemical species in and out of the cell are also important components of cell membranes.
The interior of a cell contains a variety of structures (organelles) that conduct chemical operations vital to the cells existence. Molecules bonded to the surfaces of cells serve to identify specific cells and facilitate interaction with external chemical entities. The sphingomyelins are also membrane lipids. They are the major component of the myelin sheath surrounding nerve fibers. Multiple Sclerosis is a devastating disease in which the myelin sheath is lost, causing eventual paralysis.
Prostaglandins Thromboxanes and Leukotrienes
The members of this group of structurally related natural hormones have an extraordinary range of biological effects. They can lower gastric secretions, stimulate uterine contractions, lower blood pressure, influence blood clotting and induce asthma-like allergic responses. Because their genesis in body tissues is tied to the metabolism of the essential fatty acid arachadonic acid (5,8,11,14-eicosatetraenoic acid) they are classified as eicosanoids. Many properties of the common drug aspirin result from its effect on the cascade of reactions associated with these hormones.
The metabolic pathways by which arachidonic acid is converted to the various eicosanoids are complex and will not be discussed here. A rough outline of some of the transformations that take place is provided below. It is helpful to view arachadonic acid in the coiled conformation shown in the shaded box.
Leukotriene A is a precursor to other leukotriene derivatives by epoxide opening reactions. The prostaglandins are given systematic names that reflect their structure. The initially formed peroxide \(PGH_2\) is a common intermediate to other prostaglandins, as well as thromboxanes such as \(TXA_2\). | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Lipids/Properties_and_Classification_of_Lipids/Phospholipids.txt |
Carboxylic acids and salts having alkyl chains longer than eight carbons exhibit unusual behavior in water due to the presence of both hydrophilic (CO2) and hydrophobic (alkyl) regions in the same molecule. Such molecules are termed amphiphilic (Gk. amphi = both) or amphipathic. Fatty acids made up of ten or more carbon atoms are nearly insoluble in water, and because of their lower density, float on the surface when mixed with water. Unlike paraffin or other alkanes, which tend to puddle on the waters surface, these fatty acids spread evenly over an extended water surface, eventually forming a monomolecular layer in which the polar carboxyl groups are hydrogen bonded at the water interface, and the hydrocarbon chains are aligned together away from the water. This behavior is illustrated in the diagram on the right. Substances that accumulate at water surfaces and change the surface properties are called surfactants.
Alkali metal salts of fatty acids are more soluble in water than the acids themselves, and the amphiphilic character of these substances also make them strong surfactants. The most common examples of such compounds are soaps and detergents, four of which are shown below. Note that each of these molecules has a nonpolar hydrocarbon chain, the "tail", and a polar (often ionic) "head group". The use of such compounds as cleaning agents is facilitated by their surfactant character, which lowers the surface tension of water, allowing it to penetrate and wet a variety of materials.
Very small amounts of these surfactants dissolve in water to give a random dispersion of solute molecules. However, when the concentration is increased an interesting change occurs. The surfactant molecules reversibly assemble into polymolecular aggregates called micelles. By gathering the hydrophobic chains together in the center of the micelle, disruption of the hydrogen bonded structure of liquid water is minimized, and the polar head groups extend into the surrounding water where they participate in hydrogen bonding. These micelles are often spherical in shape, but may also assume cylindrical and branched forms, as illustrated on the right. Here the polar head group is designated by a blue circle, and the nonpolar tail is a zig-zag black line.
The oldest amphiphilic cleaning agent known to humans is soap. Soap is manufactured by the base-catalyzed hydrolysis (saponification) of animal fat. Before sodium hydroxide was commercially available, a boiling solution of potassium carbonate leached from wood ashes was used. Soft potassium soaps were then converted to the harder sodium soaps by washing with salt solution. The importance of soap to human civilization is documented by history, but some problems associated with its use have been recognized. One of these is caused by the weak acidity (pKa ca. 4.9) of the fatty acids. Solutions of alkali metal soaps are slightly alkaline (pH 8 to 9) due to hydrolysis. If the pH of a soap solution is lowered by acidic contaminants, insoluble fatty acids precipitate and form a scum. A second problem is caused by the presence of calcium and magnesium salts in the water supply (hard water). These divalent cations cause aggregation of the micelles, which then deposit as a dirty scum.
These problems have been alleviated by the development of synthetic amphiphiles called detergents (or syndets). By using a much stronger acid for the polar head group, water solutions of the amphiphile are less sensitive to pH changes. Also the sulfonate functions used for virtually all anionic detergents confer greater solubility on micelles incorporating the alkaline earth cations found in hard water. Variations on the amphiphile theme have led to the development of other classes, such as the cationic and nonionic detergents shown above. Cationic detergents often exhibit germicidal properties, and their ability to change surface pH has made them useful as fabric softeners and hair conditioners. These versatile chemical "tools" have dramatically transformed the household and personal care cleaning product markets over the past fifty years. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Lipids/Properties_and_Classification_of_Lipids/Soaps_and_Detergents.txt |
The important class of lipids called steroids are actually metabolic derivatives of terpenes, but they are customarily treated as a separate group. Steroids may be recognized by their tetracyclic skeleton, consisting of three fused six-membered and one five-membered ring, as shown in the diagram to the right. The four rings are designated A, B, C & D as noted, and the peculiar numbering of the ring carbon atoms (shown in red) is the result of an earlier misassignment of the structure. The substituents designated by R are often alkyl groups, but may also have functionality. The R group at the A:B ring fusion is most commonly methyl or hydrogen, that at the C:D fusion is usually methyl. The substituent at C-17 varies considerably, and is usually larger than methyl if it is not a functional group. The most common locations of functional groups are C-3, C-4, C-7, C-11, C-12 & C-17. Ring A is sometimes aromatic. Since a number of tetracyclic triterpenes also have this tetracyclic structure, it cannot be considered a unique identifier.
Steroids are widely distributed in animals, where they are associated with a number of physiological processes. Examples of some important steroids are shown in the following diagram. Norethindrone is a synthetic steroid, all the other examples occur naturally. A common strategy in pharmaceutical chemistry is to take a natural compound, having certain desired biological properties together with undesired side effects, and to modify its structure to enhance the desired characteristics and diminish the undesired. This is sometimes accomplished by trial and error.
The generic steroid structure drawn above has seven chiral stereocenters (carbons 5, 8, 9, 10, 13, 14 & 17), which means that it may have as many as 128 stereoisomers. With the exception of C-5, natural steroids generally have a single common configuration. This is shown in the last of the toggled displays, along with the preferred conformations of the rings.
Chemical studies of the steroids were very important to our present understanding of the configurations and conformations of six-membered rings. Substituent groups at different sites on the tetracyclic skeleton will have axial or equatorial orientations that are fixed because of the rigid structure of the trans-fused rings. This fixed orientation influences chemical reactivity, largely due to the greater steric hindrance of axial groups versus their equatorial isomers. Thus an equatorial hydroxyl group is esterified more rapidly than its axial isomer.
It is instructive to examine a simple bicyclic system as a model for the fused rings of the steroid molecule. Decalin, short for decahydronaphthalene, exists as cis and trans isomers at the ring fusion carbon atoms. Planar representations of these isomers are drawn at the top of the following diagram, with corresponding conformational formulas displayed underneath. The numbering shown for the ring carbons follows IUPAC rules, and is different from the unusual numbering used for steroids. For purposes of discussion, the left ring is labeled A (colored blue) and the right ring B (colored red). In the conformational drawings the ring fusion and the angular hydrogens are black.
The trans-isomer is the easiest to describe because the fusion of the A & B rings creates a rigid, roughly planar, structure made up of two chair conformations. Each chair is fused to the other by equatorial bonds, leaving the angular hydrogens (Ha) axial to both rings. Note that the bonds directed above the plane of the two rings alternate from axial to equatorial and back if we proceed around the rings from C-1 to C-10 in numerical order. The bonds directed below the rings also alternate in a complementary fashion.
Conformational descriptions of cis- decalin are complicated by the fact that two energetically equivalent fusions of chair cyclohexanes are possible, and are in rapid equilibrium as the rings flip from one chair conformation to the other. In each of these all chair conformations the rings are fused by one axial and one equatorial bond, and the overall structure is bent at the ring fusion. In the conformer on the left, the red ring (B) is attached to the blue ring (A) by an axial bond to C-1 and an equatorial bond to C-6 (these terms refer to ring A substituents). In the conformer on the right, the carbon bond to C-1 is equatorial and the bond to C-6 is axial. Each of the angular hydrogens (Hae or Hea) is oriented axial to one of the rings and equatorial to the other. This relationship reverses when double ring flipping converts one cis-conformer into the other.
Cis-decalin is less stable than trans-decalin by about 2.7 kcal/mol (from heats of combustion and heats of isomerization data). This is due to steric crowding (hindrance) of the axial hydrogens in the concave region of both cis-conformers, as may be seen in the model display activated by the following button. This difference is roughly three times the energy of a gauche butane conformer relative to its anti conformer. Indeed three gauche butane interactions may be identified in each of the cis-decalin conformations, as will be displayed by clicking on the above conformational diagram. These gauche interactions are also shown in the model.
Steroids in which rings A and B are fused cis, such as the example on the right, do not have the sameconformational mobility exhibited by cis-decalin. The fusion of ring C to ring B in a trans configuration prevents ring B from undergoing a conformational flip to another chair form. If this were to occur, ring C would have to be attached to ring B by two adjacent axial bonds directed 180º apart. This is too great a distance to be bridged by the four carbon atoms making up ring C. Consequently, the steroid molecule is locked in the all chair conformation shown here. Of course, all these steroids and decalins may have one or more six-membered rings in a boat conformation. However the high energy of boat conformers relative to chairs would make such structures minor components in the overall ensemble of conformations available to these molecules.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Lipids/Properties_and_Classification_of_Lipids/Steroids.txt |
Compounds classified as terpenes constitute what is arguably the largest and most diverse class of natural products. A majority of these compounds are found only in plants, but some of the larger and more complex terpenes (e.g. squalene & lanosterol) occur in animals. Terpenes incorporating most of the common functional groups are known, so this does not provide a useful means of classification. Instead, the number and structural organization of carbons is a definitive characteristic. Terpenes may be considered to be made up of isoprene (more accurately isopentane) units, an empirical feature known as the isoprene rule. Because of this, terpenes usually have \(5n\) carbon atoms (\(n\) is an integer), and are subdivided as follows:
Classification Isoprene Units Carbon Atoms
monoterpenes 2 C10
sesquiterpenes 3 C15
diterpenes 4 C20
sesterterpenes 5 C25
triterpenes 6 C30
Isoprene itself, a C5H8 gaseous hydrocarbon, is emitted by the leaves of various plants as a natural byproduct of plant metabolism. Next to methane it is the most common volatile organic compound found in the atmosphere. Examples of C10 and higher terpenes, representing the four most common classes are shown in the following diagrams. Most terpenes may be structurally dissected into isopentane segments. How this is done can be seen in the diagram directly below.
Figure: Monoterpenes and diterpenes
The isopentane units in most of these terpenes are easy to discern, and are defined by the shaded areas. In the case of the monoterpene camphor, the units overlap to such a degree it is easier to distinguish them by coloring the carbon chains. This is also done for alpha-pinene. In the case of the triterpene lanosterol we see an interesting deviation from the isoprene rule. This thirty carbon compound is clearly a terpene, and four of the six isopentane units can be identified. However, the ten carbons in center of the molecule cannot be dissected in this manner. Evidence exists that the two methyl groups circled in magenta and light blue have moved from their original isoprenoid locations (marked by small circles of the same color) to their present location. This rearrangement is described in the biosynthesis section. Similar alkyl group rearrangements account for other terpenes that do not strictly follow the isoprene rule.
Figure: Triterpenes
Polymeric isoprenoid hydrocarbons have also been identified. Rubber is undoubtedly the best known and most widely used compound of this kind. It occurs as a colloidal suspension called latex in a number of plants, ranging from the dandelion to the rubber tree (Hevea brasiliensis). Rubber is a polyene, and exhibits all the expected reactions of the C=C function. Bromine, hydrogen chloride and hydrogen all add with a stoichiometry of one molar equivalent per isoprene unit. Ozonolysis of rubber generates a mixture of levulinic acid ( \(CH_3COCH_2CH_2CO_2H\) ) and the corresponding aldehyde. Pyrolysis of rubber produces the diene isoprene along with other products.
The double bonds in rubber all have a Z-configuration, which causes this macromolecule to adopt a kinked or coiled conformation. This is reflected in the physical properties of rubber. Despite its high molecular weight (about one million), crude latex rubber is a soft, sticky, elastic substance. Chemical modification of this material is normal for commercial applications. Gutta-percha (structure above) is a naturally occurring E-isomer of rubber. Here the hydrocarbon chains adopt a uniform zig-zag or rod like conformation, which produces a more rigid and tough substance. Uses of gutta-percha include electrical insulation and the covering of golf balls.
Waxes
Waxes are esters of fatty acids with long chain monohydric alcohols (one hydroxyl group). Natural waxes are often mixtures of such esters, and may also contain hydrocarbons. The formulas for three well known waxes are given below, with the carboxylic acid moiety colored red and the alcohol colored blue.
Spermaceti Beeswax Carnuba wax
CH3(CH2)14CO2-(CH2)15CH3 CH3(CH2)24CO2-(CH2)29CH3 CH3(CH2)30CO2-(CH2)33CH3
Waxes are widely distributed in nature. The leaves and fruits of many plants have waxy coatings, which may protect them from dehydration and small predators. The feathers of birds and the fur of some animals have similar coatings which serve as a water repellent. Carnuba wax is valued for its toughness and water resistance. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Lipids/Properties_and_Classification_of_Lipids/Terpenes.txt |
• Lipid Biosynthesis
The products of photosynthesis are a class of compounds called carbohydrates, the most common and important of which is glucose (C6H12O6). Subsequent reactions effect an oxidative cleavage of glucose to pyruvic acid (CH3COCO2H), and this in turn is transformed to the two-carbon building block, acetate. The multitude of lipid structures described here are constructed from acetate by enzymatic reactions that in many respects correspond to reactions used by chemists for laboratory syntheses of simi
Synthesis of Lipids
The complex organic compounds found in living organisms on this planet originate from photosynthesis, an endothermic reductive condensation of carbon dioxide requiring light energy and the pigment chlorophyll.
$x CO_2 + x H_2O + \text{energy} \rightarrow C_xH_{2x}O_x + x O_2$
The products of photosynthesis are a class of compounds called carbohydrates, the most common and important of which is glucose (C6H12O6). Subsequent reactions effect an oxidative cleavage of glucose to pyruvic acid (CH3COCO2H), and this in turn is transformed to the two-carbon building block, acetate. The multitude of lipid structures described here are constructed from acetate by enzymatic reactions that in many respects correspond to reactions used by chemists for laboratory syntheses of similar compounds. However, an important restriction is that the reagents and conditions must be compatible with the aqueous medium, neutral pH and moderate temperatures found in living cells. Consequently, the condensation, alkylation, oxidation and reduction reactions that accomplish the biosynthesis of lipids will not make use of the very strong bases, alkyl halides, chromate oxidants or metal hydride reducing agents that are employed in laboratory work.
Condensations
Claisen condensation of ethyl acetate (or other acetate esters) forms an acetoacetate ester, as illustrated by the top equation in the following diagram. Reduction, dehydration and further reduction of this product would yield an ester of butyric acid, the overall effect being the elongation of the acetate starting material by two carbons. In principle, repetition of this sequence would lead to longer chain acids, made up of an even number of carbon atoms. Since most of the common natural fatty acids have even numbers of carbon atoms, this is an attractive hypothesis for their biosynthesis.
Nature's solution to carrying out a Claisen-like condensation in a living cell is shown in the bottom equation of the diagram. Thioesters are more reactive as acceptor reactants than are ordinary esters, and preliminary conversion of acetate to malonate increases the donor reactivity of this species. The thiol portion of the thioester is usually a protein of some kind, with efficient acetyl transport occurring by way of acetyl coenzyme A. Depending on the enzymes involved, the condensation product may be reduced and then further elongated so as to produce fatty acids (as shown), or elongated by further condensations to polyketone intermediates that are precursors to a variety of natural phenolic compounds.
The reduction steps (designated by [H] in the equations) and the intervening dehydrations needed for fatty acid synthesis require unique coenzymes and phosphorylating reagents. The pyridine ring of nicotinamide adenine dinucleotide (NAD) and its 2'-phosphate derivative (NADP) function as hydride acceptors, and the corresponding reduced species (NADH & NADPH) as a hydride donors. Partial structures for these important redox reagents are shown below.
As noted earlier, the hydroxyl group is a poor anionic leaving group (hydroxide anion is a strong base). Phosphorylation converts a hydroxyl group into a phosphate (PO4) or pyrophosphate (P2O7) ester, making it a much better leaving group (the pKas at pH near 7 are 7.2 and 6.6 respectively). The chief biological phosphorylation reagents are phosphate derivatives of adenosine (a ribose compound). The strongest of these is the triphosphate ATP, with the diphosphate and monophosphate being less powerful.
The overall process of fatty acid synthesis is summarized for palmitic acid, CH3(CH2)14CO2H, in the following equation:
$8 CH_3CO-CoA + 14 NADPH + 14 H^+ + 7 ATP + H_2O \rightarrow CH_3(CH_2)_{14}CO_2H + 8 CoA + 14 NADP^+ + 7 ADP + 7 H_2PO_4^–$
Alkylations
The branched chain and cyclic structures of the terpenes and steroids are constructed by sequential alkylation reactions of unsaturated isopentyl pyrophosphate units. As depicted in the following diagram, these 5-carbon reactants are made from three acetate units by way of an aldol-like addition of a malonate intermediate to acetoacetate. Selective hydrolysis and reduction gives a key intermediate called mevalonic acid. Phosphorylation and elimination of mevalonic acid then generate isopentenyl pyrophosphate, which is in equilibrium with its double bond isomer, dimethylallyl pyrophosphate. The allylic pyrophosphate group in the latter compound is reactive in enzymatically catalyzed alkylation reactions, such as the one drawn in the green box. This provides support for the empirical isoprene rule.
The simplest fashion in which isopentane units combine is termed "head-to-tail". This is the combination displayed in the green box, and these terms are further defined in the upper equation. Non head-to-tail coupling of isopentane units is also observed, as in the chrysanthemic acid construction shown in the second equation.
On the diagram below, the series of cation-like cyclizations and rearrangements, known as the Stork-Eschenmoser hypothesis, is shown, which were identified in the biosynthesis of the triterpene lanosterol. Lanosterol is a precursor in the biosynthesis of steroids. This takes place by metabolic removal of three methyl groups and degradation of the side chain.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Lipids/Synthesis_of_Lipids/Lipid_Biosynthesis.txt |
Name the parent alkane (include the carbon atom of the nitrile as part of the parent) followed with the word -nitrile. The carbon in the nitrile is given the #1 location position. It is not necessary to include the location number in the name because it is assumed that the functional group will be on the end of the parent chain.
Cycloalkanes are followed by the word -carbonitrile. The substituent name is cyano.
• 1-butanenitrile or 1-cyanopropane
Try to name the following compounds using these conventions�
Try to draw structures for the following compounds�
• butanedinitrile J
• 2-methycyclohexanecarbonitrile J
Some common names that you should know are...
acetonitrile
benzonitrile
Try to draw a structure for the following compound�
• 2-methoxybenzonitrile J
Nomenclature of Nitriles
Nomenclature of Nitro Compounds
Name these compounds by naming the -NO2 as a nitro substituent.
nitromethane
nitrobenzene
Try to draw a structure for the following compound
• m-nitrobenzonitrile J
Nomenclature of Organic Nitrates
Name the alkyl group and follow with a space and the word nitrate.
ethyl nitrate
Try to draw a structure for the following compound
• benzyl nitrate J
Nitrile Occurrence and Uses
One of the most common occurrences of nitriles is in Nitrile rubber. Nitrile rubber is a synthetic copolymer of acrylonitrile and butadiene. This form of rubber is highly resistant to chemicals and is used to make protective gloves, hoses and seals.
Contributors
Prof. Steven Farmer (Sonoma State University)
Nitrile Properties
The electronic structure of nitriles is very similar to that of an alkyne with the main difference being the presence of a set of lone pair electrons on the nitrogen. Both the carbon and the nitrogen are sp hydridized which leaves them both with two p orbitals which overlap to form the two $\pi$ bond in the triple bond. The R-C-N bond angle in and nitrile is 180° which give a nitrile functional group a linear shape.
The lone pair electrons on the nitrogen are contained in a sp hybrid orbital which makes them much less basic and an amine. The 50% character of an sp hybrid orbital close to the nucleus and therefore less basic compared to other nitrogen containing compounds such as amines.
The presence of an electronegative nitrogen causes nitriles to be very polar molecules. Consequently, nitriles tend to have higher boiling points than molecules with a similar size.
Physical Properties of Nitriles
Nitriles contain the -CN group, and used to be known as cyanides. The smallest organic nitrile is ethanenitrile, CH3CN, (old name: methyl cyanide or acetonitrile - and sometimes now called ethanonitrile). Hydrogen cyanide, HCN, doesn't usually count as organic, even though it contains a carbon atom.
Notice the triple bond between the carbon and nitrogen in the -CN group. The three simplest nitriles are:
CH3CN ethanenitrile
CH3CH2CN propanenitrile
CH3CH2CH2CN butanenitrile
When you are counting the length of the carbon chain, don't forget the carbon in the -CN group. If the chain is branched, this carbon usually counts as the number 1 carbon.
Boiling points
The small nitriles are liquids at room temperature.
nitrile boiling point (°C)
CH3CN 82
CH3CH2CN 97
CH3CH2CH2CN 116 - 118
These boiling points are very high for the size of the molecules - similar to what you would expect if they were capable of forming hydrogen bonds. However, they don't form hydrogen bonds - they don't have a hydrogen atom directly attached to an electronegative element. They are just very polar molecules. The nitrogen is very electronegative and the electrons in the triple bond are very easily pulled towards the nitrogen end of the bond. Nitriles therefore have strong permanent dipole-dipole attractions as well as van der Waals dispersion forces between their molecules.
Solubility in water
Ethanenitrile is completely soluble in water, and the solubility then falls as chain length increases.
nitrile solubility at 20°C
CH3CN miscible
CH3CH2CN 10 g per 100 cm3 of water
CH3CH2CH2CN 3 g per 100 cm3 of water
The reason for the solubility is that although nitriles cannot hydrogen bond with themselves, they can hydrogen bond with water molecules. One of the slightly positive hydrogen atoms in a water molecule is attracted to the lone pair on the nitrogen atom in a nitrile and a hydrogen bond is formed.
There will also, of course, be dispersion forces and dipole-dipole attractions between the nitrile and water molecules. Forming these attractions releases energy. This helps to supply the energy needed to separate water molecule from water molecule and nitrile molecule from nitrile molecule before they can mix together. As chain lengths increase, the hydrocarbon parts of the nitrile molecules start to get in the way. By forcing themselves between water molecules, they break the relatively strong hydrogen bonds between water molecules without replacing them by anything as good. This makes the process energetically less profitable, and so solubility decreases. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Nitriles/Nomenclature_of_Nitriles/Nomenclature_of_Nitro_Compounds_and_Organic_Nitrates.txt |
• Conversion of nitriles to 1° amines using LiAlH4
Nitriles can be converted to 1° amines by reaction with LiAlH4. During this reaction the hydride nucleophile attacks the electrophilic carbon in the nitrile to form an imine anion. Once stabilized by a Lewis acid-base complexation the imine salt can accept a second hydride to form a dianion. The dianion can then be converted to an amine by addition of water.
• Conversion of nitriles to amides
Nitriles can be converted to amides. This reaction can be acid or base catalyzed. In the case of acid catalysis the nitrile becomes protonated. Protonation increases the electrophilicity of the nitrile so that it will accept water, a poor nucleophile.
• Conversion of nitriles to carboxylic acids
Nitriles can be converted to carboxylic acid with heating in sulfuric acid. During the reaction an amide intermediate is formed.
• Conversion to ketones using Grignard reagents
Grignard reagents can attack the electophillic carbon in a nitrile to form an imine salt. This salt can then be hydrolyzed to become a ketone.
• The Hydrolysis of Nitriles
This page looks at the hydrolysis of nitriles under either acidic or alkaline conditions to make carboxylic acids or their salts.
• The Reduction of Nitriles
This page looks at the reduction of nitriles to primary amines using either lithium tetrahydridoaluminate(III) (lithium aluminium hydride) or hydrogen and a metal catalyst.
The carbon in a nitrile is electrophilic because a resonance structure can be drawn which places a positive charge on it. Because of this the triple bond of a nitrile accepts a nucleophile in a manner similar to a carbonyl.
Contributors
Prof. Steven Farmer (Sonoma State University)
Reactivity of Nitriles
Nitriles can be converted to amides. This reaction can be acid or base catalyzed. In the case of acid catalysis the nitrile becomes protonated. Protonation increases the electrophilicity of the nitrile so that it will accept water, a poor nucleophile.
Mechanism
1) Protonation
2) Nucleophilic attack by water
3) Proton Transfer
4) Resonance
5) Deprotonation
Conversion of nitriles to carboxylic acids
Nitriles can be converted to carboxylic acid with heating in sulfuric acid. During the reaction an amide intermediate is formed.
Contributors
Prof. Steven Farmer (Sonoma State University)
Conversion to ketones using Grignard reagents
Grignard reagents can attack the electophillic carbon in a nitrile to form an imine salt. This salt can then be hydrolyzed to become a ketone.
Mechanism
1) Nucleophilic Attack by the Grignard Reagent
2) Protonation
3) Protonation
4) Nucleophilic attack by water
5) Proton Transfer
6) Leaving group removal
7) Deprotonation
Contributors
Prof. Steven Farmer (Sonoma State University)
The Hydrolysis of Nitriles
This page looks at the hydrolysis of nitriles under either acidic or alkaline conditions to make carboxylic acids or their salts.
Introduction
When nitriles are hydrolysed you can think of them reacting with water in two stages - first to produce an amide, and then the ammonium salt of a carboxylic acid. For example, ethanenitrile would end up as ammonium ethanoate going via ethanamide.
In practice, the reaction between nitriles and water would be so slow as to be completely negligible. The nitrile is instead heated with either a dilute acid such as dilute hydrochloric acid, or with an alkali such as sodium hydroxide solution. The end result is similar in all the cases, but the exact nature of the final product varies depending on the conditions you use for the reaction.
Acidic hydrolysis of nitriles
The nitrile is heated under reflux with dilute hydrochloric acid. Instead of getting an ammonium salt as you would do if the reaction only involved water, you produce the free carboxylic acid. For example, with ethanenitrile and hydrochloric acid you would get ethanoic acid and ammonium chloride.
\[ \ce{CH_3CN + 2H_2O + HCl \longrightarrow CH_3COOH + NH_4Cl}\]
Why is the free acid formed rather than the ammonium salt? The ethanoate ions in the ammonium ethanoate react with hydrogen ions from the hydrochloric acid to produce ethanoic acid. Ethanoic acid is only a weak acid and so once it has got the hydrogen ion, it tends to hang on to it.
Alkaline hydrolysis of nitriles
The nitrile is heated under reflux with sodium hydroxide solution. This time, instead of getting an ammonium salt as you would do if the reaction only involved water, you get the sodium salt. Ammonia gas is given off as well.
For example, with ethanenitrile and sodium hydroxide solution you would get sodium ethanoate and ammonia.
\[ \ce{CH_3CN + H_2O + NaOH \longrightarrow CH_3COONa^{-} + NH_3}\]
The ammonia is formed from reaction between ammonium ions and hydroxide ions. If you wanted the free carboxylic acid in this case, you would have to acidify the final solution with a strong acid such as dilute hydrochloric acid or dilute sulphuric acid. The ethanoate ion in the sodium ethanoate will react with hydrogen ions as mentioned above.
The Reduction of Nitriles
This page looks at the reduction of nitriles to primary amines using either lithium tetrahydridoaluminate(III) (lithium aluminium hydride) or hydrogen and a metal catalyst.
The reduction of nitriles using LiAlH4
Despite its name, the structure of the reducing agent is very simple. There are four hydrogens ("tetrahydido") around the aluminium in a negative ion (shown by the "ate" ending). The "(III)" shows the oxidation state of the aluminium, and is often left out because aluminium only ever shows the +3 oxidation state in its compounds. To make the name shorter, that's what I shall do for the rest of this page.
The structure of LiAlH4 is:
In the negative ion, one of the bonds is a co-ordinate covalent (dative covalent) bond using the lone pair on a hydride ion (H-) to form a bond with an empty orbital on the aluminum.
The overall reaction
The nitrile reacts with the lithium tetrahydridoaluminate in solution in ethoxyethane (diethyl ether, or just "ether") followed by treatment of the product of that reaction with a dilute acid. Overall, the carbon-nitrogen triple bond is reduced to give a primary amine. Primary amines contain the \(-NH_2\) group. For example, with ethanenitrile you get ethylamine:
\[ \ce{CH_3CN + 4[H] \longrightarrow CH_3CH_2NH_2}\]
The reduction of nitriles using hydrogen and a metal catalyst
The carbon-nitrogen triple bond in a nitrile can also be reduced by reaction with hydrogen gas in the presence of a variety of metal catalysts. Commonly used catalysts are palladium, platinum or nickel. The reaction will take place at a raised temperature and pressure, but the exact details vary from catalyst to catalyst.
For example, ethanenitrile can be reduced to ethylamine by reaction with hydrogen in the presence of a palladium catalyst.
\[ \ce{CH_3CN + 2H_2 ->[Pd] CH3CH2NH2}\] | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Nitriles/Reactivity_of_Nitriles/Conversion_of_nitriles_to_amides.txt |
Nitriles are formed by an SN2 reaction between a bromide and sodium cyanide
1o Amides can be converted to nitriles by dehydration with thionyl chloride
Addition of cyanide (-:C≡N) to an aldehyde or ketone forms a cyanohydrin
Synthesis of Nitriles
Notes
• Conditions are relatively mild: dimethylformamide at room temperature.
• Compatible with most sensitive functionalities.
• Only one third molar equivalent of cyanuric chloride is needed to go to completion.
• Cyanuric chloride should be recrystallized before use.
• The excess cyanuric chloride and the cyanuric acid by-product can be removed by 5% aq.sodium bicarbonate washings.
Preparation of Nitriles
This page looks at various ways of making nitriles - from halogenoalkanes (haloalkanes or alkyl halides), from amides, and from aldehydes and ketones. It pulls together information from pages dealing with each of these kinds of compounds
Making nitriles from halogenoalkanes
The halogenoalkane is heated under reflux with a solution of sodium or potassium cyanide in ethanol. The halogen is replaced by a -CN group and a nitrile is produced. Heating under reflux means heating with a condenser placed vertically in the flask to prevent loss of volatile substances from the mixture. The solvent is important. If water is present you tend to get substitution by -OH instead of -CN.
Example 1
using 1-bromopropane as a typical halogenoalkane:
\[ CH_3CH_2CH_2Br + CN^- \longrightarrow CH_3CH_2CH_2CN + Br^-\]
You could write the full equation rather than the ionic one, but it slightly obscures what's going on:
\[ CH_3CH_2CH_2Br + KCN \longrightarrow CH_3CH_2CH_2CN + KBr\]
The bromine (or other halogen) in the halogenoalkane is simply replaced by a -CN group - hence a substitution reaction. In this example, butanenitrile is formed.
Making a nitrile by this method is a useful way of increasing the length of a carbon chain. Having made the nitrile, the -CN group can easily be modified to make other things - as you will find if you explore the nitriles menu (link a the bottom of the page).
Making nitriles from amides
Nitriles can be made by dehydrating amides. Amides are dehydrated by heating a solid mixture of the amide and phosphorus(V) oxide, P4O10. Water is removed from the amide group to leave a nitrile group, -CN. The liquid nitrile is collected by simple distillation.
Example
Once will get ethanenitrile by dehydrating ethanamide.
Making nitriles from aldehydes and ketones
Aldehydes and ketones undergo an addition reaction with hydrogen cyanide. The hydrogen cyanide adds across the carbon-oxygen double bond in the aldehyde or ketone to produce a hydroxynitrile. Hydroxynitriles used to be known as cyanohydrins.
Example
with ethanal (an aldehyde) you get 2-hydroxypropanenitrile:
With propanone (a ketone) you get 2-hydroxy-2-methylpropanenitrile:
In every example of this kind, the -OH group will be on the number 2 carbon atom - the one next to the -CN group.
The reaction is not normally done using hydrogen cyanide itself, because this is an extremely poisonous gas. Instead, the aldehyde or ketone is mixed with a solution of sodium or potassium cyanide in water to which a little sulfuric acid has been added. The pH of the solution is adjusted to about 4 - 5, because this gives the fastest reaction. The reaction happens at room temperature. The solution will contain hydrogen cyanide (from the reaction between the sodium or potassium cyanide and the sulfuric acid), but still contains some free cyanide ions. This is important for the mechanism.
These are useful reactions because they not only increase the number of carbon atoms in a chain, but also introduce another reactive group as well as the -CN group. The -OH group behaves just like the -OH group in any alcohol with a similar structure.
For example, starting from a hydroxynitrile made from an aldehyde, you can quite easily produce relatively complicated molecules like 2-amino acids - the amino acids which are used to construct proteins.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Nitriles/Synthesis_of_Nitriles/Amide_to_Nitrile_Reduction.txt |
Nucleophilicity of Phosphorus Compounds
Phosphorous analogs of amines are called phosphines. The chemistry of phosphines and the related phosphite esters is dominated by their strong nucleophilicity and reducing character. The nucleophilicity of trivalent phosphorus results in rapid formation of phosphonium salts when such compounds are treated with reactive alkyl halides. For example, although resonance delocalization of the nitrogen electron pair in triphenylamine, (C6H5)3N, renders it relatively unreactive in SN2 reactions, the corresponding phosphorus compound, triphenylphosphine, undergoes a rapid and exothermic reaction to give a phosphonium salt, as shown below in the first equation. Phosphite esters react in the same manner, but the resulting phosphonium salts (shaded box) are often unstable, and on heating yield dialkyl phosphonate esters by way of a second SN2 reaction (equation 2 below).
Oxidation States of Phosphorus Compounds
The difference in oxidation states between nitrogen and phosphorus is less pronounced than between oxygen and sulfur. Organophosphorus compounds having phosphorus oxidation states ranging from –3 to +5, as shown in the following table, are well known (some simple inorganic compounds are displayed in green). As in the case of sulfur, the P=O double bonds drawn in some of the formulas do not consist of the customary sigma & pi-orbitals found in carbon double bonds. Phosphorus is a third row element, and has five empty 2d-orbitals that may be used for p-d bonding in a fashion similar to p-p (π) bonding. In this way phosphorus may expand an argon-like valence shell octet by two electrons (e.g. phosphine oxides).
Phosphorus Compounds as Reducing Agents
Trivalent phosphorus is easily oxidized. In contrast with ammonia and amines, phosphine and its mono and dialkyl derivatives are pyrophoric, bursting into flame on contact with the oxygen in air. The affinity of trivalent phosphorus for oxygen (and sulfur) has been put to use in many reaction systems, three of which are shown here. The triphenylphosphine oxide produced in reactions 1 & 3 is a very stable polar compound, and in most cases it is easily removed from the other products. Reaction 2 is a general formulation of the useful Corey-Winter procedure for converting vicinal glycols to alkenes.
Triphenylphosphine is also oxidized by halogens, and with bromine yields dibromotriphenylphosphorane, a crystalline salt-like compound, useful for converting alcohols to alkyl bromides. As in a number of earlier examples, the formation of triphenylphosphine oxide in the irreversible SN2 step provides a thermodynamic driving force for the reaction.
Reactions of Ylides
The most important use of ylides in synthesis comes from their reactions with aldehydes and ketones, which are initiated in every case by a covalent bonding of the nucleophilic $\alpha$-carbon to the electrophilic carbonyl carbon.
Reactivity of Phosphorus Compounds
The most important use of ylides in synthesis comes from their reactions with aldehydes and ketones, which are initiated in every case by a covalent bonding of the nucleophilic alpha-carbon to the electrophilic carbonyl carbon. Alkylidenephosphorane ylides react to give substituted alkenes in a transformation called the Wittig reaction. This reaction is illustrated by the first three equations below. In each case the new carbon-carbon double bond is colored blue, and the oxygen of the carbonyl reactant is transferred to the phosphorus. The Wittig reaction tolerates epoxides and many other functional groups, as demonstrated by reaction # 1. The carbanionic center may also be substituted, as in reactions # 2 & 3. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration. With simple substituted ylides Z-alkenes are favored (reaction # 2).
The fourth equation shows a characteristic reaction of a sulfur ylide. Again, the initial carbon-carbon bond is colored blue, but subsequent steps lead to an epoxide product rather than an alkene.
Reaction # 5 illustrates a double Wittig reaction, using a dialdehyde reactant (colored orange). Because of the additional allylic stabilization of the ylide group, the new double bonds (colored blue) have an E-configuration, in contrast to the Z-configuration favored by unstabilized ylides (equation 2). Reaction # 6 shows a related synthesis that employs a phosphonate enolate base as the nucleophile. This is known as the Horner-Wadsworth-Emmons reaction. Here, as with the Wittig reaction, the formation of a stable phosphorus oxygen bond in the phosphate product provides a driving force for the transformation. Again, stabilization of the ylide-like carbanion leads to an E-configuration of the product double bond. These remarkable and useful changes can be explained by the mechanisms displayed below.
Following the initial carbon-carbon bond formation, two intermediates have been identified for the Wittig reaction, a dipolar charge-separated species called a betaine and a four-membered heterocyclic structure referred to as an oxaphosphatane. Cleavage of the oxaphosphatane to alkene and phosphine oxide products is exothermic and irreversible. Depending on the stability of the starting ylide, the betaine may be formed reversibly and this will ultimately influence the stereochemistry of the alkene product.
In contrast to the phosphorus ylides and related reagents, reactions of sulfur ylides with carbonyl compounds do not usually lead to four-membered ring species analogous to oxaphosphatanes. The favored reaction path is therefore an internal SN2 process that leads to an epoxide product. The sulfur leaves as dimethyl sulfide. Additional examples of sulfur ylide reactions, illustrating differences in the reactivity of dimethylsulfonium methylide and dimethyloxosulfonium methylide, are given in the following diagram. Of the two, the oxosulfonium ylide is less reactive and is thought to add reversibly to carbonyl groups, eventually forming the thermodynamically favored product.
Preparation of Ylides
It has been noted that dipolar phosphorus and sulfur oxides are stabilized by p-d bonding. This may be illustrated by a resonance description, as shown here.
This bonding stabilization extends to carbanions alpha to phosphonium and sulfonium centers, and the zwitterionic conjugate bases derived from such cations are known as ylides. Approximate pKa's for some ylide precursors and related compounds are provided in the following table. The acidic hydrogen atoms are colored red. By convention, pKa's are usually adjusted to conform to the standard solvent water; however, in practice, measurements of very weak acids are necessarily made in non-aqueous solvents such as DMSO (dimethyl sufoxide). The green numbers in the table represent DMSO measurements, and although these are larger than the aqueous approximations, the relative order is unchanged. Note that DMSO itself is the weakest acid of those shown.
Compound (CH3)3S(+) I(–) (CH3)3S(+)=O I(–) (CH3)2S=O (CH3)2S(=O)2 CH3-P(OCH3)2=O
Name trimethylsulfonium
iodide
trimethylsulfoxonium
iodide
dimethyl sulfoxide dimethyl sulfone dimethyl methylphosphonate
pKa
DMSO
20
25
15
18
30
35
26
31
25
31
Some characteristic preparations of ylide reagents are shown below. Very strong bases, such as butyl lithium, are required for complete formation of ylides. Sodium hydride (NaH), another powerful base, is insoluble in most solvents, but its reaction with DMSO (the weakest acid in the table) generates a strong conjugate base, CH3)S(=O)CH2(–) Na(+), known as dimsyl sodium. This soluble base is widely used for the generation of ylides in DMSO solution.
The ylides shown here are all strong bases. Like other strongly basic organic reagents, they are protonated by water and alcohols, and are sensitive to oxygen. Water decomposes alkylidenephosphoranes to hydrocarbons and phosphine oxides, as shown. Oxygen cleaves these ylides in a similar fashion, the alkylidene moiety being converted to a carbonyl compound.
R3P=CR'2 + H2O [ R3P(OH)–CHR'2 ] R3P=O + R'2CH2
R3P=CR'2 + O2 R3P=O + R'2C=O | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Organo-phosphorus_Compounds/Properties_of_Phosphorus_Compounds.txt |
Compounds like alcohols and phenol which contain an -OH group attached to a hydrocarbon are very weak acids. Alcohols are so weakly acidic that, for normal lab purposes, their acidity can be virtually ignored. However, phenol is sufficiently acidic for it to have recognizably acidic properties - even if it is still a very weak acid. A hydrogen ion can break away from the -OH group and transfer to a base.
Why is phenol acidic?
Compounds like alcohols and phenol which contain an -OH group attached to a hydrocarbon are very weak acids. Alcohols are so weakly acidic that, for normal lab purposes, their acidity can be virtually ignored. However, phenol is sufficiently acidic for it to have recognizably acidic properties - even if it is still a very weak acid. A hydrogen ion can break away from the -OH group and transfer to a base. For example, in solution in water:
Phenol is a very weak acid and the position of equilibrium lies well to the left. Phenol can lose a hydrogen ion because the phenoxide ion formed is stabilised to some extent. The negative charge on the oxygen atom is delocalised around the ring. The more stable the ion is, the more likely it is to form. One of the lone pairs on the oxygen atom overlaps with the delocalised electrons on the benzene ring.
This overlap leads to a delocalization which extends from the ring out over the oxygen atom. As a result, the negative charge is no longer entirely localized on the oxygen, but is spread out around the whole ion.
Spreading the charge around makes the ion more stable than it would be if all the charge remained on the oxygen. However, oxygen is the most electronegative element in the ion and the delocalized electrons will be drawn towards it. That means that there will still be a lot of charge around the oxygen which will tend to attract the hydrogen ion back again. That is why phenol is only a very weak acid.
Why is phenol a much stronger acid than cyclohexanol? To answer this question we must evaluate the manner in which an oxygen substituent interacts with the benzene ring. As noted in our earlier treatment of electrophilic aromatic substitution reactions, an oxygen substituent enhances the reactivity of the ring and favors electrophile attack at ortho and para sites. It was proposed that resonance delocalization of an oxygen non-bonded electron pair into the pi-electron system of the aromatic ring was responsible for this substituent effect. A similar set of resonance structures for the phenolate anion conjugate base appears below the phenol structures.
The resonance stabilization in these two cases is very different. An important principle of resonance is that charge separation diminishes the importance of canonical contributors to the resonance hybrid and reduces the overall stabilization. The contributing structures to the phenol hybrid all suffer charge separation, resulting in very modest stabilization of this compound. On the other hand, the phenolate anion is already charged, and the canonical contributors act to disperse the charge, resulting in a substantial stabilization of this species. The conjugate bases of simple alcohols are not stabilized by charge delocalization, so the acidity of these compounds is similar to that of water. An energy diagram showing the effect of resonance on cyclohexanol and phenol acidities is shown on the right. Since the resonance stabilization of the phenolate conjugate base is much greater than the stabilization of phenol itself, the acidity of phenol relative to cyclohexanol is increased. Supporting evidence that the phenolate negative charge is delocalized on the ortho and para carbons of the benzene ring comes from the influence of electron-withdrawing substituents at those sites.
Properties of phenol as an acid
With indicators
The pH of a typical dilute solution of phenol in water is likely to be around 5 - 6 (depending on its concentration). That means that a very dilute solution isn't really acidic enough to turn litmus paper fully red. Litmus paper is blue at pH 8 and red at pH 5. Anything in between is going to show as some shade of "neutral". Phenol reacts with sodium hydroxide solution to give a colourless solution containing sodium phenoxide.
In this reaction, the hydrogen ion has been removed by the strongly basic hydroxide ion in the sodium hydroxide solution.
With sodium carbonate or sodium hydrogencarbonate
Phenol isn't acidic enough to react with either of these. Or, looked at another way, the carbonate and hydrogencarbonate ions aren't strong enough bases to take a hydrogen ion from the phenol. Unlike the majority of acids, phenol does not give carbon dioxide when you mix it with one of these. This lack of reaction is actually useful. You can recognise phenol because:
• It is fairly insoluble in water.
• It reacts with sodium hydroxide solution to give a colourless solution (and therefore must be acidic).
• It does not react with sodium carbonate or hydrogencarbonate solutions (and so must be only very weakly acidic).
With metallic sodium
Acids react with the more reactive metals to give hydrogen gas. Phenol is no exception - the only difference is the slow reaction because phenol is such a weak acid. Phenol is warmed in a dry tube until it is molten, and a small piece of sodium added. There is some fizzing as hydrogen gas is given off. The mixture left in the tube will contain sodium phenoxide.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Phenols/Properties_of_Phenols/Acidity_of_Phenols.txt |
Substitution of the hydroxyl hydrogen atom is even more facile with phenols, which are roughly a million times more acidic than equivalent alcohols. This phenolic acidity is further enhanced by electron-withdrawing substituents ortho and para to the hydroxyl group, as displayed in the following diagram. The alcohol cyclohexanol is shown for reference at the top left. It is noteworthy that the influence of a nitro substituent is over ten times stronger in the para-location than it is meta, despite the fact that the latter position is closer to the hydroxyl group. Furthermore additional nitro groups have an additive influence if they are positioned in ortho or para locations. The trinitro compound shown at the lower right is a very strong acid called picric acid.
Why is phenol a much stronger acid than cyclohexanol? To answer this question we must evaluate the manner in which an oxygen substituent interacts with the benzene ring. As noted in our earlier treatment of electrophilic aromatic substitution reactions, an oxygen substituent enhances the reactivity of the ring and favors electrophile attack at ortho and para sites. It was proposed that resonance delocalization of an oxygen non-bonded electron pair into the pi-electron system of the aromatic ring was responsible for this substituent effect. Formulas illustrating this electron delocalization will be displayed when the "Resonance Structures" button beneath the previous diagram is clicked. A similar set of resonance structures for the phenolate anion conjugate base appears below the phenol structures.
The resonance stabilization in these two cases is very different. An important principle of resonance is that charge separation diminishes the importance of canonical contributors to the resonance hybrid and reduces the overall stabilization. The contributing structures to the phenol hybrid all suffer charge separation, resulting in very modest stabilization of this compound. On the other hand, the phenolate anion is already charged, and the canonical contributors act to disperse the charge, resulting in a substantial stabilization of this species. The conjugate bases of simple alcohols are not stabilized by charge delocalization, so the acidity of these compounds is similar to that of water. An energy diagram showing the effect of resonance on cyclohexanol and phenol acidities is shown on the right. Since the resonance stabilization of the phenolate conjugate base is much greater than the stabilization of phenol itself, the acidity of phenol relative to cyclohexanol is increased. Supporting evidence that the phenolate negative charge is delocalized on the ortho and para carbons of the benzene ring comes from the influence of electron-withdrawing substituents at those sites.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Physical Properties of Phenol
This page looks at the structure and physical properties of phenol (very old name: carbolic acid). Phenol is the simplest member of a family of compounds in which an -OH group is attached directly to a benzene ring.
The structure of phenol
The simplest way to draw the structure of phenol is:
. . . but to understand phenol properly, you need to dig a bit deeper than this.
There is an interaction between the delocalized electrons in the benzene ring and one of the lone pairs on the oxygen atom. This has an important effect on both the properties of the ring and of the -OH group. One of the lone pairs on the oxygen overlaps with the delocalized ring electron system . . .
. . . giving a structure rather like this:
The donation of the oxygen's lone pair into the ring system increases the electron density around the ring. That makes the ring much more reactive than it is in benzene itself. That is explored in another page in this phenol section. It also helps to make the -OH group's hydrogen a lot more acidic than it is in alcohols. That will also be explored elsewhere in this section.
Physical properties
Pure phenol is a white crystalline solid, smelling of disinfectant. It has to be handled with great care because it causes immediate white blistering to the skin. The crystals are often rather wet and discolored.
Melting and boiling points
It is useful to compare phenol's melting and boiling points with those of methylbenzene (toluene). Both molecules contain the same number of electrons and are a very similar shape. That means that the intermolecular attractions due to van der Waals dispersion forces are going to be very similar.
melting point (°C) boiling point (°C)
C6H5OH 40 - 43 182
C6H5CH3 -95.0 111
The reason for the higher values for phenol is in part due to permanent dipole-dipole attractions due to the electronegativity of the oxygen - but is mainly due to hydrogen bonding. Hydrogen bonds can form between a lone pair on an oxygen on one molecule and the hydrogen on the -OH group of one of its neighbors.
Solubility in water
Phenol is moderately soluble in water - about 8 g of phenol will dissolve in 100 g of water. If you try to dissolve more than this, you get two layers of liquid. The top layer is a solution of phenol in water, and the bottom one a solution of water in phenol. Phenol is somewhat soluble in water because of its ability to form hydrogen bonds with the water.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Phenols/Properties_of_Phenols/Acidity_of_Substituted_Phenols.txt |
Compounds in which a hydroxyl group is bonded to an aromatic ring are called phenols. The chemical behavior of phenols is different in some respects from that of the alcohols, so it is sensible to treat them as a similar but characteristically distinct group. A corresponding difference in reactivity was observed in comparing aryl halides, such as bromobenzene, with alkyl halides, such as butyl bromide and tert-butyl chloride. Thus, nucleophilic substitution and elimination reactions were common for alkyl halides, but rare with aryl halides. This distinction carries over when comparing alcohols and phenols, so for all practical purposes substitution and/or elimination of the phenolic hydroxyl group does not occur.
Reactivity of Phenols
The facility with which the aromatic ring of phenols and phenol ethers undergoes electrophilic substitution has been noted. Two examples are shown in the following diagram. The first shows the Friedel-Crafts synthesis of the food preservative BHT from para-cresol. The second reaction is interesting in that it further demonstrates the delocalization of charge that occurs in the phenolate anion. Carbon dioxide is a weak electrophile and normally does not react with aromatic compounds; however, the negative charge concentration on the phenolate ring enables the carboxylation reaction shown in the second step. The sodium salt of salicylic acid is the major product, and the preference for ortho substitution may reflect the influence of the sodium cation. This is called the Kolbe-Schmidt reaction, and it has served in the preparation of aspirin, as the last step illustrates.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Making an Azo Dye from Phenol
This page looks at some typical reactions of diazonium ions, including examples of both substitution reactions and coupling reactions. If you have come straight to this page from a search engine and want to know about the preparation of the diazonium ions, you will find a link at the bottom of the page.
Substitution reactions of diazonium ions
Diazonium ions are present in solutions such as benzenediazonium chloride solution. They contain an -N2+ group. In the case of benzenediazonium chloride, this is attached to a benzene ring. Benzenediazonium chloride looks like this:
In this set of reactions of the diazonium ion, the -N2+ group is replaced by something else. The nitrogen is released as nitrogen gas.
Substitution by an -OH group
To get this reaction, all you need to do is warm the benzenediazonium chloride solution. The diazonium ion reacts with the water in the solution and phenol is formed - either in solution or as a black oily liquid (depending on how much is formed). Nitrogen gas is evolved.
This is the same reaction that you get if you react phenylamine with nitrous acid in the warm. The diazonium ion is formed first and then immediately reacts with the water in the solution to give phenol.
Substitution by an iodine atom
This is a good example of the use of diazonium salts to substitute things into a benzene ring which are otherwise quite difficult to attach. (That's equally true of the previous reaction, by the way.) If you add potassium iodide solution to the benzenediazonium chloride solution in the cold, nitrogen gas is given off, and you get oily droplets of iodobenzene formed. There is a simple reaction between the diazonium ions and the iodide ions from the potassium iodide solution.
Coupling reactions of diazonium ions
In the substitution reactions above, the nitrogen in the diazonium ion is lost. In the rest of the reactions on this page, the nitrogen is retained and used to make a bridge between two benzene rings.
The reaction with phenol
Phenol is dissolved in sodium hydroxide solution to give a solution of sodium phenoxide.
The solution is cooled in ice, and cold benzenediazonium chloride solution is added. There is a reaction between the diazonium ion and the phenoxide ion and a yellow-orange solution or precipitate is formed. The product is one of the simplest of what are known as azo compounds, in which two benzene rings are linked by a nitrogen bridge.
The reaction with naphthalen-2-ol
Naphthalen-2-ol is also known as 2-naphthol or beta-naphthol. It contains an -OH group attached to a naphthalene molecule rather than to a simple benzene ring. Naphthalene has two benzene rings fused together. The reaction is done under exactly the same conditions as with phenol. The naphthalen-2-ol is dissolved in sodium hydroxide solution to produce an ion just like the phenol one. This solution is cooled and mixed with the benzenediazonium chloride solution. An intense orange-red precipitate is formed - another azo compound.
The reaction with phenylamine (aniline)
Some liquid phenylamine is added to a cold solution of benzenediazonium chloride, and the mixture is shaken vigorously. A yellow solid is produced.
These strongly colored azo compounds are frequently used as dyes known as azo dyes. The one made from phenylamine (aniline) is known as "aniline yellow" (amongst many other things - see note above). Azo compounds account for more than half of modern dyes.
The use of an azo dye as an indicator - methyl orange
Azo compounds contain a highly delocalized system of electrons which takes in both benzene rings and the two nitrogen atoms bridging the rings. The delocalization can also extend to things attached to the benzene rings as well. If white light falls on one of these molecules, some wavelengths are absorbed by these delocalized electrons. The color you see is the result of the non-absorbed wavelengths. The groups which contribute to the delocalization (and so to the absorption of light) are known as a chromophore. Modifying the groups present in the molecule can have an effect on the light absorbed, and so on the color you see. You can take advantage of this in indicators. Methyl orange is an azo dye which exists in two forms depending on the pH:
As the hydrogen ion is lost or gained there is a shift in the exact nature of the delocalization in the molecule, and that causes a shift in the wavelength of light absorbed. Obviously that means that you see a different color. When you add acid to methyl orange, a hydrogen ion attaches to give the red form. Methyl orange is red in acidic solutions (in fact solutions of pH less than 3.1). If you add an alkali, hydrogen ions are removed and you get the yellow form. Methyl orange is yellow at pH's greater than 4.4. In between, at some point there will be equal amounts of the red and yellow forms and so methyl orange looks orange. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Phenols/Reactivity_of_Phenols/Electrophilic_Substitution_of_the_Phenol_Aromatic_Ring.txt |
This page gives details of some reactions of phenol not covered elsewhere in this section. It deals with the combustion and esterification of phenol, and the use of iron(III) chloride solution (ferric chloride solution) as a test for phenol.
Combustion of phenol
Phenol burns in a plentiful supply of oxygen to give carbon dioxide and water.
$C_6H_5OH + 7O_2 \rightarrow 6CO_2 + 3H2O$
However, for compounds containing benzene rings, combustion is hardly ever complete, especially if they are burnt in air. The high proportion of carbon in phenol means that you need a very high proportion of oxygen to phenol to get complete combustion. Look at the equation.
As a general rule, the hydrogen in a molecule tends to get what oxygen is available first, leaving the carbon to form carbon itself, or carbon monoxide, if there isn't enough oxygen to go round. Phenol tends to burn in air with an extremely smoky flame - full of carbon particles.
Esterification of phenol
You will probably remember that you can make esters from alcohols by reacting them with carboxylic acids. You might expect phenol to be similar. However, unlike alcohols, phenol reacts so slowly with carboxylic acids that you normally react it with acyl chlorides (acid chlorides) or acid anhydrides instead.
Making esters from phenol using an acyl chloride
A typical acyl chloride is ethanoyl chloride, CH3COCl. Phenol reacts with ethanoyl chloride at room temperature, although the reaction isn't as fast as the one between ethanoyl chloride and an alcohol. Phenyl ethanoate is formed together with hydrogen chloride gas.
Sometimes it is necessary to modify the phenol first to make the reaction faster.
For example, benzoyl chloride has the formula C6H5COCl. The -COCl group is attached directly to a benzene ring. It is much less reactive than simple acyl chlorides like ethanoyl chloride. In order to get a reasonably quick reaction with benzoyl chloride, the phenol is first converted into sodium phenoxide by dissolving it in sodium hydroxide solution.
The phenoxide ion reacts more rapidly with benzoyl chloride than the original phenol does, but even so you have to shake it with benzoyl chloride for about 15 minutes. Solid phenyl benzoate is formed.
Making esters from phenol using an acid anhydride
A typical acid anhydride is ethanoic anhydride, (CH3CO)2O. The reactions of acid anhydrides are slower than the corresponding reactions with acyl chlorides, and you usually need to warm the mixture. Again, you can react the phenol with sodium hydroxide solution first, producing the more reactive phenoxide ion. If you simply use phenol and ethanoic anhydride, phenyl ethanoate is formed together with ethanoic acid.
This reaction isn't important itself, but a very similar reaction is involved in the manufacture of aspirin (covered in detail on another page - link below).
If the phenol is first converted into sodium phenoxide by adding sodium hydroxide solution, the reaction is faster. Phenyl ethanoate is again formed, but this time the other product is sodium ethanoate rather than ethanoic acid.
The reaction with iron(III) chloride solution
Iron(III) chloride is sometimes known as ferric chloride. Iron(III) ions form strongly colored complexes with several organic compounds including phenol. The colour of the complexes vary from compound to compound. The reaction with iron(III) chloride solution can be used as a test for phenol. If you add a crystal of phenol to iron(III) chloride solution, you get an intense violet-purple solution formed. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Phenols/Reactivity_of_Phenols/Other_Reactions_of_Phenol.txt |
Phenols are rather easily oxidized despite the absence of a hydrogen atom on the hydroxyl bearing carbon. Among the colored products from the oxidation of phenol by chromic acid is the dicarbonyl compound para-benzoquinone (also known as 1,4-benzoquinone or simply quinone); an ortho isomer is also known. These compounds are easily reduced to their dihydroxybenzene analogs, and it is from these compounds that quinones are best prepared. Note that meta-quinones having similar structures do not exist. The redox equilibria between the dihydroxybenzenes hydroquinone and catechol and their quinone oxidation states are so facile that milder oxidants than chromate (Jones reagent) are generally preferred. One such oxidant is Fremy's salt, shown on the right. Reducing agents other than stannous chloride (e.g. NaBH4) may be used for the reverse reaction.
The position of the quinone-hydroquinone redox equilibrium is proportional to the square of the hydrogen ion concentration, as shown by the following half-reactions (electrons are colored blue). The electrode potential for this interconversion may therefore be used to measure the pH of solutions.
Quinone + 2H(+)
2e(–)
–2e(–)
Hydroquinone
Although chromic acid oxidation of phenols having an unsubstituted para-position gives some p-quinone product, the reaction is complex and is not synthetically useful. It has been found that salcomine, a cobalt complex, binds oxygen reversibly in solution, and catalyzes the oxidation of various substituted phenols to the corresponding p-quinones. The structure of salcomine and an example of this reaction are shown in the following equation. The solvent of choice for these oxidations is usually methanol or dimethylformamide (DMF).
Ring Reactions of Phenol
This page looks at the way the reactions of the benzene ring in phenol are modified by the presence of the attached -OH group. It covers the reactions of phenol with bromine water and with nitric acid.
Activation of the ring
The -OH group attached to the benzene ring in phenol has the effect of making the ring much more reactive than it would otherwise be. For example, as you will find below, phenol will react with a solution of bromine in water (bromine water) in the cold and in the absence of any catalyst. It also reacts with dilute nitric acid, whereas benzene itself needs a nitrating mixture of concentrated nitric acid and concentrated sulfuric acid.
Figure: One of the lone pairs on the oxygen atom in the -OH group overlaps with the delocalised ring electron system giving a structure rather like the right.
The donation of the oxygen's lone pair into the ring system increases the electron density around the ring. A benzene ring undergoes substitution reactions in which the ring electrons are attacked by positive ions or the slightly positive parts of molecules. In other words, it undergoes electrophilic substitution. If you increase the electron density around the ring, it becomes even more attractive to incoming electrophiles. That's what happens in phenol.
The directing effect of the -OH group
The -OH group has more activating effect on some positions around the ring than others (for reasons which go beyond UK A level). That means that incoming groups will go into some positions much faster than they will into others.
The net effect of this is that the -OH group has a 2,4-directing effect. That means that incoming groups will tend to go into the 2- position (next door to the -OH group) or the 4- position (opposite the -OH group). You will get hardly any of the 3- isomer formed - it is produced too slowly.
Example 1: Reaction with Bromine Water
If bromine water is added to a solution of phenol in water, the bromine water is decolorized and a white precipitate is formed which smells of antiseptic. The precipitate is 2,4,6-tribromophenol.
Notice the multiple substitution around the ring - into all the activated positions. (The 6- position is, of course, just the same as the 2- position. Both are next door to the -OH group.)
Example 2: Reactions with Nitric Acid
The reactions with nitric acid are complicated because nitric acid is an oxidizing agent, and phenol is very easily oxidized to give complex tarry products. What follows misses all that complication out, and just concentrates on the ring substitution which happens as well.
With dilute nitric acid: Phenol reacts with dilute nitric acid at room temperature to give a mixture of 2-nitrophenol and 4-nitrophenol.
With concentrated nitric acid: With concentrated nitric acid, more nitro groups substitute around the ring to give 2,4,6-trinitrophenol (common name: picric acid).
Contributors
Jim Clark (Chemguide.co.uk)
Ring Substitution of Phenols
Substitution of the Hydroxyl Hydrogen
As with the alcohols, the phenolic hydroxyl hydrogen is rather easily replaced by other substituents. For example, phenol reacts easily with acetic anhydride to give phenyl acetate. Likewise, the phenolate anion is an effective nucleophile in SN2 reactions, as in the second example below.
C6H5–OH + (CH3CO)2O C6H5–O–COCH3 + CH3CO2H
C6H5–O(–) Na(+) + CH3CH2CH3–Br C6H5–O–CH2CH2CH3 + NaBr
Electrophilic Substitution of the Phenol Aromatic Ring
The facility with which the aromatic ring of phenols and phenol ethers undergoes electrophilic substitution has been noted. Two examples are shown in the following diagram. The first shows the Friedel-Crafts synthesis of the food preservative BHT from para-cresol. The second reaction is interesting in that it further demonstrates the delocalization of charge that occurs in the phenolate anion. Carbon dioxide is a weak electrophile and normally does not react with aromatic compounds; however, the negative charge concentration on the phenolate ring enables the carboxylation reaction shown in the second step. The sodium salt of salicylic acid is the major product, and the preference for ortho substitution may reflect the influence of the sodium cation. This is called the Kolbe-Schmidt reaction, and it has served in the preparation of aspirin, as the last step illustrates.
Substitution of the Hydroxyl Hydrogen
As with the alcohols, the phenolic hydroxyl hydrogen is rather easily replaced by other substituents. For example, phenol reacts easily with acetic anhydride to give phenyl acetate. Likewise, the phenolate anion is an effective nucleophile in SN2 reactions, as in the second example below.
Examples:
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Synthesis of Phenols
Under construction
Several laboratory methods for the synthesis of phenols: | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Phenols/Reactivity_of_Phenols/Oxidation_of_Quinones.txt |
This page looks at the structure and physical properties of phenylamine - also known as aniline or aminobenzene. Phenylamine has an -NH2 group attached directly to a benzene ring.
The structure of phenylamine
Phenylamine is a primary amine - a compound in which one of the hydrogen atoms in an ammonia molecule has been replaced by a hydrocarbon group. However, in comparison with simple primary amines like methylamine, the properties of phenylamine are slightly different. This is because the lone pair on the nitrogen atom interacts with the delocalized electrons in the benzene ring. The simplest way to draw the structure of phenylamine is:
There is an interaction between the delocalised electrons in the benzene ring and the lone pair on the nitrogen atom. The lone pair overlaps with the delocalised ring electron system . . .
. . . giving a structure rather like this:
The donation of the nitrogen's lone pair into the ring system increases the electron density around the ring. That makes the ring much more reactive than it is in benzene itself. It also reduces the availability of the lone pair on the nitrogen to take part in other reactions. In particular, it makes phenylamine much more weakly basic than primary amines where the -NH2 group isn't attached to a benzene ring.
Physical properties
Pure phenylamine is a colorless liquid, but it darkens rapidly on exposure to light and air. It is normally a brown oily liquid.
Melting and boiling points
It is useful to compare phenylamine's melting and boiling points with those of methylbenzene (toluene). Both molecules contain a similar number of electrons and are a very similar shape. That means that the intermolecular attractions due to van der Waals dispersion forces are going to be very similar.
melting point (°C) boiling point (°C)
C6H5NH2 -6.2 184
C6H5CH3 -95.0 111
The reason for the higher values for phenylamine is in part due to permanent dipole-dipole attractions due to the electronegativity of the nitrogen - but is mainly due to hydrogen bonding. Hydrogen bonds can form between a lone pair on a nitrogen on one molecule and the hydrogen on the -NH2 group of one of its neighbors.
Solubility in water
Phenylamine is slightly soluble in water - about 3.6 g (depending on where you get the data from!) of phenylamine will dissolve in 100 g of water at 20°C. Mixtures containing more phenylamine than this separate into two layers, with the phenylamine forming the bottom one.
Phenylamine is somewhat soluble in water because of its ability to form hydrogen bonds with the water. However, the benzene rings in the phenylamine break more hydrogen bonds between water molecules than are reformed between water and the -NH2 groups. The water molecules also disrupt fairly strong van der Waals attractions between the phenylamine molecules. Both of these effects mean that dissolving phenylamine in water isn't very energetically profitable, and so stop the phenylamine from being very soluble.
Making Diazonium Salts
This page looks at the reaction between phenylamine (also known as aniline and aminobenzene) and nitrous acid - particularly its reaction at temperatures of less than 5°C to produce diazonium salts. If you want to know about the reactions of the diazonium ions formed, you will find a link at the bottom of the page.
The reactions of phenylamine with nitrous acid
Nitrous acid (also known as nitric(III) acid) has the formula HNO2. It is sometimes written as HONO to show the way it is joined up. Nitrous acid decomposes very readily and is always made in situ. In the case of its reaction with phenylamine, the phenylamine is first dissolved in hydrochloric acid, and then a solution of sodium or potassium nitrite is added. The reaction between the hydrochloric acid and the nitrite ions produces the nitrous acid.
$H^+ (aq) + NO_2^- (aq) \rightleftharpoons HNO_2 (aq)$
Because nitrous acid is a weak acid, the position of equilibrium lies well the right. Phenylamine reacts with nitrous acid differently depending on the temperature.
The reaction on warming
If the mixture is warmed, you get a black oily product which contains phenol (amongst other things), and nitrogen gas is given off.
$CH_6H_5NH_2 + HNO_2 \rightarrow C_6H_5OH + H_2O + N_2$
The reaction at low temperatures
The solution of phenylamine in hydrochloric acid (phenylammonium chloride solution) is stood in a beaker of ice. The sodium or potassium nitrite solution is also cooled in the ice. The solution of the nitrite is then added very slowly to the phenylammonium chloride solution - so that the temperature never goes above 5°C. You end up with a solution containing benzenediazonium chloride:
The positive ion, containing the -N2+ group, is known as a diazonium ion. The "azo" bit of the name refers to nitrogen. The ionic equation for the reaction is:
Notice that the chloride ions from the acid aren't involved in this in any way. If you use hydrochloric acid, the solution will contain benzenediazonium chloride. If you used a different acid, you would just get a different salt - a sulphate or hydrogensulphate, for example, if you used sulfuric acid.
The reactions of a diazonium salt are always done with a freshly prepared solution made in this way since the solutions do not keep. Diazonium salts are very unstable and tend to be explosive as solids.
Contributors
Jim Clark (Chemguide.co.uk)
Preparation of Phenylamine Compounds
This page looks in outline at the preparation of phenylamine (also known as aniline or aminobenzene) starting from benzene. The benzene is first converted to nitrobenzene which is in turn reduced to phenylamine.
Benzene to nitrobenzene
Benzene is nitrated by replacing one of the hydrogen atoms on the benzene ring by a nitro group, NO2. The benzene is treated with a mixture of concentrated nitric acid and concentrated sulphuric acid at a temperature not exceeding 50°C. The mixture is held at this temperature for about half an hour. Yellow oily nitrobenzene is formed.
You could write this in a more condensed form as:
$C_6H_6 + HNO_3 \rightarrow C_6H_5NO_2 + H_2O$
The concentrated sulfuric acid is acting as a catalyst and so isn't written into the equations. The temperature is kept relatively low to prevent more than one nitro group being substituted onto the ring.
Nitrobenzene to phenylamine
The conversion is done in two main stages:
Stage 1: conversion of nitrobenzene into phenylammonium ions
Nitrobenzene is reduced to phenylammonium ions using a mixture of tin and concentrated hydrochloric acid. The mixture is heated under reflux in a boiling water bath for about half an hour. Under the acidic conditions, rather than getting phenylamine directly, you instead get phenylammonium ions formed. The lone pair on the nitrogen in the phenylamine picks up a hydrogen ion from the acid.
The electron-half-equation for this reaction is:
The nitrobenzene has been reduced by gaining electrons in the presence of the acid. The electrons come from the tin, which forms both tin(II) and tin(IV) ions.
$Sn \rightarrow Sn^{2+} + 2e^-$
$Sn^{2+} \rightarrow Sn^{4+} + 2e^-$
Stage 2: conversion of the phenylammonium ions into phenylamine
All you need to do is to remove the hydrogen ion from the -NH3+ group. Sodium hydroxide solution is added to the product of the first stage of the reaction.
The phenylamine is formed together with a complicated mixture of tin compounds from reactions between the sodium hydroxide solution and the complex tin ions formed during the first stage. The phenylamine is finally separated from this mixture. The separation is long, tedious and potentially dangerous - involving steam distillation, solvent extraction and a final distillation.
Summary
You are almost bound to need the mechanism for the nitration reaction as well.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Phenylamine_and_Diazonium_Compounds/Introduction_to_Phenylamine_Compounds.txt |
This page looks at some typical reactions of diazonium ions, including examples of both substitution reactions and coupling reactions. If you have come straight to this page from a search engine and want to know about the preparation of the diazonium ions.
Substitution reactions of diazonium ions
Diazonium ions are present in solutions such as benzenediazonium chloride solution. They contain an -N2+ group. In the case of benzenediazonium chloride, this is attached to a benzene ring. Benzenediazonium chloride looks like this:
In this set of reactions of the diazonium ion, the -N2+ group is replaced by something else. The nitrogen is released as nitrogen gas.
Substitution by an -OH group
To get this reaction, all you need to do is warm the benzenediazonium chloride solution. The diazonium ion reacts with the water in the solution and phenol is formed - either in solution or as a black oily liquid (depending on how much is formed). Nitrogen gas is evolved.
This is the same reaction that you get if you react phenylamine with nitrous acid in the warm. The diazonium ion is formed first and then immediately reacts with the water in the solution to give phenol.
Substitution by an iodine atom
This is a good example of the use of diazonium salts to substitute things into a benzene ring which are otherwise quite difficult to attach. (That's equally true of the previous reaction, by the way.) If you add potassium iodide solution to the benzenediazonium chloride solution in the cold, nitrogen gas is given off, and you get oily droplets of iodobenzene formed.
There is a simple reaction between the diazonium ions and the iodide ions from the potassium iodide solution.
Coupling reactions of diazonium ions
In the substitution reactions above, the nitrogen in the diazonium ion is lost. In the rest of the reactions on this page, the nitrogen is retained and used to make a bridge between two benzene rings.
The reaction with phenol
Phenol is dissolved in sodium hydroxide solution to give a solution of sodium phenoxide.
The solution is cooled in ice, and cold benzenediazonium chloride solution is added. There is a reaction between the diazonium ion and the phenoxide ion and a yellow-orange solution or precipitate is formed. The product is one of the simplest of what are known as azo compounds, in which two benzene rings are linked by a nitrogen bridge.
The reaction with naphthalen-2-ol
Naphthalen-2-ol is also known as 2-naphthol or beta-naphthol. It contains an -OH group attached to a naphthalene molecule rather than to a simple benzene ring. Naphthalene has two benzene rings fused together. The reaction is done under exactly the same conditions as with phenol. The naphthalen-2-ol is dissolved in sodium hydroxide solution to produce an ion just like the phenol one. This solution is cooled and mixed with the benzenediazonium chloride solution.
An intense orange-red precipitate is formed - another azo compound.
The reaction with phenylamine (aniline)
Some liquid phenylamine is added to a cold solution of benzenediazonium chloride, and the mixture is shaken vigorously. A yellow solid is produced.
These strongly coloured azo compounds are frequently used as dyes known as azo dyes. The one made from phenylamine (aniline) is known as "aniline yellow" (amongst many other things - see note above). Azo compounds account for more than half of modern dyes.
The use of an azo dye as an indicator - methyl orange
Azo compounds contain a highly delocalized system of electrons which takes in both benzene rings and the two nitrogen atoms bridging the rings. The delocalization can also extend to things attached to the benzene rings as well. If white light falls on one of these molecules, some wavelengths are absorbed by these delocalized electrons. The colour you see is the result of the non-absorbed wavelengths. The groups which contribute to the delocalization (and so to the absorption of light) are known as a chromophore.
Modifying the groups present in the molecule can have an effect on the light absorbed, and so on the color you see. You can take advantage of this in indicators. Methyl orange is an azo dye which exists in two forms depending on the pH:
As the hydrogen ion is lost or gained there is a shift in the exact nature of the delocalisation in the molecule, and that causes a shift in the wavelength of light absorbed. Obviously that means that you see a different colour. When methyl orange is added, a hydrogen ion attaches to give the red form. Methyl orange is red in acidic solutions (in fact solutions of pH less than 3.1). If an alkali is added or hydrogen ions are removed, then the yellow form if generated. Methyl orange is yellow at pH's greater than 4.4.
In between, at some point there will be equal amounts of the red and yellow forms and so methyl orange looks orange.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Phenylamine_and_Diazonium_Compounds/Reactions_of_Diazonium_Salts.txt |
This page looks at reactions of phenylamine (also known as aniline or aminobenzene) where it behaves as a fairly straightforward primary amine. It explains why phenylamine is a weaker base than other primary amines, and summarises its reactions with acyl chlorides (acid chlorides), acid anhydrides and halogenoalkanes (haloalkanes or alkyl halides). Before you read each section on this page, you should follow the link to the corresponding page about aliphatic amines (those not based on benzene rings). In most cases, the reactions are the same, and this page only really looks in detail at the differences in the phenylamine case.
Amines are bases because the lone pair of electrons on the nitrogen atom can accept a hydrogen ion - in other words, for exactly the same reason that ammonia is a base. With phenylamine, the only difference is that it is a much weaker base than ammonia or an amine like ethylamine - for reasons that we will explore later.
Reaction of Phenylamine with Acids
Phenylamine reacts with acids like hydrochloric acid in exactly the same way as any other amine. Despite the fact that the phenylamine is only a very weak base, with a strong acid like hydrochloric acid the reaction is completely straightforward. Phenylamine is only very slightly soluble in water, but dissolves freely in dilute hydrochloric acid. A solution of a salt is formed - phenylammonium chloride.
To show the formation of the salt, you could write:
$C_6H_5NH_2 + HCl \rightarrow C_6H_5NH_3^+ Cl^-$
. . . or if you want to emphasize the fact that the phenylamine is acting as a base, you could most simply use:
$C_6H_5NH_2 (l) + H^+ (aq) \rightarrow C_6H_5NH_3^+$
To get the phenylamine back from the phenylammonium ion present in the salt, all you have to do is to take the hydrogen ion away again. You can do that by adding any stronger base. Normally, you would choose sodium hydroxide solution:
$C_6H_5NH_3^+ (aq) + OH^- (aq) \rightarrow C_6H_5NH_2 + H_2O (l)$
The phenylamine is formed first as an off-white emulsion - tiny droplets of phenylamine scattered throughout the water. This then settles out to give an oily bottom layer of phenylamine under the aqueous layer.
Reactions of Phenylamine with Ethers
This is where it is possible to tell that phenylamine is a much weaker base than ammonia and the aliphatic amines like methylamine and ethylamine. Phenylamine reacts reversibly with water to give phenylammonium ions and hydroxide ions.
$C_6H_5NH_2 (aq) + H_2O (l) \rightleftharpoons C_6H_5NH_3^+ + OH^- (aq)$
The position of equilibrium lies well to the left of the corresponding ammonia or aliphatic amine equilibria - which means that not many hydroxide ions are formed in the solution. The effect of this is that the pH of a solution of phenylamine will be quite a bit lower than a solution of ammonia or one of the aliphatic amines of the same concentration. For example, a 0.1 M phenylamine solution has a pH of about 9 compared to a pH of about 11 for 0.1 M ammonia solution.
Why is phenylamine such a weak base?
Amines are bases because they pick up hydrogen ions on the lone pair on the nitrogen atom. In phenylamine, the attractiveness of the lone pair is lessened because of the way it interacts with the ring electrons. The lone pair on the nitrogen touches the delocalized ring electrons . . .
. . . and becomes delocalized with them:
That means that the lone pair is no longer fully available to combine with hydrogen ions. The nitrogen is still the most electronegative atom in the molecule, and so the delocalized electrons will be attracted towards it, but the electron density around the nitrogen is nothing like it is in, say, an ammonia molecule. The other problem is that if the lone pair is used to join to a hydrogen ion, it is no longer available to contribute to the delocalization. That means that the delocalization would have to be disrupted if the phenylamine acts as a base. Delocalization makes molecules more stable, and so disrupting the delocalization costs energy and won't happen easily. Taken together - the lack of intense charge around the nitrogen, and the need to break some delocalization - means that phenylamine is a very weak base indeed.
Reactions of Phenylamine with Acyl Chlorides and with Acid Anhydrides
These are reactions in which the phenylamine acts as a nucleophile. There is no essential difference between these reactions and the same reactions involving any other primary amine. You will find a summary of the reactions below, but all the detailed explanations are on other pages. We'll take ethanoyl chloride as a typical acyl chloride, and ethanoic anhydride as a typical acid anhydride. The important product of the reaction of phenylamine with either of these is the same. Phenylamine reacts vigorously in the cold with ethanoyl chloride to give a mixture of solid products - ideally white, but usually stained brownish. A mixture of N-phenylethanamide (old name: acetanilide) and phenylammonium chloride is formed.
The overall equation for the reaction is:
$CH_3COCl + 2C_6H_5NH_2 \rightarrow CH_3CONHC_6H_5 + C_6H_5NH_3^+ Cl^-$
With ethanoic anhydride, heat is needed. In this case, the products are a mixture of N-phenylethanamide and phenylammonium ethanoate.
$(CH_3CO)_2O + 2C_6H_5NH_2 \rightarrow CH_3CONHC_6H_5 + CH_3COO^-NH_3C_6H_5^+$
The main product molecule (the N-phenylethanamide) is often drawn looking like this:
If you stop and think about it, this is obviously the same molecule as in the equation above, but it stresses the phenylamine part of it much more. Looking at it this way, notice that one of the hydrogens of the -NH2 group has been replaced by an acyl group - an alkyl group attached to a carbon-oxygen double bond. Tthe phenylamine has been acylated or has undergone acylation. Because of the nature of this particular acyl group, it is also described as ethanoylation. The hydrogen is being replaced by an ethanoyl group, CH3CO-.
Reactions of Phenylamine with Halogenoalkanes
This is another reaction of phenylamine as a nucleophile, and again there is no essential difference between its reactions and those of aliphatic amines. Taking bromoethane as a typical halogenoalkane, the reaction with phenylamine happens in the same series of complicated steps as with any other amine. We'll just look at the first step. On heating, the bromoethane and phenylamine react to give a mixture of a salt of a secondary amine and some free secondary amine. In this case, you would first get N-ethylphenylammonium bromide:
This would instantly be followed by a reversible reaction in which some unreacted phenylamine would take a hydrogen ion from the salt to give some free secondary amine: N-ethylphenylamine.
The reaction would not stop there. You will get further reactions to produce a tertiary amine and its salt, and eventually a quaternary ammonium compound. If you want to explore this further, refer to the last link just up the page, and trace the sequence of equations through using phenylamine rather than ethylamine.
Reactions of Phenylamine with Bromine Water
This page looks at the reaction of the benzene ring in phenylamine (aniline) with bromine water. It explains why the amino group activates the ring.
Activation of the ring
The -NH2 group attached to the benzene ring in phenylamine has the effect of making the ring much more reactive than it would otherwise be. This is exactly the same as the effect of the -OH group in phenol if you have already come across that. For example, phenylamine will react with an aqueous solution of bromine (bromine water) in the cold and in the absence of any catalyst. Unactivated rings will only react with bromine in the presence of a catalyst.
The lone pair on the nitrogen touches the delocalized ring electrons . . .
. . . and becomes delocalized with them:
A benzene ring undergoes substitution reactions in which the ring electrons are attacked by positive ions or the slightly positive parts of molecules. In other words, it undergoes electrophilic substitution. If you increase the electron density around the ring by involving extra electrons from the -NH2 group, it becomes even more attractive to incoming electrophiles. That's what happens in phenylamine.
The directing effect of the -NH2 group
The -NH2 group has more activating effect on some positions around the ring than others (for reasons which go beyond UK A level). That means that incoming groups will go into some positions much faster than they will into others. The net effect of this is that the -NH2 group has a 2,4-directing effect. That means that incoming groups will tend to go into the 2- position (next door to the -NH2 group) or the 4- position (opposite the -NH2 group). You will get hardly any of the 3- isomer formed - it is produced too slowly.
The reaction with bromine water
If bromine water is added to phenylamine, the bromine water is decolourized and a white precipitate is formed. This is exactly like the reaction which happens with phenol. The precipitate is 2,4,6-tribromophenylamine.
Notice the multiple substitution around the ring - into all the activated positions. (The 6- position is, of course, just the same as the 2- position. Both are next door to the -NH2 group.) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Phenylamine_and_Diazonium_Compounds/Reactions_of_Phenylamine_as_a_Primary_Amine.txt |
Polymers are long chain, giant organic molecules are assembled from many smaller molecules called monomers. Polymers consist of many repeating monomer units in long chains, sometimes with branching or cross-linking between the chains.
• Addition Polymers
An addition polymer is a polymer which is formed by an addition reaction, where many monomers bond together via rearrangement of bonds without the loss of any atom or molecule under specific conditions of heat, pressure, and/or the presence of a catalyst.
• Condensation Polymers
Condensation polymers are any kind of polymers formed through a condensation reaction—where molecules join together—losing small molecules as byproducts such as water or methanol, as opposed to addition polymers which involve the reaction of unsaturated monomers.
• Introduction to Polymers
Polymers are substances containing a large number of structural units joined by the same type of linkage. These substances often form into a chain-like structure. Polymers in the natural world have been around since the beginning of time. Starch, cellulose, and rubber all possess polymeric properties. Man-made polymers have been studied since 1832. Today, the polymer industry has grown to be larger than the aluminum, copper and steel industries combined.
• Molecular Weights of Polymers
Most polymers are not composed of identical molecules. The HDPE molecules, for example, are all long carbon chains, but the lengths may vary by thousands of monomer units. Because of this, polymer molecular weights are usually given as averages.
• Polyethylene
Polyethylene is the most popular plastic in the world. This is the polymer that makes grocery bags, shampoo bottles, children's toys, and even bullet proof vests. For such a versatile material, it has a very simple structure, the simplest of all commercial polymers. A molecule of polyethylene is nothing more than a long chain of carbon atoms, with two hydrogen atoms attached to each carbon atom.
• Rubber Polymers
Rubber is an example of an elastomer type polymer, where the polymer has the ability to return to its original shape after being stretched or deformed. The rubber polymer is coiled when in the resting state. The elastic properties arise from the its ability to stretch the chains apart, but when the tension is released the chains snap back to the original position. The majority of rubber polymer molecules contain at least some units derived from conjugated diene monomers.
Thumbnail: Space-filling model of a section of the polyethylene terephthalate polymer, also known as PET and PETE, a polyester used in most plastic bottles. Color code: Carbon, C (black), Hydrogen, H (white), and Oxygen, O (red). (Public Domain; Jynto).
Polymers
Polymers are long chain giant organic molecules are assembled from many smaller molecules called monomers. Polymers consist of many repeating monomer units in long chains. A polymer is analogous to a necklace made from many small beads (monomers). Many monomers are alkenes or other molecules with double bonds which react by addition to their unsaturated double bonds.
Introduction
The electrons in the double bond are used to bond two monomer molecules together. This is represented by the red arrows moving from one molecule to the space between two molecules where a new bond is to form. The formation of polyethylene from ethylene (ethene) may be illustrated in the graphic on the left as follows. In the complete polymer, all of the double bonds have been turned into single bonds. No atoms have been lost and you can see that the monomers have just been joined in the process of addition. A simple representation is -[A-A-A-A-A]-. Polyethylene is used in plastic bags, bottles, toys, and electrical insulation.
• LDPE - Low Density Polyethylene: The first commercial polyethylene process used peroxide catalysts at a temperature of 500 C and 1000 atmospheres of pressure. This yields a transparent polymer with highly branched chains which do not pack together well and is low in density. LDPE makes a flexible plastic. Today most LDPE is used for blow-molding of films for packaging and trash bags and flexible snap-on lids. LDPE is recyclable plastic #4.
• HDPE - High Density Polyethylene: An alternate method is to use Ziegler-Natta aluminum titanium catalysts to make HDPE which has very little branching, allows the strands to pack closely, and thus is high density. It is three times stronger than LDPE and more opaque. About 45% of the HDPE is blow molded into milk and disposable consumer bottles. HDPE is also used for crinkly plastic bags to pack groceries at grocery stores. HDPE is recyclable plastic #2.
Other Addition Polymers
• PVC (polyvinyl chloride), which is found in plastic wrap, simulated leather, water pipes, and garden hoses, is formed from vinyl chloride (H2C=CHCl). The reaction is shown in the graphic on the left. Notice how every other carbon must have a chlorine attached.
• Polypropylene: The reaction to make polypropylene (H2C=CHCH3) is illustrated in the middle reaction of the graphic. Notice that the polymer bonds are always through the carbons of the double bond. Carbon #3 already has saturated bonds and cannot participate in any new bonds. A methyl group is on every other carbon.
• Polystyrene: The reaction is the same for polystrene where every other carbon has a benzene ring attached. Polystyrene (PS) is recyclable plastic #6. In the following illustrated example, many styrene monomers are polymerized into a long chain polystyrene molecule. The squiggly lines indicate that the polystyrene molecule extends further at both the left and right ends.
• Blowing fine gas bubbles into liquid polystyrene and letting it solidify produces expanded polystyrene, called Styrofoam by the Dow Chemical Company.
• Polystyrene with DVB: Cross-linking between polymer chains can be introduced into polystyrene by copolymerizing with p-divinylbenzene (DVB). DVB has vinyl groups (-CH=CH2) at each end of its molecule, each of which can be polymerized into a polymer chain like any other vinyl group on a styrene monomer.
Table 1: Various Polymers
Monomer Polymer Name Trade Name Uses
F2C=CF2 polytetrafluoroethylene Teflon
Non-stick coating for cooking utensils, chemically-resistant specialty plastic parts, Gore-Tex
H2C=CCl2 polyvinylidene dichloride Saran Clinging food wrap
H2C=CH(CN) polyacrylonitrile Orlon, Acrilan, Creslan Fibers for textiles, carpets, upholstery
H2C=CH(OCOCH3) polyvinyl acetate Elmer's glue - Silly Putty Demo
H2C=CH(OH) polyvinyl alcohol Ghostbusters Demo
H2C=C(CH3)COOCH3 polymethyl methacrylate Plexiglass, Lucite Stiff, clear, plastic sheets, blocks, tubing, and other shapes
Addition polymers from conjugated dienes
Polymers from conjugated dienes usually give elastomer polymers having rubber-like properties.
Table 2. Addition homopolymers from conjugated dienes
Monomer Polymer name Trade name Uses
H2C=CH-C(CH3)=CH2 polyisoprene natural or some synthetic rubber applications similar to natural rubber
H2C=CH-CH=CH2 polybutadiene polybutadiene synthetic rubber select synthetic rubber applications
H2C=CH-CCl=CH2 polychloroprene Neoprene chemically-resistant rubber
Ring opening polymerization
In this kind of polymerization, molecular rings are opened in the formation of a polymer. Here epsilon-caprolactam, a 6-carbon cyclic monomer, undergoes ring opening to form a Nylon 6 homopolymer, which is somewhat similar to but not the same as Nylon 6,6 alternating copolymer.
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Addition_Polymers.txt |
A large number of important and useful polymeric materials are not formed by chain-growth processes involving reactive species such as radicals, but proceed instead by conventional functional group transformations of polyfunctional reactants. These polymerizations often (but not always) occur with loss of a small byproduct, such as water, and generally (but not always) combine two different components in an alternating structure. The polyester Dacron and the polyamide Nylon 66, shown here, are two examples of synthetic condensation polymers, also known as step-growth polymers. In contrast to chain-growth polymers, most of which grow by carbon-carbon bond formation, step-growth polymers generally grow by carbon-heteroatom bond formation (C-O & C-N in Dacron & Nylon respectively). Although polymers of this kind might be considered to be alternating copolymers, the repeating monomeric unit is usually defined as a combined moiety.
Examples of naturally occurring condensation polymers are cellulose, the polypeptide chains of proteins, and poly(β-hydroxybutyric acid), a polyester synthesized in large quantity by certain soil and water bacteria. Formulas for these will be displayed below by clicking on the diagram.
Characteristics of Condensation Polymers
Condensation polymers form more slowly than addition polymers, often requiring heat, and they are generally lower in molecular weight. The terminal functional groups on a chain remain active, so that groups of shorter chains combine into longer chains in the late stages of polymerization. The presence of polar functional groups on the chains often enhances chain-chain attractions, particularly if these involve hydrogen bonding, and thereby crystallinity and tensile strength. The following examples of condensation polymers are illustrative.
Note that for commercial synthesis the carboxylic acid components may actually be employed in the form of derivatives such as simple esters. Also, the polymerization reactions for Nylon 6 and Spandex do not proceed by elimination of water or other small molecules. Nevertheless, the polymer clearly forms by a step-growth process. Some Condensation Polymers
The difference in Tg and Tm between the first polyester (completely aliphatic) and the two nylon polyamides (5th & 6th entries) shows the effect of intra-chain hydrogen bonding on crystallinity. The replacement of flexible alkylidene links with rigid benzene rings also stiffens the polymer chain, leading to increased crystalline character, as demonstrated for polyesters (entries 1, 2 &3) and polyamides (entries 5, 6, 7 & 8). The high Tg and Tm values for the amorphous polymer Lexan are consistent with its brilliant transparency and glass-like rigidity. Kevlar and Nomex are extremely tough and resistant materials, which find use in bullet-proof vests and fire resistant clothing.
Many polymers, both addition and condensation, are used as fibers The chief methods of spinning synthetic polymers into fibers are from melts or viscous solutions. Polyesters, polyamides and polyolefins are usually spun from melts, provided the Tm is not too high. Polyacrylates suffer thermal degradation and are therefore spun from solution in a volatile solvent. Cold-drawing is an important physical treatment that improves the strength and appearance of these polymer fibers. At temperatures above Tg, a thicker than desired fiber can be forcibly stretched to many times its length; and in so doing the polymer chains become untangled, and tend to align in a parallel fashion. This cold-drawing procedure organizes randomly oriented crystalline domains, and also aligns amorphous domains so they become more crystalline. In these cases, the physically oriented morphology is stabilized and retained in the final product. This contrasts with elastomeric polymers, for which the stretched or aligned morphology is unstable relative to the amorphous random coil morphology.
This cold-drawing treatment may also be used to treat polymer films (e.g. Mylar & Saran) as well as fibers.
Step-growth polymerization is also used for preparing a class of adhesives and amorphous solids called epoxy resins. Here the covalent bonding occurs by an SN2 reaction between a nucleophile, usually an amine, and a terminal epoxide. In the following example, the same bisphenol A intermediate used as a monomer for Lexan serves as a difunctional scaffold to which the epoxide rings are attached. Bisphenol A is prepared by the acid-catalyzed condensation of acetone with phenol. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Condensation_Polymers.txt |
The synthesis of macromolecules composed of more than one monomeric repeating unit has been explored as a means of controlling the properties of the resulting material. In this respect, it is useful to distinguish several ways in which different monomeric units might be incorporated in a polymeric molecule. The following examples refer to a two component system, in which one monomer is designated A and the other B.
• Statistical Copolymers: Also called random copolymers. Here the monomeric units are distributed randomly, and sometimes unevenly, in the polymer chain: ~ABBAAABAABBBABAABA~.
• Alternating Copolymers: Here the monomeric units are distributed in a regular alternating fashion, with nearly equimolar amounts of each in the chain: ~ABABABABABABABAB~.
• Block Copolymers: Instead of a mixed distribution of monomeric units, a long sequence or block of one monomer is joined to a block of the second monomer: ~AAAAA-BBBBBBB~AAAAAAA~BBB~.
• Graft Copolymers: As the name suggests, side chains of a given monomer are attached to the main chain of the second monomer: ~AAAAAAA(BBBBBBB~)AAAAAAA(BBBB~)AAA~.
Addition Copolymerization
Most direct copolymerizations of equimolar mixtures of different monomers give statistical copolymers, or if one monomer is much more reactive a nearly homopolymer of that monomer. The copolymerization of styrene with methyl methacrylate, for example, proceeds differently depending on the mechanism. Radical polymerization gives a statistical copolymer. However, the product of cationic polymerization is largely polystyrene, and anionic polymerization favors formation of poly(methyl methacrylate). In cases where the relative reactivities are different, the copolymer composition can sometimes be controlled by continuous introduction of a biased mixture of monomers into the reaction.
Formation of alternating copolymers is favored when the monomers have different polar substituents (e.g. one electron withdrawing and the other electron donating), and both have similar reactivities toward radicals. For example, styrene and acrylonitrile copolymerize in a largely alternating fashion.
Some Useful Copolymers
Monomer A
Monomer B
Copolymer
Uses
H2C=CHCl
H2C=CCl2
Saran
films & fibers
H2C=CHC6H5
H2C=C-CH=CH2
SBR
styrene butadiene rubber
tires
H2C=CHCN
H2C=C-CH=CH2
Nitrile Rubber
adhesives
hoses
H2C=C(CH3)2
H2C=C-CH=CH2
Butyl Rubber
inner tubes
F2C=CF(CF3)
H2C=CHF
Viton
gaskets
A terpolymer of acrylonitrile, butadiene and styrene, called ABS rubber, is used for high-impact containers, pipes and gaskets.
Block Copolymerization
Several different techniques for preparing block copolymers have been developed, many of which use condensation reactions (next section). At this point, our discussion will be limited to an application of anionic polymerization. In the anionic polymerization of styrene described above, a reactive site remains at the end of the chain until it is quenched. The unquenched polymer has been termed a living polymer, and if additional styrene or a different suitable monomer is added a block polymer will form. This is illustrated for methyl methacrylate in the following diagram.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Introduction to Polymers
Prior to the early 1920's, chemists doubted the existence of molecules having molecular weights greater than a few thousand. This limiting view was challenged by Hermann Staudinger, a German chemist with experience in studying natural compounds such as rubber and cellulose. In contrast to the prevailing rationalization of these substances as aggregates of small molecules, Staudinger proposed they were made up of macromolecules composed of 10,000 or more atoms. He formulated a polymeric structure for rubber, based on a repeating isoprene unit (referred to as a monomer). For his contributions to chemistry, Staudinger received the 1953 Nobel Prize. The terms polymer and monomer were derived from the Greek roots poly (many), mono (one) and meros (part).
Recognition that polymeric macromolecules make up many important natural materials was followed by the creation of synthetic analogs having a variety of properties. Indeed, applications of these materials as fibers, flexible films, adhesives, resistant paints and tough but light solids have transformed modern society. Some important examples of these substances are discussed in the following sections.
Molecular Weights of Polymers
Unlike simpler pure compounds, most polymers are not composed of identical molecules. The HDPE molecules, for example, are all long carbon chains, but the lengths may vary by thousands of monomer units. Because of this, polymer molecular weights are usually given as averages. Two experimentally determined values are common: \(M_n\), the number average molecular weight, is calculated from the mole fraction distribution of different sized molecules in a sample, and \(M_w\), the weight average molecular weight, is calculated from the weight fraction distribution of different sized molecules. These are defined below. Since larger molecules in a sample weigh more than smaller molecules, the weight average Mw is necessarily skewed to higher values, and is always greater than \(M_n\). As the weight dispersion of molecules in a sample narrows, \(M_w\) approaches \(M_n\), and in the unlikely case that all the polymer molecules have identical weights (a pure mono-disperse sample), the ratio \(M_w\)/ \(M_n\) becomes unity.
Polyethylene
Polyethylene is probably the polymer you see most in daily life. Polyethylene is the most popular plastic in the world. This is the polymer that makes grocery bags, shampoo bottles, children's toys, and even bullet proof vests. For such a versatile material, it has a very simple structure, the simplest of all commercial polymers. A molecule of polyethylene is nothing more than a long chain of carbon atoms, with two hydrogen atoms attached to each carbon atom. That's what the picture at the top of the page shows, but it might be easier to draw it like the picture below, only with the chain of carbon atoms being many thousands of atoms long.
Sometimes it's a little more complicated. Sometimes some of the carbons, instead of having hydrogens attached to them, will have long chains of polyethylene attached to them. This is called branched, or low-density polyethylene, or LDPE. When there is no branching, it is called linear polyethylene, or HDPE. Linear polyethylene is much stronger than branched polyethylene, but branched polyethylene is cheaper and easier to make.
Linear polyethylene is normally produced with molecular weights in the range of 200,000 to 500,000, but it can be made even higher. Polyethylene with molecular weights of three to six million is referred to as ultra-high molecular weight polyethylene, or UHMWPE. UHMWPE can be used to make fibers which are so strong they replaced Kevlar for use in bullet proof vests. Large sheets of it can be used instead of ice for skating rinks. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Copolymers.txt |
Polymers are long chain, giant organic molecules are assembled from many smaller molecules called monomers. Polymers consist of many repeating monomer units in long chains, sometimes with branching or cross-linking between the chains. A polymer is analogous to a necklace made from many small beads (monomers). A chemical reaction forming polymers from monomers is called polymerization, of which there are many types. A common name for many synthetic polymer materials is plastic, which comes from the Greek word "plastikos", suitable for molding or shaping.
In the following illustrated example, many monomers called styrene are polymerized into a long chain polymer called polystyrene. The squiggly lines indicate that the polymer molecule extends further at both the left and right ends. In fact, polymer molecules are often hundreds or thousands of monomer units long.
Introduction
Many objects in daily use from packing, wrapping, and building materials include half of all polymers synthesized. Other uses include textiles, many electronic appliance casings, CD's, automobile parts, and many others are made from polymers. A quarter of the solid waste from homes is plastic materials - some of which may be recycled as shown in the table below.
Some products, such as adhesives, are made to include monomers which can be polymerized by the user in their application.
Types of Polymers
There are many types of polymers including synthetic and natural polymers.
Synthetic polymers
• Plastics
• Elastomers - solids with rubber-like qualities
• Rubber (carbon backbone often from hydrocarbon monomers)
• silicones (backbone of alternating silicon and oxygen atoms).
• Fibers
• Solid materials of intermediate characteristics
• Gels or viscous liquids
Classification of Polymers
• Homopolymers: These consist of chains with identical bonding linkages to each monomer unit. This usually implies that the polymer is made from all identical monomer molecules. These may be represented as : -[A-A-A-A-A-A]- Homopolymers are commonly named by placing the prefix poly in front of the constituent monomer name. For example, polystyrene is the name for the polymer made from the monomer styrene (vinylbenzene).
• Copolymers: These consist of chains with two or more linkages usually implying two or more different types of monomer units. These may be represented as : -[A-B-A-B-A-B]-
Polymers classified by mode of polymerization
• Addition Polymers: The monomer molecules bond to each other without the loss of any other atoms. Addition polymers from alkene monomers or substituted alkene monomers are the biggest groups of polymers in this class. Ring opening polymerization can occur without the loss of any small molecules.
• Condensation Polymers: Usually two different monomer combine with the loss of a small molecule, usually water. Most polyesters and polyamides (nylon) are in this class of polymers. Polyurethane Foam in graphic above.
Polymers classified by Physical Response to Heating
Thermoplastics
Plastics that soften when heated and become firm again when cooled. This is the more popular type of plastic because the heating and cooling may be repeated and the thermoplastic may be reformed.
Thermosets
These are plastics that soften when heated and can be molded, but harden permanently. They will decompose when reheated. An example is Bakelite, which is used in toasters, handles for pots and pans, dishes, electrical outlets and billiard balls.
Recycled Plastics
Recycle Code Abbreviation and Chemical Name of Plastic Types of Uses and Examples
1 PET - polyethylene terephthalate Many types of clear plastic consumer bottles, including clear, 2-liter beverage bottles
2 HDPE - High density polyethylene Milk jugs, detergent bottles, some water bottles, some grocery plastic bags
3 PVC - Polyvinyl chloride Plastic drain pipe, shower curtains, some water bottles
4 LDPE - Low density polyethylene Plastic garbage and other bags, garment bags, snap-on lids such as coffee can lids
5 PP - Polypropylene Many translucent (or opaque) plastic containers; containers for some products such as yogurt, soft butter, or margarine; aerosol can tops; rigid bottle caps; candy wrappers; bottoms of bottles
6 PS - Polystyrene Hard clear plastic cups, foam cups, eating utensils, deli food containers, toy model kits, some packing popcorn
7 Other Polycarbonate is a common type, Biodegradable, Some packing popcorn
Contributors
• Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Polymer_Fundamentals.txt |
A comparison of the properties of polyethylene (both LDPE & HDPE) with the natural polymers rubber and cellulose is instructive. As noted above, synthetic HDPE macromolecules have masses ranging from 105 to 106 amu (LDPE molecules are more than a hundred times smaller). Rubber and cellulose molecules have similar mass ranges, but fewer monomer units because of the monomer's larger size. The physical properties of these three polymeric substances differ from each other, and of course from their monomers.
• HDPE is a rigid translucent solid which softens on heating above 100º C, and can be fashioned into various forms including films. It is not as easily stretched and deformed as is LDPE. HDPE is insoluble in water and most organic solvents, although some swelling may occur on immersion in the latter. HDPE is an excellent electrical insulator.
• LDPE is a soft translucent solid which deforms badly above 75º C. Films made from LDPE stretch easily and are commonly used for wrapping. LDPE is insoluble in water, but softens and swells on exposure to hydrocarbon solvents. Both LDPE and HDPE become brittle at very low temperatures (below -80º C). Ethylene, the common monomer for these polymers, is a low boiling (-104º C) gas.
• Natural (latex) rubber is an opaque, soft, easily deformable solid that becomes sticky when heated (above. 60º C), and brittle when cooled below -50º C. It swells to more than double its size in nonpolar organic solvents like toluene, eventually dissolving, but is impermeable to water. The C5H8 monomer isoprene is a volatile liquid (b.p. 34º C).
• Pure cellulose, in the form of cotton, is a soft flexible fiber, essentially unchanged by variations in temperature ranging from -70 to 80º C. Cotton absorbs water readily, but is unaffected by immersion in toluene or most other organic solvents. Cellulose fibers may be bent and twisted, but do not stretch much before breaking. The monomer of cellulose is the C6H12O6 aldohexose D-glucose. Glucose is a water soluble solid melting below 150º C.
To account for the differences noted here we need to consider the nature of the aggregate macromolecular structure, or morphology, of each substance. Because polymer molecules are so large, they generally pack together in a non-uniform fashion, with ordered or crystalline-like regions mixed together with disordered or amorphous domains. In some cases the entire solid may be amorphous, composed entirely of coiled and tangled macromolecular chains. Crystallinity occurs when linear polymer chains are structurally oriented in a uniform three-dimensional matrix. In the diagram on the right, crystalline domains are colored blue.
Increased crystallinity is associated with an increase in rigidity, tensile strength and opacity (due to light scattering). Amorphous polymers are usually less rigid, weaker and more easily deformed. They are often transparent.
Three factors that influence the degree of crystallinity are:
i) Chain length
ii) Chain branching
iii) Interchain bonding
The importance of the first two factors is nicely illustrated by the differences between LDPE and HDPE. As noted earlier, HDPE is composed of very long unbranched hydrocarbon chains. These pack together easily in crystalline domains that alternate with amorphous segments, and the resulting material, while relatively strong and stiff, retains a degree of flexibility. In contrast, LDPE is composed of smaller and more highly branched chains which do not easily adopt crystalline structures. This material is therefore softer, weaker, less dense and more easily deformed than HDPE. As a rule, mechanical properties such as ductility, tensile strength, and hardness rise and eventually level off with increasing chain length.
The nature of cellulose supports the above analysis and demonstrates the importance of the third factor (iii). To begin with, cellulose chains easily adopt a stable rod-like conformation. These molecules align themselves side by side into fibers that are stabilized by inter-chain hydrogen bonding between the three hydroxyl groups on each monomer unit. Consequently, crystallinity is high and the cellulose molecules do not move or slip relative to each other. The high concentration of hydroxyl groups also accounts for the facile absorption of water that is characteristic of cotton.
Natural rubber is a completely amorphous polymer. Unfortunately, the potentially useful properties of raw latex rubber are limited by temperature dependence; however, these properties can be modified by chemical change. The cis-double bonds in the hydrocarbon chain provide planar segments that stiffen, but do not straighten the chain. If these rigid segments are completely removed by hydrogenation (H2 & Pt catalyst), the chains lose all constrainment, and the product is a low melting paraffin-like semisolid of little value. If instead, the chains of rubber molecules are slightly cross-linked by sulfur atoms, a process called vulcanization which was discovered by Charles Goodyear in 1839, the desirable elastomeric properties of rubber are substantially improved. At 2 to 3% crosslinking a useful soft rubber, that no longer suffers stickiness and brittleness problems on heating and cooling, is obtained. At 25 to 35% crosslinking a rigid hard rubber product is formed. The following illustration shows a cross-linked section of amorphous rubber. By clicking on the diagram it will change to a display of the corresponding stretched section. The more highly-ordered chains in the stretched conformation are entropically unstable and return to their original coiled state when allowed to relax
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Properties_of_Polymers.txt |
Most plastics crumble into ever-tinier fragments as they are exposed to sunlight and the elements. Except for the small amount that's been incinerated–and it's a very small amount–every bit of plastic ever made still exists, unless the material's molecular structure is designed to favor biodegradation. Unfortunately, cleaning up the garbage patch is not a realistic option, and unless we change our disposal and recycling habits, it will undoubtedly get bigger. One sensible solution would require manufacturers to use natural biodegradable packaging materials whenever possible, and consumers to conscientiously dispose of their plastic waste. Thus, instead of consigning all plastic trash to a land fill, some of it may provide energy by direct combustion, and some converted for reuse as a substitute for virgin plastics. The latter is particularly attractive since a majority of plastics are made from petroleum, a diminishing resource with a volatile price.
The energy potential of plastic waste is relatively significant, ranging from 10.2 to 30.7MJ kg Ã, suggesting application as an energy source and temperature stabilizer in municipal incinerators, thermal power plants and cement kilns. The use of plastic waste as a fuel source would be an effective means of reducing landfill requirements while recovering energy. This, however, depends on using appropriate materials. Inadequate control of combustion, especially for plastics containing chlorine, fluorine and bromine, constitutes a risk of emitting toxic pollutants.
Whether used as fuels or a source of recycled plastic, plastic waste must be separated into different categories. To this end, an identification coding system was developed by the Society of the Plastics Industry (SPI) in 1988, and is used internationally. This code, shown on the right, is a set of symbols placed on plastics to identify the polymer type, for the purpose of allowing efficient separation of different polymer types for recycling. The abbreviations of the code are explained in the following table.
PETE
HDPE
V
LDPE
polyethylene
terephthalate
high density
polyethylene
polyvinyl chloride
low density
polyethylene
PP
PS
OTHER
polypropylene
polystyrene
polyesters, acrylics
polyamides, teflon etc.
Despite use of the recycling symbol in the coding of plastics, there is consumer confusion about which plastics are readily recyclable. In most communities throughout the United States, PETE and HDPE are the only plastics collected in municipal recycling programs. However, some regions are expanding the range of plastics collected as markets become available. (Los Angeles, for example, recycles all clean plastics numbered 1 through 7) In theory, most plastics are recyclable and some types can be used in combination with others. In many instances, however, there is an incompatibility between different types that necessitates their effective separation. Since the plastics utilized in a given manufacturing sector (e.g. electronics, automotive, etc.) is generally limited to a few types, effective recycling is often best achieved with targeted waste streams.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Regio and Stereoisomerization in Polymers
Symmetrical monomers such as ethylene and tetrafluoroethylene can join together in only one way. Monosubstituted monomers, on the other hand, may join together in two organized ways, described in the following diagram, or in a third random manner. Most monomers of this kind, including propylene, vinyl chloride, styrene, acrylonitrile and acrylic esters, prefer to join in a head-to-tail fashion, with some randomness occurring from time to time. The reasons for this regioselectivity will be discussed in the synthetic methods section.
If the polymer chain is drawn in a zig-zag fashion, as shown above, each of the substituent groups (Z) will necessarily be located above or below the plane defined by the carbon chain. Consequently we can identify three configurational isomers of such polymers. If all the substituents lie on one side of the chain the configuration is called isotactic. If the substituents alternate from one side to another in a regular manner the configuration is termed syndiotactic. Finally, a random arrangement of substituent groups is referred to as atactic. Examples of these configurations are shown here.
Many common and useful polymers, such as polystyrene, polyacrylonitrile and poly(vinyl chloride) are atactic as normally prepared. Customized catalysts that effect stereoregular polymerization of polypropylene and some other monomers have been developed, and the improved properties associated with the increased crystallinity of these products has made this an important field of investigation. The following values of Tg have been reported.
Polymer
Tg atactic
Tg isotactic
Tg syndiotactic
PP
–20 ºC
0 ºC
–8 ºC
PMMA
100 ºC
130 ºC
120 ºC
The properties of a given polymer will vary considerably with its tacticity. Thus, atactic polypropylene is useless as a solid construction material, and is employed mainly as a component of adhesives or as a soft matrix for composite materials. In contrast, isotactic polypropylene is a high-melting solid (ca. 170 ºC) which can be molded or machined into structural components.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Recycling_and_Disposal_of_Polymers.txt |
Rubber is an example of an elastomer type polymer, where the polymer has the ability to return to its original shape after being stretched or deformed. The rubber polymer is coiled when in the resting state. The elastic properties arise from the its ability to stretch the chains apart, but when the tension is released the chains snap back to the original position. The majority of rubber polymer molecules contain at least some units derived from conjugated diene monomers (see Polymerization of Conjugated Dienes). Such conjugated diene monomers have a constructive backbone of at least four carbon atoms with a double-single-double bond reactive core (C=C-C=C ). Most if not practically all such dienes undergo 1,4-addition to the polymer chain, where 1 and 4 refer to the 1st and 4th carbons of the backbone unit, which become single-bonded to the rest of the polymer chain. The diene's double bonds turn into single bonds, and the single bond between them turns into a Z or E configured double bond, depending on the polymerization conditions. The unit's backbone thus becomes like this (-C-C=C-C-). Rubber gets its elasticity when the formed double bond gets the Z configuration. For 1,3-butadiene, Z is equivalent to a cis and E is equivalent to a trans configuration.
Natural Rubber
Natural rubber is an addition polymer that is obtained as a milky white fluid known as latex from a tropical rubber tree. Natural rubber is from the monomer isoprene (2-methyl-1,3-butadiene), which is a conjugated diene hydrocarbon as mentioned above. In natural rubber, most of the double fonds formed in the polymer chain have the Z configuration, resulting in natural rubber's elastomer qualities.
Charles Goodyear accidentally discovered that by mixing sulfur and rubber, the properties of the rubber improved in being tougher, resistant to heat and cold, and increased in elasticity. This process was later called vulcanization after the Roman god of fire. Vulcanization causes shorter chains to cross link through the sulfur to longer chains. The development of vulcanized rubber for automobile tires greatly aided this industry.
Synthetic Rubber
Important conjugated dienes used in synthetic rubbers include isoprene (2-methyl-1,3-butadiene), 1,3-butadiene, and chloroprene (2-chloro-1,3-butadiene). Polymerized 1,3-butadiene is mostly referred to simply as polybutadiene. Polymerized chloroprene was developed by DuPont and given the trade name Neoprene. In a number of cases, monomers which are not dienes are also used for certain types of synthetic rubber, often copolymerized with dienes. Some of the most commercially important addition polymers are the copolymers. These are polymers made by polymerizing a mixture of two or more monomers. An example is styrene-butadiene rubber (SBR) - which is a copolymer of 1,3-butadiene and styrene which is mixed in a 3 to 1 ratio, respectively.
SBR rubber was developed during World War II when important supplies of natural rubber were cut off. SBR is more resistant to abrasion and oxidation than natural rubber and can also be vulcanized. More than 40% of the synthetic rubber production is SBR and is used in tire production. A tiny amount is used for bubble-gum in the unvulcanized form. Nitrile rubber is copolymerized from butadiene and acrylonitrile (H2C=CH-CN). Butyl rubber is copolymerized from isobutylene [which is methylpropene H2C=C(CH3)2 ] and a small percentage of isoprene. Silicone rubber and other compounds, chemically called polysiloxanes, are not from conjugated dienes but have repeating units like -O-SiR2- where R is some organic radical group like methyl.
Needle Through a Balloon
The polymer rubber chains exist in random loose clumps in the unstretched state. At the nipple end of the balloon, there is lots of rubber and therefore many, many polymer chains - still loosely coiled. These chains can be pierced without popping the balloon because the the chains can still be stretched. This is because they allow the skewer in between the chains without breaking the chains or the bonds that connect them. But on the sides of the balloon, these chains are stretched almost to their limit and very far apart. The piercing is too much for the stretched chains and they break apart, and the balloon pops.
Contributors
• Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook
• Henry A. Padleckas | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Rubber_Polymers.txt |
Silicone polymers, more properly called polysiloxanes, do not have carbon as part of the backbone structure. Although silicon is in the same group as carbon in the periodic table, it has quite different chemistry. Many silanes are known which are analogous to the hydrocarbons with Si-Si bonds. These compounds are not very stable and hence not very useful. Silicones on the other hand have an alternating -Si-O- type structure. This basic structural unit is found in many rocks and minerals in nature including common sand.
Various organic groups such as methyl or the benzene ring may be bonded to the silicon as shown below. Silicones are water repellent, heat stable, and very resistant to chemical attack. They find many uses in oils, greases, and rubberlike materials. Silicone oils are very desirable since they do not decompose at high temperature and do not become viscous. Other silicones are used in hydraulic fluids, electrical insulators and moisture proofing agent in fabrics.
The preparation of dimethyl silicon dichloride, or dimethyldichlorosilane, is the first step in the production of modern dimethylsilicone (polydimethylsiloxane) products. As explained in US Patent 2,380,995, Eugene G. Rochow describes the reaction of elemental silicon with gaseous methyl chloride within a tube furnace at 300˚C.
2 CH3Cl + Si → (CH3)2SiCl2
Silicones have a number of medical applications because they are chemically inert. Medical devices composed of silicone may be approved by the FDA for permanent or temporary implantation. Catheters, tubing, gastric bags, drains, and endoscopic windows are examples of consumable medical devices that are often molded from silicone. Breast implants, stents, and prostheses are examples of permanent implants often molded from silicone. Silicone rubber approved for use in FDA devices approved for permanent implantation differs from that used in medical consumables is several important ways. Implant grade silicone is of long linear chain length often exceeding one million molecular weight. Implant grade silicone rubber has had low molecular weight silicone oils, added for improved dispersion of silica fume fillers, removed thru high temperature vacuum mixing. Implant grade silicone rubber is normally cross-linked using a platinum catalyst. Silicone rubber is often used in medical devices because it can be heat sterilized. Most silicone consumables are removed from hot press molds while hot, saving expensive chilling cycles and simplifying mold design.
Silicone Rubber tubing used in medical practice can cause some problems. Gas permeability of dimethylsilicone is high enough to cause bubbles to form in silicone tubing often used in pumps that deliver medication to patients. Some medications are sensitive to oxygen permeate. A simple way to demonstrate this effect is to fill a section of silicone tubing with water then tie off both ends excluding any air pockets. The water filled tubing may be draped over any suitable ledge for several hours. Notice air cavities form in tube lumen.
A good deal of controversy has involved the the use of silicone in polyurethane bags as breast implants. Again they were used because they were thought to be very inert and resistant to dissolving or other reactions. Reports have cited increased cancer risk and severe immune responses from possible leakage of the silicone from the implants. Some scientists dispute these findings.
Superball Demonstration
The liquids solution of sodium silicate is already in the form polymer. The silicate is alternating atoms of silicon and oxygen in long chains. When the ethanol is added, it bridges and connects the chains by cross-linking them. The analogy of a chain-link fence is a good picture of the idea of chains that are cross linked. That is what the ethanol and the silicate are doing to form this super ball. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Silicone_Polymers.txt |
All the monomers from which addition polymers are made are alkenes or functionally substituted alkenes. The most common and thermodynamically favored chemical transformations of alkenes are addition reactions. Many of these addition reactions are known to proceed in a stepwise fashion by way of reactive intermediates, and this is the mechanism followed by most polymerizations. A general diagram illustrating this assembly of linear macromolecules, which supports the name chain growth polymers, is presented here. Since a pi-bond in the monomer is converted to a sigma-bond in the polymer, the polymerization reaction is usually exothermic by 8 to 20 kcal/mol. Indeed, cases of explosively uncontrolled polymerizations have been reported.
It is useful to distinguish four polymerization procedures fitting this general description.
• Radical Polymerization The initiator is a radical, and the propagating site of reactivity (*) is a carbon radical.
• Cationic Polymerization The initiator is an acid, and the propagating site of reactivity (*) is a carbocation.
• Anionic Polymerization The initiator is a nucleophile, and the propagating site of reactivity (*) is a carbanion.
• Coordination Catalytic Polymerization The initiator is a transition metal complex, and the propagating site of reactivity (*) is a terminal catalytic complex.
Radical Chain-Growth Polymerization
Virtually all of the monomers described above are subject to radical polymerization. Since this can be initiated by traces of oxygen or other minor impurities, pure samples of these compounds are often "stabilized" by small amounts of radical inhibitors to avoid unwanted reaction. When radical polymerization is desired, it must be started by using a radical initiator, such as a peroxide or certain azo compounds. The formulas of some common initiators, and equations showing the formation of radical species from these initiators are presented below.
By using small amounts of initiators, a wide variety of monomers can be polymerized. One example of this radical polymerization is the conversion of styrene to polystyrene, shown in the following diagram. The first two equations illustrate the initiation process, and the last two equations are examples of chain propagation. Each monomer unit adds to the growing chain in a manner that generates the most stable radical. Since carbon radicals are stabilized by substituents of many kinds, the preference for head-to-tail regioselectivity in most addition polymerizations is understandable. Because radicals are tolerant of many functional groups and solvents (including water), radical polymerizations are widely used in the chemical industry.
In principle, once started a radical polymerization might be expected to continue unchecked, producing a few extremely long chain polymers. In practice, larger numbers of moderately sized chains are formed, indicating that chain-terminating reactions must be taking place. The most common termination processes are Radical Combination and Disproportionation. These reactions are illustrated by the following equations. The growing polymer chains are colored blue and red, and the hydrogen atom transferred in disproportionation is colored green. Note that in both types of termination two reactive radical sites are removed by simultaneous conversion to stable product(s). Since the concentration of radical species in a polymerization reaction is small relative to other reactants (e.g. monomers, solvents and terminated chains), the rate at which these radical-radical termination reactions occurs is very small, and most growing chains achieve moderate length before termination.
The relative importance of these terminations varies with the nature of the monomer undergoing polymerization. For acrylonitrile and styrene combination is the major process. However, methyl methacrylate and vinyl acetate are terminated chiefly by disproportionation.
Another reaction that diverts radical chain-growth polymerizations from producing linear macromolecules is called chain transfer. As the name implies, this reaction moves a carbon radical from one location to another by an intermolecular or intramolecular hydrogen atom transfer (colored green). These possibilities are demonstrated by the following equations
Chain transfer reactions are especially prevalent in the high pressure radical polymerization of ethylene, which is the method used to make LDPE (low density polyethylene). The 1º-radical at the end of a growing chain is converted to a more stable 2º-radical by hydrogen atom transfer. Further polymerization at the new radical site generates a side chain radical, and this may in turn lead to creation of other side chains by chain transfer reactions. As a result, the morphology of LDPE is an amorphous network of highly branched macromolecules.
Cationic Chain-Growth Polymerization
Polymerization of isobutylene (2-methylpropene) by traces of strong acids is an example of cationic polymerization. The polyisobutylene product is a soft rubbery solid, Tg = _70º C, which is used for inner tubes. This process is similar to radical polymerization, as demonstrated by the following equations. Chain growth ceases when the terminal carbocation combines with a nucleophile or loses a proton, giving a terminal alkene (as shown here).
Monomers bearing cation stabilizing groups, such as alkyl, phenyl or vinyl can be polymerized by cationic processes. These are normally initiated at low temperature in methylene chloride solution. Strong acids, such as HClO4 , or Lewis acids containing traces of water (as shown above) serve as initiating reagents. At low temperatures, chain transfer reactions are rare in such polymerizations, so the resulting polymers are cleanly linear (unbranched).
Anionic Chain-Growth Polymerization
Treatment of a cold THF solution of styrene with 0.001 equivalents of n-butyllithium causes an immediate polymerization. This is an example of anionic polymerization, the course of which is described by the following equations. Chain growth may be terminated by water or carbon dioxide, and chain transfer seldom occurs. Only monomers having anion stabilizing substituents, such as phenyl, cyano or carbonyl are good substrates for this polymerization technique. Many of the resulting polymers are largely isotactic in configuration, and have high degrees of crystallinity.
Species that have been used to initiate anionic polymerization include alkali metals, alkali amides, alkyl lithiums and various electron sources. A practical application of anionic polymerization occurs in the use of superglue. This material is methyl 2-cyanoacrylate, CH2=C(CN)CO2CH3. When exposed to water, amines or other nucleophiles, a rapid polymerization of this monomer takes place.
Ziegler-Natta Catalytic Polymerization
An efficient and stereospecific catalytic polymerization procedure was developed by Karl Ziegler (Germany) and Giulio Natta (Italy) in the 1950's. Their findings permitted, for the first time, the synthesis of unbranched, high molecular weight polyethylene (HDPE), laboratory synthesis of natural rubber from isoprene, and configurational control of polymers from terminal alkenes like propene (e.g. pure isotactic and syndiotactic polymers). In the case of ethylene, rapid polymerization occurred at atmospheric pressure and moderate to low temperature, giving a stronger (more crystalline) product (HDPE) than that from radical polymerization (LDPE). For this important discovery these chemists received the 1963 Nobel Prize in chemistry.
Ziegler-Natta catalysts are prepared by reacting certain transition metal halides with organometallic reagents such as alkyl aluminum, lithium and zinc reagents. The catalyst formed by reaction of triethylaluminum with titanium tetrachloride has been widely studied, but other metals (e.g. V & Zr) have also proven effective. The following diagram presents one mechanism for this useful reaction. Others have been suggested, with changes to accommodate the heterogeneity or homogeneity of the catalyst. Polymerization of propylene through action of the titanium catalyst gives an isotactic product; whereas, a vanadium based catalyst gives a syndiotactic product.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Synthesis_of_Addition_Polymers.txt |
This module guides you through the mechanism for the polymerisation of ethene by a free radical addition reaction. We are going to talk through this mechanism in a very detailed way so that you get a feel for what is going on.
A Free Radical Addition Reaction
You will remember that during the polymeriation of ethene, thousands of ethene molecules join together to make poly(ethene) - commonly called polythene. The reaction is done at high pressures in the presence of a trace of oxygen as an initiator.
Step 1: Chain Initiation
The oxygen reacts with some of the ethene to give an organic peroxide. Organic peroxides are very reactive molecules containing oxygen-oxygen single bonds which are quite weak and which break easily to give free radicals. You can short-cut the process by adding other organic peroxides directly to the ethene instead of using oxygen if you want to. The type of the free radicals that start the reaction off vary depending on their source. For simplicity we give them a general formula: $Ra ^{\bullet}$
Step 2: Chain Propagation
In an ethene molecule, CH2=CH2, the two pairs of electrons which make up the double bond aren't the same. One pair is held securely on the line between the two carbon nuclei in a bond called a sigma bond. The other pair is more loosely held in an orbital above and below the plane of the molecule known as a $\pi$ bond.
Note
It would be helpful - but not essential - if you read about the structure of ethene before you went on. If the diagram above is unfamiliar to you, then you certainly ought to read this background material.
Imagine what happens if a free radical approaches the $\pi$ bond in ethene.
Note
Don't worry that we've gone back to a simpler diagram. As long as you realise that the pair of electrons shown between the two carbon atoms is in a $\pi$ bond - and therefore vulnerable - that's all that really matters for this mechanism.
The sigma bond between the carbon atoms isn't affected by any of this. The free radical, Ra, uses one of the electrons in the $\pi$ bond to help to form a new bond between itself and the left hand carbon atom. The other electron returns to the right hand carbon. You can show this using "curly arrow" notation if you want to:
Note
If you aren't sure about about curly arrow notation you can follow this link.
This is energetically worth doing because the new bond between the radical and the carbon is stronger than the $\pi$ bond which is broken. You would get more energy out when the new bond is made than was used to break the old one. The more energy that is given out, the more stable the system becomes. What we've now got is a bigger free radical - lengthened by CH2CH2. That can react with another ethene molecule in the same way:
So now the radical is even bigger. That can react with another ethene - and so on and so on. The polymer chain gets longer and longer.
Step 3: Chain Termination
The chain does not, however, grow indefinitely. Sooner or later two free radicals will collide together.
That immediately stops the growth of two chains and produces one of the final molecules in the poly(ethene). It is important to realise that the poly(ethene) is going to be a mixture of molecules of different sizes, made in this sort of random way.
Summary
The over-all process is known as free radical addition.
• Chain initiation: The chain is initiated by free radicals, Ra, produced by reaction between some of the ethene and the oxygen initiator.
• Chain propagation: Each time a free radical hits an ethene molecule a new longer free radical is formed (e.g.,
• Chain termination: Eventually two free radicals hit each other producing a final molecule. The process stops here because no new free radicals are formed.
Because chain termination is a random process, poly(ethene) will be made up of chains of different lengths.
Contributors
Jim Clark (Chemguide.co.uk)
Thermosetting vs. Thermoplastic Polymers
Most of the polymers described above are classified as thermoplastic. This reflects the fact that above Tg they may be shaped or pressed into molds, spun or cast from melts or dissolved in suitable solvents for later fashioning. Because of their high melting point and poor solubility in most solvents, Kevlar and Nomex proved to be a challenge, but this was eventually solved.
Another group of polymers, characterized by a high degree of cross-linking, resist deformation and solution once their final morphology is achieved. Such polymers are usually prepared in molds that yield the desired object. Because these polymers, once formed, cannot be reshaped by heating, they are called thermosets .Partial formulas for four of these will be shown below by clicking the appropriate button. The initial display is of Bakelite, one of the first completely synthetic plastics to see commercial use (circa 1910).
A natural resinous polymer called lignin has a cross-linked structure similar to bakelite. Lignin is the amorphous matrix in which the cellulose fibers of wood are oriented. Wood is a natural composite material, nature's equivalent of fiberglass and carbon fiber composites. A partial structure for lignin is shown here:
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Writing Formulas for Polymers
The repeating structural unit of most simple polymers not only reflects the monomer(s) from which the polymers are constructed, but also provides a concise means for drawing structures to represent these macromolecules. For polyethylene, arguably the simplest polymer, this is demonstrated by the following equation. Here ethylene (ethene) is the monomer, and the corresponding linear polymer is called high-density polyethylene (HDPE). HDPE is composed of macromolecules in which n ranges from 10,000 to 100,000 (molecular weight $2 \times 10^5$ to $3 \times10^6$ ).
If Y and Z represent moles of monomer and polymer respectively, Z is approximately $10^{-5}$ Y. This polymer is called polyethylene rather than polymethylene, $\ce{(-CH_2-)_{n}}$, because ethylene is a stable compound (methylene is not), and it also serves as the synthetic precursor of the polymer. The two open bonds remaining at the ends of the long chain of carbons (colored magenta) are normally not specified, because the atoms or groups found there depend on the chemical process used for polymerization. The synthetic methods used to prepare this and other polymers will be described later in this chapter.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/The_Polymerization_of_Ethene.txt |
A secondary theme in carbonyl chemistry centers on the role played by the oxygen lone pairs. A compound with lone pairs can act as a Lewis base. Can carbonyl compounds also act as Lewis bases? The answer is yes, although it is most important to think about carbonyls primarily as Lewis acids.
One of the reasons the basicity of the lone pair matters is because of carbonyl activation. If a carbonyl donates a lone pair to a Lewis acid, forming a bond, the carbonyl gets a formal positive charge. If the carbonyl has a formal positive charge, it attracts electrons more strongly. In that case, nucleophiles react more easily with the carbonyl. The carbonyl is said to be activated.
Figure CO10.1. Activation of a carbonyl via donation to a proton.
A carbonyl can be activated by the addition of a proton donor, such as HCl or other common acids.
• An activated carbonyl has a positive charge.
• Carbonyls become activated by donating a lone pair to a Lewis acid (also called an electrophile).
• Once activated, carbonyls become more reactive.
• Activated carbonyls attract nucleophiles more strongly.
Most common mineral acids are used as aqueous solutions (the familiar HCl, H2SO4, HNO3, H3PO4 and so on). Sometimes, in a laboratory reaction, it isn't helpful to have all that water around (the reasons will become clear later). Other, organic acids are sometimes used instead, such as camphorsulfonic acid or toluenesulfonic acid; these are both solids that are easy to weight out and add to a reaction, and they don't add a bunch of water to the reaction.
Figure CO10.2. Some protic acids useful in activating carbonyls.
Carbonyls are also activated by more general Lewis acids. Often, metal chloride salts are used. These may include main group metals, such as aluminum, bismuth or indium, or transition metals such as scandium, titanium or iron.
Figure CO10.3. Activation of a carbonyl by a metal ion.
Once the carbonyl is activated, nucleophiles are more strongly attracted to the carbon. The carbon was already partially positive, but with a full positive charge on the molecule, electrons are attracted much more strongly.
It is tempting to donate electrons from a nucleophile to the positive oxygen. However, the oxygen already has three bonds and an octet. Remember, donating a lone pair from a nucleophile means the lone pair is becoming a bond between the nucleophile and the electrophile. Giving a pair of electrons directly to the oxygen would give it four bonds and more than an octet-- it would have 10 electrons. Instead, donation to the neighbouring carbon allows the C=O pi bond to move to the oxygen and become a lone pair. The positive charge on the oxygen disappears.
• The nucleophile donates to the activated carbonyl carbon
• That event lets the pi bond become a lone pair on oxygen
Figure CO10.4. Donation of nucleophile to an activated carbonyl.
Problem CO10.1.
Show, with arrows, the activation of the following carbonyls.
Problem CO10.2.
Show, with arrows, the activation of the following carbonyls, followed by donation from the nucleophile.
CO11. Addition of Neutral, Protic Nucleophiles
CO11. Addition of Neutral Nucleophiles
Nucleophiles do not have to be ionic, or even "semi-anionic." The basic requirement for a nucleophile is a lone pair. If a nucleophile has a lone pair, it can donate the lone pair to an electrophile such as a carbonyl. By donating a lone pair to a carbonyl, it can form a bond.
However, donation of a lone pair to a carbonyl is reversible. If there is something about the nucleophile/electrophile adduct that isn't very stable, the reaction may revert to reactants again. In the case of neutral nucleophiles, charge separation may destabilize the first-formed products of the reaction. Let's think about addition of water to propanone. Two neutral molecules, water and propanone, come together. The water donates a lone pair to the carbonyl carbon in propanone. That leaves the oxygen atom from the water with a positive charge, and the oxygen atom from the propanone with a negative charge.
One easy way to get rid of the charge separation is for the water to leave again. That step would just be the reverse of the first one.
On the other hand, another way to solve the charge problem is to move a proton (H+) from the positively charged oxygen to the negatively charged one. That turns out to be pretty easy to do. If that happens, a "hydrate" or a "geminal diol" forms. A geminal diol, or twin diol, has two hydroxy groups on one carbon.
ProblemCO11.1.
In the following cases, a nucleophile donates to the carbonyl, followed by a proton transfer. Show mechanisms, with curved arrows, for each of the following reactions.
The immediate products that form from the addition of neutral nucleophiles to carbonyls turn out to be a little unstable. That's partly because they can easily revert back to reactants. It's also because there are other processes that carry the reaction away from the first-formed products and turn them into other things. For example, in most cases the reaction does not involve the simple addition of one nucleophile, but involves a second molecule of the nucleophile as well.
We will keep looking at these other processes and see where these reactions lead. In the meantime, it can be useful to know the patterns that different types of nucleophiles will usually follow. For example, addition of alcohols to aldehydes or ketones leads to the formation of ketals or acetals. Ketals or acetals have the specific chain of atoms C-O-C-O-C, in which each carbon is tetrahedral or sp3.
On the other hand, the addition of amines leads to a very different kind of structure. Instead of adding two amine molecules into the final structure, only one amine is incorporated, and a double bond appears again. The C=O of the aldehyde or ketone is replaced with the C=N of an imine. Imines form important linkages in biological chemistry, especially in connecting small molecules to lysine side chains in proteins.
If the amine is "secondary", meaning the nitrogen is connected to two carbons instead of just one, the double bond ends up one position over. It is between the former carbonyl carbon and the one next to it, which is called the alpha position.
Note that all three of these reactions involve ultimate loss of water from the original carbonyl compound. That's a very important feature of reactions with neutral nucleophiles.
• Addition of a neutral nucleophile to a carbonyl almost always leads to loos of the original carbonyl oxygen in a water molecule. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO10._Activation_of_Carbonyls.txt |
CO12. Proton Transfer in Carbonyl Addition
Proton transfer is rapid, especially if it is transferred from a very acidic position. For example, a proton can easily be transferred from a positively charged oxygen atom to a neutral oxygen (resulting in a new, neutral oxygen and a new, positive oxygen). These species would be in equilibrium with each other.
• Proton transfer is rapid.
• Protons can be transferred from more acidic to less acidic position.
• Protons can be transferred from one acidic position to another of similar acidity, although the equilibrium may not be favoured.
It would not be as easy to transfer a proton from a neutral oxygen to another neutral oxygen. Sometimes, a neutral oxygen can transfer a proton to a negatively charged one, but the equilibrium will depend on the relative pKa values of the two species. In the case below, the tert-butoxide is a less stable anion than the hydroxide because the tert-butoxide is larger and requires more organization of solvent molecules around it.
It is tempting to think that a proton could be transferred directly from a cationic position to an anionic position in the same molecule. That might not occur, however. In terms of conformational analysis, the two positions might not be able to twist around and reach each other. The usual rule applies: two atoms may need to be greater than five atoms away from each other along a chain before they can reach around and make contact.
If the solvent has a lone pair, it may pick up the proton from the acidic position and drop it off on the basic position. These events are made easy by the fact that the reacting molecules are usually surrounded by many solvent molecules.
• Solvent can often act as a proton shuttle.
CO13. (pi) Donation Steps
Some nucleophiles are added to carbonyls, lose a proton to drop a positive charge, and have a lone pair again. If the (former) carbonyl oxygen also has a lone pair, a potential lone-pair/lonepair repulsion problem exists. Partly for this reason, two heteroatoms bonded to one carbon often present an inherently unstable situation. These kinds of species often decompose readily via pi donation.
In pi donation, a lone pair on one heteroatom is donated to the carbon shared by both heteroatoms. As a result, the other heteroatom is pushed off the carbon. This event is helped if one of the heteroatoms is already protonated, so that it comes off as a neutral species.
Problem CO13.1.
Fill in any missing lone pairs, provide curved arrows to show pi donation, and show the resonance structures that result.
Problem CO13.2.
Show how each of the following species might break down into two molecules via pi donation.
As a result of pi donation, neutral, protic nucleophiles often replace the carbonyl oxygen entirely. In effect, the nucleophile adds twice. The lone pair that is revealed after a deprotonation step adds again to the same carbon, pushing that carbonyl oxygen out of the molecule entirely. In the case of alcohol nucleophiles, a ketal or acetal results. This kind of molecule looks like two ethers that meet at one carbon. A ketal is a "masked" carbonyl; it still contains a carbon with two bonds to oxygen. However, a ketone is no longer an electrophile like a carbonyl compound.
Imines also result from pi donation. If an primary amine donates to a carbonyl, it can lose its first proton to reveal a lone pair. Once that lone pair donates, pushing off the carbonyl oxygen, a second proton can be dropped to allow the nitrogen to lose its positive charge. Imines are similar to carbonyls in that they contain a carbon-heteroatom double bond and so they are still good electrophiles. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO12._Proton_Transfer_Steps.txt |
When a sugar cyclizes via donation of a hydroxy lone pair to the carbonyl, it forms a "hemiacetal". We have already seen that hemiacetals are unstable with respect to further substitution. Pi donation from an oxygen in the hemiacetal can displace the other oxygen. A second nucleophile can then donate to the pseudo-carbonyl that results.
The hemiacetal position in the sugar is called the anomeric center. The anomeric center is special for two reasons. First, as you have already seen, the anomeric center is a chiral center. This new center can form with either of two configurations. The sugar, which is already chiral, can become either of two diastereomers when it cyclizes. Second, the anomeric center is a site of enhanced reactivity in the sugar, in terms of substitution of the carbonyl.
Anomeric reactivity involves pi donation from one oxygen to push off the other oxygen. This mode of reaction should be familiar. The C=O+ unit that forms resembles a carbonyl. Furthermore, the positive charge on the oxygen brings to mind an activated carbonyl. This position is especially attractive for nucleophiles.
It isn't an accident that in many sugar-containing biomolecules, substituents are found at the anomeric center. For example, nucleosides sub-units found in DNA and RNA are all substituted at this position. A number of other biological agents contain this motif as well.
Problem CO15.1.
Substitution at the anomeric position can be accelerated if a proton source is available. Show why.
Problem CO15.2.
Show the stereochemical results of substitution at the anomeric center of glucose with methanol.
Problem CO15.3.
Nucleotides, which form DNA and RNA chains, are just like nucleosides, but they all have a phosphate at a specific position. Explain what is special about this position that could make it form a phosphate more easily than the other hydroxyl sites.
Problem CO15.4.
Show the products of the following substitution reactions.
Although sugars contain a number of chiral centers, characterizing them by polarimetry is complicated. Optical rotation measurements are done in solution, in a polarimetry cell. Polarimetry is always a little bit complicated, because the optical rotation varies with the concentration of the solution and the length of the polarimetry cell. When a pure enantiomer of a sugar such as alpha-D-glucose is dissolved, usually in water, its optical rotation also varies with time. In other words, the reading keeps changing, eventually settling out far from the initial value. That means care must be taken in measuring this information, and in interpreting the data.
Problem CO15.5.
Show why alpha-D-glucose would exhibit a changing optical rotation value after being dissolved.
Problem CO15.6.
Suppose a one gram sample of alpha-D-glucopyranose is dissolved in 1 mL of water and its optical rotation is measured in a a 1 dm cell. Initially, a value of 100 degrees is recorded. After several hours, the value has stopped changing, and is 48 degrees.
The experiment is repeated with beta-D-glucopyranose.
1. What can you predict about the initial value with beta-D-glucopyranose?
2. What can you predict after several hours?
Problem CO15.7.
In water, alpha-D-glucopyranose predominates over the beta form in solution. Explain why.
Problem CO15.8.
In less polar solvents (compared to water) such as dichloromethane, beta-D-glucopyranose predominates over the alpha form. This phenomenon is thought to result from the influence of lone pair-lone pair repulsion.
1. Show why this factor might favour one isomer over the other.
2. Show why this factor is less important in water.
CO16. Biological Reduction
CO16. Biological Reduction
Addition to a carbonyl by a semi-anionic hydride, such as NaBH4, results in conversion of the carbonyl compound to an alcohol. The hydride from the BH4- anion acts as a nucleophile, adding H- to the carbonyl carbon. A proton source can then protonate the oxygen of the resulting alkoxide ion, forming an alcohol.
Formally, that process is referred to as a reduction. Reduction generally means a reaction in which electrons are added to a compound; the compound that gains electrons is said to be reduced. Because hydride can be thought of as a proton plus two electrons, we can think of conversion of a ketone or an aldehyde to an alcohol as a two-electron reduction. An aldehyde plus two electrons and two protons becomes an alcohol.
Aldehydes, ketones and alcohols are very common features in biological molecules. Converting between these compounds is a frequent event in many biological pathways. However, semi-anionic compounds like sodium borohydride don't exist in the cell. Instead, a number of biological hydride donors play a similar role.
NADH is a common biological reducing agent. NADH is an acronym for nicotinamide adenine dinucleotide hydride. Insetad of an anionic donor that provides a hydride to a carbonyl, NADH is actually a neutral donor. It supplies a hydride to the carbonyl under very specific circumstances. In doing so, it forms a cation, NAD+. However, NAD+ is stabilized by the fact that its nicotinamide ring is aromatic; it was not aromatic in NADH. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO15._The_Anomeric_Center.txt |
CO17. Organic Oxidation
You may recall that conversion of an aldehyde or ketone to an alcohol is referred to as a reduction. The hydride from an NADH molecule or a BH4- anion acts as a nucleophile, adding H- to the carbonyl carbon. A proton source can then protonate the oxygen of the resulting alkoxide ion, forming an alcohol.
In this reduction, two electrons and two protons are donated to the carbonyl compound to produce an alcohol.
The opposite process, the loss of two protons and two electrons from an alcohol to form a ketone or aldehyde, is an oxidation.
In biological pathways, oxidation is often the microscopic reverse of reduction. That means that the products of a reduction, NAD+ and an alcohol, could react together under the right circumstances to form NADH and a carbonyl. The reduction of NAD+ by a hydride donor is possible because, although the NAD+ loses the aromaticity of its nicotinamide ring upon becoming NADH, it also loses its positive charge. Charge stabilization is frequently an energetic problem for molecules.
Notice that this same argument has been used to look at biological oxidation and reduction in both directions. That's because each side of the equation has some energetic advantage. The NAD+ is aromatic. The NADH is neutral. This is a well-balanced system energetically, and the balance of the reaction can be tipped in either direction. The direction of the reaction is influenced by the surroundings.
In biology, NADH and NAD+ are just cofactors in a reaction. A reduction would take place in an enzyme that specifically carries out reductions, and an oxidation would occur in an enzyme specific for the oxidation. The enzyme is just a large protein that holds the substrate (such as the alcohol) and cofactor (such as NAD+) together in close proximity. Acidic and basic sites are provided by nearby amino acid residues, and other amino acid side chains may push the reactants into optimum position for one reaction or another.
In the laboratory, enzymes and cofactors can sometimes be added to reaction flasks in order to oxidize or reduce substrates. Sometimes these reactions are not convenient, however. In addition, there are a number of other ways to carry out oxidations and reductions. For example, addition of a hydride could be accomplished via addition of sodium borohydride. That would result in reduction of a carbonyl to an alcohol.
By analogy to the NADH / NAD+ approach in nature, the easiest way to oxidize an alcohol to a carbonyl would be to remove a hydride and a proton. Nature uses enzymes to bring reactants together for transformations like this. In the laboratory, metals are often used to bring two reactants together. This is sometimes true in enzymes, too: a metal at the enzyme active site may tether two molecules together, or even activate one so that it is ready to react. Metals can "hold onto" reactants because molecules with lone pairs will often coordinate to metals; a lone pair is shared with the electrophilic metal.
In a process known as an Oppenauer oxidation, a Lewis acidic metal such as aluminum is used for this tethering role. Aluminum tris(isopropoxide), Al(OCH(CH3)2)3 carries out the oxidation of an alcohol when dissolved in acetone or propanone, (CH3)2C=O. In this reaction, the acetone is a sacrificial oxidant. When both the alcohol and a molecule of acetone are coordinated to the aluminum, a hydride is transferred from the alcohol carbon to the carbonyl carbon of the coordinated acetone. The alcohol is converted to an aldehyde or ketone and the acetone is converted to isopropanol.
This general type of reaction, hydride transfer reduction, has been adapted by Ryoji Noyori, of Nagoya University in Japan, to produce a single enantiomer of a chiral alcohol product. Noyori's work on this reaction, and others, led to him being awarded the Nobel Prize in Chemistry in 2001.
Problem CO17.1.
Show the two enantiomers that could be produced from reduction of acetophenone, CH3(CO)C6H5.
Problem CO17.2.
Provide a mechanism with arrows for the Oppenauer oxidation of benzyl alcohol, C6H5CH2OH.
Problem CO17.3.
Explain why acetone is used as the solvent in an Oppenauer oxidation.
Problem CO17.4.
Meerwein-Ponndorf-Verley reduction of a ketone is carried out with aluminum tris(isopropoxide) in isopropanol as solvent. Provide a mechanism for the reduction of acetophenone, CH3(CO)C6H5, via this reaction.
Problem CO17.5.
Explain why isopropanol is used as the solvent in a Meerwein-Pondorf-Verley reduction.
A second general method for alcohol oxidation employs a "redox-active" transition metal to accept a pair of electrons from from an alcohol during the oxidation. Because oxidation of an alcohol formally involves the loss of two electrons and two protons, a proton acceptor is also involved in this oxidation. There are many redox-active metals, but one of the most commonly used is Cr(VI). When Cr(VI) accepts a pair of electrons, it becomes Cr(IV).
In order to look at how chromium oxidation works, we'll use chromium oxide, CrO3, as an oxidant and water as a solvent. Note that water could also act as a proton acceptor or proton shuttle, moving protons from one place to another as needed. To carry out an oxidation, a number of events need to happen.
• The alcohol needs to bind to the chromium.
• A proton needs to be removed. This event is helped by the formal positive charge on the alcohol after it donates a lone pair to the chromium.
• A second proton must be removed and a pair of electrons given to the chromium for good.
In reality, CrO3 isn't used that often as an oxidant. It tends to catch fire when mixed with organic compounds. Instead, a variety of other chromium compounds are used.
Problem CO17.6.
In determining an oxidation state, we imagine giving both electrons in a bond to the more electronegative atom and looking at the resulting charges on the ions that result. Assuming all of the oxygens in chromium oxide can be thought of as dianions, confirm that the chromium can be thought of as a Cr6+ cation (in other words, in oxidation state Cr(VI)).
Problem CO17.7.
By the reasoning used in the previous question, determine the oxidation state of the transition metal in the following compounds. Note that in some cases, there is an anion and cation in the compound.
a) KMnO4 b) NaIO4 c) Ag2O d) OsO4 d) (CH3CH2CH2)4N RuO4 | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO17._Oxidation.txt |
CO19. Conjugate Addition
Conjugated systems are structures that contain alternating double and single bonds (or, in some cases, a double bond that is next to an atom with either a lone pair or a vacant orbital). Conjugated systems are usually at lower energy than regular double bonds because the electrons involved in bonding are delocalized; they are spread out over a greater area and thus can have a longer wavelength.
For example, the -bonding system for 3-butene-2-one (or methyl vinyl ketone) is described by orbitals involving both the carbonyl group and the alkene group. These two groups become linked together so that there is not longer an independent carbonyl nor an independent alkene, but one "enone" (a term taken from the words alkene and ketone).
Because of that extra stability, it might not be surprising that conjugated carbonyls are often a little slower to react than regular carbonyls. The surprise is that conjugated carbonyls can sometimes give additional products in which addition does not take place at the carbonyl.
The product shown above is called a conjugate addition product, or a 1,4-addition product. In conjugate addition, the nucleophile does not donate to the carbonyl, but instead donates to an atom that is involved in conjugation with the carbonyl. This additional electrophilic position is sometimes called a "vinylogous" position (from the word vinyl, which refers to that CH=CH2 unit next to the carbonyl).
• Conjugate additions (or 1,4-additions) can occur when a carbonyl is attached to a C=C bond.
Problem CO19.1.
Draw a mechanism with curved arrows for the conjugate addition shown above.
Problem CO19.2.
Regular additions to carbonyls are sometimes called 1,2-additions, whereas conjugate additions are called 1,4-additions. Show why.
Remember that we can look at another resonance structure of a carbonyl, one that emphasizes the electron-poverty of the carbonyl carbon. It's not a good Lewis structure because of the lack of an octet on carbon, but it does reinforce the idea that there is at least some positive charge at that carbon because it is less electronegative than oxygen. Extending that idea, we can draw an additional resonance structure in a conjugated system. That third structure suggests there may be some positive charge two carbons away from the carbonyl, on the β position on the double bond.
The idea that there are two electrophilic positions in an enone is reinforced by the picture of the LUMO (the lowest-energy "empty" frontier orbital, the virtual place where an additional electron would probably go). When a lone pair is donated to an electrophile, the electrons are most likely to be donated into the LUMO.
Although it isn't obvious from the cartoons we often draw for molecular orbitals, quantum mechanical calculations suggest that the LUMO is "larger" at the carbonyl position as well as the β-position on the vinyl group.
How can it be larger on some atoms than others? A molecular orbital is an algebraic combination of atomic orbitals. In this case,
LUMO = apC1 + bpC2 + cpC3 + dpO
in which pC1 is the p orbital on the carbon on the left, pO is the p orbital on the oxygen, and so on. The letters a, b, c and d are just numbers; they are the coefficients in the equation. The result of the molecular orbital calculation in this case suggests that the numbers a and c are a little bigger than b and d. Incidentally, it also suggests that a and d have opposite sign from b and c (maybe a and d are positive numbers whereas b and c are negative numbers), meaning that a and d are out of phase with b and c.
In any case, we sometimes think of the large LUMO on particular atoms as being an easier "target", an easier place to throw the incoming electrons. These mathematical results really just reflect what we would expect from the resonance structures.
Problem CO19.3.
Indicate whether the following systems are capable of undergoing conjugate addition, and show why or why not.
Having two possible products of a reaction can be confusing. How do you know which one will result? Often, you don't know. Frequently, both products result, so there is a mixture of commpounds. However, one product often predominates. In conjugate addition, there are a few different factors that may tilt the reaction in one direction or another.
Possibly the simplest reason is steric effects. Maybe one of the electrophilic positions is more crowded than the other, and the nucleophile can access that position more easily.
• Addition of a nucleophile often occurs at the least crowed electrophile.
Problem CO19.4.
In each of the following cases, indicate whether the addition of a nucleophile will be via 1,2-addition, via 1,4-addition, or an equal mixture.
The hard/soft acid characteristics of the two electrophilic positions also influence the reaction. The carbonyl position is closer to the oxygen, of course, and it makes sense that the oxygen would have a greater influence on this carbon. The carbonyl position, with its more concentrated positive charge, is a harder electrophile. The vinylogous position, with less positive charge, is a softer electrophile.
• Carbonyls are hard electrophiles.
• Vinylogous positions are soft electrophiles.
Soft nucleophiles are more likely to react with soft electrophiles, and hard nucleophiles are more likely to react with hard electrophiles. The amount of negative charge concentrated at the nucleophilic atom is the biggest factor determining hardness. The lower the charge, or the more spread out the charge, the softer the nucleophile.
Problem 19.5.
Indicate whether the following nucleophiles are more likely to undergo 1,2-addition or 1,4-addition.
There are other factors that play roles in influencing the course of these reactions. Sometimes the mechanism of reaction is slightly different under different circumstances. The presence of Lewis acid catalysts can also influence reactivity in these systems, but not always in a predictable way.
• Lewis acid catalysts can influence whether a reaction proceeds via 1,4-addition. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO19._Conjugate_Addition.txt |
In the following pictures, a number of anions are added to a simple carbonyl compound, a ketone (2-propanone, or acetone). In each case, addition of the nucleophile is followed by addition of a proton source. Note that, overall, the reaction involves addition of the nucleophile to the carbonyl carbon and addition of the proton to the carbonyl oxygen.
Figure CO2.1. Addition of some anionic (and "semi-anionic") nucleophiles to a ketone.
• Addition of anionic nucleophiles to ketones or aldehydes transforms the carbonyl into an alcohol.
Look at the way the reaction is presented in each case. The organic (carbon-based) starting material is presented on the left hand side of the reaction arrow. The reagent added to this starting material is often shown over the arrow. This reagent transforms the starting material into something else. That something else, the product, is shown to the right of the arrow.
Very often, the solvent for the reaction is shown underneath the arrow. The solvent is the liquid that is used to dissolve the starting material and reagents. This is done for a number of reasons. First, reactions generally happen much more quickly in solution than they do without a solvent. When dissolved, the reactants can move around more easily and bump into each other, as if they are swimming. Also, most useful reactions generate heat, and the solvent acts as a heat sink, carrying the excess heat away. (People who have not thought about the importance of solvent sometimes accidentally start fires as a result.) However, there are exceptions, and not all reactions need solvent.
These reactions shown above do need solvent, but the solvent is not shown for other reasons. These reagents must be added in a particular order: first the nucleophile and then the base. The nucleophile and base cannot be allowed to mix before the nucleophile has a chance to react with the carbonyl.
Problem CO2.1.
1. For each of the cases shown above, use curved arrows to show the movement of electrons in the reaction between the anion and the carbonyl.
2. Show the intermediate that results.
3. Use curved arrows to show the movement of electrons in the reaction of the intermediate with acid to form the product.
Problem CO2.2.
1. For each of the cases shown above, used curved arrows to show what would happen if acid were mixed with the nucleophile.
2. Why would the nucleophile no longer be able to react with the carbonyl?
The pattern of reactivity is very different with another class of nucleophile. These could be called neutral nucleophiles (as opposed to anionic ones).
Take another look at the general pattern of reactivity for the anionic nucleophiles and the neutral nucleophiles. In the case of the anionic nucleophiles, the pattern is relatively easy to discover. The product has incorporated the nucleophile into its structure (or at least the anionic part of the nucleophile, which you will soon learn about). The nucleophile has attached at the carbonyl carbon. The carbonyl oxygen has become part of a hydroxyl group. These are very common patterns in the addition of nucleophiles to carbonyls.
In the case of the neutral nucleophiles, there are some similarities and some differences. The nucleophile is still incorporated into the product structure. It has added at the carbonyl position in the electrophile. However, the fate of the carbonyl oxygen is a little bit different with neutral nucleophiles. Generally, this atom is lost as a water molecule in these cases. If you look closely, you will be able to tell where the two protons come from in each case in order to form the water molecule. It's not really the HCl, which is only added in very tiny amounts and acts catalytically. The protons come from other positions in the nucleophile, and sometimes from the electrophile, too.
This chapter will help you to develop skills so that you can recognize where nucleophilic additions have taken place in reactions. You will also be able to predict what products may result from a nucleophilic addition.
Much of the chapter will focus on mechanisms of reaction. A reaction mechanism is, at the very least, the series of elementary steps needed to accomplish an overall reaction, and all of the intermediate structures that would be formed on the way from the reactants to the products.
• A reaction mechanism shows the structures of intermediates that occur after each elementary step.
An elementary reaction is typically a bond-forming or a bond-breaking step. In a bond-forming step, a pair of electrons are donated from one atom to another. In a bond-breaking step, a pair of electrons that were shared between two atoms are drawn to one end of the bond or the other, so that the bond breaks and the electrons end up on one atom only.
Very often, curved arrows are used to show the path that electrons take in these elementary steps. These arrows are always drawn from the source of the electrons to the place to which the electrons are attracted. These arrows help to illustrate bond-making and bond-breaking steps and also serve a book-keeping function, helping us to keep track of electrons over the course of the reaction.
• Curved arrows from the nucleophile to the electrophile show the path of electrons in the reaction.
Often, only one arrow is required in showing an elementary step, but not always. Sometimes, a bond-making step can happen at the same time as a bond-breaking step. This usually happens when an atom isn't large enough to accommodate the electrons from the new bond and sill keep the electrons from an old bond. In this case, two pairs of electrons move in the same elementary step, so two curved arrows are shown. Very rareley, more than two curved arrows are needed to show the events in one elementary step.
Sometimes other information is displayed in a reaction mechanism. Computational chemists will often leave out the curved arrow notation but will instead indicate the relative energy differences between all the intermediate structures along the reaction pathway. These energies may be experimentally determined (i.e. they may be based on the measurement of real reactions) or they may be calculated using an appropriate level of quantum theory. The energies may be displayed numerically, possibly in a table, or they may be illustrated using a picture, such as a reaction profile.
CO20. Conjugate Addition-Elimination in Aromatics
CO20. Conjugate Addition & Elimination in Aromatics
In conjugate addition, a carbonyl group turns a neighbouring alkene into an electrophilic site. An enone, such as the one below, has two electrophilic positions.
A similar situation happens when pi-acceptors such as nitro groups are attached to aromatic rings.
The key step in the mechanism is the loss of the halide ion, which allows the aromaticity to be restored.
Problem CO20.2.
The location of the halogen and the electron-withdrawing group matters. Explain why the reaction occurs if the groups are in the 1 and 2 positions (ortho to each other) or the 1 and 4 positions (para to each other), but not if they are in the 1 and 3 positions (meta to each other).
Problem CO20.3.
Explain why the reaction is faster if additional electron-withdrawing groups are present.
CO21. Carbonyl Addition Summary: Mechanistic Steps
CO21. Summary of Elementary Steps
The reactions of carbonyls can become very complicated, involving many steps. In essence, though, the steps involve only a few, different elementary reactions.
• Protons are most often transferred from a positively charged atom to a neutral atom with a lone pair.
• Protons are also easily transferred from a positively charged atom to a negatively charged atom.
• Sometimes, a protos might be transfered from a neutral atom to a negatively charged one.
In many of the reactions of anionic and semi-anionic nucleophiles, these two steps complete the entire reaction mechanism. However, if an additional lone pair can be revealed at the nucleophilic atom (often by transferring a proron away from this site), additional steps occur.
• Pi donation.
• In pi donation, two heteroatoms, both with lone pairs, are attached to the same carbon. A lone pair is donated to the carbon, and one of the heteroatoms is pushed off.
Many mechanisms involve a number of proton transfers and pi donations. These steps occur over and over, inching the molecule along step by step towards the product. Usually, each proton transfer helps to prepare an atom for eventual removal via pi donation.
Occasionally, if the nucleophile is neutral, these steps are preceded by an initial activation step.
• Carbonyl activation.
• Usually makes the reaction faster.
• Is especially helpful when the nucleophile is uncharged, and hence less reactive.
• The carbonyl is often activated by a proton (from a protic acid) but it can also be activated by a Lewis acid (such as a metal ion). | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO2._General_Reactivity_Patterns.txt |
CO23. Solutions to Selected Problems, Part A
Problem CO1.3.
a) The double bond means two pairs of electrons are shared between the carbon and oxygen, instead of just one. As a result, the oxygen is able to pull more electron density away from the carbon. The carbon becomes much more positive in this case than in the case of a double bond.
Not only that, but the second bond between the carbon and the oxygen is a pi bond. Those electrons are farther from the nucleus than a sigma bond, in which the electrons are tightly held between the atoms. That means the pi electrons are more easily drawn toward the oxygen, so the bond becomes even more polarized.
b) A C=N bond would be very similar to a C=O bond, because nitrogen is the third most electronegative element after oxygen and fluorine (ignoring noble gases).
Problem CO3.1.
The LUMO in this case is the π*, an antibonding level. If electrons populate this level, the π bond will break.
Problem CO3.3.
Both the imine (b) and nitrile (c) have a low-lying pi antibonding level (π*), similar to a carbonyl.
Problem CO4.1.
a) On the basis of steric crowding, the first one is most reactive, then the last one, then the middle.
b) On the basis of steric crowding, the last one is most reactive, then the first one, and then the middle.
Although there is a large group on the nitrogen in that last compound, the site of reactivity is the carbon, which is less crowded.
c) On the basis of electronics, the middle one is most reactive, then the last, and then the first. The fluorine atom is very electronegative and pulls electron density towards itself. That leaves more positive charge on the nearby carbonyl carbon. The more fluorines on that nearby carbon, the more positive the carbon. The more positive the carbon, the more it attracts electrons from a nucleophile.
Problem CO7.1.
a) In acetylide, the lone pair is on a linear carbon or sp carbon. In methyl anion, the lone pair is on a tetrahedral carbon or sp3 carbon. The description "sp" indicates that sigma bonding to neighbours involves a 2s orbital and a 2p orbital on carbon; there is a 50% contribution from the s orbital.
The description "sp3", on the other hand, indicates that sigma bonding to neighbours involves a 2s orbital and three 2p orbitals; there is a 25% contribution from the s orbital.
The 2s orbital is lower in energy than the 2p orbital. The greater the s orbital contribution to the bond (or in this case to the lone pair), the lower it is in energy. Thus, a lone pair on an sp carbon is lower in enrgy than a lone pair on an sp3 carbon.
b) In cyanide, the same argument outlined in pary (a) hold true. In addition, the nearby electronegative nitrogen stabilizes the charge by drawing electron density toward itself.
Problem CO7.2.
1. CH3OK, because of the ionic O-K bond. This is an anionic nucleophile. It is more reactive and nucleophilic than the corresponding neutral nucleophile.
2. CH3NH2, because nitrogen is less electronegative than oxygen. Its lone pair is held less tightly and is more easily donated to the electrophile.
3. NaCCH, because the neighbouring carbon in this case does not have the inductive electron-withdrawing effect that the nitrogen does in the case of NaCN. In that case, the lone pair is stabilized and made less reactive.
4. c-C6H11ONa, because the negative charge is localized on one atom. In c-C6H5ONa, the negative charge is delocalized over four different positions in the molecule. Delocalization of charge stabilizes the anion and makes it less reactive.
CO24. Solutions to Selected Problems, CO10-18
CO24. Solutions to Selected Problems, Part B
Problem CO18.4.
One of the keys in this problem is recognizing that in some steps, two different reactions are involved. For example, in the first box, there is an addition of a diol to a carbonyl followed by an ylide addition.
CO3. MO Picture of Carbonyls
Often it is useful to look at the molecular orbital picture of a molecule to learn something about its reactivity. In the case of carbonyls, frontier orbital ideas tell us to look at the lowest unoccupied molecular orbital (LUMO) and the highest occupied molecular orbital (HOMO).
Figure CO3.1. Frontier orbitals in a carbonyl compound.
When two different atoms bond together, the molecular orbitals that they form are not evenly distributed between the atoms. Instead, the new molecular orbital is closest in space to the atom to which it is closest in energy.
In the case of carbon and oxygen, oxygen is more electronegative than carbon. That means its electrons are more tightly held than carbon’s. That means its electrons are lower in energy than carbon’s.
When carbon and oxygen combine, a bonding orbital and an antibonding orbital result. The bonding orbital is lower in energy than the orbitals on either carbon or oxygen. However, it is closer in energy to oxygen. Thus, the orbital itself is more centered on oxygen. In other words, the electrons in the bond are closer to oxygen than to carbon.
The antibonding orbital, on the other hand, is closer to carbon in energy, although it is higher in energy than either carbon or oxygen. It is more centered on the carbon than the oxygen. That means the “target” for the electron donation is mostly found on the carbon. The carbonyl carbon is the electrophilic position.
If electrons are going to be donated to the molecule, the lowest energy position available for electrons in the molecule is described by the LUMO. The LUMO in this case is the C=O pi* or pi antibonding orbital.
If the carbonyl is going to donate electrons, the electrons will come from the HOMO. In this case, that refers to the non-bonding electrons. These electrons are found on the oxygen, and are equivalent to the lone pairs in the Lewis structure.
Problem CO3.1.
Explain what happens to bonding in the molecule if an electron pair is donated to the LUMO of H2C=O.
Problem CO3.2.
Draw the frontier-orbital diagram (including orbital pictures) for the following compounds.
a) CH3OH b) H2C=NH c) CH3CN
Problem CO3.3.
In the previous problem, which MO pictures most closely resemble the one shown for H2C=O? Which compounds will behave most like a carbonyl compound? Explain. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO23._Solutions_to_Selected_Problems%2C_CO1-9.txt |
CO4. Relative Reactivity of Carbonyls
Sterics
Steric hindrance, or crowdedness around the electrophile, is an important factor that influences reactivity.
• The less crowded the electrophile, the more easily it will react.
• Aldehydes are more reactive than ketones.
Crowdedness affects reactivity simply by preventing nucleophiles from easily approaching the electrophilic site in the carbonyl. If the nucleophile hits something other than the carbonyl carbon, it will probably just bounce off. It needs to collide with the carbonyl carbon in order to deliver its electrons to the right place.
Charge
Amount of positive charge on the electrophile is an important factor that influences reactivity.
• The more positive the electrophile, the more easily it will react.
Factors that place more positive charge on the carbonyl (electron withdrawing groups nearby) make the carbonyl more positive and more reactive. Factors that place additional electron density on the carbonyl (electron donors nearby) make the carbonyl less reactive.
There is another resonance structure that we can think about that illustrates the electrophilicity of a carbonyl. That structure places a full negative charge on the oxygen and a full positive charge on the carbon. This isn’t a good Lewis structure because the carbon does not have an octet. Nevertheless, when taken together with the regular Lewis structure, it suggests something real about the nature of the carbonyl: there is partial positive charge on the carbon and partial negative charge on the oxygen.
There is a general rule about cation stability on carbon atoms: a carbocation with more carbons attached to it is more stable than a carbocation with more hydrogens attached to it.
This observation is sometimes explained as an inductive effect. The positively charged carbon is more electronegative than the uncharged carbons, so it draws electrons away from them. It can polarize the neighbouring carbons, drawing some negative charge towards itself and leaving some positive charge on the other carbons. In that way, it s charge is delocalized and stabilized.
In a more sophisticated explanation, the cation becomes stabilized by a molecular orbital interaction involving the empty p orbital on the carbocation and C-H bonds on the neighbouring carbons.
A similar situation results in the partially positive carbon in the carbonyl. The carbonyl carbon in the ketone is a little more stable than the carbonyl carbon in the aldehyde.
• The partial positive charge on an aldehyde carbonyl carbon is less stable than the partial positive charge on a ketone carbonyl carbon.
• Again, aldehydes are more reactive than ketones.
Problem CO4.1.
Rank the following carbonyl compounds from most reactive to least reactive towards nucleophilic addition. Explain your reasoning.
CO5. Nucleophilic Addition
CO5. Introduction to Carbonyl Addition
Carbonyls act most importantly as electrophiles. They attract a pair of electrons from a nucleophile. When that happens, a bond forms between the nucleophile and the carbonyl carbon.
At the same time, the carbon-oxygen bond breaks. We think of that as a consequence of donating a pair of electrons into the LUMO of the carbonyl. The LUMO on the carbonyl is the C-O pi antibonding orbital. When that orbital is populated, there is no longer a net lowering in energy due to the pi interaction between the carbon and oxygen. The pi bond breaks. The electron pair from the pi bond goes to the oxygen, the more electronegative of the two atoms in the original bond. It becomes a lone pair.
After the pi bond breaks, the reaction reaches a branching point or decision point. The reaction may go forwards or backwards.
In other words, this reaction can occur in equilibrium.
To go backwards, the reaction simply slides into reverse. A lone pair on the oxygen donates to the carbon, forming a pi bond again, and pushes the nucleophile off. Whether the reaction ends up going forward or sliding backward depends partly on the relative stability of those two ends of the reaction. That’s often very difficult to assess qualitatively, because there are too many factors involved. However, one factor that plays a role is charge stability. Because an “O minus” or alkoxide is produced in this reaction, if the original nucleophile was a more reactive ion than an alkoxide, the reaction probably goes to the right. If the nucleophile was less reactive than alkoxide, the reaction can easily go to the left again.
Problem CO5.1.
Provide curved arrows and predict the direction of equilibrium in the following cases.
What happens after the initial equilibrium? In most cases, the alkoxide that is formed will become protonated. It will pick up a proton to become an alcohol. The source of the proton may be an acid, deliberately added to provide the H+. Alternatively, it may just be a very slightly acidic molecule such as water or another alcohol. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO4._Relative_Reactivity_of_Carbonyls.txt |
CO6. Protonation of the Alkoxide Anion
After the addition of an anionic nucleophile, reaction mixtures are usually treated with or water or dilute aqueous acid. The acid provides a proton that can be picked up by the alkoxide ion formed in the nucleophilic addition. An alcohol is formed as a result. Overall, carbonyl addition reactions usually involve addition of a nucleophile to the carbonyl carbon and addition of a proton to the carbonyl oxygen.
The order of these steps can be very important. On paper, a carbonyl could be turned into an alcohol by adding a proton to the carbonyl oxygen, and then adding a nucleophile to the carbonyl carbon. However, things might not work out that way in reality. The potential problem lies in the fact that anionic nucleophiles can be pretty basic. If protons are added first, the anionic nucleophile is likely to pick up a proton rather than donate to the carbonyl. Once the nucleophile has picked up a proton, it is no longer anionic, and is less attracted to the partially positive carbonyl. Furthermore, it may have donated its only lone pair to the proton, leaving it completely unable to donate to the carbonyl.
Problem CO6.1.
Provide curved arrows and predict the direction of equilibrium in the following cases.
Problem CO6.2.
Suppose nucleophilic addition was performed in methanol (10 mL) as a solvent, using 25 microliters of cyclopentanone and an equimolar amount as the of sodium cyanide (that means the same number of moles of sodium cyanide as cyclopentanone).
1. Show the mechanism for the reaction, using curved arrows in each step.
2. Comment on how the use of methanol as a solvent (rather than just adding an equimolar amount of methanol) would influence the direction of equilibrium for the final protonation step.
CO7. What is a Good Nucleophile?
CO7. What is a Nucleophile?
Lots of things can be nucleophiles. In principle, a nucleophile only needs a lone pair. However, some nucleophiles are better than others.
You already know something about nucleophiles if you know something about acidity and basicity. Nucleophiles are really Lewis bases. Some of the factors that account for basicity also account for nucleophilicity.
Halides are not very good nucleophiles for carbonyls. The negative charge on a halide is pretty stable, either because of electronegativity or polarizability. If a halide donates to a carbonyl, producing an oxygen anion, the reaction is uphill.
Hydroxide and alkoxide anions (such as CH3O-) are more reactive than halides. They are better nucleophiles. The sulfur analogues are similarly good nucleophiles (such as CH3S-). In addition, water, alcohols and thiols are nucleophilic, because they all have lone pairs that could be donated to an electrophile.
Nitrogen also has a lone pair in most compounds. That means amines are good nucleophiles, too.
Carbon does not normally have a lone pair, unless it is a carbanion. Carbanions are usually not very stable. As a result, they are not very common, except for cyanide (CN-) and acetylides (RCC-, in which R is a hydrogen or an alkyl group).
Problem CO7.1.
Carbanions such as CH3- (methyl anion) are very unstable and highly reactive. Explain why the following anions are more stable than a methyl anion.
1. Acetylide, HCC-
2. cyanide, CN-
Problem CO7.2.
Nucleophilicity is the degree of attraction of a nucleophile to a positive charge (or partial positive charge). It is related to basicity. Choose the most nucleophilic item from each of the following pairs, and explain your answer.
1. CH3OK or CH3OH
2. CH3OH or CH3NH2
3. NaCN or NaCCH
4. c-C6H11ONa or c-C6H5ONa (c- in this case means "cyclo")
Problem CO7.3.
Carbonyl compounds such as aldehydes and ketones contain a very slightly acidic hydrogen next to the carbonyl. Some nucleophiles are basic enough to remove that proton instead of donating to the carbonyl. Show why the resulting anion is stable, using cyclopentanone as an example.
Problem CO7.4.
Accidental deprotonation (proton removal) alpha to a carbonyl (one carbon away from the carbonyl) can occur when a nucleophile is added to a ketone. In the following cases, explain which nucleophile is more likely to add to the carbonyl in cyclohexanone and which is more likely to deprotonate it. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO6._Protonation_of_Oxygen.txt |
CO8. Semi-Anionic Nucleophiles
Some nucleophiles are added to carbonyls in the form of salts, such as sodium cyanide. In a salt, there is an anion and a cation. The anion can act as a nucleophile, donating a lone pair to the carbonyl. The cation is just a counterion; it is there to balance the charge but doesn't usually play an active role.
Some anions are too unstable and reactive to be used as salts. This is especially true with a number of carbon nucleophiles. C-H bonds are not usually acidic enough to deprotonate with a strong base. That makes it hard to make a simple salt containing such nucleophiles. There are exceptions, such as acetylide or alkynyl protons like CH3CCH. In that case, the resulting anion is relatively stable because the lone pair is in a lower-energy orbital with more s character, so it is held more tightly to the nucleus.
Less stable carbon anions can be be stabilized through a covalent bond. If the carbon is covalently attached to a less electronegative atom, the carbon has a partially negative charge. It can still act as though it were an anion. However, the covalent bond stabilizes the would-be "lone pair." Compounds like this can be considered to be "semi-anionic." Frequently, they are described as polar covalent compounds, although that is really a much more general term.
These polar covalent bonds can be found any time a carbon atom is bound to a metal. Remember, the metals are the aroms in the coloured boxes in the periodic table below.
One of the most common classes of this type of compounds is the family of organomagnesium halides or Grignard reagents (Green-yard reagents). Victor Grignard was awarded the Nobel Prize in Chemistry for his development of these reagents. These compounds are made by reacting an alkyl halide (such as CH3CH2CH2Cl) with magnesium metal. The metal undergoes an insertion into the C-Cl bond, forming CH3CH2CH2MgCl. (You don't need to worry about how this happens.) Because magnesium is less electronegative than carbon, the C-Mg bond acts as though it were a lone pair on the carbon and the magnesium acts as though it were a cation.
What's striking about Grignard formation is that polarity is reversed in this reaction. In the alkyl halide, the carbon attached to the halogen has a partially positive charge, because carbon is further to the left than halogens in the periodic table. After magnesium insertion, this same carbon has a partial negative charge, because carbon is farther to the right in the periodic table than magnesium. This sort of reversal in reactivity is sometimes called "umpolung chemistry."
Propylmagnesium chloride and other Grignard reagents can deliver alkyl nucleophiles to carbonyls. Just like with simple anionic nucleophiles, an alkoxide ion results. Subsequent treatment of the alkoxide with acid provides a proton, resulting in an alcohol.
Remember, the order of these two steps is very important. Adding the acid before the Grignard reagent would not work, because the Grignard reagent would become protonated at the carbon. Although the Mg-C bond is covalent, it is still polar enough so that the carbon can act as a nucleophile or as a base. Once propylmagnesium has become protonated, it forms propane, which isn't likely to act as a nucleophile.
Problem CO8.1.
Show the products of the following reactions. Assume workup with aqueous acid after each reaction.
The situation with Grignard reagents is even more delicate than that. Solvents must be chosen very carefully for Grignard reactions. Grignard reagents are basic enough that they can't tolerate protic solvents. Protic solvents are solvents that are capable of hydrogen-bonding. Although they don't seem very acidic, they can still give up a proton to a strong enough base. A Grignard reagent is a strong enough base to take that proton.
Problem CO8.2.
Show why Grignard reagents cannot be used with ethanol as a solvent.
In fact, Grignard reagents are even fussier than that. Not only do they not get along well with acidic or even semi-acidic protons, but they tend to need coordinating solvents to help support the magnesium atom and keep the complex stable. The most common solvents for this use are (diethyl) ether and tetrahydrofuran (THF). Because coordination of magnesium by these weakly donating solvents is crucial, Grignard reagents can't generally be isolated. Instead, they are sold and used as solutions in ethereal solvents.
Problem CO8.3.
There are plenty of other semi-anionic nucleophiles. For example, alkyl lithium reagents are also very common, and they are prepared by treatment of alkyl halides with finely divided lithium metal. The reaction produces lithium chloride as a side product.
1. Show an equation, with structures, for the preparation of butyllithium from 1-bromobutane.
2. Explain what happens to polarity at carbon number one before and after this reaction.
3. Why would the amount of charge on carbon number one be somewhat similar in butyllithium and butylmagnesium bromide?
Problem CO8.4.
Another class of semi-anionic nucleophiles is the family of complex metal hydrides. Examples include sodium borohydride, NaBH4, and lithium aluminum hydride (LAH), LiAlH4. There are many other variations.
1. Draw a Lewis structure for lithium aluminum hydride.
2. Explain why LAH functions as a source of the hydride nucleophile, H-.
3. LAH is much more reactive that sodium borohydride; it can reduce compounds that sodium borohydride will not. For example, it can reduce a nitrile such as CH3CN to an amine such as CH3CH2NH2 (after an aqueous workup). Explain why LAH is so much more reactive than NaBH4.
4. Sodium borohydride is sometimes used in methanol, but care must be taken in dissolving the NaBH4. It doesn't just dissolve; it quickly reacts with the methanol to produce a flammable gas and NaBH3OCH3. Provide a mechanism for this reaction with arrows.
5. Although NaBH3OCH3 also reacts with methanol, it does so much more slowly than NaBH4, and so it is still able to reduce aldehydes and ketones in methanol. Explain the difference between sodium borohydride and sodium methoxyborohydride in terms of reactivity with methanol.
6. LAH cannot be used in protic solvents such as methanol. Explain why.
Problem CO8.5.
Barbier reactions are a general class of reactions involving metal alkyls and carbonyls. Treatment of a halide such as propargyl bromide (HCCCH2Br) with zinc metal in the presence of an aldehyde such as benzaldehyde (C6H5CHO) results in nucleophilic addition of the propargyl group to the aldehyde.
1. Zinc can insert into a carbon-halogen bond, just like magnesium. Show the product of the insertion described above.
2. This reaction is usually performed in water with some ammonium chloride, NH4Cl, in solution. Show a mechanism, with curved arrows, for the reaction of the alkylzinc species with the aldehyde to yield an alcohol.
3. Explain why this alkylzinc reaction can be conducted in the presence of water, but a Grignard reaction cannot.
4. "Green chemistry" refers to the intentional use of processes that are better for the environment, by minimizing the use of toxic reagents and solvents. Compare the zinc-mediated Barbier reaction with the Grignard reaction in terms of "greenness."
Problem CO8.6.
Semi-anionic nucleophiles do not just react with carbonyls. They are also frequently used to prepare organometallic compounds via "transmetallation." For example, treatment of tantalum pentachloride, TaCl5, with dimethylzinc, (CH3)2Zn, affords trimethyltantalum dichloride, (CH3)3TaCl2.
1. Assume for the moment that tantalum pentachloride and dimethylzinc are both covalently-bonded molecules. What would you say about bond polarity in each case?
2. What is the side-product of the reaction (i.e. what else must be produced given the production of trimethyltantalum dichloride from these reactants?)
3. In what ratio would you mix the two reactants to get these products?
4. Show a mechanism, with curved arrows, for the formation of trimethyltantalum dichloride.
5. Like some of the other compounds on this page, trimethyltantalum dichloride is remarkably pyrophoric: it catches fire upon contact with air. This behaviour often depends on weather and humidity. Show a mechanism, with curved arrows, for what happens when this compound is exposed to air. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO8._Semi-Anionic_Nucleophiles.txt |
CO9. Enolate Addition
Enolate ions are just another example of anionic nucelophiles. The reason they get a page to themselves is that they are especially important, especially in biological chemistry. They are also important in the synthesis of organic compounds, such as in the pharmaceutical industry.
An enolate ion is the anion that forms when a proton is removed next to a carbonyl. The carbon next to the carbonyl is called the α-position (alpha position). The alpha position is acidic both because of the amount of positive charge on a proton in that position and because of the stability of the anion that results if that proton is removed.
Problem CO9.1.
Show why an enolate ion, such as the one formed from 2-propanone, above, is particularly stable.
Problem CO9.2.
Show a mechanism, with curved arrows, for the formation of the enolate ion from 2-propanone, above.
In the example above, 2-propanone is deprotonated at the α position to form the corresponding enolate ion. Note that sodium hydroxide is not a strong enough base to convert all of the 2-propanone to its enolate. The resulting enolate is basic enough to pull a proton from a water molecule, so an equilibrium results.
Negative charges are fairly stable on oxygen atoms. That allows this particular reaction to shift back to the left again. To make the reaction go all the way to the right, we would need a less stable anion on the left. That would make that anion more basic. Can you think of atoms that would be less stable as anions than oxygen?
The most commonly used very strong bases in synthetic chemistry involve anions of carbon, nitrogen or hydrogen. Some examples of compounds used as very strong bases are sodium hydride (NaH), sodium amide (NaNH2), lithium diisopropylamide (LiN[CH(CH3)2]2), and butyllithium (CH3CH2CH2CH2Li).
• Enolate ions form in equilibrium with their paent carbonyl compounds if a moderately strong base like sodium hydroxide is used.
• A very strong base, like sodium amide (NaNH2) or sodium hydride (NaH), would result in complete enolate formation.
However, it is sometimes really useful to have an equilibrium between a carbonyl compound and its enolate. That situation allows both a ketone (the 2-propanone, left) and its enolate (right) to be present at the same time. That means there is both a nucleophile and an electrophile (the ketone and the enolate). They will be able to react together.
• Simple carbonyls are electrophiles.
• The enolate ions that form from simple carbonyls are nucleophiles.
• Carbonyls react with enolate ions.
The reaction of an enolate nucleophile with another carbonyl compound is called an aldol reaction. A simple example of this reaction is shown here. This example involves the reaction of 2-propanone with its enolate.
Problem CO9.3.
Provide a mechanism, with curved arrows, for the aldol reaction of 2-propanone, above.
The biosynthesis of sugars, such as fructose, involves coupling smaller sugars together. If one sugar is converted into a nucleophile, it can donate electrons to the carbonyl on the other sugar, forming a new C-C bond. The carbonyl on the second sugar becomes a hydroxyl group in the new, larger sugar.
In the cell, sugars are typically in a phosphorylated form when they react in this way. Phosphorylation is often an important step in activating molecules for biochemical reactions.
Problem CO9.4.
Show the mechanism for the formation of the phosphorylated fructose shown above.
Sometimes, aldol reactions are followed by a subsequent reaction, called an elimination reaction. That reaction formally produces a molecule of water. Early studies of this reaction would result in droplets of condensation on the glassware in which the reaction occured; hence, it is sometimes called a condensation reaction.
Problem CO9.5.
Provide a mechanism for the aldol condensation shown above.
It can be hard to predict the outcome of an aldol reaction because of the fact that there are two possible products from an aldol reaction (one with a new hydroxyl and one with a new double bond). A chemist might try to make one product in the laboratory, and end up with the other. This process can be difficult to control. However, in general, the elimination reaction is encouraged by heating the reaction. The reaction sometimes occurs without elimination if the reaction is kept cool. However, there are also other factors that may come into play.
Problem CO9.6.
Predict the products of the following aldol reactions.
Problem CO9.7.
The following compound would give multiple products through different aldol condensation reactions. Show the products.
Problem CO9.8.
Sometimes, two different compounds may react together in an aldol reaction. The following compounds would give multiple products through different aldol condensation reactions. Show the products.
Problem CO9.9.
Only some of the following compounds may undergo aldol reactions. Select which ones may not undergo the reaction, and explain what factor prevents them from reacting.
Problem CO9.10.
Fill in the products of the following aldol condensations.
Aldol reactions do not just occur with enolate anions, however. Enols are the neutral form of enolates, protonated on the oxygen instead of the alpha carbon. Enols are also good nucleophiles. In an enol nucleophile, the pi bond acts as the electron source, rather than the lone pair. However, the pi bond gets a boost from the lone pair on the oxygen.
Enols are always present in equilibrium with aldehydes and ketones. An enol is a simple tautomer of a carbonyl compound. To get from one to the other, a proton is simply transferred from one position to the other.
Enamines are also good nucleophiles for aldol-type reactions.
Problem CO9.11.
Show the subsequent protonation step in the reactions involving the enol and the enamine above.
Problem CO9.12.
An enamine reaction is usually followed by hydrolysis of the C=N bond in the iminium ion. Show the mechanism for conversion of the iminium ion to the carbonyl. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO9._Enolate_Addition_and_Homologation.txt |
Carbonyl compounds are very common in biological chemistry. Classes of compounds that contain the C=O bond include amides (found in proteins & peptides, used as signaling molecules and to help catalyze and guide reactions), aldehydes and ketones (found in carbohydrates, which play structural roles in cellulose, starch and DNA, for example) and esters (found in fats that form cell membranes, among other things). Understanding the reactivity of these bonds will help you to learn about many biological processes, as well as other transformations that are important in human society.
Figure CO1.1. A tripeptide, composed of three amino acids. See if you can identify the three separate sub-units.
Figure CO1.2. A simple carbohydrate, glyceraldehyde.
Figure CO1.3. A triglyceride.
The carbonyl bond is very polar. There is a partial positive charge on the carbon and a partial negative charge on the oxygen, because oxygen is more electronegative than carbon. This charge separation is intensified because of the double bond between the carbon and oxygen. Rather than just pulling one pair of bonding electrons towards itself, the oxygen pulls two pairs of electrons towards itself.
• The C=O bond is very polar.
• The carbonyl carbon is very positive.
Sometimes, a resonance structure is drawn to emphasize the charge separation in the carbonyl. The structure has only one bond between the carbon and oxygen. In this structure, oxygen has an octet but carbon does not. This is not really a good Lewis structure, because the other resonance structure satisfies octets on all the atoms. However, this Lewis structure emphasizes the polarity of the bond and is sometimes drawn to reinforce that idea.
Figure CO1.4. Thinking about charge distribution in a carbonyl group.
Because of the positive charge on the carbonyl carbon, the most important theme in carbonyl chemistry is reaction of the carbonyl as a Lewis acid. Reactions of carbonyls almost always involve addition of an electron donor to the carbonyl carbon.
• Electrophile is another term for Lewis acid.
• Lewis acids attract electrons.
• Lewis acids have a positive charge on an atom, a partial positive charge on an atom, or an atom lacking an octet.
• Carbonyl compounds are good electrophiles.
The electrophilicity of carbonyls is very important in their reactivity. The goal of this chapter is to develop an understanding of how carbonyls react. We will learn about a few key factors that will be used in different combinations under different circumstances. Eventually, you will build an understanding that will allow you to follow both biological reactions and modern synthetic reactions.
Figure CO1.5. Reactivity in carbonyl compounds. The carbonyl in the lower sugar on the left has reacted with the neighboring molecule.
It is important to realize that biological reactions, such as carbohydrate synthesis, are very complex and can involve many, many steps. For example, the carbohydrate synthesis shown above involves additional acid-base steps as well as a reaction of a carbonyl. The additional acid base steps may involve proton donors and acceptors as well as more general Lewis acids.
Problem CO1.1.
Locate the carbonyls in the following biological molecules and identify the functional groups that contain carbonyls in each case.
Ginkgolides are biologically active terpenoids from Ginkgo trees. They are thought to have medicinal properties.
Okundoperoxide is isolated from a type of sedge in Cameroon. It has modest anti-malarial properties.
D-erythrose is a typical carbohydrate.
Problem CO1.2.
Frequently, the carbonyls in carbohydrates are "masked," as in deoxyribose (below).
By moving one proton from one position to another, and then breaking a single C-O bond, discover where the carbonyl is hiding.
Problem CO1.3.
1. Explain why the carbon in a C=O unit is very electrophilic, but the carbon in a C-O unit is much less so.
2. Propose other carbon-heteroatom bonds that may make the carbon electrophilic (heteroatom means not carbon or hydrogen). | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/Carbonyls_are_Electrophiles.txt |
Carbohydrates are an important class of biological molecules. Although their best-known role is in energy storage in the form of glucose and starch, carbohydrates play a number of other roles. For example, they lend structural support in the backbone of DNA. They aid in homebody security and defense operations by forming molecular codes on the surfaces of cells that identify whether the cell is one of our own or an intruder. They carry specialized chemical reagents into enzymes where reactions essential to life are carried out. Some of them are sweet.
The term "sugar" is often applied to any simple carbohydrate. One strict definition of a carbohydrate is a polyhydroxylated aldehyde or ketone. In general, a sugar is a molecule that contains an aldehyde or a ketone, and every carbon other than the carbonyl carbon has a hydroxyl group attached to it.
This sort of structure presents lots of possibilities for reactions. In just one molecule, we have both an electrophile (the carbonyl) and a number of nucleophiles (the hydroxyls). Can sugars react with each other? Yes. Can a sugar react with itself? Of course.
In fact, if you have seen drawings of sugars before, you might not have noticed the carbonyl. That's because the carbonyl is usually "masked" as a hemiacetal. The hemiacetal forms when a hydroxyl group along the carbon chain reaches back and bonds to the electrophilic carbonyl carbon. As a result, five- and six-membered rings are very common in sugars. Five-membered rings are called "furanoses" and six-membered rings are called "pyranoses".
The most common way of drawing these rings are in "Haworth projections." Haworth projections don't reflect the real shape of the ring. For example, in a six-membered ring, the atoms in the ring adopt a zig-zag, up-and-down pattern in order to optimize bond angles. The chair drawing shows that relationship, but in a Haworth projection, the ring is drawn as though it were flat. Also, substituents on the atoms in the ring can be found above the ring, below the ring, or sticking out around the edge of the ring. The chair drawing or "diamond lattice projection" shows these relationships pretty well. A Haworth projection doesn't try to do that. Instead, it tries to depict stereochemical relationships: whether two substituents are on the same face of the ring or opposite faces of the ring. In a diamond lattice projection, we have to keep track of whether a given substituent is in the upper or lower position at its particular site on the ring, and that requires careful attention. In a Haworth projection, substituents are simply drawn straight up or straight down. Because there are many chiral canters in sugars, and becuase two sugars can differ by just one chiral center, Haworth projections make it easier to tell different sugars apart.
When an open-chain sugar cyclizes by forming a hemiacetal, it forms a new stereocenter. Because the carbonyl carbon is trigonal planar, the hydroxyl group can approach it from either face. There is nothing to distinguish one face from the other, and so approach from either face is equally likely. That means that two different stereochemical configurations can form at the hemiacetal carbon: R and S.
In sugar chemistry, these two isomers are named a different way: alpha and beta. To distinguish these two designations, you need to look at the Haworth projection. In a Haworth projection, the lower edge of the ring is read as being nearer to you. The upper edge is read as being farther away. Remember, that's how we usually read a chair structure, too. However, in a Haworth projection, we have to orient the ring in a specific way. The hemiacetal carbon is always placed at the right edge of the drawing. In addition, we always keep the oxygen atom on the back edge of the ring (i.e. the upper edge of the drawing). That means the ring oxygen in a Haworth projection is always found in the upper or upper right part of the drawing, with the hemiacetal carbon directly beside it to the right.
If we have the Haworth projection, we can designate whether we have the alpha or beta from by seeing whether the hydroxy part of the hemiacetal points up or down. If it is down like the ants, we have an alpha isomer. If it is up like the butterflies, we have a beta isomer.
Because the hemiacetal carbon can adopt either of two configurations in a ring, it is given a special name. It is called the anomeric position. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/Pyranose_and_Furanose_Forms.txt |
Sometimes, nucleophiles adding to a carbonyl do not follow the normal reactivity patterns that have been common so far. This is often the case with the addition of ylides. Ylides are compounds that are often depicted with a positive charge one one atom and a negative charge on the next atom. They are examples of zwitterions, compounds that contain both positive and negative charges within the same molecule. What distinguishes them from other zwitterions is the proximity of the opposite charges.
The classic example of an ylide addition to a carbonyl is the Wittig reaction, which involves the addition of a phosphorus ylide to an aldehyde or ketone. Rather than producing an alcohol, the reaction produces and alkene. The reaction is driven by formation of a phosphorus oxide "side product".
This is a special case. The phosphorus-oxygen bond is strong enough to change the course of this reaction away from the normal pattern, and it isn't something you would have been able to predict based on related reactions.
An ylide is an example of a molecular compound that contains both a positive and a negative formal charge on two adjacent atoms. The charges are right beside each other: in this case, there is a positive charge on the phosphorus and a negative charge on the carbon.
Ylides are specific examples of zwitterions, which are molecules that contain positive and negative charges. The most common example of a zwitterion is probably an amino acid, which contains a positive ammonium ion and a negative carboxylate ion, within the same molecule.
Phosphorus ylides are made one charge at a time. A phosphonium ion must first be assembled, containing the positive charge on phosphorus. This event occurs via a nucleophilic substitution reaction, in which a phosphorus nucleophile displaces a halogen from an alkyl halide.
Problem CO18.1
Show, with reaction arrows, formation of the three alkyltriphenyl phosphonium bromide salts shown below.
In most cases, the source of the phosphorus is triphenylphosphine. Triphenylphosphine is used for several practical reasons. First of all, it is a solid, so it is easy to weigh out the right amount of it and add it to a reaction. Secondly, organophosphorus compounds are often very toxic and smelly, but triphenylphosphine is less offensive. Thirdly, in the Wittig reaction, the original phosphorus compound is eventually discarded as waste, and the more useful alkene is kept. Since the phosphorus part doesn't matter that much, the most convenient possible phosphine is generally used. However, there are other variations of this reaction that use other phosphorus compounds.
Once the phosphonium salt has been made, the phosphorus ylide can then be obtained via deprotonation of a phosphonium ion. The hydrogens on a carbon next to a phosphorus cation are a little bit acidic because of the positive charge on the phosphorus. One of these hydrogens is easily removed via adddition of a very strong base such as sodium hydride.
Problem CO18.2
Show, with reaction arrows, formation of ylides from the three alkyltriphenyl phosphonium bromide salts shown above in Problem CO18.1.
The reaction of a phosphorus ylide with a carbonyl compound does begin like other nucleophilic additions. The ylide donates its nucleophilic lone pair to the carbonyl and the carbonyl pi bond breaks. However, the strong P-O bond then takes over the reaction. To begin, a lone pair on the resulting alkoxide ion is donated to the positively charged phosphonium ion.
Wait! That violates one of our mechanistic rules. Usually, we don't have an atom donate to a positively charged atom that already has an octet; if we do so, the atom will have too many electrons. However, the octet rule doesn't strictly apply to sulfur and phosphorus. These atoms are larger than second-row atoms like nitrogen and oxygen, and they are often observed to "exceed the octet rule". Sulfur and phosphorus are frequently observed with trigonal bipyramidal or octahedral molecular geometries, meaning they may have up to 12 electrons in their valence shells.
So go ahead! Donate a pair of electrons to the phosphorus. It can't help itself, because of the strength of the P-O bond that forms.
Problem CO18.3.
Show the products of the reactions of each of the ylides you made in Problem CO18.2. with the following electrophiles:
a) butanal b) benzaldehyde c) 4-methylpentanal
This is when things really get interesting. It turns out that one P-O bond just isn't enough. The phosphorus is so oxophilic that it takes the oxygen atom all to itself, pulling it right out of the molecule. It probably doesn't hurt that the four-membered ring is pretty strained, so it is motivated to decompose (but be careful: there are plenty of stable four- and even three-membered rings in nature).
The arrows shown in the decomposition of the four-membered ring (called a betaine) are just meant to keep track of electrons; there isn't a true nucleophile and electrophile in this step. Instead this step may resemble a pericyclic reaction, which is covered in another section.
Exactly how to draw the P=O bond is debatable. There isn't much doubt that it is a double bond; it is stronger and shorter than a P-O single bond. However, quantum mechanical calculations indicate that the phosphorus can't form a pi bond. This double bond is different than other double bonds you have seen. For that reason, some people prefer to draw this compound as an ylide, too, with a positive charge on the phosphorus, a single bond, and a negative charge on the oxygen.
The phosphorus oxide compound forms, leaving behind an alkene. Alkenes are very common in nature, and this reaction has frequently been used to make interesting alkene-containing compounds for further use or study.
Problem CO18.4.
The juvenile hormone of the cecropia moth caterpillar (JH-1, below) is a regulatory hormone used to control the organism's development by preventing it from pupating until conditions are right.
Synthesis of insect hormones is often undertaken in order to control insect populations. The following synthesis of JH-1 was developed by Barry Trost (Stanford) in the 1960's. Fill in the missing reagents and reaction products.
Sulfur ylides are also good nucleophiles for aldehydes and ketones. However, the unusual stability of the phosphorus-oxygen bond does not have a similar analogue in sulfur chemistry.
Sulfur ylides are formed in a manner very similar to phosphorus ylides.
Problem CO18.5.
Show, with arrows, the mechanism for formation of the sulfur ylide above.
Once formed, sulfur ylides react with aldehydes or ketones. Like phosphorus ylides, the reaction starts out just like any other nucleophile, but a second step takes a very different direction. Epoxides are formed in these reactions, and the original sulfur compound (a thioether) is regenerated.
Problem CO18.6.
Show, with arrows, the mechanism for the epoxide-forming reaction above. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/Ylide_Addition.txt |
Electrophilic addition is a reaction between an electrophile and nucleophile, adding to double or triple bonds. An electrophile is defined by a molecule with a tendency to react with other molecules containing a donatable pair of electrons. Thus, it is an "electron lover." A nucleophile is one that possesses a lone pair of electrons that can be easily shared. In essence, all nucleophiles are Lewis bases that attack nonhydrogen atoms (Lewis acids).
Electrophilic Addition Reactions
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic addition reactions between bromine (and the other halogens) and alkenes like ethene and cyclohexene
The electrophilic addition of bromine to ethene
Alkenes react in the cold with pure liquid bromine, or with a solution of bromine in an organic solvent like tetrachloromethane. The double bond breaks, and a bromine atom becomes attached to each carbon. The bromine loses its original red-brown color to give a colorless liquid. In the case of the reaction with ethene, 1,2-dibromoethane is formed.
This decoloration of bromine is often used as a test for a carbon-carbon double bond. If an aqueous solution of bromine is used ("bromine water"), you get a mixture of products. The other halogens, apart from fluorine, behave similarly. (Fluorine reacts explosively with all hydrocarbons - including alkenes - to give carbon and hydrogen fluoride.) If you are interested in the reaction with, say, chlorine, all you have to do is to replace Br by Cl.
Mechanism
The reaction is an example of electrophilic addition.The bromine is a very "polarizable" molecule and the approaching pi bond in the ethene induces a dipole in the bromine molecule. If you draw this mechanism in an exam, write the words "induced dipole" next to the bromine molecule - to show that you understand what's going on.
Figure: A simplified version of the mechanism
In the first stage of the reaction, one of the bromine atoms becomes attached to both carbon atoms, with the positive charge being found on the bromine atom. A bromonium ion is formed.
Figure: Step 1 in mechanism of addition of Bromine to ethene
The bromonium ion is then attacked from the back by a bromide ion formed in a nearby reaction.
Figure: Step 2 in mechanism of addition of Bromine to ethene
Electrophilic addition of bromine to cyclohexene
Cyclohexene reacts with bromine in the same way and under the same conditions as any other alkene. 1,2-dibromocyclohexane is formed.
Mechanism
The reaction is an example of electrophilic addition. Again, the bromine is polarized by the approaching $\pi$ bond in the cyclohexene. Do not forget to write the words "induced dipole" next to the bromine molecule.
Figure: A simplified version of the mechanism
In the first stage of the reaction, one of the bromine atoms becomes attached to both carbon atoms, with the positive charge being found on the bromine atom. A bromonium ion is formed.
Figure: Step 1 in mechanism of addition of Bromine to cyclohexane
The bromonium ion is then attacked from the back by a bromide ion formed in a nearby reaction.
Figure: Step 2 in mechanism of addition of Bromine to cyclohexane
Contributors
Jim Clark (Chemguide.co.uk)
Reactions of Alkenes with Hydrogen Halides
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic addition reactions between the hydrogen halides and alkenes like ethene and cyclohexene. Hydrogen halides include hydrogen chloride and hydrogen bromide.
Electrophilic addition reactions involving hydrogen bromide
Alkenes react with hydrogen bromide in the cold. The double bond breaks and a hydrogen atom ends up attached to one of the carbons and a bromine atom to the other. In the case of ethene, bromoethane is formed.
$\ce{CH_2=CH_2 + HBr \rightarrow CH_3CH_2Br}$
With cyclohexene you get bromocyclohexane.
The structures of the cyclohexene and the bromocyclohexane are often simplified:
Be sure that you understand the relationship between these simplified diagrams and the full structures.
The mechanisms
The reactions are examples of electrophilic addition. With ethene and HBr:
and with cyclohexene:
Electrophilic addition reactions involving the other hydrogen halides
Hydrogen chloride and the other hydrogen halides add on in exactly the same way. For example, hydrogen chloride adds to ethene to make chloroethane:
$\ce{CH_2=CH_2 + HCl \rightarrow CH_3CH_2Cl}$
The only difference is in how fast the reactions happen with the different hydrogen halides. The rate of reaction increases as you go from HF to HCl to HBr to HI.
HF > HCl > HBr > HI
The reason for this is that as the halogen atoms get bigger, the strength of the hydrogen-halogen bond falls. Bond strengths (measured in kiloJoules per mole) are:
H-F (569 kJ) > HCl (432 kJ) > HBr (366 kJ) > HI (298 kJ)
As you have seen in the HBr case, in the first step of the mechanism the hydrogen-halogen bond gets broken. If the bond is weaker, it will break more readily and so the reaction is more likely to happen.
The mechanisms
The reactions are still examples of electrophilic addition. With ethene and HCl, for example:
This is exactly the same as the mechanism for the reaction between ethene and HBr, except that we've replaced Br by Cl. All the other mechanisms for symmetrical alkenes and the hydrogen halides would be done in the same way.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Electrophilic_Addition_Reactions/Reactions_of_Alkenes_with_Bromine.txt |
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic addition reactions between sulfuric acid and alkenes like ethene and cyclohexene.
The electrophilic addition reaction between ethene and sulfuric acid
Alkenes react with concentrated sulfuric acid in the cold to produce alkyl hydrogensulphates. For example, ethene reacts to give ethyl hydrogensulphate.
$\ce{CH_2=CH_2 + H_2SO_4 \rightarrow CH_3CH_2OSO_2OH}$
The structure of the product molecule is sometimes written as $CH_3CH_2HSO_4$, b ut the version in the equation is better because it shows how all the atoms are linked up. You may also find it written as $CH_3CH_2OSO_3H$.
All you need to do is to learn the structure of sulfuric acid, and after that the mechanism is exactly the same as the one with hydrogen bromide. As you will find out, the formula of the product follows from the mechanism in an inevitable way.
The mechanism for the reaction between ethene and sulfuric acid
The hydrogen atoms are attached to very electronegative oxygen atoms which means that the hydrogens will have a slight positive charge while the oxygens will be slightly negative. In the mechanism, we just focus on one of the hydrogen to oxygen bonds, because the other one is too far from the carbon-carbon double bond to be involved in any way.
Look carefully at the structure of the product so that you can see how it relates to the various formulae given earlier (CH3CH2OSO2OH etc).
The electrophilic addition reaction between cyclohexene and sulfuric acid
This time we are going straight for the mechanism without producing an initial equation. This is to show that you can work out the structure of obscure products provided you can write the mechanism.
Having worked out the structure of the product, you could then write a simple equation for the reaction if you wanted to.
Contributors
Jim Clark (Chemguide.co.uk)
The Generalized Electrophilic Addition
Electrophilic addition is a reaction between an electrophile and nucleophile, adding to double or triple bonds. An electrophile is defined by a molecule with a tendency to react with other molecules containing a donatable pair of electrons. Thus, it is an "electron lover." A nucleophile is one that possesses a lone pair of electrons that can be easily shared. In essence, all nucleophiles are Lewis bases that attack nonhydrogen atoms (Lewis acids).
Introduction
In a general electrophilic addition reaction, one of the pi bonds is removed and creates two new sigma bonds. Another electrophilic addition reaction known as halohydrination includes bromoalcohol, more commonly known as bromohydrin. There are a variety of electrophilic reactions, and therefore a variety of different products that are very useful. It simply depends on which reagents are used to determine the final product.
Step 1
In the first step there is an electrophilic addition of bromine to the cyclopentene, forming a cyclic bromonium ion or also known as an open-chained carbenium ion (Troll, T). For more information about the cyclic Bromonium Ion please look at Electrophilic Addition of Halogens to Alkenes.
Step 2
Now the nucleophilic water molecule attacks the back of the more substituted carbon and pushes the bromonium ion onto the less substituted carbon. In general the regiochemistry of this reaction follows Markovnikov’s rule (Troll, T). The stereochemistry of the reaction is anti-addition because of better orbital overlap from backside attack, which means that the Br and the H2O are on opposite sides of the double bond. The electrophilic bromide in the product becomes linked to the less substituted carbon. The nucleophile attacks the more substituted carbon, because the carbon is more positively polarized than the other carbon.
Step 3
The negatively charged bromonium ion that was not used in the reaction is still floating freely in the water. The bromine atom attacks one of the H’s located on the water molecule. The Hydrogen drops off its electrons on the oxygen molecule making the oxygen neutral.
Final Product
Finally we are left with the trans-2-Bromocyclopentanol and hydrobromination. Here's the entire mechanism in gumdrop form. (Orange = C, Yellow = H, Red = Br, White = O)
Problems
Just for some extra practice try and answer the following questions
1. What is the product for the following reaction
2. What is the product for the following reaction
3. Write the products (hint what acts as an electrophile and what acts as the nucleophile)
4. What is the stereochemistry and the Regiochemistry of Bromoalcohol?
5. In Markovnikov Addition the electrophile attacks the more substituted carbon and the nucleophile attacks the less substituted carbon?
A. True
B. False
Answers
1.
2.
3.
4. Stereochemistry: Anti-Addition & Regiochemistry: Markovnikov
5.B. False
Contributors
• Simarjit Batth (UCD) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Electrophilic_Addition_Reactions/Reactions_of_Alkenes_with_Sulfuric_Acid.txt |
Electrophilic addition happens in many of the reactions of compounds containing carbon-carbon double bonds - the alkenes.
The structure of ethene
We are going to start by looking at ethene, because it is the simplest molecule containing a carbon-carbon double bond. What is true of C=C in ethene will be equally true of C=C in more complicated alkenes.
Ethene, C2H4, is often modeled as shown on the right. The double bond between the carbon atoms is, of course, two pairs of shared electrons. What the diagram doesn't show is that the two pairs aren't the same as each other. One of the pairs of electrons is held on the line between the two carbon nuclei as you would expect, but the other is held in a molecular orbital above and below the plane of the molecule. A molecular orbital is a region of space within the molecule where there is a high probability of finding a particular pair of electrons.
In this diagram, the line between the two carbon atoms represents a normal bond - the pair of shared electrons lies in a molecular orbital on the line between the two nuclei where you would expect them to be. This sort of bond is called a sigma bond.
The other pair of electrons is found somewhere in the shaded part above and below the plane of the molecule. This bond is called a pi bond. The electrons in the pi bond are free to move around anywhere in this shaded region and can move freely from one half to the other. The pi electrons are not as fully under the control of the carbon nuclei as the electrons in the sigma bond and, because they lie exposed above and below the rest of the molecule, they are relatively open to attack by other things.
Electrophiles
An electrophile is something which is attracted to electron-rich regions in other molecules or ions. Because it is attracted to a negative region, an electrophile must be something which carries either a full positive charge, or has a slight positive charge on it somewhere.
Ethene and the other alkenes are attacked by electrophiles. The electrophile is normally the slightly positive (+) end of a molecule like hydrogen bromide, HBr.
Electrophiles are strongly attracted to the exposed electrons in the pi bond and reactions happen because of that initial attraction - as you will see shortly.
You might wonder why fully positive ions like sodium, Na+, don't react with ethene. Although these ions may well be attracted to the pi bond, there is no possibility of the process going any further to form bonds between sodium and carbon, because sodium forms ionic bonds, whereas carbon normally forms covalent ones.
Addition reactions
In a sense, the pi bond is an unnecessary bond. The structure would hold together perfectly well with a single bond rather than a double bond. The pi bond often breaks and the electrons in it are used to join other atoms (or groups of atoms) onto the ethene molecule. In other words, ethene undergoes addition reactions.
For example, using a general molecule X-Y . . .
Summary: electrophilic addition reactions
An addition reaction is a reaction in which two molecules join together to make a bigger one. Nothing is lost in the process. All the atoms in the original molecules are found in the bigger one.
An electrophilic addition reaction is an addition reaction which happens because what we think of as the "important" molecule is attacked by an electrophile. The "important" molecule has a region of high electron density which is attacked by something carrying some degree of positive charge.
Understanding the electrophilic addition mechanism
The mechanism for the reaction between ethene and a molecule X-Y.
It is very unlikely that any two different atoms joined together will have the same electronegativity. We are going to assume that Y is more electronegative than X, so that the pair of electrons is pulled slightly towards the Y end of the bond. That means that the X atom carries a slight positive charge.
The slightly positive X atom is an electrophile and is attracted to the exposed pi bond in the ethene. Now imagine what happens as they approach each other.
You are now much more likely to find the electrons in the half of the pi bond nearest the XY. As the process continues, the two electrons in the pi bond move even further towards the X until a covalent bond is made. The electrons in the X-Y bond are pushed entirely onto the Y to give a negative Y- ion.
Important term
An ion in which the positive charge is carried on a carbon atom is called a carbocation or a carbonium ion (an older term).
In the final stage of the reaction the electrons in the lone pair on the Y- ion are strongly attracted towards the positive carbon atom. They move towards it and form a co-ordinate (dative covalent) bond between the Y and the carbon.
How to write this mechanism
The movements of the various electron pairs are shown using curly arrows.
Don't leave this page until you are sure that you understand how this relates to the electron pair movements drawn in the previous diagrams.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Electrophilic_Addition_Reactions/What_is_Electrophilic_Addition%3F.txt |
Addition reactions of carbonyl compounds such as ethanal and propanone.
Nucleophilic Addition Reactions
This page gives you the facts and simple, uncluttered mechanisms for the nucleophilic addition reactions between carbonyl compounds (specifically aldehydes and ketones) and hydrogen cyanide, HCN.
The reaction of aldehydes and ketones with hydrogen cyanide
Hydrogen cyanide adds across the carbon-oxygen double bond in aldehydes and ketones to produce compounds known as hydroxynitriles. For example, with ethanal (an aldehyde) you get 2-hydroxypropanenitrile:
With propanone (a ketone) you get 2-hydroxy-2-methylpropanenitrile:
The reaction isn't normally done using hydrogen cyanide itself, because this is an extremely poisonous gas. Instead, the aldehyde or ketone is mixed with a solution of sodium or potassium cyanide in water to which a little sulphuric acid has been added. The pH of the solution is adjusted to about 4 - 5, because this gives the fastest reaction. The solution will contain hydrogen cyanide (from the reaction between the sodium or potassium cyanide and the sulphuric acid), but still contains some free cyanide ions. This is important for the mechanism.
The mechanisms
These are examples of nucleophilic addition. The carbon-oxygen double bond is highly polar, and the slightly positive carbon atom is attacked by the cyanide ion acting as a nucleophile.
The mechanism for the addition of HCN to propanone
In the first stage, there is a nucleophilic attack by the cyanide ion on the slightly positive carbon atom.
The negative ion formed then picks up a hydrogen ion from somewhere - for example, from a hydrogen cyanide molecule.
The hydrogen ion could also come from the water or the H3O+ ions present in the slightly acidic solution. You don't need to remember all of these. One equation is perfectly adequate.
The mechanism for the addition of HCN to ethanal
As before, the reaction starts with a nucleophilic attack by the cyanide ion on the slightly positive carbon atom.
It is completed by the addition of a hydrogen ion from, for example, a hydrogen cyanide molecule.
Optical isomerism in 2-hydroxypropanenitrile
When 2-hydroxypropanenitrile is made in this last mechanism, it occurs as a racemic mixture - a 50/50 mixture of two optical isomers. It is possible that you might be exected to explain how this arises. Optical isomerism occurs in compounds which have four different groups attached to a single carbon atom. In this case, the product molecule contains a CH3, a CN, an H and an OH all attached to the central carbon atom.
The reason for the formation of equal amounts of two isomers lies in the way the ethanal gets attacked. Ethanal is a planar molecule, and attack by a cyanide ion will either be from above the plane of the molecule, or from below. There is an equal chance of either happening.
Attack from one side will lead to one of the two isomers, and attack from the other side will lead to the other.
All aldehydes will form a racemic mixture in this way. Unsymmetrical ketones will as well. (A ketone can be unsymmetrical in the sense that there is a different alkyl group either side of the carbonyl group.) What matters is that the product molecule must have four different groups attached to the carbon which was originally part of the carbon-oxygen double bond. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Nucleophilic_Addition_Reactions/The_Addition_of_Hydrogen_Cyanide_to_Aldehydes_and_Ketones.txt |
This page gives you the facts and mechanisms for the reduction of carbonyl compounds (specifically aldehydes and ketones) using sodium tetrahydridoborate (sodium borohydride) as the reducing agent.
The reduction of aldehydes and ketones by sodium tetrahydridoborate
Sodium tetrahydridoborate (previously known as sodium borohydride) has the formula NaBH4, and contains the BH4- ion. That ion acts as the reducing agent.
There are several quite different ways of carrying out this reaction. Two possible variants (there are several others!) are:
• The reaction is carried out in solution in water to which some sodium hydroxide has been added to make it alkaline. The reaction produces an intermediate which is converted into the final product by addition of a dilute acid like sulphuric acid.
• The reaction is carried out in solution in an alcohol like methanol, ethanol or propan-2-ol. This produces an intermediate which can be converted into the final product by boiling it with water.
In each case, reduction essentially involves the addition of a hydrogen atom to each end of the carbon-oxygen double bond to form an alcohol. Reduction of aldehydes and ketones lead to two different sorts of alcohol.
The reduction of an aldehyde
For example, with ethanal you get ethanol:
Notice that this is a simplified equation - perfectly acceptable to examiners. The H in square brackets means "hydrogen from a reducing agent". In general terms, reduction of an aldehyde leads to a primary alcohol. A primary alcohol is one which only has one alkyl group attached to the carbon with the -OH group on it. They all contain the grouping -CH2OH.
The reduction of a ketone
For example, with propanone you get propan-2-ol:
Reduction of a ketone leads to a secondary alcohol. A secondary alcohol is one which has two alkyl groups attached to the carbon with the -OH group on it. They all contain the grouping -CHOH.
The simplified mechanisms
The BH4- ion is essentially a source of hydride ions, H-. The simplification used is to write H- instead of BH4-. Doing this not only makes the initial attack easier to write, but avoids you getting involved with some quite complicated boron compounds that are formed as intermediates.
The reduction is an example of nucleophilic addition. The carbon-oxygen double bond is highly polar, and the slightly positive carbon atom is attacked by the hydride ion acting as a nucleophile. A hydride ion is a hydrogen atom with an extra electron - hence the lone pair.
The mechanism for the reduction of ethanal
In the first stage, there is a nucleophilic attack by the hydride ion on the slightly positive carbon atom. The lone pair of electrons on the hydride ion forms a bond with the carbon, and the electrons in one of the carbon-oxygen bonds are repelled entirely onto the oxygen, giving it a negative charge.
What happens now depends on whether you add an acid or water to complete the reaction.
Adding an acid:
When the acid is added, the negative ion formed picks up a hydrogen ion to give an alcohol.
Adding water:
This time, the negative ion takes a hydrogen ion from a water molecule.
The mechanism for the reduction of propanone
As before, the reaction starts with a nucleophilic attack by the hydride ion on the slightly positive carbon atom.
Again, what happens next depends on whether you add an acid or water to complete the reaction.
Adding an acid:
The negative ion reacts with a hydrogen ion from the acid added in the second stage of the reaction.
Adding water:
This time, the negative ion takes a hydrogen ion from a water molecule.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Nucleophilic_Addition_Reactions/The_Reduction_of_Aldehydes_and_Ketones.txt |
• E1 Reactions
Unimolecular Elimination (E1) is a reaction in which the removal of an HX substituent results in the formation of a double bond. It is similar to a unimolecular nucleophilic substitution reaction (SN1) in various ways. One being the formation of a carbocation intermediate. Also, the only rate determining (slow) step is the dissociation of the leaving group to form a carbocation, hence the name unimolecular. Thus, since these two reactions behave similarly, they compete against each other.
• E2 Reactions
E2, bimolecular elimination, was proposed in the 1920s by British chemist Christopher Kelk Ingold. Unlike E1 reactions, E2 reactions remove two subsituents with the addition of a strong base, resulting in an alkene.
• Elimination Reactions
Elimination Reactions
Unimolecular Elimination (E1) is a reaction in which the removal of an HX substituent results in the formation of a double bond. It is similar to a unimolecular nucleophilic substitution reaction (SN1) in various ways. One being the formation of a carbocation intermediate. Also, the only rate determining (slow) step is the dissociation of the leaving group to form a carbocation, hence the name unimolecular. Thus, since these two reactions behave similarly, they compete against each other. Many times, both these reactions will occur simultaneously to form different products from a single reaction. However, one can be favored over another through thermodynamic control. Although Elimination entails two types of reactions, E1 and E2, we will focus mainly on E1 reactions with some reference to E2.
General Reaction
An E1 reaction involves the deprotonation of a hydrogen nearby (usually one carbon away, or the beta position) the carbocation resulting in the formation of an alkene product. In order to accomplish this, a Lewis base is required. For a simplified model, we’ll take B to be a Lewis base, and LG to be a halogen leaving group.
As can be seen above, the preliminary step is the leaving group (LG) leaving on its own. Because it takes the electrons in the bond along with it, the carbon that was attached to it loses its electron, making it a carbocation. Once it becomes a carbocation, a Lewis Base (\(B^-\)) deprotonates the intermediate carbocation at the beta position, which then donates its electrons to the neighboring C-C bond, forming a double bond. Unlike E2 reactions, which require the proton to be anti to the leaving group, E1 reactions only require a neighboring hydrogen. This is due to the fact that the leaving group has already left the molecule. The final product is an alkene along with the HB byproduct.
Reactivity
Due to the fact that E1 reactions create a carbocation intermediate, rules present in \(S_N1\) reactions still apply.
As expected, tertiary carbocations are favored over secondary, primary and methyl’s. This is due to the phenomena of hyperconjugation, which essentially allows a nearby C-C or C-H bond to interact with the p orbital of the carbon to bring the electrons down to a lower energy state. Thus, this has a stabilizing effect on the molecule as a whole. In general, primary and methyl carbocations do not proceed through the E1 pathway for this reason, unless there is a means of carbocation rearrangement to move the positive charge to a nearby carbon. Secondary and Tertiary carbons form more stable carbocations, thus this formation occurs quite rapidly.
Secondary carbocations can be subject to the E2 reaction pathway, but this generally occurs in the presence of a good / strong base. Adding a weak base to the reaction disfavors E2, essentially pushing towards the E1 pathway. In many instances, solvolysis occurs rather than using a base to deprotonate. This means heat is added to the solution, and the solvent itself deprotonates a hydrogen. The medium can effect the pathway of the reaction as well. Polar protic solvents may be used to hinder nucleophiles, thus disfavoring E2 / Sn2 from occurring.
Acid catalyzed dehydration of secondary / tertiary alcohols
We’ll take a look at a mechanism involving solvolysis during an E1 reaction of Propanol in Sulfuric Acid.
• Step 1: The OH group on the pentanol is hydrated by H2SO4. This allows the OH to become an H2O, which is a better leaving group.
• Step 2: Once the OH has been hydrated, the H2O molecule leaves, taking its electrons with it. This creates a carbocation intermediate on the attached carbon.
• Step 3: Another H2O molecule comes in to deprotonate the beta carbon, which then donates its electrons to the neighboring C-C bond. The carbons are rehybridized from sp3 to sp2, and thus a pi bond is formed between them.
Mechanism for Alkyl Halides
This mechanism is a common application of E1 reactions in the synthesis of an alkene.
Once again, we see the basic 2 steps of the E1 mechanism.
1. The leaving group leaves along with its electrons to form a carbocation intermediate.
2. A base deprotonates a beta carbon to form a pi bond.
In this case we see a mixture of products rather than one discrete one. This is the case because the carbocation has two nearby carbons that are capable of being deprotonated, but that only one forms a major product (more stable).
How are Regiochemistry & Stereochemistry involved?
In terms of regiochemistry, Zaitsev's rule states that although more than one product can be formed during alkene synthesis, the more substituted alkene is the major product. This infers that the hydrogen on the most substituted carbon is the most probable to be deprotonated, thus allowing for the most substituted alkene to be formed.
Unlike E2 reactions, E1 is not stereospecific. Thus, a hydrogen is not required to be anti-periplanar to the leaving group.
In this mechanism, we can see two possible pathways for the reaction. One in which the methyl on the right is deprotonated, and another in which the CH2 on the left is deprotonated. Either one leads to a plausible resultant product, however, only one forms a major product. As stated by Zaitsev's rule, deprotonation of the most substituted carbon results in the most substituted alkene. This then becomes the most stable product due to hyperconjugation, and is also more common than the minor product.
Outside Sources
1. Vollhardt, K. Peter C., and Neil E. Schore. Organic Chemistry Structure and Function. New York: W. H. Freeman, 2007.
2. McMurry, J., Simanek, E. Fundamentals of Organic Chemistry, 6th edition. Cengage Learning, 2007.
Problems
1) Which of these steps is the rate determining step (A or B)?
What is the major product formed (C or D)?
2) In order to produce the most stable alkene product, from which carbon should the base deprotonate (A, B, or C)?
If the carbocation were to rearrange, on which carbon would the positive charge go onto without sacrificing stability (A, B, or C)?
3) Predict the major product of the following reaction.
4) (True or False) – There is no way of controlling the product ratio of E1 / Sn1 reactions.
5) Explain why the presence of a weak base / nucleophile favors E1 reactions over E2.
Answers
1. A , C
2. B, B
3.
4. False - They can be thermodynamically controlled to favor a certain product over another.
5. By definition, an E1 reaction is a Unimolecular Elimination reaction. This means the only rate determining step is that of the dissociation of the leaving group to form a carbocation. Since E2 is bimolecular and the nucleophilic attack is part of the rate determining step, a weak base/nucleophile disfavors it and ultimately allows E1 to dominate. (Don't forget about Sn1 which still pertains to this reaction simultaneously).
Contributors
• Satish Balasubramanian
E1 Reactions
Carbocation rearrangements are extremely common in organic chemistry reactions are are defined as the movement of a carbocation from an unstable state to a more stable state through the use of various structural reorganizational "shifts" within the molecule. Once the carbocation has shifted over to a different carbon, we can say that there is a structural isomer of the initial molecule. However, this phenomenon is not as simple as it sounds.
Introduction
Whenever alcohols are subject to transformation into various carbocations, the carbocations are subject to a phenomenon known as carbocation rearrangement. A carbocation, in brief, holds the positive charge in the molecule that is attached to three other groups and bears a sextet rather than an octet. However, we do see carbocation rearrangements in reactions that do not contain alcohol as well. Those, on the other hand, require more difficult explanations than the two listed below. There are two types of rearrangements: hydride shift and alkyl shift. These rearrangements usualy occur in many types of carbocations. Once rearranged, the molecules can also undergo further unimolecular substitution (SN1) or unimolecular elimination (E1). Though, most of the time we see either a simple or complex mixture of products. We can expect two products before undergoing carbocation rearrangement, but once undergoing this phenomenon, we see the major product.
Hydride Shift
Whenever a nucleophile attacks some molecules, we typically see two products. However, in most cases, we normally see both a major product and a minor product. The major product is typically the rearranged product that is more substituted (aka more stable). The minor product, in contract, is typically the normal product that is less substituted (aka less stable).
The reaction: We see that the formed carbocations can undergo rearrangements called hydride shift. This means that the two electron hydrogen from the unimolecular substitution moves over to the neighboring carbon. We see the phenomenon of hydride shift typically with the reaction of an alcohol and hydrogen halides, which include HBr, HCl, and HI. HF is typically not used because of its instability and its fast reactivity rate. Below is an example of a reaction between an alcohol and hydrogen chloride:
GREEN (Cl) = nucleophile BLUE (OH) = leaving group ORANGE (H) = hydride shift proton RED(H) = remaining proton
The alcohol portion (-OH) has been substituted with the nucleophilic Cl atom. However, it is not a direct substitution of the OH atom as seen in SN2 reactions. In this SN1 reaction, we see that the leaving group, -OH, forms a carbocation on Carbon #3 after receiving a proton from the nucleophile to produce an alkyloxonium ion. Before the Cl atom attacks, the hydrogen atom attached to the Carbon atom directly adjacent to the original Carbon (preferably the more stable Carbon), Carbon #2, can undergo hydride shift. The hydrogen and the carbocation formally switch positions. The Cl atom can now attack the carbocation, in which it forms the more stable structure because of hyperconjugation. The carbocation, in this case, is most stable because it attaches to the tertiary carbon (being attached to 3 different carbons). However, we can still see small amounts of the minor, unstable product. The mechanism for hydride shift occurs in multiple steps that includes various intermediates and transition states. Below is the mechanism for the given reaction above:
Hydration of Alkenes: Hydride Shift
In a more complex case, when alkenes undergo hydration, we also observe hydride shift. Below is the reaction of 3-methyl-1-butene with H3O+ that furnishes to make 2-methyl-2-butanol:
Once again, we see multiple products. In this case, however, we see two minor products and one major product. We observe the major product because the -OH substitutent is attached to the more substituted carbon. When the reactant undergoes hydration, the proton attaches to carbon #2. The carbocation is therefore on carbon #2. Hydride shift now occurs when the hydrogen on the adjacent carbon formally switch places with the carbocation. The carbocation is now ready to be attacked by H2O to furnish an alkyloxonium ion because of stability and hyperconjugation. The final step can be observed by another water molecule attacking the proton on the alkyloxonium ion to furnish an alcohol. We see this mechanism below:
Alkyl Shift
Not all carbocations have suitable hydrogen atoms (either secondary or tertiary) that are on adjacent carbon atoms available for rearrangement. In this case, the reaction can undergo a different mode of rearrangement known as alkyl shift (or alkyl group migration). Alkyl Shift acts very similarily to that of hydride shift. Instead of the proton (H) that shifts with the nucleophile, we see an alkyl group that shifts with the nucleophile instead. The shifting group carries its electron pair with it to furnish a bond to the neighboring or adjacent carbocation. The shifted alkyl group and the positive charge of the carbocation switch positions on the moleculeReactions of tertiary carbocations react much faster than that of secondary carbocations. We see alkyl shift from a secondary carbocation to tertiary carbocation in SN1 reactions:
We observe slight variations and differences between the two reactions. In reaction #1, we see that we have a secondary substrate. This undergoes alkyl shift because it does not have a suitable hydrogen on the adjacent carbon. Once again, the reaction is similar to hydride shift. The only difference is that we shift an alkyl group rather than shift a proton, while still undergoing various intermediate steps to furnish its final product.
With reaction #2, on the other hand, we can say that it undergoes a concerted mechanism. In short, this means that everything happens in one step. This is because primary carbocations cannot be an intermediate and they are relatively difficult processes since they require higher temperatures and longer reaction times. After protonating the alcohol substrate to form the alkyloxonium ion, the water must leave at the same time as the alkyl group shifts from the adjacent carbon to skip the formation of the unstable primary carbocation.
Carbocation Rearrangements for E1 Reactions
E1 reactions are also affected by alkyl shift. Once again, we can see both minor and major products. However, we see that the more substituted carbons undergo the effects of E1 reactions and furnish a double bond. See practice problem #4 below for an example as the properties and effects of carbocation rearrangements in E1 reactions are similar to that of alkyl shifts.
1,3-Hydride and Greater Shifts
Typically, hydride shifts can occur at low temperatures. However, by heating the solutionf of a cation, it can easily and readily speed the process of rearrangement. One way to account for a slight barrier is to propose a 1,3-hydride shift interchanging the functionality of two different kinds of methyls. Another possibility is 1,2 hydride shift in which you could yield a secondary carbocation intermediate. Then, a further 1,2 hydride shift would give the more stable rearranged tertiary cation.
More distant hydride shifts have been observed, such as 1,4 and 1,5 hydride shifts, but these arrangements are too fast to undergo secondary cation intermediates.
Analogy
Carbocation rearrangements happen very readily and often occur in many organic chemistry reactions. Yet, we typically neglect this step. Dr. Sarah Lievens, a Chemistry professor at the University of California, Davis once said carbocation rearrangements can be observed with various analogies to help her students remember this phenomenon. For hydride shifts: "The new friend (nucleophile) just joined a group (the organic molecule). Because he is new, he only made two new friends. However, the popular kid (the hydrogen) glady gave up his friends to the new friend so that he could have even more friends. Therefore, everyone won't be as lonely and we can all be friends." This analogy works for alkyl shifts in conjunction with hydride shift as well.
• Jeffrey Ma
Electrophilic Alkene Addition Mechanism
Mechanism Overview
The electrophilic Alkene Adddition reaction involves two sequential steps. The red arrows are curved or curly arrows showing electron movement.
Step 1 (slow)
The alkene pi (π) electrons react with a source of electron deficient H. The most stable carbocation is formed. Occasionally the first-formed carbocation undergoes a rearrangement to a more stable carbocation. Formation of the carbocation is the rate limiting (slowest) step in the mechanism.
Step 2 (fast)
The electron deficient carbocation reacts with an electron rich X- ion.
Examples
Reactant Reagent(s) and conditions
(yield: %)
Product
(fraction formed: %)
Comments.
HBr
acetic acid
(90%)
Regioselective: Markovnikov
Note the bromine attached to the C-2 position from the more stable secondary carbocation
HBr
acetic acid
(90%)
Regioselective: Markovnikov
Note the bromine attached to the C-2 position from the more stable tertiary carbocation
HCl
(100%)
Regioselective: Markovnikov
Note the chlorine attached to the C-1 position from the more stable tertiary carbocation
KI
H3PO4, 80ºC
(90%)
HCl
(40%) + (60%) Rearrangement occurs
(hydride migration)
+
The first-formed carbocation is secondary, some of this carbocation reacts with Cl- but a hydride migration converts this into a more stable tertiary carbocation from which the majority of the product is derived.
HCl
(17%)
(83%)
Rearrangement occurs
(methyd [CH3 with 2 electrons] migration)
+
The first-formed carbocation is secondary, a methide migration converts this into the more stable tertiary carbocation.
A small amount of the secondary carbocation reacts with Cl- before rearranging, but the majority rearranges before reacting.
Wagner-Meerwein Rearrangement
A Wagner-Meerwein rearrangement is any reaction in which the carbon skeleton of a reactant changes due to one or more rearrangements involving carbocations, e.g.:
mechanism: | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Elimination_Reactions/E1_Reactions/Carbocation_Rearrangements.txt |
E2 reactions are typically seen with secondary and tertiary alkyl halides, but a hindered base is necessary with a primary halide. The mechanism by which it occurs is a single step concerted reaction with one transition state. The rate at which this mechanism occurs is second order kinetics, and depends on both the base and alkyl halide. A good leaving group is required because it is involved in the rate determining step. The leaving groups must be coplanar in order to form a pi bond; carbons go from sp3 to sp2 hybridization states.
General Reaction
In this reaction Ba represents the base and X represents a leaving group, typically a halogen. There is one transition state that shows the concerted reaction for the base attracting the hydrogen and the halogen taking the electrons from the bond. The product be both eclipse and staggered depending on the transition states. Eclipsed products have a synperiplanar transition states, while staggered products have antiperiplanar transition states. Staggered conformation is usually the major product because of its lower energy confirmation.
An E2 reaction has certain requirements to proceed:
• Secondary and tertiary alkyl halides will proceed with E2 in the presence of a base (OH-, RO-, R2N-)
• Both leaving groups should be on the same plane, this allows the double bond to form in the reaction. In the reaction above you can see both leaving groups are in the plane of the carbons.
• Follows Zaitsev's rule, the most substituted alkene is usually the major product.
• Hoffman Rule, if a sterically hindered base will result in the least substituted product.
Reaction Coordinate
Problems
1.
2. What is the major product and why?
3. What is the major prodcut when 2-bromo-2-methylbutane reacts with with sodium ethoxide?
4.
5.
A. Elimination from 2-Bromopropane
The formation of alkenes from halogenoalkanes such as 2-bromopropane, and from the dehydration of alcohols.
Elimination Reactions
This page gives you the facts and a simple, uncluttered mechanism for the elimination reaction between a simple halogenoalkane like 2-bromopropane and hydroxide ions (from, for example, sodium hydroxide) to give an alkene like propene.
The elimination reaction involving 2-bromopropane and hydroxide ions
2-bromopropane is heated under reflux with a concentrated solution of sodium or potassium hydroxide in ethanol. Heating under reflux involves heating with a condenser placed vertically in the flask to avoid loss of volatile liquids. Propene is formed and, because this is a gas, it passes through the condenser and can be collected.
Everything else present (including anything formed in the alternative substitution reaction) will be trapped in the flask.
The mechanism
In elimination reactions, the hydroxide ion acts as a base - removing a hydrogen as a hydrogen ion from the carbon atom next door to the one holding the bromine. The resulting re-arrangement of the electrons expels the bromine as a bromide ion and produces propene.
B. Elimination from Unsymmetrical Halogenoalkanes
This page looks at elimination from unsymmetric halogenoalkanes such as 2-bromobutane.
2-bromobutane is an unsymmetric halogenoalkane in the sense that it has a CH3 group one side of the C-Br bond and a CH2CH3 group the other.
The basic facts and mechanisms for these reactions are exactly the same as with simple halogenoalkanes like 2-bromopropane. This page only deals with the extra problems created by the possibility of more than one elimination product.
Background to the mechanism
You will remember that elimination happens when a hydroxide ion (from, for example, sodium hydroxide) acts as a base and removes a hydrogen as a hydrogen ion from the halogenoalkane.
For example, in the simple case of elimination from 2-bromopropane:
The hydroxide ion removes a hydrogen from one of the carbon atoms next door to the carbon-bromine bond, and the various electron shifts then lead to the formation of the alkene - in this case, propene.
With an unsymmetric halogenoalkane like 2-bromobutane, there are several hydrogens which might possibly get removed. You need to think about each of these possibilities.
Where does the hydrogen get removed from?
The hydrogen has to be removed from a carbon atom adjacent to the carbon-bromine bond. If an OH- ion hit one of the hydrogens on the right-hand CH3 group in the 2-bromobutane (as we've drawn it), there's nowhere for the reaction to go.
To make room for the electron pair to form a double bond between the carbons, you would have to expel a hydrogen from the CH2 group as a hydride ion, H-. That is energetically much too difficult, and so this reaction doesn't happen.
That still leaves the possibility of removing a hydrogen either from the left-hand CH3 or from the CH2 group.
If it was removed from the CH3 group:
The product is but-1-ene, CH2=CHCH2CH3.
If it was removed from the CH2 group:
This time the product is but-2-ene, CH3CH=CHCH3.
In fact the situation is even more complicated than it looks, because but-2-ene exhibits geometric isomerism. You get a mixture of two isomers formed - cis-but-2-ene and trans-but-2-ene.
Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an explanation of the two ways of naming these two compounds, follow the link in the box below.
Which isomer gets formed is just a matter of chance.
The overall result
Elimination from 2-bromobutane leads to a mixture containing:
• but-1-ene
• cis-but-2-ene (also known as (Z)-but-2-ene)
• trans-but-2-ene (also known as (E)-but-2-ene) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Elimination_Reactions/E2_Reactions.txt |
This page discusses the factors that decide whether halogenoalkanes undergo elimination reactions or nucleophilic substitution when they react with hydroxide ions from, say, sodium hydroxide or potassium hydroxide.
The reactions
Both reactions involve heating the halogenoalkane under reflux with sodium or potassium hydroxide solution.
Nucleophilic substitution
The hydroxide ions present are good nucleophiles, and one possibility is a replacement of the halogen atom by an -OH group to give an alcohol via a nucleophilic substitution reaction.
In the example, 2-bromopropane is converted into propan-2-ol.
Elimination
Halogenoalkanes also undergo elimination reactions in the presence of sodium or potassium hydroxide.
The 2-bromopropane has reacted to give an alkene - propene.
Notice that a hydrogen atom has been removed from one of the end carbon atoms together with the bromine from the centre one. In all simple elimination reactions the things being removed are on adjacent carbon atoms, and a double bond is set up between those carbons.
What decides whether you get substitution or elimination?
The reagents you are using are the same for both substitution or elimination - the halogenoalkane and either sodium or potassium hydroxide solution. In all cases, you will get a mixture of both reactions happening - some substitution and some elimination. What you get most of depends on a number of factors.
The type of halogenoalkane
This is the most important factor.
type of halogenoalkane substitution or elimination?
primary mainly substitution
secondary both substitution and elimination
tertiary mainly elimination
For example, whatever you do with tertiary halogenoalkanes, you will tend to get mainly the elimination reaction, whereas with primary ones you will tend to get mainly substitution. However, you can influence things to some extent by changing the conditions.
The solvent
The proportion of water to ethanol in the solvent matters.
• Water encourages substitution.
• Ethanol encourages elimination.
The temperature
Higher temperatures encourage elimination.
Concentration of the sodium or potassium hydroxide solution
Higher concentrations favor elimination.
In summary
For a given halogenoalkane, to favour elimination rather than substitution, use:
• heat
• a concentrated solution of sodium or potassium hydroxide
• pure ethanol as the solvent
The role of the hydroxide ions
The role of the hydroxide ion in a substitution reaction
In the substitution reaction between a halogenoalkane and OH- ions, the hydroxide ions are acting as nucleophiles. For example, one of the lone pairs on the oxygen can attack the slightly positive carbon. This leads on to the loss of the bromine as a bromide ion, and the -OH group becoming attached in its place.
The role of the hydroxide ion in an elimination reaction
Hydroxide ions have a very strong tendency to combine with hydrogen ions to make water - in other words, the OH- ion is a very strong base. In an elimination reaction, the hydroxide ion hits one of the hydrogen atoms in the CH3 group and pulls it off. This leads to a cascade of electron pair movements resulting in the formation of a carbon-carbon double bond, and the loss of the bromine as Br-.
Contributors
Jim Clark (Chemguide.co.uk)
D. The Dehydration of Propan-2-ol
This page looks at the mechanism for the acid catalysed dehydration of propan-2-ol.
The dehydration of propan-2-ol is taken as an simple example of the way that secondary and tertiary alcohols dehydrate.
Primary alcohols like ethanol use a different mechanism, and ethanol is discussed separately on another page. You will find a link to this from the elimination mechanisms menu.
You will also find a link there to a page on the dehydration of more complicated alcohols where more than one product may be formed.
The dehydration of propan-2-ol
The facts
Propan-2-ol can be dehydrated to give propene by heating it with an excess of concentrated sulphuric acid at about 170°C.
Concentrated phosphoric(V) acid, H3PO4, can be used instead.
The acids aren't written into the equation because they serve as catalysts. If you like, you could write, for example, "conc H2SO4" over the top of the arrow.
Notice that the -OH group is lost, together with a hydrogen from a next-door carbon - it doesn't matter which one. If you chose the other one, you would get CH3CH=CH2. That's the same molecule flipped over.
The mechanism - the full version
We are going to discuss the mechanism using sulphuric acid. Afterwards, we'll describe how you can use a simplified version which will work for any acid, including phosphoric(V) acid.
In the first stage, one of the lone pairs of electrons on the oxygen picks up a hydrogen ion from the sulphuric acid. The alcohol is said to be protonated.
The negative ion produced is the hydrogensulphate ion, HSO4-.
Notice that the oxygen atom in the alcohol has gained a positive charge. That charge has to be there for two reasons:
• On the left hand side of the equation you start with two overall neutral molecules. Assuming you forgot about the positive charge, you would end up with a neutral species and a negative ion on the right. Charges must balance in equations, so something is wrong.
• The oxygen looks wrong! The oxygen atom is joined to 3 things rather than its usual 2. Oxygen can only join to 3 things if it carries a positive charge.
In the second stage of the reaction the protonated propan-2-ol loses a water molecule to leave a carbocation (previously known as a carbonium ion) - an ion with a positive charge on a carbon atom.
The carbon atom is positive because it has lost the electron that it originally contributed to the carbon-oxygen bond. Both of the electrons in that bond have moved onto the oxygen atom, neutralising the oxygen's charge.
Finally, a hydrogensulphate ion (from the sulphuric acid) pulls off a hydrogen ion from the carbocation, and a double bond forms.
The mechanism - a simplified version
People normally quote a simplified version of this mechanism. Instead of showing the full structure of the sulphuric acid, you write it as if it were simply a hydrogen ion, H+. That leaves the full mechanism:
An advantage of this (apart from the fact that it doesn't require you to draw the structure of sulphuric acid) is that it can be used for any acid catalyst without changing it at all. For example, if you use this version, you wouldn't need to worry about the structure of phosphoric(V) acid. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Elimination_Reactions/Elimination_Reactions/C._Elimination_vs._Substitution.txt |
This page looks at the mechanism for the acid catalysed dehydration of a simple primary alcohol like ethanol to give an alkene like ethene.
The dehydration of ethanol
The facts
Ethanol can be dehydrated to give ethene by heating it with an excess of concentrated sulphuric acid at about 170°C.
Concentrated phosphoric(V) acid, H3PO4, can be used instead.
The acids aren't written into the equation because they serve as catalysts. If you like, you could write, for example, "conc H2SO4" over the top of the arrow.
The mechanism
A problem!
You will find two versions of the mechanism for the dehydration of primary alcohols on the web and in various textbooks.
One of these is exactly the same as the mechanism for the reaction involving propan-2-ol and other secondary or tertiary alcohols (known technically as an E1 mechanism), but the other is different (known as an E2 mechanism).
The more reliable sources give the E2 mechanism for the dehydration of primary alcohols including ethanol. I am going to treat this as the "correct" version.
The correct version in full
If you have read the page on the dehydration of propan-2-ol, you will know that it involves the formation of a carbocation (a carbonium ion).
If ethanol used the same mechanism, you would get a primary carbocation formed, CH3CH2+, but this is much less stable than a secondary or tertiary carbocation.
That would lead to a very high activation energy for the reaction.
The alternative mechanism avoids the formation of the carbocation, and so avoids the high activation energy.
We are going to discuss the mechanism using sulphuric acid.
In the first stage, one of the lone pairs of electrons on the oxygen picks up a hydrogen ion from the sulphuric acid. The alcohol is said to be protonated. That is exactly the same as happens with propan-2-ol and the other secondary and tertiary alcohols.
In the mechanism we have already looked at with propan-2-ol, the next thing to happen was loss of water to form a carbocation, followed by removal of a hydrogen ion from the carbocation and the formation of a double bond.
In this case, instead of happening in two separate steps, this all happens at the same time in one smooth operation. By doing that, you avoid the formation of an unstable primary carbocation.
No simplified version of this!
I am not giving a simplified version of this mechanism just in terms of hydrogen ions. If you don't show something removing the hydrogen ion from the protonated alcohol, you are really missing an important feature of the reaction.
The other mechanism that you might come across
This is the mechanism that looks justs like the one involving propan-2-ol, and involves the formation of an unstable primary carbocation.
This is the version that you would have found on Chemguide up to December 2012, and was originally included because it was the one wanted by one of the UK Exam Boards at the time I wrote the page in 2000.
It is possible that your examiners may still want it. As far as I am aware, the only syllabus which I track which still asks this topic is IB, and one of their mark schemes gave the version below. If you need to know about this, you need to check what your examiners are currently expecting.
F. The Dehydration of Butan-2-ol
This page builds on your understanding of the acid catalysed dehydration of alcohols.
You have to be wary with more complicated alcohols in case there is the possibility of more than one alkene being formed. Butan-2-ol is a good example of this, with no less than three different alkenes being formed when it is dehydrated.
Background
To make the diagrams less cluttered, we'll use the simplified version of the mechanism showing gain and loss of H+.
Remember that the mechanism takes place in three stages:
• The alcohol is protonated by the acid catalyst.
• The protonated alcohol loses a water molecule to give a carbocation (carbonium ion).
• The carbocation formed loses a hydrogen ion and forms a double bond.
So, in the case of the dehydration of propan-2-ol:
The dehydration of butan-2-ol
The first two stages
There is nothing new at all in these stages.
In the first stage, the alcohol is protonated by picking up a hydrogen ion from the sulphuric acid.
In the second stage, the positive ion then sheds a water molecule and produces a carbocation.
The complication arises in the next step. When the carbocation loses a hydrogen ion, where is it going to come from?
Where does the hydrogen get removed from?
So that a double bond can form, it will have to come from one of the carbons next door to the one with the positive charge.
If a hydrogen ion is lost from the CH3 group
But-1-ene is formed.
If a hydrogen ion is lost from the CH2 group
This time the product is but-2-ene, CH3CH=CHCH3.
In fact the situation is even more complicated than it looks, because but-2-ene exhibits geometric isomerism. You get a mixture of two isomers formed - cis-but-2-ene and trans-but-2-ene.
Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an explanation of the two ways of naming these two compounds, follow the link in the box below.
The overall result
Dehydration of butan-2-ol leads to a mixture containing:
• but-1-ene
• cis-but-2-ene (also known as (Z)-but-2-ene)
• trans-but-2-ene (also known as (E)-but-2-ene) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Elimination_Reactions/Elimination_Reactions/E._The_Dehydration_of_Ethanol.txt |
Free radical substitution reactions in alkanes and alkyl groups. Free radical addition during the polymerization of ethene and the reaction between HBr and alkenes in the presence of organic peroxides.
• Bromination of Methane
This page gives you the facts and a simple, uncluttered mechanism for the free radical substitution reaction between methane and bromine. This reaction between methane and bromine happens in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction - a reaction brought about by light.
• Chlorination of Methane
This page gives you the facts and a simple, uncluttered mechanism for the free radical substitution reaction between methane and chlorine.
• Methylbenzene and Chlorine
This page gives you the facts and a simple, uncluttered mechanism for the free radical substitution reaction between methylbenzene (previously known as toluene) and chlorine.
• What is Free Radical Substitution?
Substitution reactions These are reactions in which one atom in a molecule is replaced by another atom or group of atoms. Free radical substitution often involves breaking a carbon-hydrogen bond in alkanes
Free Radical Reactions
A Free Radical Substitution Reaction
This page gives you the facts and a simple, uncluttered mechanism for the free radical substitution reaction between methane and bromine. This reaction between methane and bromine happens in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction - a reaction brought about by light.
$CH_4 + Br_2 \rightarrow CH_3Br + HBr$
The organic product is bromomethane. One of the hydrogen atoms in the methane has been replaced by a bromine atom, so this is a substitution reaction. However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by bromine atoms.
The mechanism
The mechanism involves a chain reaction. During a chain reaction, for every reactive species you start off with, a new one is generated at the end - and this keeps the process going.The over-all process is known as free radical substitution, or as a free radical chain reaction.
• Chain initiation: The chain is initiated (started) by UV light breaking a bromine molecule into free radicals.
Br22Br
• Chain propagation reactions: These are the reactions which keep the chain going.
CH4 + BrCH3 + HBr
CH3 + Br2CH3Br + Br
• Chain termination reactions: These are reactions which remove free radicals from the system without replacing them by new ones.
2BrBr2
CH3 + BrCH3Br
CH3 + CH3CH3CH3
Contributors
Jim Clark (Chemguide.co.uk)
Chlorination of Methane
If a mixture of methane and chlorine is exposed to a flame, it explodes - producing carbon and hydrogen chloride. This is not a very useful reaction! The reaction we are going to explore is a more gentle one between methane and chlorine in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction - a reaction brought about by light.
$\ce{CH4 + Cl_2 -> CH_3Cl + HCl} \nonumber$
The organic product is chloromethane. One of the hydrogen atoms in the methane has been replaced by a chlorine atom, so this is a substitution reaction. However, the reaction does not stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms. Multiple substitution is dealt with on a separate page, and you will find a link to that at the bottom of this page.
The mechanism
The mechanism involves a chain reaction. During a chain reaction, for every reactive species you start off with, a new one is generated at the end - and this keeps the process going. The over-all process is known as free radical substitution, or as a free radical chain reaction.
• Chain initiation: The chain is initiated (started) by UV light breaking a chlorine molecule into free radicals.
Cl2 $\rightarrow$ 2Cl
• Chain propagation reactions : These are the reactions which keep the chain going.
CH4 + Cl$\rightarrow$CH3 + HCl
CH3 + Cl2$\rightarrow$CH3Cl + Cl
• Chain termination reactions: These are reactions which remove free radicals from the system without replacing them by new ones.
2Cl$\rightarrow$Cl2
CH3 + Cl $\rightarrow$ CH3C l
CH3 + CH3$\rightarrow$CH3CH3
Methylbenzene and Chlorine
A Free Radical Substitution Reaction
This page gives you the facts and a simple, uncluttered mechanism for the free radical substitution reaction between methylbenzene (previously known as toluene) and chlorine.
Methylbenzene has a methyl group attached to a benzene ring. The hexagon with the circle inside is the standard symbol for this ring. There is a carbon atom at each corner of the hexagon, and a hydrogen atom on each carbon apart from the one with the methyl group attached.
The facts
The reaction we are going to explore happens between methylbenzene and chlorine in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction - a reaction brought about by light.
The organic product is (chloromethyl)benzene. The brackets in the name emphasizes that the chlorine is part of the attached methyl group, and isn't on the ring.
One of the hydrogen atoms in the methyl group has been replaced by a chlorine atom, so this is a substitution reaction. However, the reaction doesn't stop there, and all three hydrogens in the methyl group can in turn be replaced by chlorine atoms. Multiple substitution is dealt with on a separate page, and you will find a link to that at the bottom of this page.
The mechanism
The mechanism involves a chain reaction. During a chain reaction, for every reactive species you start off with, a new one is generated at the end - and this keeps the process going. The over-all process is known as free radical substitution, or as a free radical chain reaction.
• Chain initiation: The chain is initiated (started) by UV light breaking a chlorine molecule into free radicals.
Cl22Cl
• Chain propagation reactions: These are the reactions which keep the chain going.
• Chain termination reactions: These are reactions which remove free radicals from the system without replacing them by new ones. If any two free radicals collide, they will join together without producing any new radicals. The simplest example of this is a collision between two chlorine radicals.
2ClCl2
What is Free Radical Substitution?
Substitution reactions
These are reactions in which one atom in a molecule is replaced by another atom or group of atoms. Free radical substitution often involves breaking a carbon-hydrogen bond in alkanes such as
methane CH4
ethane CH3CH3
propane CH3CH2CH3
A new bond is then formed to something else. It also happens in alkyl groups like methyl, ethyl (and so on) wherever these appear in more complicated molecules.
methyl CH3
ethyl CH3CH2
For example, ethanoic acid is CH3COOH and contains a methyl group. The carbon-hydrogen bonds in the methyl group behave just like those in methane, and can be broken and replaced by something else in the same way. A simple example of substitution is the reaction between methane and chlorine in the presence of UV light (or sunlight).
$CH_4 + Cl_2\rightarrow CH_3Cl + HCl$
Notice that one of the hydrogen atoms in the methane has been replaced by a chlorine atom. That's substitution.
Free radical reactions
Free radicals are atoms or groups of atoms which have a single unpaired electron. A free radical substitution reaction is one involving these radicals. Free radicals are formed if a bond splits evenly - each atom getting one of the two electrons. The name given to this is homolytic fission.
To show that a species (either an atom or a group of atoms) is a free radical, the symbol is written with a dot attached to show the unpaired electron. For example:
a chlorine radical Cl
a methyl radical CH3
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Free_Radical_Reactions/Bromination_of_Methane.txt |
Bioconjugation is a chemical technique used to couple two molecules together, at least one of which is a biomolecule, such as a carbohydrate, nucleic acid, or protein. Proteins are especially diverse biomolecules due to the variety of amino acids available and are thus important substrates in bioconjugation reactions. Bioconjugation reactions play a critical role in the modification of proteins. Because of recent advances in the study of biomolecules, proteins can be modified to perform a variety of functions, including cellular tracking, imaging biomarkers, and target drug delivery.
Synthesis and Strategies
Though the chemistry behind conjugation is simple, its execution is not. There are several obstacles that can hinder efficient bioconjugation reactions. For example, because certain amino acid residues are more prevalent than others, some reactions cannot be selective and are therefore inefficient. Another obstacle is polar molecules on the surface of proteins that can interfere with a reaction. In addition, selectively modifying more than one site on a protein presents considerable challenges.
Classical approaches to protein modification are typically second-order reactions that target side chains of specific amino acids such as cysteine and lysine. Cysteine and lysine side chains contain thiol and amino groups, respectively, which allow them to undergo modification with a wide variety of reagents. The table below summarizes common reactions involving these amino acids.
However, with the advent of new technologies and advancements in biochemistry, strategies have been developed to make bioconjugation reactions more efficient.
Choosing a strategy is largely dependent on the protein of interest. If the protein is present in a mixture and cannot be isolated, alternative techniques will have to be used. If the protein is present in its purified form, the next question to consider is whether site specificity is needed. From these initial criteria, a flowchart, developed by Nicholas Stephanopolous and Matthew Francis, can be used to determine efficient synthesis techniques on a case-by-case basis.
Some common bioconjugation techniques introduced in this flowchart are discussed below.
Bioorthogonal Reactions
One subset of bioconjugation reactions which has become increasingly important is bioorthogonal reactions, which are reactions in living systems that do not interfere with native processes. They provide a mechanism for targeting and modifying a specific site on a protein. Applications include live cell surface labeling, enhanced fluorescence, and photochemical switching behavior in proteins.
Ketone and aldehyde modification reactions are a common example of bioorthogonal reactions. These functional groups are effective because they are not present on cell surfaces and can thus be uniquely identified when attached. In this scenario, a ketone or aldehyde functional group on a biomolecule is coupled to a protein using aminooxy or hydrazide compounds to form stable oxime or hydrazone linkages between the biomolecule and protein. The figure below summarizes these reactions.
This reaction is efficient because the unique functional group is targeted. It also does not compete with innate processes.
N- and C- Termini Modification
Because natural amino acid residues are prevalent in proteins, it is difficult to selectively target a single residue. Therefore, techniques have been developed to target residues on the N- and C-termini because of enhanced site selectivity in these locations.
One example of N-terminal modification involves the oxidation of the serine or threonine residues to form an N-terminal aldehyde, which can undergo bioorthogonal reactions similar to the one described earlier.
An example of C-terminal modification is native chemical ligation (NCL), in which a thioester residue on the C-terminal is coupled to a thiol group on a cysteine residue on the N-terminal of another protein. This technique allows for the construction of large polypeptides by assembling smaller ones. NCL is especially powerful because the first step, the interaction between the thiol and thioester groups, is reversible, whereas the second step, which forms an amide, is irreversible. This leads to high yields of the final ligation product.
Labeling-Site Selectivity
The alkylation of cysteine residues is used to selectively modify a protein site. This technique is typically effective for naturally expressed proteins and also depends on the low abundance of cysteine residues. However, many of the reagents used in the alkylation process, such as iodoacetamides and vinyl sulfones, have been shown to modify other amino acids in the protein, which lowers selectivity.
Purification
Because most bioconjugation reactions do not go to full completion, excess reagent is often added. This makes it more difficult to extract the product out of the resulting mixture. In addition, very few general techniques have been developed to purify products of bioconjugation reactions. A simple technique such as size-exclusion chromatography can be used to isolate bioconjugated molecules if the product of interest is sufficiently large to elute faster than the other molecules present in the mixture. In other instances, purification methods unique to the bioconjugation reaction must be developed. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Introduction_to_Bioconjugation.txt |
Mechanism
Appel, R. Angew. Chem. Int. Ed. Engl. 1975, 14, 801–811.
Baeyer-Villiger oxidation
Baeyer-Villiger oxidation is the oxidation of a ketone to a carboxylic acid ester using a peroxyacid as the oxidizing agent. eg. 1:
eg. 2:
Mechanism
When the two ligands on the carbonyl carbon in the ketone are different, Baeyer-Villiger oxidation is regioselective. Of the two alpha carbons in the ketone, the one that can stabilize a positive charge more effectively, which is the more highly substituted one, migrates from carbon to oxygen preferentially.
eg. 1:
eg. 2:
Beckmann Rearrangement
Introduction
The Beckmann rearrangement is a reaction employed in many sectors to convert oximes to amides. The reaction has greatly been improved since its discovery in the sense of safety and viability. This work focuses on the history of the Beckmann rearrangement and improvements applied to current syntheses of mass-produced, widely available compounds that previously utilized expensive, toxic, and difficult to synthesize or hard to obtain reagents.
The Beckmann rearrangement is a reaction discovered in the mid-1880’s by the chemist Ernst Otto Beckmann. The reaction converts oximes into their corresponding amides1 allowing the insertion of the nitrogen atom from the C=N bond into the carbon chain forming a C–N bond. Depending on the starting material, it could also produce nitriles from aldehydes.2 Traditional methods for the rearrangement involve harsh reaction conditions, such as a strongly acidic medium and high temperatures that can lead to undesired side products and unsuitable for sensitive substrates.3 Further historical accounts on the reaction’s discovery shall be discussed shortly, along with the current advancements for reactions that employ the Beckmann rearrangement for mass-produced compounds in the chemical, pharmaceutical, and agricultural sectors
Mechanism
In 1883, a chemist by the name of Alois Janny was working on the reaction of acetoxime with phosphorus pentachloride but could not identify the products.6 It was not until 1886, when Beckmann published his work identifying the products that were being produced by a ketoxime (benzophenone oxime) reacting with phosphorus pentachloride or phosphorus oxychloride, and determined that the product of his reaction was benzanilide.1 Originally, Beckmann did not intend to identify the products that Janny failed to. Beckmann was working on a method to distinguish between aldehydes and ketones.1 Just like many other discoveries, Beckmann’s was predominantly accidental. Beckmann determined as well that a ketoxime could be reacted with other reagents, such as sulfuric acid.1,7 Beckmann presented his so-called “Beckmann mixture” which consisted of hydrochloric acid, acetic anhydride, and acetic acid.8 His traditional method calls for a highly acidic medium and high temperatures. The highly acidic medium and high temperatures are what can be considered harsh, as these conditions are not suitable for sensitive substrates. Methods employed today vary from “one-pot” reactions to a few step reactions, low-to-mild temperatures, have low reaction times, and use readily available, less toxic, cost-effective reagents. Within the last 25 years, green chemistry studies have focused on employing these methods in effort to minimize the cost of production, waste, and use of toxic reagents.
In 1900, Otto Wallach discovered the compound caprolactam, and the commercial synthesis consists of the acid-catalyzed Beckmann rearrangement of cyclohexanone oxime.9 Today, many reagents and methods can lead to the synthesis of caprolactam. Some reactions employ Ga(OTf)3, HgCl2, and P2O5 or Eaton's reagent (7.7% P2O5 in methanesulfonic acid) as catalysts. The Ga(OTf)3 and HgCl2 catalysts were put in a CH3CN mixture with the cyclohexanone oxime. The mixture utilizing Ga(OTf)3 was reacted at 40ºC for 20 min, which led to a 92% conversion.10 The mixture with HgCl2 was heated at 80ºC and allowed to react for 8 h.11 The mixture utilizing P2O5 or Eaton’s reagent used an ionic liquid, bmiPF6 (1-n-butyl-3-methylimidazoliumhexafluorophosphate). The optimum temperature for the P2O5 or Eaton’s reagent mixture was 75ºC, and ran for 16 h and 21 h, respectively.12
Paracetamol was discovered in the late 1800s and its use as a drug did not come about until the mid-1900s. Celanese Corp., USA patented their synthesis of paracetamol in 1985. Their work states that paracetamol is produced by reacting a hydroxy aromatic ketone (4-hydroxyacetophenone) with a hydroxylamine salt to form the ketoxime and exposing the ketoxime to a catalyst which induces the Beckmann rearrangement forming the N-acyl-hydroxy aromatic amine. For the Beckmann rearrangement to take place, 4-hydroxyacetophenone oxime was allowed to react in a mixture of Amberlyst 15 (catalytic resin) and acetic acid, which was then refluxed under N2 for 2 h, resulting in a good yield (66.7%).14 The Beckmann rearrangement was utilized for its synthesis more recently by reacting 4-hydroxyacetophenone oxime with ammonium persulphate and dimethyl sulfoxide in 1,4-dioxane. This mixture was heated to 100ºC and allowed to react 45 min, and results in a good yield.15
Contributors and Attributions
• Jorge Calderon Moreno, Department of Chemistry, Sonoma State University
References
[1] E. Beckmann, Ber. Dtsch. Chem. Ges. 1886, 19, 988-993.
[2] A. Martínez-Asencio, M. Yus, D. J.Ramon, Tetrahedron, 2012, 68, 3948-3951.
[3] S. Srivastava and K. Kaur, New J. Chem., 2020, 3.
[4] B. Waskow, et al., Tetrahedron Lett., 2016, 57, 5575-5580.
[5] Beckmann Rearrangement https://www.alfa.com/en/beckmann-rearrangement/ (accessed April 20, 2021)
[6] A. Janny, Ber. Dtsch. Chem. Ges., 1883, 16, 172.
[7] E. Beckmann, Ber. Dtsch. Chem. Ges., 1887, 20, 1507.
[8] E. Beckmann, Ber. Dtsch. Chem. Ges., 1887, 20, 2580-2585.
[9] Molecule of the Week Archive Caprolactam, https://www.acs.org/content/acs/en/molecule-of-the-week/archive/c/caprolactam.html (accessed Mar 30, 2021).
[10] P. Yan, P. Batamack, G. K. S. Prakash and G. A. Olah, Catal. Lett., 2005, 103, 165-168.
[11] C. Ramalingan and Y. -T. Park, J. Org. Chem., 2007, 72, 4536-4538.
[12] R. X. Ren, L. D. Zueva, W. Ou, Tetrahedron Lett., 2001, 42, 8441-8443.
[13] Acetominophen, https://.ncbi.nlm.nih.gov/compound/A...inophenpubchem (accessed May 11, 2021).
[14] K. G. Davenport, C. B. Hilton, Process For Producing N-acyl-hydroxy Aromatic Amines,
https://worldwide.espacenet.com/pate...n%3DUS4524217A (accessed Apr 18, 2021).
[15] S. B. Mhaske, P. S. Mahajan, A Process For Synthesis Of Amides Via Radical-mediated Beckmann Rearrangement, 2016, 1-22.
Birch Reduction
The Birch reduction is the dissolving-metal reduction of aromatic rings in the presence of an alcohol, eg:
Chugaev Reaction
Recent Examples
Padwa's Synthesis of (±)-lycoricidine
For the full synthesis see: J. Org. Chem., 2007, 72, 2570-2582.
Cope Elimination
When a tertiary amine oxide bearing one or more beta hydrogens is heated, it is converted to an alkene. The reaction is known as Cope elimination or Cope reaction, not to be confused with Cope Rearrangement. For example:
The net reaction is 1,2-elimination, hence the name Cope elimination.
mechanism: Cope elimination is an intramolecular E2 reaction. It is also a pericyclic reaction.
Intermolecular E2 reactions occur preferentially from the conformation of the substrate in which the leaving group and the beta hydrogen abstracted by the base are antiperiplanar, which is not possible in intramolecular E2 reactions in which the base is built into the leaving group because the basic atom is too far away from the beta hydrogen anti to the leaving group. Intramolecular E2 reactions occur preferentially from the conformation of the substrate in which the leaving group and the beta hydrogen abstracted by the base are synperiplanar. The basic atom and the beta hydrogen abstracted by it are closest to each other in this conformation. For example:
mechanism:
Cope elimination is regioselective. Unlike intermolecular E2 reactions, it does not follow Zaitsev’s rule; the major product is always the least stable alkene, i.e., the alkene with the least highly substituted double bond. For example:
This trend is most likely due to the fact that the less highly substituted β-carbon bears more hydrogen atoms than the more highly substituted one; at a given moment, in a sample of the substrate, there are more molecules in which a hydrogen atom on the less highly substituted beta carbon is synperiplanar to the leaving group than there are in which a hydrogen atom on the more highly substituted beta carbon is.
Dakin Oxidation
Recent Examples
The Njardarson Synthesis of (±)-vinigrol
For the full synthesis see: Angew. Chem. Int. Ed. 2013, 52, 8648-8651.
Dieckmann condensation
The Dieckmann condensation is the intramolecular version of Claisen condensation: | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Named_Reactions/Appel_Reaction.txt |
The Heck reaction is a famous chemical reaction discovered by Mizoroki and Heck in 1972 through independent research. It involves the cross-coupling reaction between organohalides and alkenes, these two substances react in the presence of a palladium catalyst and a base to form a substituted alkene:
Step A is the oxidative addition of a polar substrate onto a palladium catalyst to form a tetrasubstituted complex. Step B is the migratory insertion of an olefin into the system. Step C is the β-Hydride Elimination of the alkene and Step D is the addition of base to the palladium to regenerate the starting catalyst and close the cycle[1].
Detailed Study of the Mechanism of Heck Reaction
Pre-activation of Palladium Catalyst
It is noteworthy that the first step of the catalytic cycle is actually not the oxidative addition of the substrate, as the palladium catalyst must be activated before the reaction. Therefore, a thorough study of the structure of the palladium catalyst and its properties will be important in understanding Heck reactions [2]. The catalytic precursor Pd(II)(OAc)2, associated with monodentate phosphine ligands such as PPh3 is normally used to catalyze the reaction, but this Pd(II) complex must be reduced to Pd(0) in order to enter the catalytic cycle. There are two different mechanisms involving phosphine-mediated Pd(II) reduction[2]:
The catalytic precursor Pd(II)(OAc)2, associated with monophosphine ligands, is much more efficient in catalyzing Heck reactions when compared to Pd(0)(PPh3)4 catalyst. This is because Pd(0)(PPh3)2(OAc)- can be destabilized by the interaction between OAc- and protons to readily form the unstable Pd(0)(PPh3)2 catalyst, which then enters the catalytic cycle to catalyze the reaction[3]. Since Pd(0)(PPh3)4 is a relatively stable 18-electron complex, it is unlikely that it dissociates two ligands to form an unstable 14-electron structure[3].
Furthermore, Pd(0) catalysts must possess an appropriate coordination number to enter the catalytic cycle[2]. If there are too many monophosphine ligands, it may inhibit the catalyst because a coordinatively saturated metal complex will be formed via ligand association:
The activation of Pd(II) catalysts can also be achieved without the assistance of phosphines. For example, triethylamine is a good reagent to selectively reduce Pd(II):
Oxidative Addition
Oxidative addition is the most difficult step of the entire catalytic cycle. However, the presence of electron-donating groups on the phosphine ligands can activate the Pd(0) catalyst such that the R-X bond can be easily broken along with the formation of Pd-R and Pd-X bond. The rate of oxidative addition also depends on the chemical property of halides, the following trend is generally observed [2]:
Olefin Addition to the Palladium Complex
Before migratory insertion of the olefin to the palladium-R bond, the olefin must first associate onto the palladium complex, which requires the dissociation of the existing ligands. Historically, the Heck reaction was viewed as the functionalization of olefins through aryl halides typically without ligands for aryl iodide or with monodentate phosphines for the other compounds [1]. This cycle produces a strong Pd-X bond while there is a weak Pd-PR3 attachment. The oncoming group would attack and form a net neutral square planar molecule, as is shown in Path A. This reaction mechanism was found to not be ideal as, “chelating diphosphines… in general do not produce useful catalysts [1].” This concern was addressed through the introduction of triflates as leaving groups which would allow the mechanism coordination-insertion step to follow Path B. Catalytic and stoichiometric studies have been utilized to determine these pathways.
• Path A: If a neutral ligand dissociates, then the neutral mechanism predominates the reaction. The neutral mechanism occurs when X=Cl, Br, I (i.e. strong sigma-donor).
• Path B: If an anionic halide ligand dissociates, then the cationic mechanism dominates the reaction. The cationic mechanism is believed to happen when X= OTs- or OAc- (e.g. weakly associated ligand).
The two mechanisms shown above display the possible coordination-insertion paths found in Heck reactions. It has been found that the insertion of ethylene into the Pt-H bond is critical in the reaction characterization [1]. Thorn and Hoffman conducted orbital studies to determine key information from this step. The first conclusion they reached is that there must be a coplanar assembly of the metal, hydride and ethylene for insertion to occur. This indicated that insertion is stereoselective and occurs in a syn manner. Experimental data supported this claim [1]. They also determined that the energy barrier for a pentacoordinated complex is significantly higher than that of a tetracoordinated complex, indicating pentacoordinated complexes are not involved in the mechanism. This observation was supported by kinetic and experimental data [1].
Path B was unknown until 1991, when Ozawa and Hayashi proposed the existence of a cationic form of the square planar complex. To obtain this complex, triflates must be used as the leaving group [1]. The lability of Pd-OTf bonds in the oxidation addition complex aids in this formation. Bidentate phosphorus or nitrogen ligands, along with the triflate leaving group, allow for the reaction to follow this path. It was also determined that this path can yield high asymmetric induction when the diphosphine is chiral. This effect was not seen in Path A [1].
Ligand dissociation mechanisms are also affected by different types of phosphine ligands. Monodentate phosphine ligands lead to the occurrence of both the neutral and cationic mechanisms, whereas bidentate phosphine ligands merely induce a cationic mechanism, but the neutral mechanism is still possible in the presence of a large bite angle [2].
Migratory Insertion
Migratory insertion of the olefin into the Pd-R bond is a crucial step for the catalytic cycle because it can control the stereo-selectivity and regio-selectivity of Heck reactions. For a neutral palladium complex, the regioselectivity is governed by sterics, which means nucleophilic attack happens on the less hindered site of the alkene [4][5]:
For cationic palladium complexes, the regioselectivity is governed by electronics, which implies that nucleophilic attack occurs on the site possessing the least electron density of the alkene [4][5]:
β-Hydride Elimination
β-Hydride Elimination results in the Heck reaction product, which is a new substituted alkene. In this step, the palladium and the hydride attached to it must be syn-coplanar for the initiation of elimination. The product with the Z-conformation is strongly disfavored because of the steric interaction in the transition state [2]:
Afterβ-hydride elimination, the newly formed palladium-alkeneπcomplex is subject to olefin isomerization, resulting in the formation of an undesired Heck product[4]:
These side reactions will occur since this is a reversible reaction. If the olefin dissociation rate is too slow, this problem arises [1]. Fortunately, adding bases or silver salts can significantly reduce the chance of alkene isomerization by facilitating reductive elimination to form an H-X bond [6].
Regeneration of palladium catalyst
The addition of base is necessary to reduce the L2PdHX complex back to the starting L2Pd(0)[1]. Some common bases used are trialjylamines such as Et3N or inorganic salts such as AcONa. A proton sponge or Tl(I) or Ag(I) salts may also be employed to close the cycle[1].
Intramolecular Heck Reaction
Heck reactions can also be performed in a single molecule which is quite useful for macrocyclization. This intramolecular Heck reaction was first reported by Mori and Ban in 1977:
Mori, M; Ban, K.; Tetrahedron 1977, 12, 1037
The Intramolecular Heck reaction has many advantages compared with the intermolecular Heck reaction. First of all, only mono- or disubstituted alkenes can coordinate into the palladium complex in the intermolecular Heck reaction, whereas tri- and tetrasubstituted alkenes are able to participate readily through the intramolecular mechanism. Secondly, the intramolecular Heck reaction is much more efficient than the intermolecular reaction because of entropic considerations [7]. Finally, regioselectivity and stereoselectivity are dramatically improved in the intramolecular Heck reaction. This advantage inspired both Shibasaki and Overman to explore the asymmetric effect in the intramolecular Heck reaction, and they eventually found the first asymmetric intramolecular Heck reactions. This remarkable finding has provided enlightenment for natural product synthesis [7].
Shibasaki, M.J.Org. Chem. 1984,54,4738
Overman, L. E. J. Org. Chem. 1989, 54, 5846
Regioselectivity and Stereoselectivity
Regioselectivity reactions of Path A were conducted on several classes of olefins. Different olefin families reacted with different regioselectivity[1]. Once aryl triflates were introduced as the leaving groups, reactions via Path B were also examined for regioselectivity. These reactions were carried out with aryl triflate leaving groups and aryl halides with Pd(OAc)2 with bidentate phosphorus ligands[1]. These reactions yielded branched products more readily than Path A.
In Path A, regioselectivity is related to the coordination-insertion pathway and steric factors. A migration of the R group to the less substituted carbon with formation of linear products has seen to be favored in Pathway A [1]. Path B differs in that electronic factors dominate in determining regioselectivity. The increase in polarization is determined by the coordination of a pi-system within a cationic complex. This causes selective migration of the aryl moiety onto the carbon with a lower charge density [1].
Hayashi and Ozawa have investigated stereoselective intermolecular reactions using aryl triflate as the leaving group and chiral (R)-BINAP. Several reactions were performed under analogous conditions and their products compared [1]. As is expected with the triflate leaving group, these reactions followed Path B. In these reactions, the chiral BINAP ligand binds tightly to the metal through both of its phosphorus atoms. This chiral ligand is able to transfer its chiral information from the catalyst to the substrates [1]. These reactions have been found to be most selective when using both chiral BINAP and the triflate leaving group. Of note is that this (R)-BINAP determination of selectivity will only occur in systems that are electron rich. This information aligns with the assertion that Path B relies heavily on electronic factors in that electron rich systems react much more efficiently than electron poor [1].
Overman has reported a stereoselective synthesis for quaternary carbons through asymmetric intramolecular Heck reactions. These results were of note, as they did not follow the widely held belief that long reaction times result in isomerization of the double bond and multiple product formation [1]. PMP was used as the base in these reactions without Ag(I) salt present. Good selectivity was observed in these reactions despite the slow reaction time and the difficulty in transferring chirality through Path A [1]. It was determined that with flexible substrates, the products were almost racemic. However, with rigid substrates, the single phosphorus coordination of (R)-BINAP was still able to transfer the chirality via Path A[1].
Limitations of the Heck Reaction
The Heck reaction is widely used in the pharmaceutical, medical and industrial areas because of its ability to efficiently generate large polycyclic structures. However, this synthetically useful reaction has its own disadvantages [8].One of the major disadvantages is that the Pd catalyst will be lost at the end of the catalytic cycle, therefore, it is necessary for researchers to find an effective method to recycle the palladium catalyst. Another significant weakness is that the phosphine ligands attached to the palladium catalyst can be toxic and expensive, so the phosphine-free ligands must be discovered to improve the efficiency of the reaction [8].
Applications of the Heck Reaction
Heck couplings with dehydrocostus lactone and aryl halides can produce guaianolide sesquiterpene lactones derivatives, which have been proved effective in inhibiting resistant acute leukemic cells:
Y. H. Ding, H. X. Fan, J. Long, Q. Zhang, &Y. Chen, Bioorganic and Medicinal Chemistry Letters, vol. 23, 22, 6087–6092, 2013.
The Heck reaction can also be used to synthesize the smoking cessation aid, Chantix ®
Coe, J. W.; Brooks, P.R.; Veteline, M.G.; Bashore, C.G.;Bianco, K.;Flick, A,A.C. Tetrahedron Lett. 2011, 52, 953-954
Contributors and Attributions
• Haley Merritt and Yifan Qi | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Named_Reactions/Heck_reaction.txt |
Alpha Addition - Hell Volhard Zelinsky Reaction
An aldehyde or ketone in possession of an α hydrogen will be in equilibrium with its enol tautomer. This aspect of aldehydes and ketones allows electrophilic addition to occur at the α hydrogen. Carboxylic acids, however, don't generally form stable enols, so alpha addition is more difficult to achieve with carboxylic acids than aldehydes and ketones. The Hell Volhard Zelinsky reaction demonstrates a method for alpha addition with a carboxylic acid. The gist of the method is to convert the carboxylic acid into a derivative that does undergo tautomerization and then to carry out alpha addition upon that form.
In the Hell Volhard Zelinsky reaction PBr3 is used to replace the carboxylic OH with a bromide, resulting in a carboxylic acid bromide. The acyl bromide can then tautomerize to an enol. This enol is then made to react with Br2 at the α position forming an α-bromo acyl bromide. Reaction of the α-bromo acyl bromide with the original carboxylic acid yields the α-bromo carboxylic acid product and regenerates the acyl bromide intermediate.
Hofmann Elimination
Elimination reactions of 4º-ammonium salts are termed Hofmann eliminations. Since the counter anion in most 4º-ammonium salts is halide, this is often replaced by the more basic hydroxide ion through reaction with silver hydroxide (or silver oxide). The resulting hydroxide salt must then be heated (100 - 200 ºC) to effect the E2-like elimination of a 3º-amine. Example #1 below shows a typical Hofmann elimination. Obviously, for an elimination to occur one of the alkyl substituents on nitrogen must have one or more beta-hydrogens, as noted earlier in examining elimination reactions of alkyl halides.
In example #2 above, two of the alkyl substituents on nitrogen have beta-hydrogens, all of which are on methyl groups (colored orange & magenta). The chief product from the elimination is the alkene having the more highly substituted double bond, reflecting not only the 3:1 numerical advantage of those beta-hydrogens, but also the greater stability of the double bond. Example #3 illustrates two important features of the Hofmann elimination:
1. Simple amines are easily converted to the necessary 4º-ammonium salts by exhaustive alkylation, usually with methyl iodide (methyl has no beta-hydrogens and cannot compete in the elimination reaction). Exhaustive methylation is shown again in example #4.
2. When a given alkyl group has two different sets of beta-hydrogens available to the elimination process (colored orange & magenta here), the major product is often the alkene isomer having the less substituted double bond.
The tendency of Hofmann eliminations to give the less-substituted double bond isomer is commonly referred to as the Hofmann Rule, and contrasts strikingly with the Zaitsev Rule formulated for dehydrohalogenations and dehydrations. In cases where other activating groups, such as phenyl or carbonyl, are present, the Hofmann Rule may not apply. Thus, if 2-amino-1-phenylpropane is treated in the manner of example #3, the product consists largely of 1-phenylpropene (E & Z-isomers).
To understand why the base-induced elimination of 4º-ammonium salts behaves differently from that of alkyl halides it is necessary to reexamine the nature of the E2 transition state, first described for dehydrohalogenation. The energy diagram shown earlier for a single-step bimolecular E2 mechanism is repeated on the right. The E2 transition state is less well defined than is that of SN2 reactions. More bonds are being broken and formed, with the possibility of a continuum of states in which the extent of C–H and C–X bond-breaking and C=C bond-making varies. For example, if the bond to the leaving group (X) is substantially broken relative to the other bond changes, the transition state approaches that for an E1 reaction (initial ionization followed by a fast second step). At the other extreme, if the acidity of the beta-hydrogens is enhanced, then substantial breaking of C–H may occur before the other bonds begin to be affected. For most simple alkyl halides it was proper to envision a balanced transition state, in which there was a synchronous change in all the bonds. Such a model was consistent with the Zaitsev Rule.
When the leaving group X carries a positive charge, as do the 4º-ammonium compounds discussed here, the inductive influence of this charge will increase the acidity of both the alpha and the beta-hydrogens. Furthermore, the 4º-ammonium substituent is much larger than a halide or hydroxyl group and may perturb the conformations available to substituted beta-carbons. It seems that a combination of these factors acts to favor base attack at the least substituted (least hindered and most acidic) set of beta-hydrogens. The favored anti orientation of the leaving group and beta-hydrogen, noted for dehydrohalogenation, is found for many Hofmann eliminations; but syn-elimination is also common, possibly because the attraction of opposite charges orients the hydroxide base near the 4º-ammonium leaving group.
Three additional examples of the Hofmann elimination are shown in the following diagram. Example #1 is interesting in two respects. First, it generates a 4º-ammonium halide salt in a manner different from exhaustive methylation. Second, this salt is not converted to its hydroxide analog prior to elimination. A concentrated aqueous solution of the halide salt is simply dropped into a refluxing sodium hydroxide solution, and the volatile hydrocarbon product is isolated by distillation.
Example #2 illustrates an important aspect of the Hofmann elimination. If the nitrogen atom is part of a ring, then a single application of this elimination procedure does not remove the nitrogen as a separate 3º-amine product. In order to sever the nitrogen function from the molecule, a second Hofmann elimination must be carried out. Indeed, if the nitrogen atom was a member of two rings (fused or spiro), then three repetitions of the Hofmann elimination would be required to sever the nitrogen from the remaining molecular framework.
Example #3 is noteworthy because the less stable trans-cyclooctene is the chief product, accompanied by the cis-isomer. An anti-E2-transition state would necessarily give the cis-cycloalkene, so the trans-isomer must be generated by a syn-elimination. The cis-cyclooctene produced in this reaction could also be formed by a syn-elimination. Cyclooctane is a conformationally complex structure. Several puckered conformations that avoid angle strain are possible, and one of the most stable of these is shown on the right. Some eclipsed bonds occur in all these conformers, and transannular hydrogen crowding is unavoidable. Since the trimethylammonium substituent is large (about the size of tert-butyl) it will probably assume an equatorial-like orientation to avoid steric crowding. An anti-E2 transition state is likely to require an axial-like orientation of this bulky group, making this an unfavorable path.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Named_Reactions/Hell-Volhard-Zelinsky_reaction.txt |
Hofmann rearrangement, also known as Hofmann degradation and not to be confused with Hofmann elimination, is the reaction of a primary amide with a halogen (chlorine or bromine) in strongly basic (sodium or potassium hydroxide) aqueous medium, which converts the amide to a primary amine. For example:
Mechanism:
Mitsunobu Macrolactonization
The following is a standard procedure for a Mitsunobu macrolactonization.
Procedure
To a solution of the seco-acid (0.1 mmol) and PPh3 (0.4 mmol) in PhMe (70 mL) under N2 at -10 °C was added a solution of DIAD (0.4 mmol) in PhMe (30 mL) via syringe pump over 1 h. The resulting mixture was slowly allowed to warm to room temperature. After the disappearance of the starting material by TLC, the reaction was concentrated and the crude material was purified by flash chromatography on silica gel.
Notes:
• PhMe is the most common solvent for this reaction. Other useful ones are: PhH, THF and CH2Cl2.
• Very dilute conditions and slow addition of the diazo compound are critical to avoid formation of the diolide (dimerization of the seco-acid).
• Slow addition of the seco-acid to the azo/phosphine mixture can also prevent diolide formation.
• If addition of the azo compound onto the acid occurs, other diazo compounds can be used: DEAD (1972-28-7), DBAD (212-796-9), ADDP (10465-81-3) depending on the bulkiness needed.
• Changing the triphenylphosphine to a less bulky phosphine (PBu3, PMe3) can accelerate slow reactions.
Named Reagents
Named reagent Structure Formula Other Name CAS number
Adam's catalyst
PtO2 Platinum(IV) oxide 1314-15-4
Appel's salt
C2Cl3NS2 4,5-dichloro-1,2,3-dithiazolium chloride 75318-43-3
Bestmann ylide
C20H15OP 2-(triphenylphosphoranylidene)ethenone 15596-07-3
Brady's reagent
C6H6N4O4 (2,4-dinitrophenyl)hydrazine 119-26-6
Brassard's diene
C11H22O3Si (E)-((1,3-diethoxybuta-1,3-dien-1-yl)oxy)trimethylsilane
Bredereck's reagent
C9H22N2O tert-Butoxy bis(dimethylamino)methane 5815-08-7
Burgess reagent
C8H18N2O4S 1-methoxy-N-triethylammoniosulfonyl-methanimidate 29684-56-8
Caro's acid
H2O5S Peroxymonosulfuric acid 7722-86-3
Collins reagent
C10H10N2CrO3 Dipyridine chromium(VI) oxide 20492-50-6
Comins' reagent
C7H3ClF6N2O4S2 2-[N,N-Bis(trifluoromethylsulfonyl)amino]-5-chloropyridine 145100-51-2
Corey-Bakshi-Shibata catalyst
C18H20BNO (R)-(+)-1-Methyl-3,3-diphenylhexahydropyrrolo[1,2-c][1,3,2]oxazaborole
(S)-(-)-1-Methyl-3,3-diphenylhexahydropyrrolo[1,2-c][1,3,2]oxazaborole
112022-83-0
112022-81-8
Corey-Chaykovsky reagent
C3H8OS Dimethylsulfoxonium methylide 5367-24-8
Corey lactone
C8H12O4 (3aR,4S,5R,6aS)-(−)-Hexahydro-5-hydroxy-4-(hydroxymethyl)-2H-cyclopenta[b]furan-2-one
(3aS,4R,5S,6aR)-(+)-Hexahydro-5-hydroxy-4-(hydroxymethyl)-2H-cyclopenta[b]furan-2-one
32233-40-2
76704-05-7
Cornforth reagent
C10H12N2Cr2O7 Pyridinium dichromate 20039-37-6
Danishefsky's diene
C8H16O2Si (E)-1-Methoxy-3-trimethylsilyloxy-1,3-butadiene 54125-02-9
Dess-Martin periodinane
C13H13IO8 1,1,1-Triacetoxy-1,1-dihydro-1,2-benziodoxol-3(1H)-one 87413-09-0
Echavarren's gold complex
C22H30AuF6NPSb (Acetonitrile)[(2-biphenyl)di-tert-butylphosphine]gold(I) hexafluoroantimonate 866641-66-9
Eschenmoser's salt
C3H8NI Dimethylmethylideneammonium iodide 33797-51-2
Fétizon's reagent
N/A Silver carbonate on Celite N/A
Fort's base
N/A N/A N/A
Frémy's salt
K2NO7S2 Potassium nitrosodisulfonate 14293-70-0
Garner's aldehyde
C11H19NO4 (S)-tert-butyl 4-formyl-2,2-dimethyloxazolidine-3-carboxylate
(R)-tert-butyl 4-formyl-2,2-dimethyloxazolidine-3-carboxylate
102308-32-7
95715-87-0
Glauber's salt
H20O14Na2S Sodium sulfate decahydrate 7727-73-3
Gold's reagent
C6H14ClN3 (Dimethylaminomethyleneaminomethylene)dimethylammonium chloride 20353-93-9
Grubbs catalyst (1st generation)
C43H72Cl2P2Ru Benzylidene-bis(tricyclohexylphosphine)dichlororuthenium 172222-30-9
Grubbs catalyst (2nd generation)
C46H65Cl2N2PRu (1,3-Bis(2,4,6-trimethylphenyl)-2-imidazolidinylidene)dichloro(phenylmethylene)(tricyclohexylphosphine)ruthenium 246047-72-3
Guindon's reagent
C2H6BBr Dimethylboron bromide 5158-50-9
Hajos-Parrish ketone
C10H12O2 (S)-7a-methyl-2,3,7,7a-tetrahydro-1H-indene-1,5(6H)-dione
(R)-7a-methyl-2,3,7,7a-tetrahydro-1H-indene-1,5(6H)-dione
17553-86-5
17553-89-8
Hantzsch's ester
C13H19NO4 Diethyl 1,4-dihydro-2,6-dimethyl-3,5-pyridinedicarboxylate 214-561-6
Hatakeyama's catalyst
C19H22N2O2 (9S)-3α,9-Epoxy-10,11-dihydrocinchonan-6'-ol 253430-48-7
Hermann-Beller catalyst
C46H46O4P2Pd2 trans-di(μ-acetato)bis[o-(di-o-tolylphosphino)benzyl]dipalladium(II) 172418-32-5
Hieber anion
C3FeNO4 N/A N/A
Hoveyda-Grubbs catalyst (1st generation)
C28H45Cl2OPRu Dichloro(o-isopropoxyphenylmethylene)(tricyclohexylphosphine)ruthenium(II) 203714-71-0
Hoveyda-Grubbs catalyst (2nd generation)
C31H38Cl2N2ORu (1,3-Bis-(2,4,6-trimethylphenyl)-2-imidazolidinylidene)dichloro(o-isopropoxyphenylmethylene)ruthenium 301224-40-8
Hünig's base
C8H19N N,N-Diisopropylethylamine 7087-68-5
Kagan's reagent
SmI2 Samarium diiodide 32248-43-4
Kemp's triacid
C12H18O6 cis,cis-1,3,5-Trimethylcyclohexane-1,3,5-tricarboxylic acid 79410-20-1
Koser's reagent
C13H13IO4S [Hydroxy(tosyloxy)iodo]benzene 27126-76-7
Lawesson's reagent
C14H14O2P2S4 2,4-bis(4-methoxyphenyl)-1,3,2,4-dithiadiphosphetane 2,4-disulfide 19172-47-5
Leuchs' anhydrides
N/A N/A N/A
Lindlar catalyst
N/A Pd/CaCO3 poisoned with Lead (Pb(OAc)2 or PbO) N/A
Mander's reagent
C3H3NO2 Methyl cyanoformate 17640-15-2
Martin's sulfurane
C30H20F12O2S Bis[α,α-bis(trifluoromethyl)benzyloxy]diphenylsulfur 32133-82-7
Meerwein's reagent
C3H9OBF4 Trimethyloxonium tetrafluoroborate 420-37-1
Meldrum's acid
C6H8O4 2,2-Dimethyl-1,3-dioxane-4,6-dione 2033-24-1
Mosher's acid
C10H9F3O3 (R)-3,3,3-Trifluoro-2-methoxy-2-phenylpropanoic acid
(S)-3,3,3-Trifluoro-2-methoxy-2-phenylpropanoic acid
20445-31-2
17257-71-5
Pearlman's Catalyst
N/A Palladium hydroxide on carbon 12135-22-7
Ohira-Bestmann reagent
C5H9N2O4P Dimethyl (1-diazo-2-oxopropyl)phosphonate 90965-06-3
Roche ester
C5H10O3 Methyl (S)-(+)-3-hydroxy-2-methylpropionate
Methyl (R)-(+)-3-hydroxy-2-methylpropionate
80657-57-4
72657-23-9
Schlosser's base
N/A N/A N/A
Schwartz reagent
C10H11ClZr Bis(cyclopentadienyl)zirconium(IV) chloride hydride 37342-97-5
Seyferth-Gilbert reagent
C3H7N2O3P Dimethyl (diazomethyl)phosphonate 27491-70-9
Shiina's reagent
C16H12N2O7 2-Methyl-6-nitrobenzoic anhydride 434935-69-0
Stryker's reagent
C108H96Cu6P6 (Triphenylphosphine)copper hydride hexamer 33636-93-0
Tebbe reagent
C13H18AlClTi N/A 67719-69-1
Togni's reagent
C10H10F3IO 3,3-Dimethyl-1-(trifluoromethyl)-1,2-benziodoxole 887144-97-0
Tröger's base
C17H18N2 (+)-(5R)-2,8-dimethyl-6,12-dihydro-5,11-methanodibenzo[b,f][1,5]diazocine
(-)-(5S)-2,8-dimethyl-6,12-dihydro-5,11-methanodibenzo[b,f][1,5]diazocine
529-81-7
14645-24-0
Wieland–Miescher ketone
C11H14O2 (S)-(+)-8a-Methyl-3,4,8,8a-tetrahydro-2H-naphthalene-1,6-dione
(R)-(–)-8a-Methyl-3,4,8,8a-tetrahydro-2H-naphthalene-1,6-dione
33878-99-8
100348-93-4
Wood's metal 50% Bi / 26.7% Pb / 13.3% Sn / 10% Cd
by weight
N/A N/A 8049-22-7
Woollins' reagent
C12H10P2Se4 2,4-Diphenyl-1,3,2,4-diselenadiphosphetan-2,4-diselenide 122039-27-4
Negishi cross-coupling
The Negishi reaction involves catalytic addition of alkylzinc nucleophiles to vinyl halides.
Figure 1. The Negishi cross-coupling reaction.
The proposed catalytic cycle for this reaction involves the crucial oxidative addition of the vinyl halide to the metal. The alkylzinc probably delivers the alkyl nucleophile to the metal via conventional nucleophilic substitution. Once both species are on the metal, they experience reductive elimination together.
Figure 2. Catalytic cycle for the Negishi cross-coupling reaction. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Named_Reactions/Hofmann_rearrangement.txt |
Suzuki-Miyaura coupling (or Suzuki coupling) is a metal catalyzed reaction, typically with Pd, between an alkenyl (vinyl), aryl, or alkynyl organoborane (boronic acid or boronic ester, or special cases with aryl trifluoroborane) and halide or triflate under basic conditions. This reaction is used to create carbon-carbon bonds to produce conjugated systems of alkenes, styrenes, or biaryl compounds (Scheme 1).
Scheme 1: General reaction scheme of Suzuki cross coupling reaction.
Modified conditions have demonstrated reactivity with less reactive substrates such as alkyl boranes (BR3) or aryl or alkenyl chlorides by amending the base and ligands employed.
Historical Methods: Forming carbon-carbon bonds via cross coupling reactions
The Suzuki coupling is a pioneering reaction in cross coupling, and has been thoroughly studied since. The first Suzuki-type cross coupling reaction between phenylboronic acid and haloarenes was published by Suzuki and Miyaura in 1981 (Scheme 1).1 Commonly, Suzuki coupling is compared to Stille coupling seeing that boron has a similar electronegativity to tin, which is used for transmetallation in Stille coupling. Both couplings have a similar reaction scope and proceed via a similar mechanistic cycle.
Scheme 1: Suzuki cross coupling reaction of phenylboronic acid and haloarenes.
Stille cross coupling reactions can form carbon-carbon bonds between alkenyl (vinyl), aryl, or alkynyl halides and an extended scope of organotin alkynes, alkenes, aryl, allyl benzyl, ketones, and alkyl. However, Stille coupling poses several drawbacks seeing that organotin reagents are: (1) highly toxic, (2) costly, and have a (3) lower functional group tolerance, despite Suzuki coupling not being suitable for base sensitive substrates.2 The example below demonstrates that even under conditions for either a Stille or Suzuki coupling to ensue, primarily Suzuki coupling occurs (Scheme 2).3
Scheme 2: Suzuki versus Stille cross coupling reaction.
Negishi coupling also demonstrates similar transformations to Suzuki coupling in a comparable substrate scope. It uses organo-zinc for transmetallation. However, Negishi coupling tends to occur in lower yields, with less functional group tolerance, and is water and oxygen sensitive.1
Reaction Scheme
Suzuki coupling proceeds via three basic steps: oxidative addition, transmetallation, and reductive elimination.
Reaction Mechanism and Mechanistic Studies
The general catalytic cycle for Suzuki cross coupling involves three fundamental steps: oxidative addition, transmetalation, and reductive elimination as demonstrated in Figure 1.1 The oxidative addition of aryl halides to Pd(0) complex is the initial step to give intermediate 1, a Pd(II) species. Under the participation of base, an organoborane compound reacts with intermediate 1 in transmetalation to afford intermediate 2. This is followed by reductive elimination to give the desired product and regenerate the original Pd(0) species. Depending on different catalytic systems with various catalysts, ligands, and solvents, there are additional processes in the catalytic cycle, including ligand or solvent association and dissociation.2 Undoubtedly, these three fundamental steps occur since intermediates 1 and 2 have been characterized by isolation3 or spectroscopic analyses4.
Oxidative addition is the rate determining step in the catalytic cycle and the relative reactivity decreases in the order of I > OTf > Br > Cl.4a The oxidative addition involving Pd(0) complexes is a SN2-type mechanism based on the unambiguous evidence of stereochemical inversion of configuration at alkyl halides carbon.5 Mechanistic studies have revealed that the apparently simple oxidative addition step consists of four concurrent isomerization pathways from the cis isomer, which is initially formed, to the more stable trans isomer (Figure 2).6
Transmetallation is initiated by base to encourage the transfer of the aryl or alkyl group from the organoborane to the Pd complex. The intermediates and stereochemistry of this step were determined by:
1. Low-temperature rapid injection nuclear magnetic resonance (NMR) spectroscopy & kinetic studies have recently revealed three different species containing palladium-oxygen-boron linkages were identified and characterized as the pre-transmetalation intermediates.
1. Deuterium-labeling experiments demonstrated the retention of stereochemistry in the transmetalation step (Figure 4).8 Here, smaller J-values (approximately 0-7 Hz) correspond to a syn configuration, and larger J-values (8-15 Hz) corresponds to the anti configuration.
Reductive elimination occurs from the cis complex, therefore the trans complexes isomerizes to cis isomer to undergo reductive elimination.9 The elimination was confirmed to proceed intramolecularly from the cis isomer9 based on:
Kinetic studies showed that reductive elimination obeys first order kinetics, hence the rate and reaction is dependent only on the concentration of the post-transmetallation Pd-complex.
Deuterium labeling demonstrated no cross-coupling product from crossover between two Pd complexes (Figure 5).
Catalysts
Palladium catalysts are most widely employed in Suzuki coupling. Normally, the active Pd catalyst consist of two parts: precursors (for example: Pd(OAc)2, Pd2(dba)3, or Pd(PPh3)4) and ligands. To enhance the reactivity and stability of the catalyst, they were developed to be electron-rich and spatially bulky, which affords a high turnover number (TON) and low loading.1,2 For example, palladacycles (Figure 4) were developed and exhibit thermal stability, robust reaction times, insensitivity to air and water, low cost, and environmentally friendly.3,4
In terms of separation and recycling of the catalysts, people have developed polymer-supported heterogeneous catalysis systems for Suzuki coupling reaction.5 Polymer-supported heterogeneous catalysis have tremendous values in pharmaceutical and industrial synthesis due to the advantage of preventing contamination from the ligand residue in products, fast recovery, and the simple recycling of the catalysts
Suzuki Cross Coupling Reaction Ligand Design
Since the 1980’s, Suzuki reactions have been significantly improved, which lends to why the Suzuki reaction is so powerful: the elegant ligand design for Pd catalysts. Ligands are designed with electron-rich and spatially bulky features since electron-rich ligands can facilitate the oxidative addition step, and spatially bulky structures increase the orbital overlapping on the metal which enables reductive elimination.1 Based on which atom is coordinated with the metal, ligands are categorized as Phosphine ligands, Carbon ligands, or Nitrogen ligands.
Phosphine ligands are the most popular Pd ligands in both the laboratory and industry. PPh3 was the earliest and most widely use monodentate phosphine ligand2 in Suzuki reactions. Inspired by the structure of PPh3, scientists have manipulated substitutions on the phosphor atom and aromatic ring to tune the different reactivity of the reaction. For example, Fu and other research groups3 replaced aromatic groups with more electron-rich and bulkier alkyl groups (Figure 1), and the resulting ligands gave high catalytic reactivity on less reactive substrates with lower catalyst loading.
In the late 1990’s, Buchwald’s group introduced new electronic rich and bulky phosphine ligands based on the frame skeleton of dialkylbiaryl phosphine ligands4. Understanding the impact of substitutions on dialkylbiarylphosphines on the efficacy of catalytic reactions, the Buchwald group has synthesized a series of dialkylbiaryl ligands facilitating Palladium-catalyzed C-C, C-N, and C-O bond-forming processes, as well as supporting ligands for other reactions.4 There are other monodentate phosphine ligands with unique structures, which can also promote Suzuki reaction effectively, for example, monodentate ferrocene phosphine ligands.5
Many research groups have also designed and synthesized bidentate phosphine ligands (Figure 3) for the Suzuki reaction, and those ligands showed excellent reactivity on substrates that are generally challenging for monodentate ligands.
Carbon & Nitrogen ligands: Carbon ligands mainly comprise carbene8- and olefin9-type ligands for Suzuki couping. Efficient nitrogen ligands include amines and imines.9
Suzuki Cross Coupling Reaction Substrate Scope
Electrophilic Partners1: The electrophilic partners in Suzuki coupling are: alkenyl, alkynyl, allyl, benzyl, aryl, and alkyl halides or triflates. Aryl halides are activated by electron-withdrawing (EWD) groups in the ortho or para positions or 1-alkenyl halides with EWD groups in the alpha or beta positions. This increases their reactivity in oxidative addition compared to those with electron-donating groups. The order of reactivity of the electrophilic partners based on their leaving groups is:
I >> Br > OTf >> Cl > F
Chloride electrophiles are the most nonreactive as they are reluctant to participate in oxidative addition. Reactivity with chloride electrophiles has been established by using bulky, electron-donating (electron rich) phosphine ligands on the Pd catalyst and stronger bases to encourage dissociation of Cl.2
β- alkyl Suzuki coupling is most successful between hindered electron-rich organoboranes and electron-deficient vinyl or aryl halides or triflates. This counters the trends reported above in that the electrophilic partner would be activated by EWD groups.
Bases1: Transmetallation is initiated with base, and it’s reactivity varies with solvent.
Strong bases: NaOH, TlOH, and NaOMe perform optimally in THF or H2O solvent systems
Thallium bases such as TlOH, and its derivatives Tl2CO3 and TlOEt, enable cross coupling between alkyl boranes and alkyl halides. It also allows this reaction to proceed at lower temperatures (20 °C from the usual 50 to 80 °C). Unfortunately, these bases are air and light sensitive, but are still widely used.3
Weak bases: K2CO3 and K3PO4 perform optimally in DMF
Catalysts: Pd*, Ni, Ru, Fe, Cu
Palladium catalysts are the most widely used for Suzuki coupling and perform best with electron-donating (usually phosphine) ligands.
Nickel catalysts have been recently developed and demonstrate reactivity with inert electrophiles, especially chlorides and unreactive bromides. They have also expanded the electrophile scope to include: aryl fluorides (using a directing group, usually o-NO2), carbamates, sulfamates, esters, phosphate esters and ethers. This is enabled by nickel’s variation in oxidation states (Ni0 à Ni2+ and Ni+ à Ni3+) and increased nucleophilicity due to its small size. Use of Ni in catalysis is favored due to its abundance, and thus low cost.4
Boranes: Boronic acids, esters and tri-fluoro derivatives1,2
The reactivity of the borane to conduct transmetallation depends on its Lewis acidity. Therefore, EWD substituents increase the reactivity of the borane. The general trend of borane reactivity is:
ArBF3 > RB(OH)2 > RB(OR)2 >> R3B
Alkyl boranes are the least reactive, and rarely participate in transmetallation under standard conditions; however this has been amended with the use of stronger bases.
Selecting a borane for a reaction depends on its compatibility with its electrophile coupling partner and desired borane R group.
Regio- and Stereoselectivity1,2
Each step of the catalytic cycle of Suzuki coupling can influence the regio- or stereo- configuration of the product. General rules of regio- and stereoselectivity relative to the substrates are:
Oxidative addition of alkyl and alkenyl halides retains the configuration of the electrophilic substrate, however allylic and benzylic halides invert this configuration. This step initially produces the cis complex which isomerizes to the trans complex.
Transmetallation and reductive elimination both retains the regio- and stereochemistry established in oxidative addition.
The borane substrate can also dictate selectivity of the cross coupling product. Often times the reactive borane species is formed in situ via hydroboration with 9-BBN. This occurs as a syn addition, and reductive elimination of the product retains the stereochemistry established by the hydroboration. However, reversing the order in which the reagents are added yields the anti addition product, which is also reflected in reductive elimination.
Chiral centers are formed by using chiral borane reagents. These reagents perform asymmetric cross coupling with retention of configuration i.e. (R) boranes yield (R) products, which follows suit for (S) boranes, in >90% enantiomeric excess.3
Advantages of Suzuki Coupling
Borane Substrates: Easy to synthesize, stabile, nontoxic, and cost effective1,2
The ease of synthesizing boranes provides convenient access to substrates for Suzuki coupling at a low cost and with minimal health risks. The two common methods to synthesize boronic acids are:
1. React lithium or magnesium Grignard reagents with borate esters, followed by hydrolysis.
1. React arylsilanes with boron tribromide, followed by acid hydrolysis.
Boronic esters are synthesized from boronic acids using alcohols or diols, especially pinicol.
Aryl trifluoroborates are synthesized by subjecting the corresponding alkyl boronic acid or ester to excess KHF2. Aryltrifluoroborates possess beneficial characteristics that allow for robust reactivity, easy purified, and reduced protodeboronation, which quenches the borane reagent.3
Reaction Conditions: Milder and Greener1,2,4
Typically, cross coupling reactions are run in organic conditions; however, Suzuki couplings can be performed in heterogeneous or purely aqueous conditions as organoboranes are water soluble and compatible with water soluble, inorganically supported, ligand-free Pd-catalysts. Boranes are exceptionally nucleophilic, and thus do not require extreme conditions for transmetallation which increases its functional group tolerance. Their high reactivity is also demonstrated in high turnover rates of the Pd catalyst used, which allows for a lower loading.
Organoboranes are nontoxic and stable to extreme heating and exposure to oxygen or water. Consequently, these reagents can be easily used at benchtop, and do not require special equipment or techniques, such as gloveboxes or air-sensitive and dry technique.1 This reaction has a high atom economy seeing that the byproducts are typically salts and water soluble borane byproducts.1.4 The lack of interfering byproducts allows for one-pot syntheses to further reduce waste and increase reaction efficiency.4 These attractive features of organoborane reagents increases its utility and synthetic convenience, and makes Suzuki coupling optimal for large scale and industry synthesis.
Limitations of Suzuki Coupling:
Issues with reactants:
• Borane substrates commercially available do not encompass all R groups desired
• Borane substrates are generally preinstalled to more complicated substrates synthesized in lab, which is often difficult
• Some alkyl borates (sp3) and hetero- substrates don’t show reactivity1
• Chloride substrates react slowly, and with lower yields1
• Without base there are many side products1
Side reactions:
Beta-hydride elimination competes with reductive elimination, which affords a side product that greatly reduces the yield. This is observed with reactants that have beta-hydrides, most commonly alkyl substrates, but sometimes can be avoided by using Ni catalysts or ligands with larger bite angles.2
Scheme 1: Β-hydride elimination in Suzuki cross coupling reaction.2
Synthetic Applications:
Large-scale, industry synthesis of medicinal drugs1
Scheme 1: Suzuki cross coupling reaction to synthesize the drug Linifanib.
Linifanib (Scheme 1) is a tyrosine kinase inhibitor that was FDA approved as an anti-cancer drug that is synthesized on kilogram scale in the presence of different functional groups using Suzuki coupling for its industry production.
Natural products synthesis2
Scheme 2: Suzuki cross coupling reaction application: Epothilone total synthesis.
Suzuki coupling in the total synthesis of Epothilone A (Figure 2) was successful in the presence of numerous stereocenters and functional groups.
Polymer synthesis: Suzuki Polycondensation (SPC)3
Suzuki coupling has been applied to create polymers of aromatic groups, which yielded the longest conjugated chains. These polymers are categorized by their properties and applications:
(a) Polymers with alkyl or alkoxy chains and (c) Polyelectrolytes: SPC yields poly(p-phenylene)’s (PPP’s) which are used in LED lights
(b) Amphiphilic PPPs: SPC produces colloidal structures to create rechargeable, solid state lithium batteries
(d) PPP precursors for ladder polymers: currently being developed to apply their thermal, electrical, and optical properties to conducting materials, energy sources, and instrumentation
(e) Polymers with main-chain chirality and (f) Dendronized PPPs: used in organic synthesis of optically pure compounds
(g) Poly(arylene vinylene)’s and poly(arylene ethinylene)’s: Photoluminescence (PL) and electroluminescence (EL) characteristics for materials and instrumentation
(h) Various polymers: functionalization in synthesis via radical mechanism
Contributors and Attributions
• Peiyuan Qu and Amanda Ramdular
Swern oxidation
Below is a standard procedure for the Swern oxidation of an alcohol.
Procedure
To a solution of oxalyl chloride (5 mmol) and 3 Å MS in CH2Cl2 (1 mL) at -78 °C under N2 is added dropwise a solution of DMSO (10 mmol) in CH2Cl2 (1 mL). After 15 min a solution of the alcohol in CH2Cl2 (3 mL) is slowly added dropwise. After 30 min, Et3N (15 mmol) is added dropwise. The reaction is stirred 30 min at -78 °C then slowly allowed to warm to rt.
Notes:
• Dichloromethane is usually the solvent of choice for this reaction, but Et2O or THF may also be used.
• For secondary alcohols, warming the reaction to -40 °C after the addition of the alcohol for 20 min may help.
• Bulkier amines such as diisopropylethylamine can be used instead of triethylamine in case of epimerization of the α-center
TES protection
Below is a standard procedure for the triethylsilyl (TES) protection of an alcohol.
Using triethylsilyl chloride (TESCl)
Procedure
To an ice-cold (0 °C) solution of the alcohol (1 mmol) in DMF (2 mL, 0.5 M) under N2 is added imidazole (3 mmol). TESCl (2 mmol) is then added dropwise. After complete disappearance of the starting material by TLC, the reaction mixture is quenched by addition of water (1 mL), diluted with Et2O (10 mL) and the layers are separated. The organic layer is washed with water (10 x 3 mL) and brine (2 mL), dried (Na2SO4), filtered and concentrated under reduced pressure. The residue is purified by flash chromatography on silica gel.
Notes:
• if the reaction proceeds efficiently, the equivalents of TESCl/imidazole can be reduced to 1.1/1.5
• if the reaction is sluggish, try slowly warming to rt after the addition of TESCl or use TESOTf instead of TESCl
• DCM can usually be used instead of DMF, but the reaction is generally slower
• Excess TESOH may be hard to remove as it does not stain by TLC. It can be removed by leaving on high vac overnight.
Using triethylsilyl trifluoromethanesulfonate (TESOTf)
Procedure
To a solution of the alcohol (1 mmol) in DCM (2 mL, 0.5 M) under N2 at -78 °C, is added dry 2,6-lutidine (1.5 mmol). TESOTf (1.1 mmol) is then added dropwise. After complete disappearance of the starting material by TLC, the reaction mixture is quenched by addition of sat. aq. NaHCO3 (2 mL), diluted with DCM (5 mL) and brought back to rt. The layers are separated and the aqueous layer is extracted with DCM (3 x 2 mL). The combined organic layers are washed with brine (2 mL), dried (Na2SO4), filtered and concentrated under reduced pressure. The residue is purified by flash chromatography on silica gel.
Notes:
• if the reaction is sluggish, the equivalents of TESOTf/2,6-lutidine can be increased to 2.0/3.0
• if the reaction is sluggish, try slowly warming to -40 °C for a few hours, then to 0 °C or rt
Tamao-Fleming Oxidation
The Denmark Synthesis of (+)-1-Epiaustraline
For the full synthesis see: J. Org. Chem., 2001, 66, 4276–4284. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Named_Reactions/Suzuki_cross-coupling.txt |
http://organicreactions.org/index.ph...tions_chapters
http://www.organicreactions.org/inde...ed_or_Abridged
• Aldol Condensation
An aldol condensation is a condensation reaction in organic chemistry in which an enol or an enolate ion reacts with a carbonyl compound to form a β-hydroxyaldehyde or β-hydroxyketone, followed by dehydration to give a conjugated enone.
• Cannizzaro Reaction
The Cannizzaro reaction, named after its discoverer Stanislao Cannizzaro, is a chemical reaction that involves the base-induced disproportionation of an aldehyde lacking a hydrogen atom in the alpha position.
• Claisen Condensation
The Claisen condensation is a carbon–carbon bond forming reaction that occurs between two esters or one ester and another carbonyl compound in the presence of a strong base, resulting in a β-keto ester or a β-diketone.
• Claisen Rearrangement
The Claisen rearrangement is a carbon–carbon bond-forming chemical reaction discovered by Rainer Ludwig Claisen. The heating of an allyl vinyl ether will initiate a [3,3]-sigmatropic rearrangement to give a γ,δ-unsaturated carbonyl.
• Clemmensen Reduction
Clemmensen reduction is a chemical reaction described as a reduction of carbonyls to alkanes using zinc amalgam and hydrochloric acid.
• Dieckmann Condensation
The Dieckmann condensation is the intramolecular chemical reaction of diesters with base to give β-ketoesters. It is named after the German chemist Walter Dieckmann (1869–1925). The equivalent intermolecular reaction is the Claisen condensation.
• Formation of Cyclic Ketones by Intramolecular Acylation
In chemistry, acylation (rarely, but more formally: alkanoylation) is the process of adding an acyl group to a compound. The compound providing the acyl group is called the acylating agent.
• Hydroamination Reactions of Alkenes
Hydroamination reactions of alkenes represent additions of N-H bonds across carbon-carbon double bonds. Viable substrates for these reactions include unactivated alkenes, vinyl arenes, allenes, and strained alkenes. The scope of the nitrogen-containing reactant includes amines, azoles, and N-protected substrates. A variety of catalysts have been employed including alkali metals, alkaline earth metals, transition metals, and lanthanides.
• Jacobsen Rearrangement
The Jacobsen rearrangement is a chemical reaction, commonly described as the migration of an alkyl group in a sulfonic acid derived from a polyalkyl- or polyhalobenzene
• Mannich Reaction
The Mannich reaction is an organic reaction which consists of an amino alkylation of an acidic proton placed next to a carbonyl functional group by formaldehyde and a primary or secondary amine or ammonia. The final product is a β-amino-carbonyl compound also known as a Mannich base.
• Periodic Acid Oxidation
Oxidation with Periodic Acid is used to cleave vicinal diols (a total of two alcohols, one on two adjacent carbons) into two carbonyl molecules.
• Preparation of Aliphatic Fluorine Compounds
• Preparation of Unsymmetrical Biaryls by the Diazo Reaction and the Nitrosoacetylamine Reaction
Preparation of Unsymmetrical Biaryls by the Diazo Reaction and the Nitrosoacetylamine reaction is an aryl-aryl coupling reaction via a diazonium salt. It is also known as the Gomberg–Bachmann reaction.
• Reduction with Aluminum Alkoxides (The Meerwein-Ponndorf-Verley Reduction)
Meerwein–Ponndorf–Verley (MPV) reduction in organic chemistry is the reduction of ketones and aldehydes to their corresponding alcohols utilizing aluminium alkoxidecatalysis in the presence of a sacrificial alcohol. The beauty of the MPV reduction lies in its high chemoselectivity, and its use of a cheap environmentally friendly metal catalyst.
• Replacement of the Aromatic Primary Amino Group by Hydrogen
The Sandmeyer reaction is a chemical reaction used to synthesize aryl halides from aryl diazonium salts. The reaction is a method for substitution of an aromatic amino group via preparation of its diazonium salt followed by its displacement with a nucleophile. The nucleophile in this case is hypophosphorus acid, H3PO2 .
• Schmidt Reaction
The Schmidt reaction is the reaction of hydrazoic acid or an alkyl azide with a carbonyl compound, alkene, or alcohol, often in the presence of a Brønsted or Lewis acid. Although the family of Schmidt reactions includes a number of variants, they all result in the migration of a substituent from carbon to nitrogen with loss of a molecule of dinitrogen. This reaction has considerable utility for the synthesis of hindered or cyclic amides and amines.
• Wacker Oxidation
The Wacker oxidation refers generally to the transformation of a terminal or 1,2-disubstituted alkene to a ketone through the action of catalytic palladium(II), water, and a co-oxidant.
• Wittig Reaction
The Wittig reaction is a chemical reaction of an aldehyde or ketone with a triphenyl phosphonium ylide (often called a Wittig reagent) to give an alkene and triphenylphosphine oxide.
Thumbnail: a reaction map for ketones (not comprehensive). From Mastering Chemistry (http://www.masterorganicchemistry.com)
Contributors
• This page is based on Organic Reactions. The primary collection of full Organic Reactions chapters is maintained by John Wiley & Sons.
Organic Reactions
An aldol condensation is a condensation reaction in organic chemistry in which an enol or an enolate ion reacts with a carbonyl compound to form a β-hydroxyaldehyde or β-hydroxyketone, followed by dehydration to give a conjugated enone.
Aldol condensations are important in organic synthesis, because they provide a good way to form carbon–carbon bonds. For example, the Robinson annulation reaction sequence features an aldol condensation; the Wieland-Miescher ketone product is an important starting material for many organic syntheses. Aldol condensations are also commonly discussed in university level organic chemistry classes as a good bond-forming reaction that demonstrates important reaction mechanisms. In its usual form, it involves the nucleophilic addition of a ketone enolate to an aldehyde to form a β-hydroxy ketone, or "aldol" (aldehyde + alcohol), a structural unit found in many naturally occurring molecules and pharmaceuticals.
The name aldol condensation is also commonly used, especially in biochemistry, to refer to just the first (addition) stage of the process—the aldol reaction itself—as catalyzed by aldolases. However, the aldol reaction is not formally a condensation reaction because it does not involve the loss of a small molecule.
The reaction between an aldehyde/ketone and an aromatic carbonyl compound lacking an alpha-hydrogen (cross aldol condensation) is called the Claisen-Schmidt condensation. This reaction is named after two of its pioneering investigators Rainer Ludwig Claisen and J. G. Schmidt, who independently published on this topic in 1880 and 1881. An example is the synthesis of dibenzylideneacetone. Quantitative yields in Claisen-Schmidt reactions have been reported in the absence of solvent using sodium hydroxide as the base and plus benzaldehydes.
Mechanism
The first part of this reaction is an aldol reaction, the second part a dehydration—an elimination reaction (Involves removal of a water molecule or an alcohol molecule). Dehydration may be accompanied by decarboxylation when an activated carboxyl group is present. The aldol addition product can be dehydrated via two mechanisms; a strong base like potassium t-butoxide, potassium hydroxide or sodium hydride in an enolate mechanism, or in an acid-catalyzed enol mechanism. Depending on the nature of the desired product, the aldol condensation may be carried out under two broad types of conditions: kinetic control or thermodynamic control.
Condensation Types
It is important to distinguish the aldol condensation from other addition reactions of carbonyl compounds.
• When the base is an amine and the active hydrogen compound is sufficiently activated the reaction is called a Knoevenagel condensation.
• In a Perkin reaction the aldehyde is aromatic and the enolate generated from an anhydride.
• A Claisen condensation involves two ester compounds.
• A Dieckmann condensation involves two ester groups in the same molecule and yields a cyclic molecule
• A Henry reaction involves an aldehyde and an aliphatic nitro compound.
• A Robinson annulation involves a α,β-unsaturated ketone and a carbonyl group, which first engage in a Michael reaction prior to the aldol condensation.
• In the Guerbet reaction, an aldehyde, formed in situ from an alcohol, self-condenses to the dimerized alcohol.
• In the Japp–Maitland condensation water is removed not by an elimination reaction but by a nucleophilic displacement | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Aldol_Condensation.txt |
The Cannizzaro reaction, named after its discoverer Stanislao Cannizzaro, is a chemical reaction that involves the base-induced disproportionation of an aldehyde lacking a hydrogen atom in the alpha position
By Krishnavedala (Own work) [Public domain], via Wikimedia Commons
Cannizzaro first accomplished this transformation in 1853, when he obtained benzyl alcohol and potassium benzoate from the treatment of benzaldehyde with potash (potassium carbonate). More typically, the reaction would be conducted with sodium or potassium hydroxide:
2 C6H5CHO + KOH → C6H5CH2OH + C6H5COOK
The oxidation product is a salt of a carboxylic acid and the reduction product is an alcohol.
Mechanism
The reaction involves a nucleophilic acyl substitution on an aldehyde, with the leaving group concurrently attacking another aldehyde in the second step. First, hydroxide attacks a carbonyl. The resulting tetrahedral intermediate then collapses, re-forming the carbonyl and transferring hydride to attack another carbonyl. In the final step of the reaction, the acid and alkoxide ions formed exchange a proton. In the presence of a very high concentration of base, the aldehyde first forms a doubly charged anion from which a hydride ion is transferred to the second molecule of aldehyde to form carboxylate and alkoxide ions. Subsequently, the alkoxide ion acquires a proton from the solvent.
By Krishnavedala (Own work) [Public domain], via Wikimedia Commons
Overall, the reaction follows third-order kinetics. It is second order in aldehyde and first order in base:
\[rate = k[RCHO]^2[OH^−]\]
At very high base a second path (k') becomes important that is second order in base:
\[rate = k[RCHO]^2[OH^−] + k'[RCHO]^2[OH^−]^2\]
The k' pathway implicates a reaction between the doubly charged anion (RCHO22) and the aldehyde. The direct transfer of hydride ion is evident from the observation that the recovered alcohol does not contain any deuterium attached to the α-carbon when the reaction is performed in the presence of D2O.
Scope
Due to the strongly alkaline reaction conditions, aldehydes that have alpha hydrogen atom(s) instead undergo deprotonation there, leading to enolates and possible aldol reactions. Under ideal conditions the reaction produces only 50% of the alcohol and the carboxylic acid (it takes two aldehydes to produce one acid and one alcohol). To avoid the low yields, it is more common to conduct the crossed Cannizzaro reaction, in which a sacrificial aldehyde is used in combination with a more valuable chemical. In this variation, the reductant is formaldehyde, which is oxidized to sodium formate and the other aldehyde chemical is reduced to the alcohol. In this scenario, each of the two separate aldehydes can be converted completely to its corresponding product rather than losing 50% of a single reactant to each of two different products. Thus, the yield of the valuable chemical is high, although the atom economy is still low.
A solvent-free reaction has been reported involving grinding liquid 2-chlorobenzaldehyde with potassium hydroxide in a mortar and pestle:
V8rik at English Wikipedia [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0) or GFDL (http://www.gnu.org/copyleft/fdl.html)], via Wikimedia Commons
Certain ketones can undergo a Cannizzaro-type reaction, transferring one of their two carbon groups rather than the hydride that would be present on an aldehyde.
Variations
In the Tishchenko reaction, the base used is an alkoxide rather than hydroxide, and the product is an ester rather than the separate alcohol and carboxylate groups. After the nucleophilic base attacks an aldehyde, the resulting new oxygen anion attacks another aldehyde to give a hemiacetal linkage between two of the formerly aldehyde-containing reactants rather than undergoing tetrahedral collapse. Eventually tetrahedral collapse does occur, giving the stable ester product.
By Calvero. (Selfmade with ChemDraw.) [Public domain], via Wikimedia Commons
See also
• Meerwein–Ponndorf–Verley reduction
• Oppenauer oxidation
Contributors
Wikipedia (CC-BY-SA-3.0) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Cannizzaro_Reaction.txt |
The Claisen condensation is a carbon–carbon bond forming reaction that occurs between two esters or one ester and another carbonyl compound in the presence of a strong base, resulting in a β-keto ester or a β-diketone. It is named after Rainer Ludwig Claisen, who first published his work on the reaction in 1887.
Requirements
At least one of the reagents must be enolizable (have an α-proton and be able to undergo deprotonation to form the enolate anion). There are a number of different combinations of enolizable and nonenolizable carbonyl compounds that form a few different types of Claisen condensations.
The base used must not interfere with the reaction by undergoing nucleophilic substitution or addition with a carbonyl carbon. For this reason, the conjugate sodium alkoxide base of the alcohol formed (e.g. sodium ethoxide if ethanol is formed) is often used, since the alkoxide is regenerated. In mixed Claisen condensations, a non-nucleophilic base such as lithium diisopropylamide, or LDA, may be used, since only one compound is enolizable. LDA is not commonly used in the classic Claisen or Dieckmann condensations due to enolization of the electrophilic ester.
The alkoxy portion of the ester must be a relatively good leaving group. Methyl and ethyl esters, which yields methoxide and ethoxide, respectively, are commonly used.
Types
• The classic Claisen condensation, a self-condensation between two molecules of a compound containing an enolizable ester.
• The mixed (or "crossed") Claisen condensation, where one enolizable ester or ketone and one nonenolizable ester are used.
• The Dieckmann condensation, where a molecule with two ester groups reacts intramolecularly, forming a cyclic β-keto ester. In this case, the ring formed must not be strained, usually a 5- or 6-membered chain or ring.
Mechanism
In the first step of the mechanism, an α-proton is removed by a strong base, resulting in the formation of an enolate anion, which is made relatively stable by the delocalization of electrons. Next, the carbonyl carbon of the (other) ester is nucleophilically attacked by the enolate anion. The alkoxy group is then eliminated (resulting in (re)generation of the alkoxide), and the alkoxide removes the newly formed doubly α-proton to form a new, highly resonance-stabilized enolate anion. Aqueous acid (e.g. sulfuric acid or phosphoric acid) is added in the final step to neutralize the enolate and any base still present. The newly formed β-keto ester or β-diketone is then isolated. Note that the reaction requires a stoichiometric amount of base as the removal of the doubly α-proton thermodynamically drives the otherwise endergonic reaction. That is, Claisen condensation does not work with substrates having only one α-hydrogen because of the driving force effect of deprotonation of the β-keto ester in the last step.
Stobbe condensation
The Stobbe condensation is a modification specific for the diethyl ester of succinic acid requiring less strong bases. An example is its reaction with benzophenone:
A reaction mechanism that explains the formation of both an ester group and a carboxylic acid group is centered on a lactone intermediate (5):
Claisen Condensation
The Perkin reaction is an organic reaction developed by William Henry Perkin that is used to make cinnamic acids. It gives an α,β-unsaturated aromatic acid by the aldol condensation of an aromatic aldehyde and an acid anhydride, in the presence of an alkali salt of the acid. The alkali salt acts as a base catalyst, and other bases can be used instead.
Several reviews have been written.
Reaction mechanism
The above mechanism is not universally accepted, as several other versions exist, including decarboxylation without acetic group transfer.
Applications
• One notable application for the Perkin reaction is in the laboratory synthesis of the phytoestrogenic stilbene resveratrol (c.f. Fo-ti).
Perkin Reaction
The Elbs reaction is an organic reaction describing the pyrolysis of an ortho methyl substituted benzophenone to condensed polyaromatic. The reaction is named after its inventor, the German chemist Karl Elbs also responsible for the Elbs oxidation. The reaction was published in 1884.
Elbs however did not correctly interpret the reaction product due to a lack of knowledge about naphthalene structure.
Scope
The Elbs reaction enables the synthesis of condensed aromatic systems. As already demonstrated by Elbs in 1884 it is possible to obtain anthracene through dehydration. Larger aromatic systems like pentacene are also feasible. This reaction does not take place in a single step but leads first to dihydropentacene that is dehydrogenated in a second step with copper as a catalyst.
The acyl compounds required for this reaction can be obtained through a Friedel-Crafts acylation with aluminum chloride.
Variations
It is also possible to synthesise heterocyclic compounds via the Elbs reaction. In 1956 an Elbs reaction of a thiophene derivative was published. The expected linear product was not obtained due to a change in reaction mechanism after formation of the first intermediate which caused multiple free radical reaction steps.
Elbs Reaction
The Bucherer reaction in organic chemistry is the reversible conversion of a naphthol to a naphthylamine in the presence of ammonia and sodium bisulfite. The reaction is widely used in the synthesis of dye precursors aminonaphthalenesulfonic acids.
C10H7-2-OH + NH3 ⇌ C10H7-2-NH2 + H2O
The French chemist Robert Lepetit was the first to discover the reaction in 1898. The German chemist Hans Theodor Bucherer (1869–1949) discovered (independent from Lepetit) its reversibility and its potential especially in industrial chemistry. Bucherer published his results in 1904 and his name is connected to this reaction. The organic reaction also goes by the name Bucherer-Lepetit reaction or (wrongly) the Bucherer-Le Petit reaction.
The reaction is used to convert 1,7-dihydroxynaphthalene into 7-amino-1-naphthol and 1-aminonaphthalene-4-sulfonic acid into 1-hydroxynaphthalene-4-sulfonic acid.It is also useful for transamination reactions of 2-aminonaphthalenes.
Mechanism
in the first step of the reaction mechanism a proton adds to a carbon atom with high electron density therefore by preference to C2 or C4 of naphthol (1). This leads to resonance stabilized adducts 1a-1e.
De-aromatization of the first ring of the naphthalene system occurs at the expense of 25 kcal/mol. In the next step a bisulfite anion adds to C3 through 1e. This results in the formation of 3a which tautomerizes to the more stable 3b to the sulfonic acid of tetralone. A nucleophilic addition follows of the amine with formation of 4a and its tautomer 4b loses water to form the resonance stabilized cation 5a. This compound is deprotonated to the imine 5b or the enamine 5c but an equilibrium exists between both species. The enamine eliminates sodium bisulfite with formation of naphthylamine 6.
It is important to stress that this is a reversible reaction. The reaction is summarized as follows:
The Bucherer carbazole synthesis is a related reaction. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Claisen_Condensation/Bucherer_reaction.txt |
The Claisen rearrangement (not to be confused with the Claisen condensation) is a powerful carbon–carbon bond-forming chemical reaction discovered by Rainer Ludwig Claisenin 1912. The heating of an allyl vinyl ether will initiate a [3,3]-sigmatropic rearrangement to give a γ,δ-unsaturated carbonyl.
Figure 1: Wikimedia (Own work) [GFDL CC-BY-SA-3.0; ~K)
The Claisen rearrangement is the first recorded example of a [3,3]-sigmatropic rearrangement.
The Claisen rearrangement is an exothermic, concerted (bond cleavage and recombination) pericyclic reaction. Woodward–Hoffmann rules show a suprafacial, stereospecific reaction pathway. The kinetics are of the first order and the whole transformation proceeds through a highly ordered cyclic transition state and is intramolecular. Crossover experiments eliminate the possibility of the rearrangement occurring via an intermolecular reaction mechanism and are consistent with an intramolecular process.
There are substantial solvent effects observed in the Claisen rearrangement, where polar solvents tend to accelerate the reaction to a greater extent. Hydrogen-bonding solvents gave the highest rate constants. For example, ethanol/water solvent mixtures give rate constants 10-fold higher than sulfolane. Trivalent organoaluminium reagents, such as trimethylaluminium, have been shown to accelerate this reaction.
The first reported Claisen rearrangement is the [3,3]-sigmatropic rearrangement of an allylphenyl ether to intermediate 1, which quickly tautomerizes to an ortho-substituted phenol.
Figure 2: By Kchemyoung (Own work) [CC BY-SA 4.0, Kchemyoung)
Meta-substitution affects the regioselectivity of this rearrangement. For example, electron withdrawing groups (e.g. bromide) at the meta-position direct the rearrangement to the ortho-position (71% ortho-product), while electron donating groups (e.g. methoxy), direct rearrangement to the para-position (69% para-product). Additionally, presence of ortho-substituents exclusively leads to para-substituted rearrangement products (tandem Claisen and Cope rearrangement).
Figure 3: (Public Domain; RAN 10).
If an aldehyde or carboxylic acid occupies the ortho or para positions, the allyl side-chain displaces the group, releasing it as carbon monoxide or carbon dioxide, respectively.
Bellus–Claisen rearrangement
The Bellus–Claisen rearrangement is the reaction of allylic ethers, amines, and thioethers with ketenes to give γ,δ-unsaturated esters, amides, and thioesters. This transformation was serendipitously observed by Bellus in 1979 through their synthesis of a key intermediate of an insecticide, pyrethroid. Halogen substituted ketenes (R1, R2) are often used in this reaction for their high electrophilicity. Numerous reductive methods for the removal of the resulting α-haloesters, amides and thioesters have been developed. The Bellus-Claisen offers synthetic chemists a unique opportunity for ring expansion strategies.
at the English language Wikipedia [CC-BY-SA-3.0 , GFDL, CC-BY-SA-3.0 or CC BY-SA 3.0 , from Wikimedia Commons
Eschenmoser–Claisen rearrangement
The Eschenmoser–Claisen rearrangement proceeds by heating allylic alcohols in the presence of N,N-dimethylacetamide dimethyl acetal to form γ,δ-unsaturated amide. This was developed by Albert Eschenmoser in 1964.Eschenmoser-Claisen rearrangement was used as a key step in the total synthesis of morphine.
Wikimedia CommonsBy ~K (Own work) [GFDL (http://www.gnu.org/copyleft/fdl.html) or CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0/)], via Wikimedia Commons
Ireland–Claisen rearrangement
The Ireland–Claisen rearrangement is the reaction of an allylic carboxylate with a strong base (such as lithium diisopropylamide) to give a γ,δ-unsaturated carboxylic acid. The rearrangement proceeds via silylketene acetal, which is formed by trapping the lithium enolate with chlorotrimethylsilane. Like the Bellus-Claisen (above), Ireland-Claisen rearrangement can take place at room temperature and above. The E- and Z-configured silylketene acetals lead to anti and syn rearranged products, respectively.[30] There are numerous examples of enantioselective Ireland-Claisen rearrangements found in literature to include chiral boron reagents and the use of chiral auxiliaries.
Figure 5: (CC BY-SA 4.0; Kchemyoung)
Johnson–Claisen rearrangement
The Johnson–Claisen rearrangement is the reaction of an allylic alcohol with an orthoester to yield a γ,δ-unsaturated ester. Weak acids, such as propionic acid, have been used to catalyze this reaction. This rearrangement often requires high temperatures (100 to 200 °C) and can take anywhere from 10 to 120 hours to complete. However, microwave assisted heating in the presence of KSF-clay or propionic acid have demonstrated dramatic increases in reaction rate and yields.
Mechanism:
By Self Made by RAN 10 (-) [CC0], via Wikimedia Commons
Photo-Claisen rearrangement
The photo-Claisen rearrangement is closely related to the photo-Fries rearrangement, that proceeds through a similar radical mechanism. Aryl ethers undergo the photo-Claisen rearrangement, while the photo-Fries rearrangement utilizes aryl esters.
Aza–Claisen
An iminium can serve as one of the pi-bonded moieties in the rearrangement.
By No machine-readable author provided. ~K assumed (based on copyright claims). [Public domain], via Wikimedia Commons
Chromium oxidation
Chromium can oxidize allylic alcohols to alpha-beta unsaturated ketones on the opposite side of the unsaturated bond from the alcohol. This is via a concerted hetero-Claisen reaction, although there are mechanistic differences since the chromium atom has access to d- shell orbitals which allow the reaction under a less constrained set of geometries.
By The original uploader was Takometer at English Wikipedia [CC BY 2.5, via Wikimedia Commons
Chen–Mapp reaction
The Chen–Mapp reaction also known as the [3,3]-Phosphorimidate Rearrangement or Staudinger–Claisen Reaction installs a phosphite in the place of an alcohol and takes advantage of the Staudinger reduction to convert this to an imine. The subsequent Claisen is driven by the fact that a P=O double bond is more energetically favorable than a P=N double bond.
By Howcheng at en.Wikipedia [CC BY 2.5], via Wikimedia Commons
Overman rearrangement
The Overman rearrangement (named after Larry Overman) is a Claisen rearrangement of allylic trichloroacetimidates to allylic trichloroacetamides.
By Yikrazuul (Own work) [Public domain], via Wikimedia Commons
Overman rearrangement is applicable to synthesis of vicinol diamino comp from 1,2 vicinal allylic diol.
Zwitterionic Claisen rearrangement
Unlike typical Claisen rearrangements which require heating, zwitterionic Claisen rearrangements take place at or below room temperature. The acyl ammonium ions are highly selective for Z-enolates under mild conditions
https://upload.wikimedia.org/wikiped...ent_Scheme.png
Claisen rearrangement in nature
The enzyme Chorismate mutase (EC 5.4.99.5) catalyzes the Claisen rearrangement of chorismate ion to prephenate ion, a key intermediate in the shikimic acid pathway (the biosynthetic pathway towards the synthesis of phenylalanine and tyrosine).
By No machine-readable author provided. ~K assumed (based on copyright claims). [Public domain], via Wikimedia Commons | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Claisen_Rearrangement.txt |
Clemmensen reduction is a chemical reaction described as a reduction of ketones (or aldehydes) to alkanes using zinc amalgam and hydrochloric acid. This reaction is named after Erik Christian Clemmensena Danish chemist.
The Clemmensen reduction is particularly effective at reducing aryl-alkyl ketones, such as those formed in a Friedel-Crafts acylation. With aliphatic or cyclic ketones, zinc metal reduction is much more effective.
The substrate must be unreactive to the strongly acidic conditions of the Clemmensen reduction. Acid sensitive substrates should be reacted in the Wolff-Kishner reduction, which utilizes strongly basic conditions; a further, milder method is the Mozingo reduction. The oxygen atom is lost in the form of one molecule of water.
However, the reaction is not suitable for substances sensitive to acids. Also, -COOH group can't be reduced by this method. (-COOH group can be reduced by treating it with soda lime and then heating).
In spite of the antiquity of this reaction, the mechanism of the Clemmensen reduction remains obscure. Due to the heterogeneous nature of the reaction, mechanistic studies are difficult, and only a handful of studies have been disclosed. Proposal mechanisms invoke organozinc intermediates, possibly zinc carbenoids, either as discrete species or with the organic fragment bound to the metal surface. However, the corresponding alcohol is not believed to be an intermediate, since subjection of alcohols to Clemmensen conditions generally does not afford the alkane product.
Clemmensen Reduction
The Arndt–Eistert synthesis is a series of chemical reactions designed to convert a carboxylic acid to a higher carboxylic acid homologue (i.e. contains one additional carbon atom) and is considered a homologation process. Named for the German chemists Fritz Arndt (1885–1969) and Bernd Eistert (1902–1978), Arndt–Eistert synthesis is a popular method of producing β-amino acids from α-amino acids. Acid chlorides react with diazomethane to give diazoketones. In the presence of a nucleophile (water) and a metal catalyst (Ag2O), diazoketones will form the desired acid homologue.
While the classic Arndt–Eistert synthesis uses thionyl chloride to convert the starting acid to an acid chloride, any procedure can be used that will generate an acid chloride.
Diazoketones are typically generated as described here, but other methods such as diazo-group transfer can also apply.
Since diazomethane is toxic and violently explosive, many safer alternatives have been developed, such as the usage of ynolates (Kowalski ester homologation) or diazo(trimethylsilyl)methane.
Reaction mechanism
The key step in the Arndt–Eistert synthesis is the metal-catalyzed Wolff rearrangement of the diazoketone to form a ketene. In the insertion homologation of t-BOC protected (S)-phenylalanine (2-amino-3-phenylpropanoic acid), t-BOC protected (S)-3-amino-4-phenylbutanoic acid is formed.
Wolff rearrangement of the α-diazoketone intermediate forms a ketene via a 1,2-rearrangement, which is subsequently hydrolysed to form the carboxylic acid. The consequence of the 1,2-rearrangement is that the methylene group alpha to the carboxyl group in the product is the methylene group from the diazomethane reagant. It has been demonstrated that the rearrangement preserves the stereochemistry of the chiral centre as the product formed from t-BOC protected (S)-phenylalanine retains the (S) stereochemistry with a reported enantiomeric excess of at least 99%.
Heat, light, platinum, silver, and copper salts will also catalyze the Wolff rearrangement to produce the desired acid homologue.
Variations
In the Newman–Beal modification, addition of triethylamine to the diazomethane solution will avoid the formation of α-chloromethylketone side-products.
Blanc chloromethylation
The Blanc chloromethylation (also called the Blanc reaction) is the chemical reaction of aromatic rings with formaldehyde and hydrogen chloride catalyzed by zinc chloride or other Lewis acid to form chloromethyl arenes. The reaction was discovered by Gustave Louis Blanc (1872-1927) in 1923. The reaction is performed with care as, like most chloromethylation reactions, it produces highly carcinogenic bis(chloromethyl) ether as a by-product.
Mechanism
The reaction is carried out under acidic conditions and with a ZnCl2 catalyst. These conditions protonate the formaldehyde carbonyl making the carbon much more electrophilic. The aldehyde is then attacked by the aromatic pi-electrons, followed by rearomatization of the aromatic ring. The benzyl alcohol thus formed is quickly converted to the chloride under the reaction conditions.
Although the reaction is an efficient means of introducing a chloromethyl group, the production of small amounts of highly carcinogenic bis(chloromethyl) ether is a disadvantage.
Related chloromethylations
Chloromethylation can also be effected using chloromethyl methyl ether:
ArH + CH3OCH2Cl → ArCH2Cl + CH3OH
This reaction is employed in the chloromethylation of styrene in the production of ion-exchange resins and Merrifield resins. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Clemmensen_Reduction/Arndt-Eister_reaction.txt |
The Chichibabin reaction is a method for producing 2-aminopyridine derivatives by the reaction of pyridine with sodium amide. It was reported by Aleksei Chichibabin in 1914. The following is the overall form of the general reaction:
The direct amination of pyridine with sodium amide takes place in liquid ammonia. Following the addition elimination mechanism first a nucleophilic NH2 is added while a hydride (H) is leaving. Ciganek describes an example of an intramolecular Chichibabin reaction in which a nitrile group on a fused ring is the source of nitrogen in amination.
Mechanism
It is widely accepted that the Chichibabin reaction mechanism is an addition-elimination reaction that proceeds through an σ-adduct (Meisenheimer adduct) intermediate (the third structure). First, the nucleophilic NH2 group adds to the δ+ ring carbon pushing electrons onto the ring nitrogen and forming the anionic σ-adduct, which is stabilized by sodium. Electrons from the nitrogen are then pushed towards the ring forming a C=N bond and kicking out a hydride ion. The hydride ion abstracts a hydrogen from the positively charged nitrogen, forming hydrogen gas. The ring nitrogen then pushes electrons back into the ring, regaining aromaticity, the now negatively charged NH group abstracts a proton from water giving us the product.
Reaction progress can be measured by the formation of hydrogen gas and red color from σ-adduct formation. Sodium amide is a handy reagent for the Chichibabin reaction but handling it can be dangerous and caution is advised.
• σ-adduct (Meisenheimer adduct) formation
Evidence indicates that before addition of the amino group, the ring nitrogen is sorbed onto the surface of sodium amide and the sodium cation forms a coordination complex. This increases the δ+ on the α-carbon, thus 1,2-addition of sodium amide is favored over 1,4-addition. The proximity of the amino group to the α-carbon once the coordination complex is formed also makes the 1,2-addition more likely to occur.
Some data exists that supports a single electron transfer as the proposed pathway for σ-adduct formation.
In most cases, the anionic σ-adduct is unstable making its formation the rate determining step.
• Hydride ion elimination
In addition to the mechanism shown above, other pathways have been proposed for the elimination step. The mechanism above, loss of the hydride ion followed by abstraction of a proton, is supported by the fact that the nucleophile needs at least one hydrogen atom for the reaction to proceed. Another competing pathway could be the elimination of hydride by sodium to form sodium hydride.
Factors influencing reaction
Different aromatic nitrogen heterocyclic compounds proceed through the Chichibabin reaction in a matter of minutes and others can take hours. Factors that influence the reaction rate include:
• Basicity - The ideal pKa range is 5-8 and the reaction either does not proceed, or proceeds poorly outside of this range. The reaction occurs faster under more basic conditions but only up to a point because when electron density builds up on the α-carbon, it makes it less electrophilic. The strongest base known to aminate is 4-dimethylaminopyridine (pKa 9.37).
• δ+ on α-carbon - For kinetically controlled additions, the rate of amination is related to the magnitude of the partial positive charge on the carbon next to the ring nitrogen. For thermodynamically controlled additions, the rate of amination is related to the stability of the σ-adduct.
• Ease of hydride elimination - Success of this reaction is also dependent on the ease at which the hydride ion leaves and the ring regains aromaticity. The rate of amination for three azoles proceeds quickest to slowest as follows: 1-methylbenzimidazole > 1-methylnaphth-[2,3-d]imidazole > 3-methylnaphth[1,2-d]imidazole. Since the addition of the amide ion proceeds quickly with these substrates, the differences in reaction rates is most likely their propensity for hydride elimination and reformation of an aromatic ring.
• Substituents - Electron-withdrawing groups inhibit the Chichibabin reaction. Three proposed ideas of why this is are (1) they decrease the basicity of the ring nitrogen and slow down the sorption on sodium amide, (2) these electron-withdrawing groups can also form complexes with sodium amide, and (3) for single electron transfer pathway, altering the distribution of spin density of the intermediate radical anion.
Substrates with σ-dimethoxy groups don't aminate because they form a stable complex with sodium amide.
Electron-donating groups also inhibit the Chichibabin reaction because of their deactivating effects.
• Benzo annelation - Since the hydride ion is a poor leaving group, benzo annelation increases reactivity of the substrate in the Chichibabin reaction. This is demonstrated by the fact that 1-methylimidazole does not work as a substrate, but 1-methylbenzimidazole reacts easily.
• Solvent - The ability of the polar anionic σ-adduct to form will depend on the solvating capacity and the dielectric constant of the solvent.
• Temperature - The rule of thumb in aprotic solvents (where σ-adduct formation is the rate determining step) is to run the reaction at the lowest temperature for good hydrogen evolution to avoid the decomposition that occurs at high temperatures.
Side reaction
Dimerization is a side reaction that can occur. When heated in xylene and sodium amide at atmospheric pressure, the substrate 4-tert-butylpyridine produces 89% of the dimer product (4,4'-di-tert-butyl-2,2'-bipyridine) and only 11% of the aminated Chichibabin product (2-amino-4-tert-butylpyridine). When subjected to 350 psi nitrogen pressure and the same conditions, the yields are 74% of the aminated Chichibabin product and 26% of the dimer product. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Clemmensen_Reduction/Chichibabin_Reaction.txt |
The Reformatsky reaction (sometimes spelled Reformatskii reaction) is an organic reaction which condenses aldehydes or ketones, with α-halo esters, using a metallic zinc to form β-hydroxy-esters:
The organozinc reagent, also called a 'Reformatsky enolate', is prepared by treating an alpha-halo ester with zinc dust. Reformatsky enolates are less reactive than lithium enolates or Grignard reagents and hence nucleophilic addition to the ester group does not occur. The reaction was discovered by Sergey Nikolaevich Reformatsky.
Structure of the reagent
The crystal structures of the THF complexes of the Reformatsky reagents tert-butyl bromozincacetate and ethyl bromozincacetate have been determined. Both form cyclic eight-membered dimers in the solid state, but differ in stereochemistry: the eight-membered ring in the ethyl derivative adopts a tub-shaped conformation and has cis bromo groups and cis THF ligands, whereas in the tert-butyl derivative, the ring is in a chair form and the bromo groups and THF ligands are trans.
ethyl bromozincacetate dimer tert-butyl bromozincacetate dimer
Reaction Mechanism
Zinc metal is inserted into the carbon-halogen bond of the α-haloester by oxidative addition 1. This compound dimerizes and rearranges to form two zinc enolates 2. The oxygen on an aldehyde or ketone coordinates to the zinc to form the six-member chair like transition state 3. A rearrangement occurs in which zinc switches to the aldehyde or ketone oxygen and a carbon-carbon bond is formed 4. Acid workup 5,6 removes zinc to yield zinc(II) salts and a β-hydroxy-ester 7.
Variations
In one variation of the Reformatsky reaction an iodolactam is coupled with an aldehyde with triethylborane in toluene at -78 °C.
Dieckmann Condensation
The Dieckmann condensation is the intramolecular chemical reaction of diesters with base to give β-ketoesters. It is named after the German chemist Walter Dieckmann (1869–1925). The equivalent intermolecular reaction is the Claisen condensation.
Reaction Mechanism
Deprotonation of an ester at the α-position generates an enolate ion which then undergoes a 5-exo-trig nucleophilic attack to give a cyclic enol. Protonation with a Bronsted-Lowry acid (H3O+ for example) re-forms the β-keto ester.[6]
Owing to the steric stability of five- and six-membered ring structures, these will preferentially be formed. So 1,6 diesters will form five-membered cyclic β-keto esters, while 1,7 diesters will form six-membered β-keto esters.[7]
Animation of the reaction mechanism
See also
• Claisen condensation
• Gabriel-Colman rearrangement
Contributors
• Wikipedia (CC-BY-SA-3.0)
Formation of Cyclic Ketones by Intramolecular Acylation
Acylation is the process of adding an acyl group to a compound and the compound providing the acyl group is called the acylating agent. Acyl halides are commonly used as acylating agents because they form a strong electrophile when treated with some metal catalysts. For example, Friedel-Crafts acylation uses acetyl chloride (ethanoyl chloride), CH3COCl, as the agent and aluminum chloride (AlCl3) as a catalyst to add an ethanoyl (acetyl) group to benzene:
By Arrowsmaster (Own work) [Public domain], via Wikimedia Commons
The mechanism of this reaction is electrophilic aromatic substitution. Acyl halides and anhydrides of carboxylic acids are also commonly used acylating agents to acylate amines to form amides or acylate alcohols to form esters. The amines and alcohols are nucleophiles; the mechanism is a nucleophilic acyl substitution. Succinic acid is also commonly used in a specific type of acylation called succination. Oversuccination occurs when more than one succinate adds to a single compound.
Acylation can be used to prevent rearrangement reactions that would normally occur in alkylation. To do this an acylation reaction is performed, then the carbonyl is removed by clemmensen reduction or a similar process.
Acylation in biology
Protein acylation is the post-translational modification of proteins via the attachment of functional groups through acyl linkages. One prominent type is fatty acylation, the addition of fatty acids to particular amino acids (e.g. myristoylation or palmitoylation). Protein acylation has been observed as a mechanism of biological signaling.
Contributors
• Wikipedia (CC-BY-SA-3.0) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Clemmensen_Reduction/Reformatsky_Reaction.txt |
Hydroamination reactions of alkenes represent additions of N-H bonds across carbon-carbon double bonds. Viable substrates for these reactions include unactivated alkenes, vinyl arenes, allenes, and strained alkenes. The scope of the nitrogen-containing reactant includes amines, azoles, and N-protected substrates. A variety of catalysts have been employed including alkali metals, alkaline earth metals, transition metals, and lanthanides.[1]
Introduction
As a consequence of the importance of nitrogen-containing compounds in biological and pharmaceutical applications, methods for the construction of carbon-nitrogen bonds have been heavily studied for at least 150 years. Hydroamination, the addition of HNR2 across a carbon-carbon π bond, is in principle one of the most atom-economical methods for the synthesis of this class of compounds. Catalytic hydroaminations of alkenes, allenes, and dienes (which may be inter- or intramolecular) afford amines, imines, and enamines (Eq. 1).
The scope of amine substrates includes ammonia, primary and secondary aliphatic and aromatic amines, azoles, hydrazines, and N-protected amines. Transition metal, rare earth, and alkali metal catalysts have been applied successfully, although the appropriate choice of catalyst depends on the unsaturated and amine substrates.
In general, hydroaminations present issues of site selectivity (Markovnikov/anti-Markovnikov) and enantioselectivity. Site selectivity is a function of the unsaturated substrate as well as the catalyst. The scope of enantioselective hydroamination is very broad; however, a particular catalyst system is typically limited to a fairly narrow range of substrates.
Mechanism and Stereochemistry
Prevailing Mechanisms
Mechanisms of alkene hydroaminations depend strongly on the catalyst system employed. Catalysts that include electropositive elements such as alkali, alkaline earth, and rare earth metals typically operate through a metal-amido species that undergoes nucleophilic addition to the alkene (Eq. 2).[2] The site selectivity of this addition depends on the substitution of the alkene: whereas aliphatic alkenes typically undergo Markovnikov addition, aromatic alkenes more commonly engage in anti-Markovnikov addition due to stabilization of the resulting benzylic metal intermediate.
Hydroaminations of allenes catalyzed by Group 4 metals proceed by a different mechanism involving a metal imido intermediate (Eq. 3).[3] After formation of a bis-amido precursor, α-elimination generates the metal imido species, which engages in [2+2] cycloaddition with the allene substrate. Protonation then generates an enamido complex with an additional amido ligand; α-elimination then regenerates the metal-imido species.
Mechanisms of hydroaminations catalyzed by late transition metals are not as well understood as the mechanisms discussed above. Nonetheless, it is generally believed that these reactions involve either amine activation via N-H insertion or alkene activation via π-coordination. Intermolecular hydroamination of norbornene catalyzed by an iridium(I) complex provides an example of the former (Eq. 4).[4] Oxidative addition of the amine is followed by insertion of the alkene and reductive elimination.
Group 9 and 10 metal complexes tend to react via an alkene activation pathway involving coordination of the alkene to the metal center and external attack of the amine on the coordinated alkene (Eq. 5).[5] Product formation may occur either by proton transfer to the metal center followed by reductive elimination or direct protonation of the β-aminoalkyl ligand.
Scope and Limitations
The scope of alkene hydroaminations is very broad and includes unactivated and activated alkenes as well as primary and secondary alkyl- and arylamines. However, this broad substrate scope is accompanied by an almost equally broad array of potential catalyst systems. Careful consideration must thus be given to the nature of the alkene and amine substrates in choosing the ideal catalyst. For example, late transition metal catalysts are generally incompatible with basic alkylamines and require either N-protected or aniline substrates.
Small unactivated alkenes generally require harsh conditions to undergo intermolecular hydroaminations, particularly when ammonia is used.[6] Catalyst decomposition is often a significant concern in reactions of alkylamines with unactivated alkenes. On the other hand, less basic anilines are more amenable to reaction with unactivated alkenes; for example, a catalyst system based on rhodium trichloride is effective in the reaction of ethylene with anilines (Eq. 6).[7]
(6)
Unactivated alkenes are significantly more reactive in intramolecular hydroaminations and the scope of available catalysts is accordingly much broader. Rare earth metal catalysts are among the most active for intramolecular reactions of amino alkenes (Eq. 7).[8] Exclusive exo selectivity is observed in these reactions and in related reactions employing alkali or alkaline earth metal catalysts.
(7)
Early transition metal catalysts are also effective in intramolecular hydroaminations of amino alkenes (Eq. 8).[9] In this case, cyclization occurs efficiently even in the absence of geminal substituents in the tethering alkyl chain.
(8)
Conjugation significantly increases the reactivity of the alkene in vinyl arenes, although control of site selectivity can be difficult. For example, sodium metal catalyzes the addition of primary or secondary aliphatic amines to styrene with anti-Markovnikov selectivity (Eq. 9).[10]
(9)
Intramolecular reactions of vinyl arenes proceed similarly to reactions of unactivated amino alkenes described above. Catalysts based on rare earth metals are among the most active and react with exo selectivity (Eq. 10).[11]
(10)
Although 1,3-dienes display comparable reactivity to vinyl arenes in hydroaminations, the possibility of 1,2- or 1,4-addition complicates this class of reactions. Late transition metal catalysts are known to catalyze the 1,4-addition of anilines to acyclic 1,3-dienes (Eq. 11).[12] Related reactions of alkylamines show a lack of site selectivity and issues with diene oligomerization.[13]
(11)
Allenes may undergo hydroamination to give enamines, imines, or allyl amines. Early transition metal catalysts such as the titanium half-sandwich imido complex 1 promote formation of imines exclusively (Eq. 12).[14]
(12)
Late transition metal catalysts are more active in the formation of allyl amines. For example, the gold(I) N-heterocyclic carbene complex 2 catalyzes the addition of N-protected amines to substituted allenes at room temperature (Eq. 13).[15]
(13)
Along with conjugated alkenes and cumulenes, strained alkenes represent a third class of substrates activated toward hydroamination. Like allenes, methylenecyclopropanes may afford imines, enamines, or allyl amines upon hydroamination. For instance, rare earth metal complexes catalyze the hydroamination of methylenecyclopropane to afford imines with ring opening (Eq. 14).[2]
(14)
In analogy to reactions of allenes, late transition metal catalysts react with substituted methylenecyclopropanes to afford allylamines (Eq. 15).[16]
(15) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Hydroamination_Reactions_of_Alkenes.txt |
The Jacobsen rearrangement is a chemical reaction, commonly described as the migration of an alkyl group in a sulfonic acid derived from a polyalkyl- or polyhalobenzene:
https://upload.wikimedia.org/wikiped...e/Jacobsen.png
The exact reaction mechanism is not completely clear, but evidence indicates that the rearrangement occurs intermolecularly and that the migrating group is transferred to a polyalkylbenzene, not to the sulfonic acid (sulfonation only takes place after migration). The intermolecular mechanism is partially illustrated by the side products found in the following example:
Sergei en at the English language Wikipedia [GFDL (http://www.gnu.org/copyleft/fdl.html), CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0/) or Public domain], via Wikimedia Commons
Furthermore, the reaction is limited to benzene rings with at least four substituents (alkyl and/or halogen groups). The sulfo group is easily removed, so the Jacobsen rearrangement can also be considered as a rearrangement of polyalkylbenzenes. It was Herzig who described this type of rearrangement for the first time in 1881 using polyhalogenated benzenesulfonic acids, but the reaction took the name of the German chemist Oscar Jacobsen, who described the rearrangement of polyalkylbenzene derivatives in 1886.
Mannich Reaction
The Mannich reaction is an organic reaction which consists of an amino alkylation of an acidic proton placed next to a carbonyl functional group by formaldehyde and a primary or secondary amine or ammonia. The final product is a β-amino-carbonyl compound also known as a Mannich base. Reactions between aldimines and α-methylene carbonyls are also considered Mannich reactions because these imines form between amines and aldehydes. The reaction is named after chemist Carl Mannich.
The Mannich reaction is an example of nucleophilic addition of an amine to a carbonyl group followed by dehydration to the Schiff base. The Schiff base is an electrophile which reacts in the second step in an electrophilic addition with a compound containing an acidic proton (which is, or had become an enol). The Mannich reaction is also considered a condensation reaction.
In the Mannich reaction, primary or secondary amines or ammonia, are employed for the activation of formaldehyde. Tertiary amines lack an N–H proton to form the intermediate enamine. α-CH-acidic compounds (nucleophiles) include carbonyl compounds, nitriles, acetylenes, aliphatic nitro compounds, α-alkyl-pyridines or imines. It is also possible to use activated phenyl groups and electron-rich heterocycles such as furan, pyrrole, and thiophene. Indole is a particularly active substrate; the reaction provides gramine derivatives.
Reaction Mechanism
The mechanism of the Mannich reaction starts with the formation of an iminium ion from the amine and the formaldehyde.
The compound with the carbonyl functional group (in this case a ketone) can tautomerize to the enol form, after which it can attack the iminium ion.
Asymmetric Mannich reactions
Progress has been made towards asymmetric Mannich reactions. When properly functionalized the newly formed ethylene bridge in the Mannich adduct has two prochiral centers giving rise to two diastereomeric pairs of enantiomers. The first asymmetric Mannich reaction with an unmodified aldehyde was carried with (S)-proline as a naturally occurring chiral catalyst.
The reaction taking place is between a simple aldehyde, such as propionaldehyde, and an imine derived from ethyl glyoxylate and p-methoxyaniline (PMP = paramethoxphenyl) catalyzed by (S)-proline in dioxane at room temperature. The reaction product is diastereoselective with a preference for the syn-Mannich reaction 3:1 when the alkyl substituent on the aldehyde is a methyl group or 19:1 when the alkyl group the much larger pentyl group. Of the two possible syn adducts (S,S) or (R,R) the reaction is also enantioselective with a preference for the (S,S) adduct with enantiomeric excess larger than 99%. This stereoselectivity is explained in the scheme below.
Proline enters a catalytic cycle by reacting with the aldehyde to form an enamine. The two reactants (imine and enamine) line up for the Mannich reaction with Si facial attack of the imine by the Si-face of the enamine-aldehyde. Relief of steric strain dictates that the alkyl residue R of the enamine and the imine group are antiperiplanar on approach which locks in the syn mode of addition. The enantioselectivity is further controlled by hydrogen bonding between the proline carboxylic acid group and the imine. The transition state for the addition is a nine-membered ring with chair conformation with partial single bonds and double bonds. The proline group is converted back to the aldehyde and a single (S,S) isomer is formed.
By modification of the proline catalyst to it is also possible to obtain anti-Mannich adducts.
An additional methyl group attached to proline forces a specific enamine approach and the transition state now is a 10-membered ring with addition in anti-mode. The diastereoselectivity is at least anti:syn 95:5 regardless of alkyl group size and the (S,R) enantiomer is preferred with at least 97% ee.
Applications
The Mannich-Reaction is employed in the organic synthesis of natural compounds such as peptides, nucleotides, antibiotics, and alkaloids (e.g. tropinone). Other applications are in agro chemicals such as plant growth regulators, paint- and polymer chemistry, catalysts and main mechanism of formalin tissue crosslinking. The Mannich reaction is also used in the synthesis of medicinal compounds e.g. rolitetracycline (Mannich base of tetracycline), fluoxetine (antidepressant), tramadol, and tolmetin (anti-inflammatory drug)and azacyclophanes. The Mannich reaction is employed to synthesize alkyl amines, converting non-polar hydrocarbons into soap or detergents. This is used in a variety of cleaning applications, automotive fuel treatments, and epoxy coatings. Similar methods of substituted branched chain alkyl ethers into polyetheramines are achieved via a number of reactions. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Jacobsen_Rearrangement.txt |
Periodic acid is the highest oxoacid of iodine, in which the iodine exists in oxidation state VII. Like all periodates it can exist in two forms: orthoperiodic acid, with the chemical formula H5IO6 and metaperiodic acid, which has the formula HIO4. Periodic acid was discovered by Heinrich Gustav Magnus and C. F. Ammermüller in 1833.
Orthoperiodic Acid Metaperiodic Acid
By NEUROtiker (Own work) [Public domain], via Wikimedia Commons
By Benjah-bmm27 (Own work) [Public domain], via Wikimedia Commons
Synthesis
Modern industrial scale production involves the electrochemical oxidation of iodic acid, on a PbO2anode, with the following standard electrode potential:
H5IO6 + H+ + 2 e → IO−3 + 3 H2O E° = 1.6 V
Orthoperiodic acid can be dehydrated to give metaperiodic acid by heating to 100 °C
HIO4 + 2 H2O ⇌ H5IO6
Further heating to around 150 °C gives iodine pentoxide (I2O5) rather than the expected anhydride diiodine heptoxide (I2O7). Metaperiodic acid can also be prepared by from various orthoperiodates by treatment with dilute nitric acid.
H5IO6 → HIO4 + 2 H2O
Properties
Orthoperiodic acid has a number of acid dissociation constants.The pKa of meta periodic acid has not been determined.
H5IO6 ⇌ H4IO6 + H+, pKa = 3.29
H4IO6 ⇌ H3IO26 + H+, pKa = 8.31
H3IO26 ⇌ H2IO36 + H+, pKa = 11.60
There being two forms of periodic acid, it follows that two types of periodate salts are formed. For example, sodium metaperiodate, NaIO4, can be synthesised from HIO4 while sodium orthoperiodate, Na5IO6 can be synthesised from H5IO6.
Structure
Orthoperiodic acid forms monoclinic crystals (space group P21/n) consisting of slightly deformed IO6 octahedron interlinked via bridging hydrogens. The crystal structure of metaperiodic acid also includes IO6 octahedron, however these are connected via cis-edge-sharing with bridging oxygens to form one-dimensional infinite chains.
Reactions
Like all periodates, periodic acid can be used to cleave various 1,2-difunctional compounds. Most notably, periodic acid will cleave vicinal diols into two aldehyde or ketone fragments.
By Project Osprey (Own work) [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)], via Wikimedia Commons
This can be useful in determining the structure of carbohydrates as periodic acid can be used to open saccharide rings. This process is often used in labeling saccharides with fluorescent molecules or other tags such as biotin. Because the process requires vicinal diols, periodate oxidation is often used to selectively label the 3′-termini of RNA (ribose has vicinal diols) instead of DNA as deoxyribose does not have vicinal diols.
Periodic acid is also used in as an oxidising agent of moderate strength.
Other oxyacids
Periodate is part of a series of oxyacids in which iodine can assume oxidation states of −1, +1, +3, +5, or +7. A number of neutral iodine oxides are also known.
Iodine Oxidation State -1 +1 +3 +5 +7
Name
Hydrogen iodide Hypoiodous acid Iodous acid Iodic acid Periodic acid
Formula HI HIO HIO2 HIO3 HIO4 or H5IO5
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• Wikipedia (CC-BY-SA-3.0)
Preparation of Unsymmetrical Biaryls by the Diazo Reaction and the Nitrosoacetylamine Reaction
Preparation of Unsymmetrical Biaryls by the Diazo Reaction and the Nitrosoacetylamine reaction is an aryl-aryl coupling reaction via a diazonium salt. It is also known as the Gomberg–Bachmann reaction.
V8rik at the English language Wikipedia [GFDL (http://www.gnu.org/copyleft/fdl.html) or CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0/)], via Wikimedia Commons
The arene compound 1 (here benzene) is coupled with base with the diazonium salt 2 to the biaryl 3 through an intermediate aryl radical. For example, p-bromobiphenyl may be prepared from 4-bromoaniline and benzene:
BrC6H4NH2 + C6H6 → BrC6H4−C6H5
The reaction offers a wide scope for both diazonium component and arene component but yields are generally low following the original procedure (less than 40%), given the many side-reactions of diazonium salts. Several improvements have been suggested. One possibility is to employ diazonium tetrafluoroborates in arene solvent together with a phase-transfer catalyst, another is to use 1-aryl-3,3-dialkyltriazenes.
Pschorr reaction
One intramolecular variation which gives better results is the Pschorr reaction:
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The group Z can be CH2, CH2CH2, NH and CO (to fluorenone) to name just a few.
See also
• Graebe–Ullmann synthesis
• Meerwein arylation
• Sandmeyer reaction
Contributors
• Wikipedia (CC-BY-SA-3.0) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Periodic_Acid_Oxidation.txt |
Meerwein–Ponndorf–Verley (MPV) reduction in organic chemistry is the reduction of ketones and aldehydes to their corresponding alcohols utilizing aluminium alkoxide catalysis in the presence of a sacrificial alcohol. The beauty of the MPV reduction lies in its high chemoselectivity, and its use of a cheap environmentally friendly metal catalyst.
Figure 1, Exchange of carbonyl oxidation states in the presence of aluminium isopropoxide.
By Lankaluf (Own work) [Public domain], via Wikimedia Commons
The MPV reduction was discovered byMeerwein and Schmidt, and separately by Verley in 1925. They found that a mixture of aluminium ethoxide and ethanol could reduce aldehydes to their alcohols. Ponndorf applied the reaction to ketones and upgraded the catalyst to aluminium isopropoxide in isopropanol.
Mechanism
The MPV reduction is believed to go through a catalytic cycle involving a six-member ring transition state as shown in Figure 2. Starting with the aluminium alkoxide 1, a carbonyl oxygen is coordinated to achieve the tetra coordinated aluminium intermediate 2. Between intermediates 2 and 3 the hydride is transferred to the carbonyl from the alkoxy ligand via a pericyclic mechanism. At this point the new carbonyl dissociates and gives the tricoordinated aluminium species 4. Finally, an alcohol from solution displaces the newly reduced carbonyl to regenerate the catalyst 1.
Figure 2, Catalytic cycle of Meerwein–Ponndorf–Verley reduction
By Jkolev (Own work) [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0) or GFDL (http://www.gnu.org/copyleft/fdl.html)], via Wikimedia Commons
Each step in the cycle is reversible and the reaction is driven by the thermodynamic properties of the intermediates and the products. This means that given time the more thermodynamically stable product will be favored.
Several other mechanisms have been proposed for this reaction, including a radical mechanism as well as a mechanism involving an aluminium hydride species. The direct hydride transfer is the commonly accepted mechanism recently supported by experimental and theoretical data.
Chemoselectivity
One of the great draws of the Meerwein–Ponndorf–Verley reduction is its chemoselectivity. Aldehydes are reduced before ketones allowing for a measure of control over the reaction. If it is necessary to reduce one carbonyl in the presence of another, the common carbonyl protecting groups may be employed. Groups, such as alkenes and alkynes, that normally pose a problem for reduction by other means have no reactivity under these conditions.
Stereoselectivity
The aluminium based Meerwein–Ponndorf–Verley reduction can be performed on prochiral ketones leading to chiralalcohols. The three main ways to achieve the asymmetric reduction is by use of a chiral alcohol hydride source, use of an intramolecular MPV reduction, or use of a chiral ligand on the aluminium alkoxide.
One method of achieving the asymmetric MPV reduction is with the use of chiral hydride donating alcohols. The use of chiral alcohol (R)-(+)-sec-o-bromophen-ethyl alcohol gave 82% e.e. (percent enantiomeric excess) in the reduction of 2-chloroacetophenone. This enantioselection is due to the sterics of the two phenol groups in the six membered transition state as shown in Figure 3. In Figure 3, 1 is favored over 2 due to the large steric effect in 2 from the two phenyl groups.
Figure 3, Transition states of MPV reduction with a chiral alcohol
By Jkolev (Own work) [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0) or GFDL (http://www.gnu.org/copyleft/fdl.html)], via Wikimedia Commons
The use of an intramolecular MPV reduction can give good enantiopurity. By tethering the ketone to the hydride source only one transition state is possible (Figure 4) leading to the asymmetric reduction. This method, however, has the ability to undergo the reverse Oppenauer oxidation due to the proximity of the two reagents. Thus the reaction runs under thermodynamic equilibrium with the ratio of the products related to their relative stabilities. After the reaction is run the hydride-source portion of the molecule can be removed.
Figure 4, Transition state of intramolecular MPV reduction
By Jkolev (Own work) [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0) or GFDL (http://www.gnu.org/copyleft/fdl.html)], via Wikimedia Commons
Chiral ligands on the aluminium alkoxide can affect the stereochemical outcome of the MPV reduction. This method lead to the reduction of substituted acetophenones in up to 83% e.e. (Figure 5). The appeal of this method is that it uses a chiral ligand as opposed to a stoiciometric source of chirality. It has been recently shown that the low selectivity of this method is due to the shape of the transition state. It has been shown that the transition state is a planar six member transition state. This is different than the believed Zimmerman-Traxler model like transition state.
Figure 5, MPV reaction with chiral ligand
By Jkolev (Own work) [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0) or GFDL (http://www.gnu.org/copyleft/fdl.html)], via Wikimedia Commons
Scope
Several problems restrict the use of the Meerwein–Ponndorf–Verley reduction compared to the use of other reducing agents. The stereochemical control is seriously limited. Often a large amount of aluminium alkoxide is needed when using commercial reagent, and there are several known side reactions.
While commercial aluminim isopropoxide is available, the use of it often requires catalyst loadings of up to 100-200 mol%. This hinders the use of the MPV reduction on scale. Recent work has shown that aluminium alkoxides made in situ from trimethyl aluminium reagents have far better activity requiring as little as 10% loading. The activity difference is believed to be due to the large aggregation state of the commercially available product.
Several side reactions are known to occur. In the case of ketones and especially aldehydes aldol condensations have been observed. Aldehydes with no α-hydrogens can undergo the Tishchenko reaction. Finally, in some cases the alcohol generated by the reduction can be dehydrated giving an alkyl carbon.
Variations
The Meerwein–Ponndorf–Verley reduction has been recently used in the synthesis of chiral amines from ketimines using a chiral alkoxide. The addition of a phosphinoyl group to the nitrogen of the ketimine allowed for high enantioselectivity up to 98% e.e.
Work has been done in the use of lanthanides and transition metals for the Meerwein–Ponndorf–Verley reduction. Both Ruthenium and Samarium have shown high yields and high stereoselectivity in the reduction of carbonyls to alcohols. The Ruthenium catalyst has been shown, however, to go through a Ruthenium hydride intermediate. The Meerwein–Ponndorf–Verley reduction has also been effected with synthetically useful yield by plutonium (III) isopropoxide.
The standard MPV reduction is a homogeneous reaction several heterogeneous reactions have been developed.
See also
• Oppenauer oxidation
• Carbonyl reduction
Contributors
• Wikipedia (CC-BY-SA-3.0)
Replacement of the Aromatic Primary Amino Group by Hydrogen
Introduction
The Sandmeyer reaction is a chemical reaction used to synthesize aryl halides from aryl diazonium salts. The reaction is a method for substitution of an aromatic amino group via preparation of its diazonium salt followed by its displacement with a nucleophile. The nucleophile in this case is hypophosphorus acid, H3PO2 .
Mechanism
The nitrous acid is usually prepared in situ from sodium nitrite and an acid. Following a 2nd protonation step, one equivalent of water is lost to form nitrogen monoxide cation i.e. the "nitrosonium ion" electrophile. An aromatic (or heterocyclic) amine quickly reacts with a nitrite to form an aryl diazonium salt.
By Nuklear (Own work) [CC BY-SA 4.0 (http://creativecommons.org/licenses/by-sa/4.0)], via Wikimedia Commons
Variations
The majority of variations of the Sandmeyer reactions consist of using various copper salts. For example, using cuprous cyanide produces benzonitriles.[ Substituting thiols or water for the copper salts generates thioethers or phenols, respectively.
The Schiemann reaction uses tetrafluoroborate and delivers the halide-substituted product, fluorobenzene, which is not obtained by the use of copper fluorides.
Sandmeyer reactions with copper salts used in catalytic amounts are also known. One bromination protocol employs a 0.2 equivalent Cu(I)/Cu(II) mixture with additional amounts of the bidentate ligand phenanthroline and phase-transfer catalyst dibenzo-18-crown-6:
V8rik at English Wikipedia [GFDL (http://www.gnu.org/copyleft/fdl.html), CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0/) or CC BY-SA 2.5-2.0-1.0 (http://creativecommons.org/licenses/by-sa/2.5-2.0-1.0)], via Wikimedia Commons
Amyl nitrites are also useful as reagents in a modification of the Sandmeyer reaction. The reaction of the alkyl nitrite with an aromatic amine in a halogenated solvent produces a radicalaromatic species, this then abstracts a halogen atom from the solvent. For the synthesis of aryl iodides diiodomethane is used, whereas bromoform is the solvent of choice for the synthesis of aryl bromides.
Aliphatic variation
William Reusch at Michigan State University states:
“The distinct behavior of 1°, 2° & 3°-aliphatic amines is an instructive challenge to our understanding of their chemistry, but is of little importance as a synthetic tool. The SN1 product mixtures from 1°-amines are difficult to control, and rearrangement is common when branched primary alkyl groups are involved. The N-nitrosamines formed from 2°-amines are carcinogenic, and are not generally useful as intermediates for subsequent reactions.”
The aliphatic version of the Sandmeyer reaction was used to prepare Batimastat and marimastat though (from 1° amino acids).
Contributors
• Wikipedia (CC-BY-SA-3.0) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Reduction_with_Aluminum_Alkoxides_%28The_Meerwein-Ponndorf-Verley_Reduction%29.txt |
The Schmidt reaction is the reaction of hydrazoic acid or an alkyl azide with a carbonyl compound, alkene, or alcohol, often in the presence of a Brønsted or Lewis acid. Although the family of Schmidt reactions includes a number of variants, they all result in the migration of a substituent from carbon to nitrogen with loss of a molecule of dinitrogen. This reaction has considerable utility for the synthesis of hindered or cyclic amides and amines.[1]
Introduction
Azides are nucleophilic at their terminal nitrogen atoms, and may add to suitably activated electrophiles in the presence of a Brønsted or Lewis acid. Upon addition, the newly bound nitrogen atom becomes electron-deficient and is subject to 1,2-migration of a carbon or hydrogen substituent with loss of a molecule of dinitrogen. Historically, carbonyl compounds were the first electrophiles successfully employed in this context.[2] Since the initial discovery of the Schmidt reaction, many variants employing alternative electrophiles and hydrazoic acid have been developed (Eq. 1). Related reactions of alkyl azides may yield substituted amides, lactams, or amines (after reduction of iminium ions). However, the scope of alkyl azides in the Schmidt reaction is limited compared to hydrazoic acid.[3]
(1)
Generally, intramolecular Schmidt reactions are more useful than their intermolecular counterparts, which are limited by poor site selectivity and sensitivity to steric hindrance. The Schmidt reaction of carboxylic acids, which produces amines, is in direct competition with the milder Curtius rearrangement,[4] and is rarely used in practice. Nonetheless, the Schmidt reaction has been applied extensively for the synthesis of medium-sized lactams and hindered amides. For these applications, the Schmidt reaction exhibits advantageous site selectivity and atom economy.
Mechanism and Stereochemistry
Reactions of Ketones
The mechanisms of reactions of ketones with hydrazoic acid begin with addition of HN3 to the carbonyl group to form adizohydrin intermediate I. At this stage, one can envision two mechanistic pathways that lead to amides. In a process analogous to the mechanism of the Baeyer-Villiger reaction (Eq. 2, blue pathway), the azidohydrin may rearrange to afford amides directly after loss of a proton.[5] A second possibility (Eq. 2, green pathway) is analogous to the mechanism of the Beckmann rearrangement. Elimination of water from the azidohydrin affords isomeric diazoiminium ions II and III, which undergo synchronous, antiperiplanar 1,2-rearrangement to yield nitrilium ions IVand V. Subsequent addition of water and tautomerization lead to the amide products.[6] The observation of tetrazole side products at high concentrations of hydrazoic acid[7] and similar product ratios in conceptually related Beckmann and Schmidt reactions[8] provide evidence of the Beckmann-type mechanism in reactions of hydrazoic acid. However, it should be noted that alkyl azides must react via the Baeyer-Villiger mechanism to avoid the formation of adjacent positive charges.
(2)
The product of the migration step is determined by the configuration of the diazoiminium ion, and it has thus been proposed that the observed amide product ratio is directly related to the ratio of diazoiminium ions. In support of this idea, calculations have shown that the barrier to direct conversion between II and III is quite high.[9] However, II and III may interconvert rapidly via the addition of water followed by elimination, and a Curtin-Hammett situation cannot be ruled out. Under Curtin-Hammett conditions, the amide product ratio is not determined by the ratio of diazoiminium ions, but by their relative rates of conversion to nitrilium ions. In any case, preferential migration of the larger R group is overwhelmingly observed, suggesting the predominant intermediacy of the less hindered diazoiminium ion.[8] Stereogenic centers migrate with retention of configuration.[10]
Reactions of Carboxylic Acids
The mechanism of the reaction of carboxylic acids with hydrazoic acid is well understood. Initial formation of an acylium ion is followed by the addition of hydrazoic acid to form an acyl azide intermediate. The only possible migration event produces an isocyanate, which undergoes hydrolysis to yield an amine and carbon dioxide (Eq. 3).[1]
(3)
Reactions of Other Electrophiles
Electrophiles capable of forming carbocationic intermediates may also participate in the Schmidt reaction. After nucleophilic addition of hydrazoic acid to the carbocation, alkyl or aryl group migration affords an imine (Eq. 2). Subsequent hydrolysis and tautomerization of the imine are both possible.[11]
(4)
The relative electron density of the R groups is likely the primary determinant of site selectivity in these Baeyer-Villiger-type reactions. More electron-rich, nucleophilic R groups migrate to a greater extent.[12]
Enantioselective Variants
The Schmidt reaction has been applied for the desymmetrization of symmetric ketones containing enantiotopic α-carbon atoms. Employing a chiral hydroxyalkyl azide resulted in highly diastereoselective migration, and subsequent removal of the nitrogen substituent produced the lactam as a single enantiomer.[13]
(5)
Scope and Limitations
Reactions of Hydrazoic Acid
The Schmidt reaction is most commonly used to convert differentially substituted ketones to amides or lactams. The most common problems with this reaction are site selectivity and tetrazole formation, although the latter can be controlled by changing reaction conditions. Steric hindrance to nucleophilic addition lowers yields somewhat, but site selectivity is improved when the substituents of the carbonyl are of different sizes. The less substituted carbon rarely migrates (Eq. 6).[14]
(6)
Schmidt reactions of aldehydes are often problematic because of amide and nitrile formation (C-migration and H-migration, respectively).[15]
Aromatic groups often migrate in preference to alkyl groups in reactions of ketones, particularly if the ring is electron rich (Eq. 7).[16] The behavior of heteroaromatic ketones is more difficult to predict.[17]
(7)
A variety of alternative electrophiles react with hydrazoic acid to afford products of Schmidt-type reactions. For instance, dithioketals activated by iodoazide form α-azido sulfides, which may be converted to amides via acidic hydrolysis or to thioimino ethers with stannic tetrachloride (Eq. 8).[18]
(8)
Reactions of Alkyl Azides
The scope of reactions of alkyl azides is generally more limited than the scope of reactions of hydrazoic acid. Intermolecular Schmidt reactions of alkyl azides are plagued by poor site selectivity and strong sensitivity to steric effects. However, intramolecular reactions of alkyl azides are among the most synthetically useful applications of the Schmidt reaction. The range of ketones that engage in the intramolecular Schmidt reactions is exceptionally broad. For these reactions, the azide and carbonyl carbon must be separated by four or five atoms to facilitate cyclization, with four preferred (Eq. 9). Substrates containing five-carbon tethers react only in the presence of relatively strong Lewis acids.[19]
(9)
One useful intermolecular method involves the use of hydroxyalkyl azides, whose pendant hydroxy groups may attack the intermediate oxocarbenium ion present after migration to form iminium ethers. Subsequent basic hydrolysis leads to the hydroxyalkyl-substituted amide, and nucleophiles other than hydroxide have been employed with some success.[20]
(10)
Many Schmidt reactions of cationic electrophiles are subject to carbocationic rearrangements. However, a judicious choice of reaction conditions (and utilizing an intramolecularly bound azide) can steer the reaction toward a single product (Eq. 11).[21]. Intermolecular reactions of benzylic and tertiary cations with alkyl azides are also known.[22]
(11)
Epoxides are a useful class of cation equivalents that react rapidly with alkyl azides in the presence of a Lewis acid to afford hydroxymethyl iminium ions. Subsequent reduction with sodium borohydride yields the corresponding amine (Eq. 12).[23]
(12)
Synthetic Applications
The Schmidt reaction has been applied to the synthesis of disubstituted amino acids. After asymmetric alkylation of a β-ketoester, treatment with trifluoroacetic acid (TFA) and sodium azide yields the N-acetyl amino ester, which may be converted to the corresponding amino acid through functional group manipulations (Eq. 13).[24]
(13)
The Schmidt reaction has also been applied extensively to access crowded or otherwise difficult amines and amides in alkaloid natural products. Efforts toward the homoerythrina spirocyclic ring system contain a nice application of the intramolecular Schmidt reaction of alkyl azides. The use of Beckmann conditions and hydrazoic acid both failed for the sterically hindered ketone in Eq. 14; however, an appropriately positioned alkyl azide was smoothly converted to the lactam in the presence of TFA.[25]
(14)
Comparison to Other Methods
The Beckmann rearrangement is conceptually very similar to the Schmidt reaction, but involves the discrete formation of an oxime (analogous to the posited diazoiminium intermediate of the Schmidt reaction) prior to the rearrangement step. Like the Schmidt reaction, the site selectivity of the Beckmann rearrangement is generally controlled by stereoelectronic factors and the relative prevalence or reactivity of geometric isomers. The site selectivity of related Schmidt and Beckmann reactions are generally similar (migration of the larger group is favored, for instance).[26]
(15)
The sense of site selectivity of the photochemical oxaziridine rearrangement is opposite that of the Beckmann and Schmidt reactions. In fact, this reaction is one of a select few that result in the selective insertion of nitrogen into the less hindered of two carbon-carbon bonds (Eq. 16).[27]
(16)
Although Schmidt reactions of carboxylic acids are certainly possible and have been studied in detail, the Curtius rearrangement of carboxylic acids to isocyanates is a much more mild and practical method for the synthesis of amines from carboxylic acids (Eq. 17).[28] In addition, the initial isocyanate products can easily be converted to urethanes and ureas through treatment with alcohols or amines, an impossible transformation under the conditions of the Schmidt reaction.
(17)
Experimental Conditions and Procedure
Typical Conditions
Hydrazoic acid is typically prepared in situ for immediate use via treatment of sodium azide with hydrochloric or sulfuric acid. In some cases an additional Lewis acid is added. Hydrazoic acid is both explosive and toxic, and should be handled with extreme care in a well-ventilated fume hood with a safety shield. Alkyl azides are in many cases explosive and toxic as well (particularly alkyl azides of low molecular weight). While intermolecular reactions typically employ titanium tetrachloride as a Lewis acid, either Lewis or Brønsted acids (most commonly, trifluoroacetic acid) may be used for intramolecular reactions.
An important caution concerns the use of sodium azide in methylene chloride solutions for the synthesis of alkyl azides. The formation of extremely explosive diazidomethane under these conditions has been observed.[29]Alternative methods for synthesizing alkyl azides from alcohols using Mitsunobu conditions[30] and from alkyl halides using azidotrimethylsilane[31] are available.
Example Procedure
(18)
Sodium azide (0.60 g, 9.1 mmol) was added portionwise over 20 min to a solution of cis-bicyclo[3.3.0]octane-3,7-dione (1.0 g, 7.2 mmol) in 36% aqueous HCl (20 mL), while keeping the temperature below 35 °C. The mixture was stirred for 3 h at rt and then brought to pH 10 with 20% aqueous NaOH at 0 °C. Precipitated NaCl was filtered off and the aqueous layer was extracted continuously with CHCl3 for 48 h. Drying the organic phase over MgSO4, followed by removal of solvent, furnished a residue that was separated on a silica gel column (30 g). Elution with CH2Cl2/acetone (85:15) gave unreacted diketone (240 mg) and the desired product (576 mg, 52%): mp 120–122 °C; 1H NMR (CDCl3) δ 2.10–2.30 (m, 3H), 2.40–2.90 (m, 5H), 3.21 (ddd, J = 3.0, 6.8, 13.0 Hz, 1H), 3.52 (ddd, J = 3.5, 5.8, 13.0 Hz, 1H), 6.3 (br s, 1H); 13C NMR (D2O) δ 30.7 (d), 31.3 (d), 31.9 (t), 40.9 (t), 41.9 (t), 43.6 (t), 175.4 (s), 223.7 (s); IR (neat) 3246, 2935, 1735, 1667, 1638 cm–1; MS m/z: M+ 153 (100), 125 (10), 112 (33), 96 (35), 82 (53), 68 (33), 54, (82), 41 (78); Anal. Calcd for C8H11NO2: C, 62.73; H, 7.24; N, 9.14. Found: C, 62.76; H, 7.21; N, 9.11. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Schmidt_Reaction.txt |
The Wacker oxidation refers generally to the transformation of a terminal or 1,2-disubstituted alkene to a ketone through the action of catalytic palladium(II), water, and a co-oxidant. Variants of the reaction yield aldehydes, allylic/vinylic ethers, and allylic/vinylic amines. Because of the ease with which terminal alkenes may be prepared and the versatility of the methyl ketone group installed by the reaction, the Wacker oxidation has been employed extensively in organic synthesis.[1]
Introduction
The stoichiometric conversion of ethylene to acetaldehyde by an acidic, aqueous solution of PdCl2 was discovered over a century ago,[2] but fifty years passed between the discovery of this reaction and the development of a catalytic method. In 1959, researchers at Wacker Chemie reported that a similar transformation takes place in an aqueous, acidic solution of catalytic PdCl2 and a stoichiometric amount of CuCl2 through which oxygen is bubbled (Eq. 1).[3]
(1)
Since this initial report, the Wacker process has been widely applied in organic synthesis and has been extended to other classes of substrates and products. To encourage mixing of the organic reactants with the aqueous phase, a co-solvent is generally employed along with water. Dimethylformamide (DMF) is a common choice; when DMF is used as a co-solvent with a stoichiometric amount of CuCl under balloon pressure of oxygen, the reaction is called the "Tsuji-Wacker oxidation."[4] Applications of the Wacker oxidation to organic synthesis typically involve the installation of a methyl ketone moiety, which may subsequently undergo nucleophilic addition or deprotonation to form an enolate.
Mechanism and Stereochemistry
Prevailing Mechanism: Water Nucleophile
The mechanism of the Wacker oxidation has been studied both experimentally and theoretically (Eq. 2). The first step of the Wacker oxidation involves coordination of the the alkene to the palladium center to form π-complex 2. Evidence for this step is provided by the relative sluggishness of electron-poor alkenes, which generally require higher catalyst loadings than unactivated alkenes. Hydroxypalladation then occurs to yield either zwitterionic complex 3 or neutral complex 4 depending on the mode of hydroxypalladation (see below). Studies employing deuterated substrates suggest that β-hydride elimination then occurs to afford enol complex 5, which re-inserts into the Pd-H bond to afford complex 6.[5] Computational studies support the involvement of chloride-assisted deprotonation in the subsequent step,[6] which affords the product and palladium(0). Oxidation of palladium(0) by copper(II) then occurs, regenerating palladium(II) species 1. The role of copper(II) in the mechanism is poorly understood at present.
(2)
The mode of hydroxypalladation is an important issue for the Wacker oxidation. Hydroxypalladation may occur either in a syn fashion through an inner-sphere mechanism or in an anti fashion via nucleophilic attack on the coordinated alkene (Eq. 3). Although the stereocenter in 4 is ultimately destroyed upon elimination, the mode of hydroxypalladation can influence the site selectivity of the reaction. Markovnikov-type addition of water to the more substituted carbon of the alkene forms a methyl ketone, while attack of water at the less substituted position ultimately yields an aldehyde. The mode of hydroxypalladation has been shown to affect the distribution of ketone and aldehyde products, and is also important in stereoselective Wacker cyclization reactions.
(3)
At low concentrations of chloride ion, syn-hydroxypalladation appears to be the norm.[7] Water coordinates to the metal and migratory insertion into the Pd-OH bond occurs to generate 4 directly. At high concentrations of chloride ion, chloride competes with hydroxide for binding at palladium, and thus anti-hydroxypalladation may occur.[8]
Prevailing Mechanism: Alcohol Nucleophiles
Alcohols may also be employed as nucleophiles in the Wacker process. The initial steps of the mechanism are similar to those for the Wacker oxidation with water, but the mechanisms diverge at alkoxy complex 11 (Eq. 4). Elimination of palladium occurs to generate an oxocarbenium ion, which may be captured by solvent to generate an acetal. Small amounts of vinyl ether products support the intermediacy of complex 10.[9]
(4)
Enantioselective Variants
When the substrate contains a tethered nucleophile, such as a hydroxyl group or protected amine, the nucleophile may react to form a cyclic, allylic or vinylic ether or amine after β-hydride elimination. This process is known as the "Wacker cyclization," and when the product is allylic, a stereocenter may be established. Chiral ligands have been used to render the Wacker cyclization enantioselective. For example, use of tetra(dihydrooxazoline) ligand 12 and benzoquinone (BQ) as a co-oxidant resulted in formation of allylic ether product in an enantiomeric ratio of 98:2 (Eq. 5). [10]
(5)
The Wacker process may also establish a stereocenter in 1,1-disubstituted alkenes via acetal formation. A chiral auxiliary in the alkene has been used successfully to generate chiral acetals with high diastereomeric ratio. The acetal forms at the more electron-deficient site of the alkene (Eq. 6).[11]
(6)
Scope and Limitations
Terminal alkenes are the most commonly employed substrates for the Wacker oxidation. Oxygen-containing functionality far from the reactive alkene is well tolerated by the reaction because palladium reacts selectively with the alkene in preference to lone pairs on oxygen. Wacker oxidation followed by intramolecular aldol condensation is a convenient method for the rapid synthesis of carbocycles (Eq. 7).[12]
(7)
The original Wacker process and Tsuji-Wacker conditions can present problems for substrates containing acid-sensitive functionality, such as acetals and silyl ethers. Use of copper(II) chloride leads to the generation of strongly acidic hydrochloric acid. To mitigate this issue, a method employing copper(II) acetate has been developed. Some acid-sensitive groups are able to withstand the milder acetic acid generated by this method (Eq. 8).[13] Under similar conditions involving copper(I) chloride, a yield of only 56% was obtained.
(8)
Unprotected amines coordinate strongly to palladium and thus cause problems for Wacker oxidations. Amines protected with electron-withdrawing substituents often do not interfere in Wacker oxidations,[14] although they may participate in aza-Wacker cyclizations if appropriately positioned in the substrate (see below).
Tri-substituted alkenes are completely inert to the conditions of the Wacker oxidation; only terminal and 1,1-disubstituted alkenes react. Terminal alkenes are usually oxidized selectively over internal alkenes, except under rather specialized circumstances.[15] Eq. 9 represents a typical case of selectivity for terminal alkenes.[14]
(9)
Lewis-basic functionality in the allylic or homoallylic position relative to the reactive alkene may cause unpredictable site selectivity in Wacker oxidations. Under standard Tsuji-Wacker conditions, unprotected allylic alcohols form mixtures of methyl ketone and aldehyde products.[16] The catalyst Pd(Quinox)Cl2 was developed to address this problem; allylic alcohols react selectively in the presence of this catalyst to afford methyl ketones (Eq. 10).[17] tert-Butyl hydroperoxide (TBHP) is used as the oxygen source in this reaction.
(10)
Protected, homoallylic alcohols generally form the methyl ketone selectively, but results are more unpredictable for unprotected homoallylic alcohols.[18]
Protected allylic amines generally give higher yields of aldehyde than comparable protected allylic alcohols; these results suggest that coordination of palladium to the Lewis base is responsible for the formation of aldehydes. Using a phthalimide group in the allylic position, the Wacker oxidation may be rendered completely selective for aldehydes (Eq. 11).[19]
(11)
Site selectivity in reactions of internal alkenes is low unless an allylic or homoallylic Lewis base is present in the substrate. Electron-withdrawing groups can direct oxidation to the β position of the alkene (Eq. 12), but relatively high catalyst loadings are required.[20]
(12)
When an appropriately situated nucleophile is present in the substrate, Wacker cyclization may occur to yield an allylic or vinylic ether (Eq. 13).[21] The aza-Wacker cyclization involves protected amino nucleophiles such as sulfonamides.[22]
(13)
When an alcohol is used in stoichiometric or solvent quantities or a diol is present in the substrate, acetal formation may occur. Acetal formation, like ketone formation, generally occurs with Markovnikov selectivity. For example, a Michael acceptor forms an acetal at the β position in the presence of (R,R)-2,4-pentanediol (Eq. 14).[23]
(14)
Synthetic Applications
The Wacker oxidation has been used extensively in organic synthesis. A number of methods exist for the installation of terminal alkenes and the methyl ketone products of the reaction can be elaborated easily to more complex compounds. In addition, the reaction is aerobic and operationally simple.
The Wacker oxidation was employed in the course of synthesis of the macrolide elaiolide. Oxidation at two terminal alkenes in a C2-symmetric intermediate afforded a diketone, which was subsequently elaborated to the symmetric natural product (Eq. 15).[24] Notably, the internal alkenes and esters in the intermediate appeared not to affect the oxidation.
(15)
A modification of Tsuji-Wacker conditions employing copper(II) acetate was used in the course of a synthesis of hennoxazole A (Eq. 16).[25] After the oxidation, deprotection under acidic conditions led to the intramolecular formation of an acetal present in the target compound.
(16)
Comparison to Other Methods
Few true alternatives to the Wacker oxidation (that is, other hydration-oxidation reactions) are known. One rather harsh method involves oxymercuration followed by transmetalation to palladium (Eq. 17).[26]
(17)
Experimental Conditions and Procedure
Typical Conditions
Tsuji-Wacker conditions represent a good starting point for experimental conditions for this reaction. Water is mixed with an organic co-solvent (often DMF) and 10 mol% PdCl2, stoichiometric CuCl, and 1 atmosphere of oxygen gas. Copper(I) salts are generally superior to copper(II) salts because the former minimize the concentration of chloride in solution, encouraging syn-hydroxypalladation. Copper(I) is rapidly oxidized to copper(II) by dissolved oxygen; a period of 30 minutes prior to the introduction of the substrate is used to ensure that this oxidation is complete. For situations in which Tsuji-Wacker conditions fail, a wide variety of modifications of the reaction conditions may be employed.
Using high oxygen pressure and a judicious choice of solvent, Wacker oxidations can be carried out without the need for a co-oxidant. Certain ligands also promote direct O2-coupled oxidations.[27]
Peroxides may also be used as the oxygen source in Wacker oxidations; these reactions do not require a co-oxidant. In the Pd(Quinox)/TBHP system, silver hexafluoroantimonate serves as a scavenger of chloride ions.
Experimental Procedure[4]
(18)
A 100-mL, 3-necked, round-bottomed flask was fitted with a magnetic stirring bar and a pressure-equalizing dropping funnel containing 1-decene (4.2 g, 30 mmol, 1.0 equiv). The flask was charged with a mixture of PdCl2 (0.53 g, 3 mmol, 0.1 equiv), CuCl (2.97 g, 30 mmol, 1.0 equiv) and aqueous DMF (DMF/H 2O = 7:1, 24 mL). With the other outlets securely stoppered and wired down, an oxygen-filled balloon was placed over one neck, and the mixture was stirred at rt to allow oxygen uptake. After 1 h, the 1-decene (4.2 g, 30 mmol) was added over 10 min, and the solution was stirred vigorously at rt under an oxygen balloon. The color of the solution turned from green to black within 15 min and returned gradually to green. After 24 h, the mixture was poured into cold 3 N HCl (100 mL) and extracted with five, 50 mL portions of ether. The extracts were combined and washed successively with 50 mL of saturated aqueous NaHCO3 solution, 50 mL of brine, and then dried over anhydrous magnesium sulfate. The solvent was removed by evaporation and the residue was distilled using a 15 cm Vigreux column to give 2-decanone as a colorless oil (3.0–3.4 g, 65–73%): bp 43.5 °C (1 mm Hg); IR (neat) 1722 cm–1; 1H NMR (CCl4) δ 2.37 (t, J = 7 Hz, 2H), 2.02 (s, 3H), 0.7–1.8 (m, 15H).
References
1. Michel, B. W.; Steffens, L. D.; Sigman, M. S. Org. React. 2014, 84, 2. (doi: 10.1002/0471264180.or084.02)
2. Phillips, F. C. Am. Chem. J. 1894, 16, 255.
3. Smidt, J.; Hafner, W.; Jira, R.; Sedlmeier, J.; Sieber, R.; Ruttinger, R.; Kojer, H. Angew. Chem. 1959, 71, 176.
4. a b Tsuji, J.; Nagashima, H.; Nemoto, H. Org. Synth. 1984, 62, 9.
5. Smidt, J.; Hafner, W.; Jira, R.; Sieber, R.; Sedlmeier, J.; Sabel, A. Angew. Chem. 1962, 74, 93.
6. Eshtiagh-Hosseini, H.; Beyramabadi, S. A.; Morsali, A.; Housaindokht, M. R. J. Mol. Struct.: THEOCHEM 2010, 941, 138.
7. Henry, P. M. J. Am. Chem. Soc. 1964, 86, 3246.
8. Bäckvall, J. E.; Akermark, B.; Ljunggren, S. O. J. Chem. Soc., Chem. Commun. 1977, 264.
9. Balija, A. M.; Stowers, K. J.; Schultz, M. J.; Sigman, M. S. Org. Lett. 2006, 8, 1121.
10. Zhang, Y. J.; Wang, F.; Zhang, W. J. Org. Chem. 2007, 72, 9208.
11. Hosokawa, T.; Yamanaka, T.; Itotani, M.; Murahashi, S.-I. J. Org. Chem. 1995, 60, 6159.
12. Srikrishna, A.; Kumar, P. P. J. Indian Chem. Soc. 1999, 76, 521.
13. Smith, A. B., III; Cho, Y. S.; Friestad, G. K. Tetrahedron Lett. 1998, 39, 8765.
14. a b Reginato, G.; Mordini, A.; Verrucci, M.; Degl'Innocenti, A.; Capperucci, A. Tetrahedron: Asymmetry 2000, 11, 3759.
15. Ho, T.-L.; Chang, M. H.; Chen, C. Tetrahedron Lett. 2003, 44, 6955.
16. Muzart, J. Tetrahedron 2007, 63, 7505.
17. Michel, B. W.; Camelio, A. M.; Cornell, C. N.; Sigman, M. S. J. Am. Chem. Soc. 2009, 131, 6076.
18. Kang, S.-K.; Jung, K.-Y.; Chung, J.-U.; Namkoong, E.-Y.; Kim, T.-H. J. Org. Chem. 1995, 60, 4678.
19. Weiner, B.; Baeza, A.; Jerphagnon, T.; Feringa, B. L. J. Am. Chem. Soc. 2009, 131, 9473.
20. Yan, B.; Spilling, C. D. J. Org. Chem. 2008, 73, 5385.
21. Mitsudome, T.; Umetani, T.; Nosaka, N.; Mori, K.; Mizugaki, T.; Ebitani, K.; Kaneda, K. Angew. Chem., Int. Ed. 2006, 45, 481.
22. Hegedus, L. S.; McKearin, J. M. J. Am. Chem. Soc. 1982, 104, 2444.
23. Hosokawa, T.; Ataka, Y.; Murahashi, S. Bull. Chem. Soc. Jpn. 1990, 63, 166.
24. Barth, R.; Mulzer, J. Tetrahedron 2008, 64, 4718.
25. Yokokawa, F.; Asano, T.; Shioiri, T. Tetrahedron 2001, 57, 6311.
26. Ellis, J. M.; Crimmins, M. T. Chem. Rev. 2008, 108, 5278.
27. Cornell, C. N.; Sigman, M. S. Org. Lett. 2006, 8, 4117.
Contributors
• This page is based on Organic Reactions. The primary collection of full Organic Reactions chapters is maintained by John Wiley & Sons. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Wacker_Oxidation.txt |
The Wittig reaction or Wittig olefination is a chemical reaction of an aldehyde or ketone with a triphenyl phosphonium ylide (often called a Wittig reagent) to give an alkene and triphenylphosphine oxide.
The Wittig reaction was discovered in 1954 by Georg Wittig, for which he was awarded the Nobel Prize in Chemistry in 1979. It is widely used in organic synthesis for the preparation of alkenes. It should not be confused with the Wittig rearrangement.
Wittig reactions are most commonly used to couple aldehydes and ketones to singly substituted phosphine ylides. With unstabilised ylides this results in almost exclusively the Z-alkene product. In order to obtain the E-alkene, stabilised ylides are used or unstabilised ylides using the Schlosser modification of the Wittig reaction can be performed.
Reaction mechanism
Classical mechanism
The steric bulk of the ylide 1 influences the stereochemical outcome of nucleophilic addition to give a predominance of the betaine 3 (cf. Bürgi–Dunitz angle). Note that for betaine 3 both R1 and R2 as well as PPh3+ and O− are positioned anti to one another.
Carbon-carbon bond rotation gives the betaine 4, which then forms the oxaphosphetane 5. Elimination gives the desired Z-alkene 7 and triphenylphosphine oxide 6. With simple Wittig reagents, the first step occurs easily with both aldehydes and ketones, and the decomposition of the betaine (to form 5) is the rate-determining step. However, with stabilised ylides (where R1 stabilises the negative charge) the first step is the slowest step, so the overall rate of alkene formation decreases and a bigger proportion of the alkene product is the E-isomer. This also explains why stabilised reagents fail to react well with sterically hindered ketones.
Mechanism
Mechanistic studies have focused on unstabilized ylides, because the intermediates can be followed by NMR spectroscopy. The existence and interconversion of the betaine (3a and 3b) is subject of ongoing research. Phosphonium ylides 1 react with carbonyl compounds 2 via a π²s/π²a [2+2] cycloaddition to directly form the oxaphosphetanes 4a and 4b. The stereochemistry of the product 5 is due to the addition of the ylide 1 to the carbonyl 2 and to the equilibration of the intermediates. Maryanoff and Reitz identified the issue about equilibration of Wittig intermediates and termed the process "stereochemical drift". For many years, the stereochemistry of the Wittig reaction, in terms of carbon-carbon bond formation, had been assumed to correspond directly with the Z/E stereochemistry of the alkene products. However, certain reactants do not follow this simple pattern. Lithium salts can also exert a profound effect on the stereochemical outcome.
Mechanisms differ for aliphatic and aromatic aldehydes and for aromatic and aliphatic phosphonium ylides. Evidence suggests that the Wittig reaction of unbranched aldehydes under lithium-salt-free conditions do not equilibrate and are therefore under kinetic reaction control. Vedejs has put forth a theory to explain the stereoselectivity of stabilized and unstabilized Wittig reactions.
Wittig reagents
Preparation of phosphorus ylides
Wittig reagents are usually prepared from a phosphonium salt, which is in turn prepared by the quaternization of triphenylphosphine with an alkyl halide. The alkylphosphonium salt is deprotonated with a strong base such as n-butyllithium:
[Ph3P+CH2R]X + C4H9Li → Ph3P=CHR + LiX + C4H10
One of the simplest ylides is methylenetriphenylphosphorane (Ph3P=CH2). It is also a precursor to more elaborate Wittig reagents. Alkylation of Ph3P=CH2 with a primary alkyl halide R−CH2−X, produces substituted phosphonium salts:
Ph3P=CH2 + RCH2X → Ph3P+CH2CH2R X
These salts can be deprotonated in the usual way to give Ph3P=CH−CH2R.
Structure of the ylide
Ball-and-stick model of Ph3P=CH2, as found in the crystal structure
The Wittig reagent may be described in the phosphorane form (the more familiar representation) or the ylide form:
The ylide form is a significant contributor, and the carbon is nucleophilic.
Reactivity
Simple phosphoranes are reactive. Most hydrolyze and oxidize readily. They are therefore prepared using air-free techniques. Phosphoranes are more air-stable when they contain an electron withdrawing group. Some examples are Ph3P=CHCO2R and Ph3P=CHPh. These ylides are sufficiently stable to be sold commercially
From the phosphonium salts, these reagent are formed more readily, requiring only NaOH, and they are usually more air-stable. These are less reactive than simple ylides, and so they usually fail to react with ketones, necessitating the use of the Horner–Wadsworth–Emmons reaction as an alternative. They usually give rise to an E-alkene product when they react, rather than the more usual Z-alkene.
Scope and limitations
The Wittig reaction is a popular method for the synthesis of alkene from ketones and aldehydes. The Wittig reagent can generally tolerate carbonyl compounds containing several kinds of functional groups such as OH, OR, aromatic nitro and even ester groups. There can be a problem with sterically hindered ketones, where the reaction may be slow and give poor yields, particularly with stabilized ylides, and in such cases the Horner–Wadsworth–Emmons (HWE) reaction (using phosphonate esters) is preferred. Another reported limitation is the often labile nature of aldehydes which can oxidize, polymerize or decompose. In a so-called Tandem Oxidation-Wittig Process the aldehyde is formed in situ by oxidation of the corresponding alcohol.
As mentioned above, the Wittig reagent itself is usually derived from a primary alkyl halide. Quaternization of triphenylphosphine with most secondary halides is inefficient. For this reason, Wittig reagents are rarely used to prepare tetrasubstituted alkenes. However the Wittig reagent can tolerate many other variants. It may contain alkenes and aromatic rings, and it is compatible with ethers and even ester groups. Even C=O and nitrile groups can be present if conjugated with the ylide- these are the stabilised ylides mentioned above. Bis-ylides (containing two P=C bonds) have also been made and used successfully.
One limitation relates to the stereochemistry of the product. With simple ylides, the product is usually mainly the Z-isomer, although a lesser amount of the E-isomer is often formed also – this is particularly true when ketones are used. If the reaction is performed in DMF in the presence of LiI or NaI, the product is almost exclusively the Z-isomer. If the E-isomer is the desired product, the Schlosser modification may be used. With stabilised ylides the product is mainly the E-isomer, and this same isomer is also usual with the HWE reaction.
Schlosser modification
The major limitation of the traditional Wittig reaction is that the reaction proceeds mainly via the erythro betaine intermediate, which leads to the Z-alkene. The erythro betaine can be converted to the threo betaine using phenyllithium at low temperature. This modification affords the E-alkene.
Allylic alcohols can be prepared by reaction of the betaine ylid with a second aldehyde. For example:
Examples
Because of its reliability and wide applicability, the Wittig reaction has become a standard tool for synthetic organic chemists.
The most popular use of the Wittig reaction is for the introduction of a methylene group using methylenetriphenylphosphorane (Ph3P=CH2). Using this reagent even a sterically hindered ketone such as camphor can be converted to its methylene derivative. In this case, the Wittig reagent is prepared in situ by deprotonation of methyltriphenylphosphonium bromide with potassium tert-butoxide. In another example, the phosphorane is produced using sodium amide as a base, and this reagent converts the aldehyde shown into alkene I in 62% yield. The reaction is performed in cold THF, and the sensitive nitro, azo and phenoxide groups are tolerated. The product can be used to incorporate a photostabiliser into a polymer, to protect the polymer from damage by UV radiation.
Another example of its use is in the synthesis of leukotriene A methyl ester. The first step uses a stabilised ylide, where the carbonyl group is conjugated with the ylide preventing self condensation, although unexpectedly this gives mainly the cis product. The second Wittig reaction uses a non-stabilised Wittig reagent, and as expected this gives mainly the cis product. Note that the epoxide and ester functional groups survive intact.
Methoxymethylenetriphenylphosphine is a Wittig reagent for the homologation of aldehydes. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Organic_Reactions/Wittig_Reaction.txt |
Iron smelting! Photo credit: http://sclowcountryoutdoors.blogspot...g-at-acba.html
In the beginning, the term actually made sense. When the alchemists and medieval metallurgists started doing experiments to quantify exactly how much iron, say was formed from the smelting of a given amount of iron ore, they found that the weight of the solid iron was always less than that of the ore. Given the decrease in mass, a sensible name for the process was made: reduction.
$2 Fe_2O_3 + 2C \rightarrow 4 Fe_{(s)} + 3CO_{2 \, (g)}$
That was more than 500 years ago. Later on, Scheele, Lavoisier and Priestley independently discovered that the loss of mass was due to the expulsion of an element named (by Lavoisier) oxygen, and subsequent burning the metal in air led to its recombination. Hence, “oxidation”.
$4 Fe_{(s)}+ 3 O_2 (g) \rightarrow 2 Fe_2O_3$
That was about 200 years ago. Then came a general understanding of how atoms are composed of a positively charged nucleus and negatively charged electrons, and the introduction of the formalism known as the “oxidation state“, which is the hypothetical charge that an atom would have if all bonds to atoms of different elements of different elements were 100% ionic. It also works for ions, of course. So here’s the same reaction. See how oxidation leads to removal of electrons, and reduction leads to a gain of electrons.
That was about 70 years ago. This also happens to be the first definition of oxidation and reduction I first learned in high school. When this was introduced in class, my first question – which is still asked by many students today, was:
“In what world does it make sense to call a process where electrons are gained “reduction” ? “
The answer from my high school chemistry teacher was , “well, you’re reducing the oxidation state – making it more negative”. Which was a very clever answer, completely jettisoning the inconvenient historical definition in favor of a simplistic mathematical one. Fortunately for him, I wasn’t fast or clever enough to counter with “then why isn’t “oxidation” called “addition”? (Feel free to use this yourself, however).
Result: I just memorized that “reduction” meant “adding electrons” and “oxidation” meant “removing electrons”. Which came in handy in general chemistry, with its seemingly endless balancing of complex redox reactions. Just when this seemed settled in my mind, along came organic chemistry, with what was seemingly yet another way of defining oxidation and reduction.
At first glance, this seems a long way away from the Gen chem definition of oxidation being loss of electrons and reduction a gain of electrons. However, upon review of the concept of the oxidation state, it makes more sense. If you just pay attention to what’s happening to the oxidation state of the carbons, you can follow along to see if it’s an oxidation or reduction. If the oxidation state is becoming more negative, it’s a reduction (gaining electrons). If the oxidation state is becoming more positive, it’s an oxidation (losing electrons).
Let’s look at those examples again (putting in an extra example for fun), paying attention to the change in oxidation state.
So is there a quick way to figure out if a carbon is being oxidized or reduced? Why yes there is.
• A reduction will result in a net increase in the number of C-H bonds, or a net decrease in the number of C-O bonds (or equivalent, such as C-Cl, C-Br, etc).
• An oxidation will result in a net decrease in the number of C-H bonds, or a net increase in the number of C-O bonds (or equivalent).
All of these events affect the oxidation state of the carbon, and this ties back to the concept of oxidation that I originally learned in high school: keeping track of the gaining (and losing) of electrons. When I finally understood this I was happy to note that the term “oxidation” finally made sense again. “Reduction” still didn’t, but I learned to live with it and moved on. You will too.
Contributors
James Ashenhurst (MasterOrganicChemistry.com)
Oxidation and Reduction Reactions
Potassium permanganate, KMnO4, is a powerful oxidizing agent, and has many uses in organic chemistry.
Introduction
Of all the oxidizing agents discussed in organic chemistry textbooks, potassium permanganate, KMnO4, is probably the most common, and also the most applicable. As will be shown below, KMnO4 can be utilized to oxidize a wide range of organic molecules. The products that are obtained can vary depending on the conditions, but because KMnO4 is such a strong oxidizing agent, the final products are often carboxylic acids.
The half-reaction and oxidation potential
Mn(VII) is reduced under acidic conditions to Mn(IV) or Mn(II) according to the half-reactions shown below, with the indicated cell potentials1
$MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O\;\;\;\;E^o = 1.68\,V$
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\;\;\;\;E^o = 1.5\, V$
$MnO_4^- + 2H_2O + 3e^- \rightarrow Mn^{2+} + 4OH^-\;\;\;\;E^o = 0.6\, V$
General Reactivity with Organic Molecules
KMnO4 is able to oxidize carbon atoms if they contain sufficiently weak bonds, including
1. Carbon atoms with $\pi$ bonds, as in alkenes and alkynes
2. Carbon atoms with weak C-H bonds, such as
1. Carbons with exceptionally weak C-C bonds such as
• C-C bonds in a glycol
• C-C bonds next to an aromatic ring AND an oxygen
KMnO4 also oxidizes phenol to para-benzoquinone.
Example
Examples of carbons that are not oxidized
1. Aliphatic carbons (except those alpha to an aromatic ring, as above)
2. Aromatic carbons (except phenol, as above)
3. Carbons without a C-H bond, except as in (3) above
Exhaustive oxidation of organic molecules by KMnO4 will proceed until the formation of carboxylic acids. Therefore, alcohols will be oxidized to carbonyls (aldehydes and ketones), and aldehydes (and some ketones, as in (3) above) will be oxidized to carboxylic acids.
Reactions with Specific Functional Groups
Using the principles above, we expect KMno4 to react with alkenes, alkynes, alcohols, aldehydes and aromatic side chains. Examples are provided below. It is easiest to start at the top.
Aldehydes
Aldehydes RCHO are readily oxidized to carboxylic acids.
Unless great efforts are taken to maintain a neutral pH, KMnO4 oxidations tend to occur under basic conditions. In fact, the most effective conditions for aldehyde oxidation by KMnO4 involves t-butanol as solvent with a NaH2PO4 buffer.2 The reactions above are deliberately not balanced equations. Balancing the reactions would involve using the methods learned in general chemistry, requiring half reactions for all processes.
Alcohols
Primary alcohols such as octan-1-ol can be oxidized efficiently by KMnO4, in the presence of basic copper salts.3 However, the product is predominantly octanoic acid, with only a small amount of aldehyde, resulting from overoxidation.
Although overoxidation is less of a problem with secondary alcohols, KMnO4 is still not considered generally well-suited for conversions of alcohols to aldehydes or ketones.
Alkenes4
Under mild conditions, potassium permanganate can effect conversion of alkenes to glycols. It is, however, capable of further oxidizing the glycol with cleavage of the carbon-carbon bond, so careful control of the reaction conditions is necessary. A cyclic manganese diester is an intermediate in these oxidations, which results in glycols formed by syn addition.
With addition of heat and/or more concentrated KMnO4, the glycol can be further oxidized, cleaving the C-C bond.
More substituted olefins will terminate at the ketone
Oxidative cleavage of the diol can be carried out more mildly by using IO4 as the oxidant.
The cleavage of alkenes to ketones/carboxylic acids can be used to determine the position of double bonds in organic molecules.5
Alkynes4
Instead of bis-hydroxylation that occurs with alkenes, permanganate oxidation of alkynes initially leads to the formation of diones.
Under harsher conditions, the dione is cleaved to form two carboxylic acids.
Aromatic side-chains6
Treatment of an alkylbenzene with potassium permanganate results in oxidation to give the benzoic acid.
Note
The position directly adjacent to an aromatic group is called the “benzylic” position.
The reaction only works if there is at least one hydrogen attached to the carbon. However, if there is at least one hydrogen, the oxidation proceeds all the way to the carboxylic acid.
Examples:
Notes: Note that in example 2 the extra carbons are cleaved to give the same product as in example 1. And in example 3, two benzoic acids are formed. Finally, when no hydrogens are present on the benzylic carbon, no reaction occurs (example 4).
The oxidation of alkyl side-chains to form benzoic acids was historically used in qualitative analysis to determine the positions of alkyl groups in substituted aromatic systems. Alkyl-substituted rings can be coverted to poly-acids, which can be distinguished on the basis of their pKas
Additional Reading
• Oxidation by Chromic Acid (H2CrO4)
• Ozonolysis
• Oxidative cleavage of double bonds
• Oxidation of alkenes
Sources
1. http://www.epa.gov/ogwdw/mdbp/pdf/alter/chapt_5.pdf
2. Abiko, Atsushi; Roberts, John C.; Takemasa, Toshiro; Masamune, Satoru, Tetrahedron Letters (1986), 27(38), 4537-40
3. Jefford, Charles W.; Wang, Ying, Journal of the Chemical Society, Chemical Communications (1988), (10), 634-5.
4. Carey, F.A.; Sundberg, R. J. Advanced Organic Chemistry
5. Downing, Donald T.; Greene, Richard S., Journal of Investigative Dermatology (1968), 50(5), 380-6.
6. http://www.masterorganicchemistry.co...boxylic-acids/ | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Oxidation_and_Reduction_Reactions/Oxidation_of_Organic_Molecules_by_KMnO4.txt |
A pericyclic reaction is a concerted reaction in which the number of rings in the transition state is greater than the total number of rings in the reactant molecules.
• Introduction to Pericyclic Reactions
An important body of chemical reactions, differing from ionic or free radical reactions in a number of respects, has been recognized and extensively studied. Since reactions of this kind often proceed by nearly simultaneous reorganization of bonding electron pairs by way of cyclic transition states, they have been termed pericyclic reactions. The four principle classes of pericyclic reactions are termed: Cycloaddition, Electrocyclic, Sigmatropic, and Ene Reactions.
• A Useful Mnemonic Rule
Before pericyclic reactions can be put to use in a predictable and controlled manner, a broad mechanistic understanding of the factors that influence these concerted transformations must be formulated. The simplest, albeit least rigorous, method for predicting the configurational path favored by a proposed pericyclic reaction is based upon a transition state electron count.
• Cope Rearrangement
The Cope rearrangement, not to be confused with the Cope elimination, is the conversion of a 1,5-diene to a more stable, constitutionally isomeric 1,5-diene at a very high temperature:
• Cycloaddition Reactions
A concerted combination of two π-electron systems to form a ring of atoms having two new σ bonds and two fewer π bonds is called a cycloaddition reaction. The number of participating π-electrons in each component is given in brackets preceding the name, and the reorganization of electrons may be depicted by a cycle of curved arrows - each representing the movement of a pair of electrons.
• Electrocyclic Reactions
An electrocyclic reaction is the concerted cyclization of a conjugated π-electron system by converting one π-bond to a ring forming σ-bond. The reverse reaction may be called electrocyclic ring opening.
• Ene Reactions
The joining of a double or triple bond to an alkene reactant having a transferable allylic hydrogen is called an ene reaction. The reverse process is called a retro ene reaction.
• Sigmatropic Rearrangements
Molecular rearrangements in which a σ-bonded atom or group, flanked by one or more π-electron systems, shifts to a new location with a corresponding reorganization of the π-bonds are called sigmatropic reactions. The total number of σ-bonds and π-bonds remain unchanged. These rearrangements are described by two numbers set in brackets, which refer to the relative distance (in atoms) each end of the σ-bond has moved.
• Stereochemical Notations
One characteristic shared by most pericyclic reactions, and noted in many cases described above, is their stereospecificity. This is not the first class of reactions for which a characteristic stereospecificity has been noted. Substitution reactions may proceed randomly or by "inversion" or "retention" of configuration. Elimination reactions may occur in an "anti" or "syn" fashion, or may be configurationally random. The terms "syn" and "anti" have also been applied to 1,2-addition reactions.
• Theoretical Models for Pericyclic Reactions
Concerted reactions proceed most readily when there is congruence between the orbital symmetries of the reactants and products. In other words, when the bonding character of all occupied molecular orbitals is preserved at all stages of a concerted molecular reorganization, that reaction will most likely take place. The greater the degree of bonding found in the transition state for the reaction, the lower will be its activation energy and the greater will be the reaction rate.
• Woodward Hoffmann rules
Robert Burns Woodward and Roald Hoffmann devised these set of rules to explain the stereochemistry of pericyclic reactions based on the orbital symmetry.
Pericyclic Reactions
An important body of chemical reactions, differing from ionic or free radical reactions in a number of respects, has been recognized and extensively studied. Among the characteristics shared by these reactions, three in particular set them apart:
1. They are relatively unaffected by solvent changes, the presence of radical initiators or scavenging reagents, or (with some exceptions) by electrophilic or nucleophilic catalysts.
2. They proceed by a simultaneous (concerted) series of bond breaking and bond making events in a single kinetic step, often with high stereospecificity.
3. In agreement with 1 & 2, no ionic, free radical or other discernible intermediates lie on the reaction path.
Since reactions of this kind often proceed by nearly simultaneous reorganization of bonding electron pairs by way of cyclic transition states, they have been termed pericyclic reactions. The four principle classes of pericyclic reactions are termed: Cycloaddition, Electrocyclic, Sigmatropic, and Ene Reactions. The cycloaddition and ene reactions are shown in their intermolecular format. Corresponding intramolecular reactions, which create an additional ring, are well known.
All these reactions are potentially reversible (note the gray arrows). The reverse of a cycloaddition is called cycloreversion and proceeds by a ring cleavage and conversion of two sigma-bonds to two pi-bonds. The electrocyclic reaction shown above is a ring forming process. The reverse electerocyclic ring opening reaction proceeds by converting a sigma-bond to a pi-bond. As shown, the retro ene reaction cleaves an unsaturated compound into two unsaturated fragments. Finally, sigmatropic bond shifts may involve a simple migrating group, as shown in the example above, or may take place between two pi-electron systems (e.g. the Cope rearrangement).
The general descriptions shown above provide a basis for reaction classification, but care must be taken to assure that a given transformation is truly concerted. Unfortunately, this is not a trivial determination, often requiring a combination of isotope labeling and stereochemical studies to arrive at a plausible conclusion. There is also a subtle distinction to be made between a synchronous reaction in which all bond-making and bond-breaking events take place in unison, and a multi-stage concerted process in which some events precede others without generating an intermediate state.
Although some pericyclic reactions occur spontaneously, most require the introduction of energy in the form of heat or light, with a remarkable product dependence on the source of energy used. An appreciation of the stereoselective structural changes these reactions promote is best achieved by inspecting some individual examples. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Pericyclic_Reactions/1Introduction_to_Pericyclic_Reactions.txt |
Before pericyclic reactions can be put to use in a predictable and controlled manner, a broad mechanistic understanding of the factors that influence these concerted transformations must be formulated. The simplest, albeit least rigorous, method for predicting the configurational path favored by a proposed pericyclic reaction is based upon a transition state electron count. In most of the earlier examples, pericyclic reactions were described by a cycle of curved arrows, each representing a pair of bonding electrons. The total number of electrons undergoing reorganization is always even, and is either a 4n+2 or 4n number (where n is an integer). Once this electron count is made, the following table may be used for predictions.
Thermal
Reactions
Transition State Class
Configurational Preference
4n + 2 (aromatic)
Suprafacial or Disrotatory
4n (antiaromatic)
Antarafacial or Conrotatory
Photochemical
Reactions
Transition State Class
Configurational Preference
4n + 2 (aromatic)
Antarafacial or Conrotatory
4n (antiaromatic)
Suprafacial or Disrotatory
Although this modest mnemonic does not make explicit use of molecular orbitals, more rigorous methods that are founded on the characteristics of such orbitals have provided important insight into these reactions. Since pericyclic reactions proceed by a cyclic reorganization of bonding electron pairs, it is necessary to evaluate changes in the associated molecular orbitals that take place in going from reactants to products. The following section describes approaches of this kind.
Cope Rearrangement
The Cope rearrangement, not to be confused with the Cope elimination, is the conversion of a 1,5-diene to a more stable, constitutionally isomeric 1,5-diene at a very high temperature, eg:
Mechanism
The Cope rearrangement is a pericyclic reaction
see also Claisen rearrangement
Cycloaddition Reactions
A concerted combination of two π-electron systems to form a ring of atoms having two new σ bonds and two fewer π bonds is called a cycloaddition reaction. The number of participating π-electrons in each component is given in brackets preceding the name, and the reorganization of electrons may be depicted by a cycle of curved arrows - each representing the movement of a pair of electrons. These notations are illustrated in the drawing on the right. The ring-forming cycloaddition reaction is described by blue arrows, whereas the ring-opening cycloreversion process is designated by red arrows. Note that the number of curved arrows needed to show the bond reorganization is half the number total in the brackets.
The most common cycloaddition reaction is the [4π+2π] cyclization known as the Diels-Alder reaction. In Diels-Alder terminology the two reactants are referred to as the diene and the dienophile. The following diagram shows two examples of [4π+2π] cycloaddition, and in the second equation a subsequent light induced [2π+2π] cycloaddition. In each case the diene reactant is colored blue, and the new σ-bonds in the adduct are colored red. The stereospecificity of these reactions should be evident. In the first example, the acetoxy substituents on the diene have identical E-configurations, and they remain cis to each other in the cyclic adduct. Likewise, the ester substituents on the dienophile have a trans-configuration which is maintained in the adduct. The reactants in the second equation are both monocyclic, so the cycloaddition adduct has three rings. The orientation of the quinone six-membered ring with respect to the bicycloheptane system (colored blue) is endo, which means it is oriented cis to the longest or more unsaturated bridge. The alternative configuration is called exo.
Since the dienophile (quinone) has two activated double bonds, a second cycloaddition reaction is possible, provided sufficient diene is supplied. The second cycloaddition is slower than the first, so the monoadduct shown here is easily prepared in good yield. Although this [4+2] product is stable to further heating, it undergoes a [2+2] cycloaddition when exposed to sunlight. Note the loss of two carbon-carbon π-bonds and the formation of two σ-bonds (colored red) in this transformation. Also note that the pi-subscript is often omitted from the [m+n] notation for the majority of cycloadditions involving π-electron systems.
Reaction 3 is an intramolecular Diels-Alder reaction. Since the diene and dienophile are joined by a chain of atoms, the resulting [4+2] cycloaddition actually forms two new rings, one from the cycloaddition and the other from the linking chain. Once again the addition is stereospecific, ignoring the isopropyl substituent, the ring fusion being cis and endo. The fourth reaction is a [6+4] cycloaddition.
Electrocyclic Reactions
An electrocyclic reaction is the concerted cyclization of a conjugated π-electron system by converting one π-bond to a ring forming σ-bond. The reverse reaction may be called electrocyclic ring opening. Two examples are shown on the right. The electrocyclic ring closure is designated by blue arrows, and the ring opening by red arrows. Once again, the number of curved arrows that describe the bond reorganization is half the total number of electrons involved in the process.
In the first case, trans,cis,trans-2,4,6-octatriene undergoes thermal ring closure to cis-5,6-dimethyl-1,3-cyclohexadiene. The sterospecificity of this reaction is demonstrated by closure of the isomeric trans,cis,cis-triene to trans-5,6-dimethyl-1,3-cyclohexadiene, as noted in the second example.
By clicking on this diagram two examples of thermal electrocyclic opening of cyclobutenes to conjugated butadienes will be displayed. This mode of reaction is favored by relief of ring strain, and the reverse ring closure (light blue arrows) is not normally observed. Photochemical ring closure can be effected, but the stereospecificity is opposite to that of thermal ring opening. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Pericyclic_Reactions/A_Useful_Mnemonic_Rule.txt |
The joining of a double or triple bond to an alkene reactant having a transferable allylic hydrogen is called an ene reaction. The reverse process is called a retro ene reaction. In the bonding direction the ene reaction is characterized by the redistribution of three pairs of bonding electrons. and may be described by a cycle of three curved arrows. As noted earlier, this bond reorganization involves the overall conversion of a π-bond to a σ-bond (or the opposite in the case of retro ene fragmentation). This is the same bond bookkeeping change exhibited by electrocyclic reactions, but no rings are formed or broken in an ene reaction unless it is intramolecular. The following examples illustrate some typical ene reactions, with equation 3 being an intramolecular ene reaction. Ene reactions are favored when the hydrogen accepting reagent, the "enophile", is electrophilic. This is the case for reactions 1 and 2, which proceed under milder conditions than 3, despite the latter's intramolecular nature.
Hydrogen is the most common atom transferred in an ene reaction. Indeed, all the examples shown above involve hydrogen shifts. Other atoms or groups may, however, participate in ene-like transformations. Two such cases will be displayed above by clicking on the diagram. Reaction 4 is drawn as a retro ene reaction, although this has not been demonstrated to be general for all reactions of allylic alcohols with thionyl chloride. Equation 5 illustrates an unusual "magnesium ene reaction" in which a Grignard function moves to a new location before reacting with an electrophilic reagent such as CO2. Because this is an intramolecular ene reaction a new ring is formed. Clicking on the diagram a second time will display two additional examples. Equation 6 demonstrates that an enol tautomer, even in low concentration, may function as the hydrogen donor in the ene reaction. Equation 7 is one of many examples of Lewis acid catalysis in the ene reaction. A similar acid-catalyzed reaction of simple aldehydes with alkenes to give allylic alcohols, 1,3-diols or 1,3-dioxanes is known as the Prins reaction.
Certain retro ene reactions have proven useful as synthetic transformations.
Sigmatropic Rearrangements
Molecular rearrangements in which a σ-bonded atom or group, flanked by one or more π-electron systems, shifts to a new location with a corresponding reorganization of the π-bonds are called sigmatropic reactions. The total number of σ-bonds and π-bonds remain unchanged. These rearrangements are described by two numbers set in brackets, which refer to the relative distance (in atoms) each end of the σ-bond has moved, as illustrated by the first equation in the diagram below. The most common atom to undergo sigmatropic shifts is hydrogen or one of its isotopes. The second equation in the diagram shows a facile [1,5] hydrogen shift which converts a relatively unstable allene system into a conjugated triene. Note that this rearrangement, which involves the relocation of three pairs of bonding electrons, may be described by three curved arrows.
These reactions are particularly informative in that [1,3] hydrogen shifts are not observed. The reactant in the first equation is a deuterium labeled 1,3,5-cyclooctatriene. On heating, this compound equilibrates with its 1,3,6-triene isomer, and the two deuterium atoms are scrambled among the four locations noted. If [1,3] or [1,7] hydrogen shifts were taking place, the deuterium atoms would be distributed equally among all eight carbon atoms. On prolonged heating, or at higher temperatures these cyclooctatrienes undergo electrocyclic ring opening to 1,3,5,7-octatetraene and reclosure to vinyl-1,3-cyclohexadienes.
The second example shows another [1,5] hydrogen shift, from the proximal methyl group to the carbonyl oxygen atom. The resulting dienol rapidly exchanges OH for OD before the [1,5] shift reverses. In this manner the reactive methyl is soon converted to CD3. Since hydrogens alpha to a carbonyl group are known to undergo acid or base catalyzed exchange by way of enol intermediates, we might expect the α'-CH2 group to exchange as well. However, if care is taken to remove potential acid or base catalysts, the thermal [1,3] shift necessary for the exchange is found to be very slow.
Cope and Claisen rearrangements
The [3,3] sigmatropic rearrangement of 1,5-dienes or allyl vinyl ethers, known respectively as the Cope and Claisen rearrangements, are among the most commonly used sigmatropic reactions. Three examples of the Cope rearrangement are shown in the following diagram. Reactions 1 and 2 (top row) demonstrate the stereospecificity of this reaction. The light blue σ-bond joins two allyl groups, oriented so their ends are near each other. Since each allyl segment is the locus of a [1,3] shift, the overall reaction is classified as a [3,3] rearrangement. The three pink colored curved arrows describe the redistribution of three bonding electron pairs in the course of this reversible rearrangement. The diene reactant in the third reaction is drawn in an extended conformation. This molecule must assume a coiled conformation (as above) before the [3,3] rearrangement can take place. The product of this rearrangement is an enol which immediately tautomerizes to its keto form. Such variants are termed the oxy-Cope rearrangement, and are useful because the reverse rearrangement is blocked by rapid ketonization. If the hydroxyl substituent is converted to an alkoxide salt, the activation energy of the rearrangement is lowered significantly.
The degenerate or self-replicating Cope rearrangement has been a fascinating subject of research. For examples .
Two examples of the Claisen Rearrangement may be seen by clicking on the above diagram. Reaction 4 is the classic rearrangement of an allyl phenyl ether to an ortho-allyl phenol. The methyl substituent on the allyl moiety serves to demonstrate the bonding shift at that site. The initial cyclohexadienone product immediately tautomerizes to a phenol, regaining the stability of the aromatic ring. Reaction 5 is an aliphatic analog in which a vinyl group replaces the aromatic ring. In both cases three pairs of bonding electrons undergo a reorganization.
By clicking on the above diagram a second time two examples of [2,3] sigmatropic rearrangements will be displayed. The allylic sulfoxide in reaction 6 rearranges reversibly to a less stable sulfenate ester. The weak S-O bond may be reductively cleaved by trimethyl phosphite to an allylic alcohol and a thiol (not shown). Reaction 7 shows a similar rearrangement of a sulfur ylide to a cyclic sulfide. The [2,3]-Wittig rearrangement is yet another example. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Pericyclic_Reactions/Ene_Reactions.txt |
One characteristic shared by most pericyclic reactions, and noted in many cases described above, is their stereospecificity. This is not the first class of reactions for which a characteristic stereospecificity has been noted. Substitution reactions may proceed randomly or by "inversion" or "retention" of configuration. Elimination reactions may occur in an "anti" or "syn" fashion, or may be configurationally random. The terms "syn" and "anti" have also been applied to 1,2-addition reactions.
Since these configurational change notations are not appropriate for pericyclic reactions, new designations are needed. Cycloaddition reactions and sigmatropic rearrangements both involve pairs of σ-bond-making events (or a coupled bond-making & bond-breaking) associated with a π-electron system. If all the bonding events take place on the same face of the π-system the configuration of the reaction is termed suprafacial. If the bonding events occur on opposite sides or faces of the π-system the reaction is termed antarafacial. Suprafacial examples of these pericyclic transformations are shown below. The bracketed numbers that designate reactions of this kind sometimes carry subscripts (s or a) that specify their configuration. Thus the cycloaddition on the left may be termed a [4s + 2s] process.
Although cycloaddition reactions are concerted (no intermediate species are formed), the two new bonds are not necessarily formed in a synchronous fashion. Depending on partial charge distribution in the diene and dienophile reactants, the formation of one bond may lead the development of the other. Such unsymmetrical transition state bonding is termed asynchronous.
A Suprafacial [4+2] Cycloaddition
A Suprafacial [3,3] Sigmatropic Rearrangement
An example of an antarafacial [1,7] hydrogen shift is shown in the following diagram. The conjugated triene assumes a nearly planar coiled conformation in which a methyl hydrogen is oriented just above the end carbon atom of the last double bond. A [1s,7a] sigmatropic hydrogen shift may then take place, as described by the four curved arrows. With reference to the approximate plane of this π-electron system (defined by the green bonds), the hydrogen atom departs from the bottom face and bonds to the top face, so the transfer is antarafacial.
A different notation for configurational change is required for electrocyclic reactions. In these cases a σ-bond between the ends of a conjugated π-electron system is either made or broken with a corresponding loss or gain of a π-bond. For this to happen, the terminal carbon atoms of the conjugated π-electron system must be rehybridized with an accompanying rotation or twisting of roughly 90º. When viewed along the axis of rotation, the two end groups may turn in the same direction, termed conrotatory, or in opposite directions, termed disrotatory. The prefixes con and dis may be remembered by association with their presence in the words concur & disagree. These two modes of electrocyclic reaction are shown in the following diagram in the general form in which they are most commonly observed. Specific examples of these electrocyclic reactions were given earlier
.
A Disrotatory Electrocyclic Closure
A Conrotatory Electrocyclic Opening
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Theoretical Models for Pericyclic Reactions
In 1965 R. B. Woodward and Roald Hoffmann of Harvard University proposed and demonstrated that concerted reactions proceed most readily when there is congruence between the orbital symmetries of the reactants and products. In other words, when the bonding character of all occupied molecular orbitals is preserved at all stages of a concerted molecular reorganization, that reaction will most likely take place. The greater the degree of bonding found in the transition state for the reaction, the lower will be its activation energy and the greater will be the reaction rate.
A general introduction to molecular orbitals was presented earlier. The simple compound ethene is made up of six atoms held together by six covalent bonds, as described in the following illustration. A molecular orbital diagram of ethene is created by combining the twelve atomic orbitals associated with four hydrogen atoms and two sp2 hybridized carbons to give twelve molecular orbitals. Six of these molecular orbitals (five sigma & one pi-orbital) are bonding, and are occupied by the twelve available valence shell electrons. The remaining six molecular orbitals are antibonding, and are empty.
Proper molecular orbitals are influenced by all the nuclei in a molecule, and require consideration of the full structure and symmetry of a molecule for their complete description. For most purposes, this level of treatment is not needed, and more localized orbitals serve well. In the case of ethene and other isolated double bonds, descriptions of the localized π orbitals will be displayed by clicking on the above diagram. Several important characteristics of molecular orbitals need to be pointed out, and this diagram will serve to illustrate them.
1. The spatial distribution of electron density for most occupied molecular orbitals is discontinuous, with regions of high density separated by regions of zero density, e.g. a nodal plane. The π-orbital on the left has one nodal plane (colored light blue), and the π*-orbital on the right has a second nodal plane (colored yellow). As a rule, higher energy molecular orbitals have a larger number of nodal surfaces or nodes.
2. The wave functions that describe molecular orbitals undergo a change in sign at nodal surfaces. This phase change is sometimes designated by plus and minus signs associated with discrete regions of the orbital, but this notation may sometimes be confused for an electric charge. In the above diagram, regions having one phase sign are colored blue, while those having an opposite sign are colored red.
3. These localized orbitals may be classified by two independent symmetry operations ; a mirror plane perpendicular to the functional plane and bisecting the the molecule (colored yellow above), and a two-fold axis of rotation (C2) created by the intersection of this mirror plane with the common nodal plane (colored light blue). The π-orbital on the left is symmetric (S) with respect to the mirror plane, but antisymmetric (A) when rotated 180º, a C2 operation. The opposite is true for the π*-orbital on the right, which has a mirror plane symmetry of A and a C2 symmetry of S. Such symmetry characteristics play an important role in creating the orbital diagrams used by Woodward and Hoffmann to rationalize pericyclic reactions.
The original approach of Woodward and Hoffmann involved construction of an "orbital correlation diagram" for each type of pericyclic reaction. The symmetries of the appropriate reactant and product orbitals were matched to determine whether the transformation could proceed without a symmetry imposed conversion of bonding reactant orbitals to antibonding product orbitals. If the correlation diagram indicated that the reaction could occur without encountering such a symmetry-imposed barrier, it was termed symmetry allowed. If a symmetry barrier was present, the reaction was designated symmetry-forbidden. Two related methods of analyzing pericyclic reactions are the transition state aromaticity approach, and the frontier molecular orbital approach. Each of these methods has merit, and a more detailed description of each may be examined by clicking the appropriate button below.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Pericyclic_Reactions/Stereochemical_Notations.txt |
Robert Burns Woodward and Roald Hoffmann devised these set of rules to explain the stereochemistry of pericyclic reactions based on the orbital symmetry. There are several ways to explain the Woodward Hoffmann rules and are discussed in separate topics. Hoffmann received the 1981 Nobel prize in chemistry along with Fukui (Frontier molecular orbital approach). Woodward had died two years before and therefore was not considered for what could have been his second Nobel prize.
General rule
A general way of stating the Woodward Hoffmann rules for thermal pericyclic reactions:
A ground-state pericyclic change is symmetry-allowed when the total number of (4q + 2)s and (4r)a components is odd.
The subscripts "s" and "a" stand for suprafacial and antarafacial respectively, "q" and "r" stand for integers (0, 1, 2, 3, ...). In the following discussion we will explain how this general rule is applicable in all the pericyclic reactions.
Notes to keep in mind
1. Orbital symmetry is conserved in concerted reactions.
2. In this page, molecular orbitals are symbolized in terms of the atomic orbitals whose interaction give the actual molecular orbital. The relative sizes of the resultant atomic orbital contributions are not identical but are ignored as the nodal properties are the subjects of interest here.
3. It is important to identify the σ and π levels, their corresponding anti-bonding levels while drawing a correlation diagram. The number of levels of each symmetry type on the right hand side of the correlation diagram should match those on the left side.
4. All bonding σ orbitals which are not symmetric or anti-symmetric with respect to every molecular symmetry element must be mixed until they become so.
5. All orbitals of a given symmetry can be mixed with each other for convenience.
Precautions
1. Each basic process must be isolated and analyzed separately. Otherwise the superposition of two forbidden but independent processes may lead one to the erroneous conclusion that the combined process is symmetry-allowed.
2. Only symmetry elements that are present in reactants, products and transition state are useful. Additionally, the symmetry elements chosen for analysis must bisect bonds made or broken in the process. Two corollaries which follow:
• a symmetry element of no use in analyzing a reaction is one with respect to which the orbitals considered are either all symmetric or all anti-symmetric.
• if the only symmetry element is one which does not bisect any bonds made or broken, then the correlation diagram constructed based on this element can lead to false conclusion.
3. Each case must be reduced to its highest inherent symmetry. Thus, if there are heteroatoms in a polyene component, they are to be replaced by their isoelectronic carbon groupings. If there are substituents with trivial electronic demands they should be replaced by hydrogens. Heteroatoms can offer new reactivity patterns by the inclusion of the lone pairs or by the availability of low lying unoccupied orbitals. These cases have to be dealt with care.
Side Notes
Robert Burns Woodward was a professor at Harvard University when Roald Hoffmann was a research fellow there. Woodward's lectures often continued for hours. As he was rather slow with chalk drawing, it was easy to take class notes. See the bottom of this page for the videos of his lectures.
• Suprafacial: If a component undergoes addition (forms bond) on the same face, it is called a suprafacial component.
• Antarafacial: If a component undergoes addition on opposite faces, it is called an antarafacial component.
• In a conrotatory mode of an electrocyclic reaction the substituents located at the termini of a conjugated double bond system move in the same (clockwise or counter clockwise) direction during ring opening or ring closure. In a disrotatory mode, they move in opposite directions.
• Correlation diagram: A diagram which shows the relative energies of orbitals, configurations, valence bond structures, or states of reactants and products of a reaction, as a function of the molecular geometry, or another suitable parameter. An example involves the interpolation between the energies obtained for the united atoms and the values for the separated atom limits.
• Conservation of orbital symmetry: IUPAC definition: The orbital symmetry control of concerted reactions; this requires transformation of the molecular orbitals of reactants into those of products to proceed continuously by following a reaction path along which the symmetry of these orbitals remains unchanged. Reactions which adhere to this requirement are classified as symmetry-allowed reactions, and those which do not as symmetry-forbidden reactions.
• Forbidden or allowed? A reaction may be allowed but may also require high energy or in other case, a reaction may be forbidden but may also often occur relatively easily. By the term "forbidden" it is meant that the reaction has a high symmetry imposed barrier along a specific geometry of approach apart from the other factors such as sterics. This does not mean that the reactions cannot happen. Similarly, a reaction may need to overcome high activation energy even though it is symmetry allowed.
Corey's claim on WH rules and Hoffmann's rebuttal: E. J. Corey's claim and Roald Hoffmann's rebuttal are here and you should judge yourself.
Cycloaddition Reactions
Here we will see how Woodward Hoffmann selection rules for cycloaddition are derived. Before doing that let's see how to draw a correlation diagram for a cycloaddition reaction. An example of cycloaddition reaction is the formation of cyclobutane from two ethylene molecules. A simple orbital correlation diagram or orbital symmetry diagram is shown below to explain the outcome of of this reaction. The symmetry element chosen are the mirror planes (plane 1 and plane 2 in section 1.2.)bisecting the bonds made or broken (depicted in blue dotted lines in the correlation diagram, see section 1.3.).
Any other symmetry elements passing through the two ethylene molecules or cyclobutane will not suffice for the discussion of the bond formation of two ethylene molecules (a pictorial description is given below).
Which plane of symmetry to choose
For a simple example, the approach of the two ethylene molecules are considered as parallel. Any other approach is not considered for now. Two mirror planes 1 and 2 are considered as the choice of symmetry elements for the purpose of discussion as they satisfy the required conditions as mentioned in the "precautions" section.
Orbital Correlation Diagram
In the orbital symmetry diagram or the orbital correlation diagram shown in the left, we consider the combinations (π1 + π2) and (π1 - π2) of the two ethylenes and (σ1 + σ2), (σ1 - σ2) of cyclobutane. The energy levels of either of these combinations should be degenerate when the two ethylenes are not interacting or the cyclobutane is not being broken apart. Any weak interactions break the degeneracy of these levels.
The symmetry labels of the combinations are with respect to the two mirror planes considered here. (SS) represents the orbital symmetry where the π orbitals are symmetric under reflection on both the mirror planes, (AA) represents where they are anti-symmetric, (SA) and (AS) represent levels where the π orbitals are symmetric under one mirror plane but anti-symmetric under the other mirror plane.
Applying the concept of "conservation of orbital symmetry" introduced by Woodward and Hoffmann, it is shown in the diagram how the levels transform along the course of the reaction. The ground state of ethylene becomes a doubly excited state of cyclobutane. The reverse is also true and the ground state of cyclobutane becomes a doubly excited state of ethylene. This is highly undesirable outcome of the reaction and termed as forbidden under thermal conditions.
State Correlation Diagram
For all the pericyclic reactions, orbital correlation diagrams are sufficient to predict the reaction outcome. However, a deeper understanding arises from the corresponding state correlation diagrams. State correlation diagrams deal with the actual quantum mechanical states and not the orbitals. As quantum mechanical states are definite solutions of the Schrödinger equation, they relate directly with the reality. State correlation diagrams can always be used to verify the outcome predicted by the orbital correlation diagram. A simple procedure to assign the state symmetry is shown below.
The figure in the right is showing an example of how to derive the state symmetry based on the individual electron configurations in the orbitals. The symmetry of the states is obtained by multiplying the symmetry labels of each electrons following the rules:
$S \times S = S = A \times A$
$S \times A = A = A \times S$
(S = symmetric; A = anti-symmetric)
In this diagram, several other excited states are possible which are not relevant for the discussion. The correlated states should have same orbital occupancies in the reactant and the product. The correlated states should also have the same symmetry with respect to each mirror planes.
An important difference between the state correlation and orbital correlation is that in the state correlation states of the same symmetry do not cross. This is known as non-crossing rule. As the energies of the two states with same symmetry approach each other, they mix and diverge. Greater the mixing, higher is the energy gap. Thus the avoided crossing means the ground state of the ethylenes correlates with the ground state of cyclobutane. However, there is a very large symmetry imposed barrier for this reaction and at the first excited state, this barrier is elevated and the reaction becomes symmetry-allowed.
The state correlation diagram corresponds to the reaction co-ordinate and this symmetry imposed barrier could be additional to other energy barriers caused by sterics, etc. Interestingly, the first excited state surface comes very close to the thermal barrier and because of this situation, funnel forms and reaction becomes allowed in the first excited state. However, this does not mean that the photochemical reaction will always follow this concerted pathway. It is often found that an allowed pericyclic photochemical reaction chooses stepwise pathway.
An interesting situation arises when one considers the cycloaddition reaction of two propylene molecules. An extra methyl group on each of ethylene molecule has in an absolute sense reduced the symmetry of all the orbitals to the same level. However, one must restrain from concluding that the reaction has become symmetry allowed. The perturbation due to a mere substitution is very little and the reaction remains symmetry forbidden.
Orthogonal Approach of Two Ethylenes
We have so far seen the parallel approach for [π2s + π2s] cycloaddition is forbidden. What about the orthogonal approach? This can be called [π2s + π2a] reaction. Let us consider the symmetry element involved. Here one of the ethylenes is suprafacial component and the other is the antarafacial component. There is a two fold axis of symmetry (C2) passing through the bonds of two ethylenes perpendicular to the plane of this screen and right through the "green dot" in the figure on the right. This also bisects the bonds to be broken. The orthogonal approach will eventually flatten out and become much flatter cyclobutane. To consider the orthogonal approach we have exaggerated the "twistedness" of cyclobutane.
With respect to the C2 axis, the π bonds are well defined i.e., they are either symmetric (S) or anti-symmetric (A). Therefore there s no need to mix them. However, for the σ bonds as they are not independently well defined with respect to C2 axis, we have to mix them as shown in the figure.
We can now see that the ground states of the two ethylenes directly correlate with the ground state of cyclobutane. The reaction is symmetry allowed and if we consider the corresponding state correlation diagram, we will see that there is no symmetry imposed barrier.
The reaction outcome for the first excited state can also be rationalized from the state correlation diagram.
[4 + 2] cycloaddition
A very common example of [4 + 2] cycloaddition would be Diels Alder reaction between cyclobutadiene and ethylene. Here the element of symmetry chosen is the mirror plane bisecting cyclobutadiene and ethylene as shown in the top figure. The orbitals on cyclobutadiene and ethylene are well defined with respect to the plane of symmetry but the σ orbitals on the product cyclohexene are not. In this case, the σ bonds has to be mixed as shown in the left figure.
The orbital correlation diagram clearly shows there is no crossing of levels between bonding and antibonding levels. The ground states of the reactants are correlating with the ground state of cyclohexene. The reaction is symmetry allowed.
Under photochemical conditions, this situation changes. The corresponding state correlation diagram will show that there is large symmetry imposed barrier to make this reaction possible and therefore, the reaction is forbidden in the first excited state.
Selection Rules for Cycloaddition
We just saw that the total number of (4q + 2) suprafacial components and (4r) antarafacial components is even for [π2s + π2s]. Here the (4q + 2) suprafacial components are the two ethylenes and there are no (4r) antarafacial component which implies that the total number is 2 (an even number). According to the general rule, the reaction is forbidden. Using the same rule, [π2s + π2a] cycloaddition should be allowed because we now have one (4q + 2) component and no (4r) component which implies the total number of (4q + 2)s and (4r)a components is 1 (an odd number).
For a two component cycloaddition, the maximum number of modes of addition is 22 = 4 (for n components, it is 2n): (s,s), (s,a), (a,s), (a,a) where s = suprafacial and a = antarafacial.
(m + n) Thermal Thermal Photochemical Photochemical
Allowed Forbidden Allowed Forbidden
4q + 2 (s,s) and (a,a) (s,a) and (a,s) (s,a) and (a,s) (s,s) and (a,a)
4q (s,a) and (a,s) (s,s) and (a,a) (s,s) and (a,a) (s,a) and (a,s)
Electrocyclic Reactions
Woodward Hoffmann general rules for the electrocyclic reactions are as follows:
The thermal electrocyclic reactions of a k π electron system will be disrotatory for k = 4q + 2, conrotatory for k = 4q (q = 0, 1, 2, ...). In the first excited state, these relationships are reversed. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Pericyclic_Reactions/Woodward_Hoffmann_rules.txt |
In organic chemistry, it is important to understand the concept of electron flow. In polar reaction mechanisms, such as the nucleophilic substitution reactions of haloalkanes, electron flow will be designated by arrows indicating the movement of electrons from electron rich regions to electron poor regions.
Introduction
In considering this concept, we must look at the two types of arrows provided in the mechanisms shown below. The curved arrows indicate the movement of electrons. The first type of arrow, shown in pink, originates from the electron pair of the nucleophile and extends to the electrophilic carbon of the haloalkane. This type of movement does not indicate that electrons leave the nucleophile; rather, it means that electrons become shared between the nucleophile and the electrophilic atom.
The second type of curved arrow, also shown in pink, originates from the R-X bond and extends to the halogen. This indicates cleavage of the bond, whereby the electron pair becomes separated from R, the electrophilic carbon, and ends up on the halogen atom.
While we are using the concept of nucleophilic substitution mechanisms to explain electron flow, it is very important to understand that this concept will be applied in nearly all the mechanisms you learn throughout your course of study. The simplest way to think about this in any mechanism you learn is that electrons will be pushed from an electron rich species or site to an electron poor species or site, and the direction of the curved arrow will indicate this.
Contributors
• Rachael Curtis
Reaction Coordinates in Potential Energy Diagrams
Reaction potential energy diagrams are graphs that show the energy of a process as a function of the extent to which that process has occurred. As these are graphs showing mathematical functions, there must be a numerical coordinate axis that shows the independent variable. This coordinate is called the reaction coordinate, and it reflects the geometry of the system. Very often, the reaction coordinate reflects extent to which a reaction has progressed from reactants to products, starting with reactants near the y-axis (the energy coordinate) and progressing toward products.
Reaction Coordinates for Diatomic Systems
The simplest reaction coordinate is found for a diatomic system. In this case, the geometry in the molecular frame is completely described by the intermolecular distance, r. Therefore, the potential energy diagram for the dissociation of a simple diatomic molecule consists of a plot of the energy of the system as a function of separation.
Reaction Coordinates for Polyatomic Systems
For systems of 3 or more atoms, the geometry coordinates get more complicated. A non-linear molecule consisting of N atoms will have 3N-6 different geometry coordinates, and there are 3N-5 for linear molecules (this also applies to N = 2). That means that a depiction of the potential energy for a non-linear molecule isa 3N - 6 dimensional surface (3N - 5 dimensions for a linear molecule).
Formally, the coordinate axes consist of the mathematical normal coordinates that describe motion of the atoms in the molecule, although more physical meaningful coordinates can also be utilized.
Triatomic Systems
A non-linear triatomic molecule can be described with three coordinates. The best example is water, HOH. The three coordinates in water are nominally the two O-H bond lengths and the H-O-H bond angle. Mathematically, these are described as symmetric and asymmetric OH stretching motion, and the HOH bend.
Considering the challenge in representing a 4-D system (3 geometry coordinates and an energy coordinate), there are simplifications and/or approximations that can be made. For example, it is to create a plot where one or two of the variables are fixed. For example, the energy can be plotted against the HOH bond angle, holding the OH bonds at their equilibrium bond lengths. Because the optimal OH bond lengths do not vary significantly as a function of bond angle, a plot of energy vs bond angle in water will be very close to the lowest energy part of the potential energy surface. Alternatively, it is possible to fix the bond angle and one of the OH bond lengths, and look at the energy as a function of the length of the other OH bond. Although this is likely a reasonable approximation of the lowest energy path for breaking the O-H bond, it does not account for changes in the preferred bond angle that might occur, or, to a lesser extent, the change in the OH bond length between water and hydroxyl radical.
Alternatively, it is possible to plot a relaxed energy surface for O-H breaking where the other variables (HOH bond angle, other O-H bond length) are always at the optimal values as a function of the breaking O-H bond length. In this plot, the energy does account for the changes in the preferred bond angle and the change to the hydroxyl bond length. If there is a readily available numerical coordinate to use, then a relaxed surface scan is very convenient and is an accurate reflection of the surface, since it refers to the lowest energy pathway in that coordinate.
Bond Rotation
The potential energy surface for bond rotation is an example of a relaxed surface scan. The reaction coordinate is the dihedral angle between groups on the two atoms, which can be easily observed in a Newman projection. Potential energy surfaces for bond rotation are commonly used for conformational analysis of molecules like ethane and butane.
More complicated systems
Most organic chemical reactions cannot be described by a single simple coordinate. Even when they can, it may be more convenient to use a more complicated coordinate. For example, chloride undergoes SN2 reaction with methyl bromide through a 5-centered transition state.
In principle, the energy of the process can be plotted vs the Cl-C bond length. Therefore, the energy increases gradually as chloride approaches from long distance until the bond length reaches that in the transition state, which is approximately 200 pm. From there, the energy will decrease precipitously as the Cl-C bond length approaches the value in CH3Cl, 175 pm. A more convenient coordinate might utilize the Cl-C bond length until the formation of the transition state, and switch to the C-Br bond length after the transition state. Alternatively, if using the Cl-C bond length, it might be better to not to use a linear scale, or even consistent scaling.
Consequently, most potential energy surfaces in organic chemistry are not drawn with any single, numerical coordinate, but with a generalized reaction coordinate that reflects the geometry changes in the reaction. Instead of numerical labels on the axis, the positions are labeled with their structures, and the regions in-between are the paths that connect the indicated geometries. Because of this, it is possible to reduce the highly-dimensional surfaces of complicated reactions to 1-dimensional curves.
Important Things to Remember
1. Potential energy diagrams are graphs.
Energy is on the y-axis, and the x-axis indicates geometry. Energy is a function of geometry.
2. The geometry changes refer to changes in bonding (changes in atom positions).
Each point on the diagram has the same molecular formula (same atoms and electrons). The bonding can change, however. For example, in the SN2 process shown above, the reactant is Cl- + CH3Br, and the product is CH3Cl + Br-. Even if this is not explicitly shown, it is implicit in the diagram. It is not possible to compare energies of different atoms.
3. The reaction coordinate in a complex reaction is the projection of a multidimensional surface into a single coordinate
A lot of important details about the reaction can be lost in the projection. It can be helpful to think of different steps in the reaction as happening in orthogonal directions - for example, if one occurs in the plane of the drawing, the next step might come out toward you, or away from you. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reaction_Fundamentals/Pushing_Arrows.txt |
• EA1. Introduction to Electrophilic Addition
• EA2. Cations in Electrophilic Addition
• EA3. Solvent Effects in Electrophilic Addition
• EA4. Stabilized Cations
• EA5. Addition to Alkene Complexes
• EA6. Concerted Additions
• EA7. Olefin Polymerization: Cationic and Ziegler-Natta Approaches
• EA8. Solutions for Selected Problems
Electrophilic Addition to Alkenes
EA1. Introduction to Electrophilic Addition
Alkenes are found throughout nature. They form the basis of many natural products, such as terpenes, which play a variety of roles in the lives of plants and insects. The C=C bonds of alkenes are very different from the C=O bonds that are also common in nature. The C=C bonds of alkenes are electron-rich and nucleophilic, in contrast to the electron-poor C=O bonds of carbohydrates, fatty acids and proteins. That difference plays a role in how terpenes form in nature.
Alkenes, or olefins, are also a major product of the petroleum industry. Reactions of alkenes form the basis for a significant porion of our manufacturing economy. Commonly used plastics such as polyethylene, polypropylene and polystyrene are all formed through the reactions of alkenes. These materials continue to find use in our society because of their valuable properties, such as high strength, flexibility and low weight.
Alkenes undergo addition reactions like carbonyls do. Often, they add a proton to one end of the double bond and another group to the other end. These reactions happen in slightly different ways, however.
Alkenes are reactive because they have a high-lying pair of π-bonding electrons. These electrons are loosely held, being high in energy compared to σ-bonds. The fact that they are not located between the carbon nuclei, but are found above and below the plane of the double bond, also makes these electrons more accessible.
Alkenes can donate their electrons to strong electrophiles other than protons, too. Sometimes their reactivity pattern is a little different than the simple addition across the double bond, but that straightforward pattern is what we will focus on in this chapter.
Problem EA1.1.
The reaction of 2-methylpropene (or isobutylene) with HBr, as depicted above, is really a 2-step process. Draw this mechanism again and in each of the two steps label both the nucleophile and the electrophile (so, that's four labels).
Problem EA1.2.
Draw a reaction progress diagram for the reaction of 2-methylpropene with hydrogen bromide.
Problem EA1.3.
Predict the rate law for the reaction of 2-methylpropene with hydrogen bromide.
Problem EA1.4.
Based on the reaction shown above, draw products for the following reactions.
Problem EA1.5.
Acids other than HBr can add to alkenes. Based on the reaction with HBr, draw products for the following reactions.
Problem EA1.6.
In the following reaction, the addition has gone a slightly different way. Draw a mechanism for this reaction.
EA2. Cations in Electrophilic Addition
EA2. Cations in Electrophilic Addition
Many of the reactions of alkenes begin with a protonation step. The cation that forms then undergoes a second step in which it combines with the counterion from the acid.
In the first step, the alkene's π bond is the nucleophile and the proton is the electrophile. In the second step, the bromide is the nucleophile and the cation is the electrophile.
If you are familiar with nucleophilic aliphatic substitution, you will already know that the presence of a cationic intermediate signals some potential complications in this reaction.
One issue is the problem of stereochemical control. A carbocation is trigonal planar, because the carbon with the positive charge has only three groups attached to it. Because the cation is trigonal planar, the bromide ion that combines with it can approach from either side. It can come from above or below the trigonal plane.
That fact may have no effect whatsoever. However, if the alkene (and the cation it forms) is prochiral, meaning it has the potential to form a new chiral center during this reaction, then there is a choice of which enantiomer to make.
A prochiral carbocation is easy to recognize because the cationic carbon has three different groups attached to it. The fourth group added, the nucleophile, would result in four different groups attached to that carbon, making it a chiral center. In order to recognize a prochiral alkene, you can picture what the alkene would look like after the reaction has taken place: will there be four different groups?
Problem EA2.1.
Which of the following alkenes are prochiral?
Problem EA2.2.
Addition of the nucleophile to one face of the alkene will result in a stereocentre with R configuration. That face is called the re face. Adding it to the other will lead to formation of S configuration. That face is called the si face.
In the following alkenes, identify whether we are looking at the re face or the si face.
Problem EA2.3.
Draw the products of the following reactions, paying attention to stereochemistry.
In addition to the problem of stereochemistry, electrophilic additions of alkenes also present potential regiochemical complications. As in aliphatic nucleophilic substitutions, formation of a cation often opens the door to rapid rearrangement via 1,2-hydride shifts. There may be one hydride shift or there may be many of them in a row.
These hydride shifts happen pretty easily. Overlap of a hydrogen atom with the empty p orbital of the the adjacent cation leads to a short hop from one carbon to the next.
A hydride shift from one secondary carbon to the next, as illustrated in the above example, is thermodynamically pretty neutral. Because the barrier is low, it happens quickly, but there isn't a driving force fo the hydride to shift one way or the other. Instead, both cations result. There is a mixture.
However, in a case in which the cation can form in a more stable position, such as a tertary position, there is a driving force for the reaction to go one way. The barrier would be too high for it to get back.
As a result, when the counterion combines with the cation, it may do so in a position away from the original double bond.
Problem EA2.4.
Draw the products of the following reactions, paying attention to stereochemistry and regiochemistry. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Electrophilic_Addition_to_Alkenes/EA1._Introduction_to_Electrophilic_Addition.txt |
EA3. Solvent Participation in Electrophilic Addition
Alkenes can donate their electrons to strong electrophiles and the resulting carbocations combine with the counterion of the electrophile to undergo an overall addition reaction. However, there may be some cases in which the counterion does not combine with the carbocation.
Hydrobrominations of the type we have looked at only occur under certain conditions. Other conditions can lead to other products. For example, a solvent such as water can also participate in the reaction.
The oxygen-based cation (or oxonium ion) that results can easily lose its charge through loss of a proton. As a result, a molecule of water adds to the alkene overall. The alkene becomes an alcohol. This reaction is called an "acid-catalyzed hdration" of an alkene.
Problem EA3.1.
Explain how the hydration of an alkene in the presence of acid is a catalytic reaction.
Problem EA3.2.
In many cases, equilibrium mixtures of multiple products may result from the addition of acids to alkenes. Show mechanisms, with curved arrows, for the following reactions.
1. The conversion of 2-butene to 2-chlorobutane with aqueous HCl.
2. The conversion of 2-butene to 2-butanol with aqueous HCl.
3. The conversion of 2-chlorobutane to 2-butene with aqueous HCl.
4. The conversion of 2-butanol to 2-butene with aqueous HCl.
On the other hand, in the absence of any solvent, the bromide ion might still have some competition in the second step. The neat reaction (neat means "without solvent" of an alkene with a small amount of acid can result in polymerization. The alkene, which acted as a nucleophile in the first step, can also act as a nucleophile in the second step.
It is important to remember that in any reaction, millions of molecules are involved. Even if one alkene molecule reacts with acid in the first step of a reaction, there are still plenty of other alkene molecules around to act as nucleophile in the second step.
Problem EA3.3.
Provide a mechanism for the polymerization shown above. Assume there are four 2-methylpropene molecules and one hydrogen bromide molecule to begin.
Problem EA3.4.
Chain reactions involve an initiation step, in which a reactive species is generated; propagation steps, in which the reactive species reacts to make a new reactive species; and a termination step, in which the reactive species reacts to make a stable molecule. Label each of the steps in your mechanism from the previous question.
EA4. Stabilized Cations
EA4. Stabilized Cations in Electrophilic Addition
Electrophilic addition to alkenes generally takes place via donation of the π-bonding electron pair from the alkene to an electrophile. So far, we have only looked at protic electrophiles, but the reaction proceeds with others, as well. For instance, alkenes react quite easily with bromine.
Dripping a solution of bromine into a solution of alkene provides a clear sign of reaction. The red-brown colour of bromine disappears almost instantly.
Although bromine isn't an obvious electrophile, most of the common diatomic elements can behave that way; the exception is dinitrogen. A fleeting asymmetry of electrons can polarize the molecules to one end. That event leaves one atom partially positive and the other end partially negative. Because these elements tend to form somewhat stable anions, the partially negative atom can be displaced failry easily.
As before, a nucleophile connects with the cation in a second step. In this case, a dibromide compound is formed. However, it isn't formed in quite the way that is shown below.
We know that the mechanism shown above does not convey the whole picture because it isn't consistent with the stereochemistry of the reaction. The stereochemical outcome is shown below. The enantiomer is formed as well.
Problem EA4.1.
Assign configurations R or S to the product shown in the above mechanism.
However, although two enantiomers are formed in the reaction, the corresponding diastereomer is not. The following step does not occur.
Problem EA4.2.
Assign configurations R or S to the product shown in the above mechanism and explain why it is not an enantiomer to the compound in problem EA4.1.
Instead, the cation that forms in the reaction appears to be stabilized by lone pair donation from bromine. The intermediate species below is called a cyclic bromnium ion. The bromine prevents approach of the nucleophilic bromide from one side, ensuring formation of product through anti addition only. The trans product forms as a result.
Problem EA4.3.
Additional evidence of the stabilized bromonium comes from the observation of just one product in the bromination of 1-hexene. How many products would be expected in the absence of a stabilized cation? Explain with a mechanism.
Problem EA4.4.
Bromonium ions like the one shown below have been isolated and characterized by X-ray crystallography in at least one case. Explain why the intermediate was isolated in this case, rather than a dibromo product.
Problem EA4.5.
In some cases, the reaction of an alkene with bromine does not provide dibromo products. Show products of the reaction of cyclohexene under the following conditions and justify your choices with mechanisms.
a) Br2 in water b) Br2 and NH4+ Cl- in THF
Problem EA4.6.
Provide products for the reaction of bromine with each of the following compounds. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Electrophilic_Addition_to_Alkenes/EA3._Solvent_Effects_in_Electrophilic_Addition.txt |
EA5. Addition to Coordinated Alkenes
Alkenes can coordinate to transition metals to form alkene complexes. In some cases, coordination of the alkene to a metal leaves it susceptible to reaction with a nucleophile such as water.
The classic case of nucleophilic donation to a coordinated alkene occurs with mercury (II) salts such as mercuric chloride, HgCl2, or mercuric acetate, Hg(OAc)2. The reaction, or rather the sequence of reactions, is called oxymercuration - demercuration or oxymercuration - reduction.
Problem EA5.1.
Compare the product of the reaction above to that obtained from treatment of 1-pentene with aqueous sulfuric acid.
We will break the two different reactions in this sequence apart and focus only on the first one: oxymercuration. This reaction qualifies as an electrophilic addition because, as in the previous cases, it begins with donation of a π-bonding pair to an electrophile. In this case, we will consider the electrophile to be aqueous Hg2+ ion.
That electrophilic addition (from the alkene's perspective) results in the formation of an alkene complex. In reality, the mercury ion is also coordinated by several water molecules, but we will ignore them for simplicity.
You may know that alkene complexes are not observed with d0 transition metals. Although π-to-metal donation is the key event in the formation of such complexes, the alkene is just a little more sticky if the metal has d electrons. These electrons are able to "back-donate" into the alkene portion of the complex, adding extra stability to the interaction.
This situation is something like formation of a cyclic bromonium ion.
Note that the overall transfer of electrons is still from alkene to metal. That imbalance isn't apparent in a Lewis structure sense, in which case you can draw the structure so that there appears to be an equal trade. In a computational chemistry approach, in which we rely on basic principles of quantum mechanics and let computers churn out high-level calculations, we would still predict a little bit of positive charge on the alkene.
In the structure below, we have over-emphasized that charge, just to see what happens next in the reaction. When we draw it that way, it looks a lot more like simple addition of electrophile, such as H+, to alkene. We know it's more subtle than that. We'll get back to the real mechanism after a small detour.
Of course, the next step is donation of a lone pair from a nucleophile to the almost-cationic carbon.
That looks easy. After that, deprotonation would result in the formation of a hydroxy group.
Problem EA5.2.
Suppose deprotonation is carried out by the acetate ion in solution. Draw a mechanism for this step.
How do we know the reaction doesn't happen through this simple cation?
Partly we know that because we know about alkene complexes. There are thousands of examples, structurally characterized by NMR spectroscopy and X-ray crystallography. In addition, we know it isn't a simple cation because nothing like the following scenario plays out during oxymercuration.
There are no hydride shifts. The cation stays put. The hydroxy group forms right where the alkene used to be. That means there is not a full carbocation like the one shown above.
If there isn't a real carbocation, though, why does the nucleophile end up at one particular end of the alkene? The hydroxy does end up at the position that would form the more stable cation. ( In other words, this reaction results in what is called "Markovnikov addition".)
There are a couple of reasons that could play a role. Foremost, the alkene isn't bound symmetrically. One end is held a little closer to the mercury than the other. Mostly that's because of sterics. Any other ligands on the mercury (such as those water molecules) push that more crowded end away a little bit. That slight asymmetry allows a little more charge to build on the more substitutted end of the alkene, which is therefore more electrophilic.
The final part of the reaction sequence is displacement of mercury from the hydroxyalkylmercury complex, effected through addition of sodium borohydride. The details of the reaction are usually dismissed in textbooks because they have little to do with electrophilic addition, the topic we are focusing on. However, the result is that the mercury is replaced by a hydrogen atom. The metal is converted to silvery, liquid, elemental mercury.
Problem EA5.3.
Suppose the demercuration reaction takes place via addition of hydride nucleophile to mercury, followed by reductive elimination. Draw this mechanism.
Problem EA5.4.
When mercuration takes place in a ethanol instead of water, an ether product results rather than an alcohol. Work through the mechanism and show the result of mercuration-demercuration in ethanol.
Problem EA5.5.
Show the products of the following reactions. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Electrophilic_Addition_to_Alkenes/EA5._Addition_to_Alkene_Complexes.txt |
Alkenes can be treated with aqueous acids or, much more efficiently, with aqueous mercuric salts, followed by sodium borohydride, to produce alcohols. In those cases, the hydroxyl group is found in different places. Treatment with acid often results in a mixture of alcohols in which the OH groups are found in the most substituted positions in the structure, regardless of the position of the original alkene. Oxymercuration - demercuration results in the hydroxyl being fixed at the most substituted end of the former C=C bond.
Those reactions are generally called hydration reactions because they result in the overall addition of H-OH across the double bond.
• Hydration reactions place an OH on one end and a H on the other end of a former double bond.
Hydroboration - oxidation is a two-step sequence of reactions that also results in hydration of a double bond. However, this reaction is complementary to oxymercuration - demercuration.
Instead of leaving an OH group at the most substituted end of the double bond, the hydroxy group is placed at the least substituted end of the double bond.
Let's modify that statement a little bit. In reality, the reaction scheme above just shows the major product. The minor product has the hydroxy group at the more substituted position of the double bond. These two products might be found in different ratios, maybe even as close as 55:45, but the least substituted product always predominates. We will see more efficient hydroboration methods soon, leading to ratios above 95:5, or almost entirely the least substituted product.
The product of non-hydrogen addition (i.e. OH group addition) at the most substituted end of the alkene is called a Markovnikov addition product. The product of non-hydrogen addition at the least substituted end of the alkene is called an anti-Markovnikov addition product.
• Oxymercuration - demercuration results in Markovnikov hydration.
• Hydroboration - oxidation results in anti-Markovnikov hydration.
This selectivity is important in synthetic applications. We use natural products all the time as pharmaceuticals, vitamins and other health and beauty applications, but we can't always obtain these compounds directly from nature, for a number of reasons. It could be that the organism needs to be killed in order to harvest its products, or that there isn't enough of the source in nature ro meet demand. Frequently it is more economical to produce commercially useful compounds from convenient chemical feedstocks. Over the last century and a half, those feedstocks have come from coal tar and, later, petroleum. Currently, there is rapid progress underway to develop chemical feedstocks from sources such as vegetable and algal oil (i.e. oil from seaweed).
These feedstocks are just compounds that can be converted synthetically into pharmaceuticals as well as in plastics, paints, coatings and other materials. Frequently, the starting materials for these processes contain C=C bonds that can be functionalized through electrophilic addition. Thus, electrophilic addition and related reactions are among the most important in the world, economically speaking. It's very valuable to be able to control the outcome of these reactions in order to make processes more efficient, producing fewer wasteful by-products.
• Regioselectivity, or control over where a reaction occurs, is very important.
• The Markovnikov versus anti-Markovnikov additions available in hydration are good examples of regiochemical control.
Problem EA6.1.
List advantages and disadvantages of producing materials based on
1. petroleum
2. vegetable oil
3. algae
4. harvesting desired compounds directly from nature
Again, we are going to focus on the first of the two reactions in this sequence. That part is where the placement of the new substituent is decided. After the addition of the borane, an alkylborane is formed. The major isomer results from anti-Markovnikov addition.
It seems pretty clear at this point that this reaction must proceed like other electrophilic additions to alkenes. The π electrons donate to the electrophile. In this case that's boron, which is strongly Lewis acidic because it lacks an octet. The boron ends up at the least substituted end of the double bond.
That outcome would certainly be favoured over this one:
That boron is beginning to look less electrophilic and more nucleophilic. We can easily imagine a hydride nucleophile being delivered to the carbocation.
So far, the picture of how the alkylboration reaction works fits pretty well within our electrophilic addition framework. Unfortunately, there are some problems with this model.
First of all, maybe 55% of the boration takes place in a Markovnikov sense, but the other 45% is added to form an anti-Markovnikov product. Certainly the secondary cation is favoured over the primary one, but if the reaction is proceeding through a carbocation, then the primary one shouldn't happen at all.
Something is wrong with our model.
Another hole is torn in the argument when we look at the results of stereochemical studies. We could, for example, take the following deuterium-labelled hexene and treat it with borane. We could look at the products via 1H NMR spectroscopy, and if we could see the coupling constant between the two protons shown in the structure, then we would know their relative arrangement in space. We would know their stereochemistry.
If we did that experiment, then we would see that the hydrogen and the boron from the borane are added to the same face of the alkene. We don't get addition of boron to one face and hydrogen to the other. This type of addition is called syn addition; it is the opposite of anti addition.
• Hydroboration results in syn addition to the alkene.
Problem EA6.2.
Show how a cationic intermediate and conformational changes would allow both syn and anti addition of borane to propene.
That result means that, although some elements of our mechanism may reflect reality, we at least have a problem with timing. How can the hydride be delivered before the conformation has a chance to change? It has to happen pretty quickly. What if it happens at the same time as π bond donation to the boron?
This reaction would best be described as a concerted addition. Two groups are added to the two ends of the double bond at the same time. During the transition state, two bonds would be breaking and two would be forming at the same time.
Problem EA6.3.
We can further improve our model of how the alkylboration works if we consider that disiamylborane and 9-BBN are much more effecient than borane in terms of regioselectivity. These reagents can produce close to 100% anti-Markovnikov addition. Explain how with the help of drawings.
The subsequent reaction in this series involves removal of the boron and replacement with a hydroxyl group. The mechanism of this reaction may not be worth memorizing becuase it doesn't fit well within categories we have looked at so far.
The important thing to know is that the oxygen ends up in exactly the same place as the boron. There is no change in stereochemistry at that position. Overall, the hydrogen and the hydroxy effectively group undergo syn addition, although they are added in different steps.
Problem EA6.4.
Borane is frequently used in THF because borane alone is not very stable; it is quite pyrophoric, bursting into flame upon contact with air. In THF, borane forms an equilibrium with a Lewis-acid-base complex. Show this equilibrium reaction.
Problem EA6.5.
Show products of the following reactions.
Problem EA6.5.
Provide reagents for the following reactions
EA8. Solutions for Selected Problems
Problem EA4.1.
Two products are formed and they are enantiomers.
Problem EA4.2.
They are diastereomers. One chiral center has the same configuration in both compounds but the others are opposite.
Problem EA4.3.
The second bromine could occupy any of the secondary positions if there were a true carbocation. That doesn't happen; the second bromine occupies only the position next to the other bromine.
Problem EA4.5.
The nucleophile in the second step changes under different conditions.
Problem EA6.3.
Crowding is more severe in the structure on the left than in the structure on the right. The structure on the right, representing an approach to the transition state of the reaction, is more favourable than the other one. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Electrophilic_Addition_to_Alkenes/EA6._Concerted_Additions.txt |
Aromatics, or arenes, are derivatives of benzene or other compounds with aromatic ring systems. That is, they are cyclic, planar, fully conjugated and have an odd number of π-electron pairs. Like alkenes, aromatics have π-electrons that are loosely held and are easily attracted to electrophiles. However, aromatics don't undergo the typical reactions of alkenes.
For example, bromine will not add across the double bond of benzene.
Instead, a bromine atom replaces one of the hydrogen atoms on the benzene. This reaction is greatly accelerated in the presence of Lewis acids, such as ferric chloride.
A similar reaction happens with chlorine. If treated with chlorine gas and a metal catalyst, a chlorine atom from chlorine gas can replace a hydrogen atom on benzene. However, the same thing doesn't work as smoothly with the other halogens, iodine and fluorine.
The reactions of chlorine and bromine with benzene and other aromatics can be catalysed by a variety of Lewis acidic metal catalysts. So can the reactions of alkyl halides and acyl halides, which we don't normally think of as electrophiles for alkene addition.
There are some limitations on what kind of groups can be added in this way. The carbon attached to the halide should be tetrahedral. Typically, it is much easier to add secondary or tertiary alkyls than primary ones. That is, the carbon attached to the halogen had best be attached to two or three other carbons as well. Methyls are very, very difficult to add in this way.
There is an exception. The carbon attached to the halogen need not be tetrahedral, provided it is a carbonyl carbon. That reaction is called an acylation.
In these cases, it is the alkyl or acyl, rather than the halogen, that replaces a hydrogen atom on the benzene. Remember, benzene is most likely acting as a nucleophile in this reaction, even though it is following a different pathway than an alkene would. It is reacting with the most electrophilic part of the alkyl halide or acyl halide.
Aromatics have a limited repertoire of electrophiles with which they commonly undergo reaction. In addition to these Lewis acid-catalysed reactions, there are also reactions strong acidic media, such as a mixture of nitric and sulfuric acid.
Another acidic medium, referred to as "fuming sulfuric acid," is really a mixture of sulfuric acid and sulfur trioxide.
Just as with the acid-catalysed reactions, the nitro group and the sulfonate group just replace a hydrogen atom on the benzene ring. The overall reaction involves bond formation between a benzene carbon and the electrophile, and bond cleavage between the same carbon and a proton.
Problem AR1.1.
Fill in the missing reagents in the following reactions.
AR2. Mechanism of EAS
Bromine will not add across the double bond of benzene.
Instead, a bromine atom can replace one of the hydrogen atoms on the benzene. This reaction is especially easy in the presence of a catalyst.
How does that outcome happen? Why does that outcome happen?
There has been a good deal of study of these reactions and there is strong evidence of the steps through which they occur. As expected, the reaction involves donation of π electrons from the benzene. For the moment, we'll assume the electrophile is a bromine cation; we will deal with its exact structure later.
The problem is, that initial step results in the loss of aromaticity. The aromatic system confers a little extra stability on the π system, so the molecule is motivated to restore the aromaticity. The easiest way to do that, and get rid of a positive charge at the same time, would be to deprotonate the cation. Some base will pick up the proton; it is likely a bromide ion in this case. We will see later where that bromide comes from.
• Electrophilic aromatic substitution proceeds through a cationic intermediate.
• The intermediate forms via donation of π electrons from the arene to an electrophile.
• Aromaticity is restored through loss of a proton from this cation.
How do we know that the mechanism unfolds this way? There are three basic steps that are clearly accomplished during the course of the reaction: the C-H bond is broken, the C-Br bond is formed, and the Br-Br bond is broken.
When is the C-H bond broken? That question can be answered by looking for what is called an "isotope effect." The most common isotope of hydrogen is 1H, or protium, but 2H is also available; it is called "deuterium." Deuterium is often represented by the symbol D and protium by the symbol H. Deuterium is twice as heavy as the common protium. That mass difference leads to a lower vibrational frequency of a C-D bond than a C-H bond. The C-H bond vibrates more rapidly and energetically than a C-D bond; as a consequence, the C-H bond is more easily broken than the C-D bond.
If we take a sample of ordinary benzene, C6H6, and a sample of deuterated benzene, C6D6, we can measure how quickly they each undergo a bromination reaction. Very often, a reaction that involves C-H bond cleavage will slow down if a C-D bond is involved. This outcome is observed in E2 eliminations, for instance. This slowing of the reaction with the heavier isotope is called the deuterium isotope effect.
However, no deuterium isotope effect is observed during bromination, or other aromatic electrophilic substitution reactions. That absence of an isotope effect usually means the C-H bond cleavage is a sort of an afterthought. The hard part of the reaction is already done. Both the C-H and C-D bonds are broken so quickly and easily, by comparison, that we don't really notice the difference between them.
There is even more evidence. In a few exceptional cases, the cationic intermediate in this reaction is stable enough to be isolated and crystallized. X-ray diffraction shows that there is a tetrahedral carbon in the ring, indicating that the C-H bond has not broken yet.
The C-H bond is broken at the end of the reaction. When is the Br-Br bond broken?
That question is a little harder to answer. We can't use the same isotope strategy that we used with the C-H bond. Although deuterium is twice as heavy as protium, producing a substantial isotope effect, 81Br is only 2.5% more massive than 79Br. Any difference in rates involving these isotopes is undetectable. The exact nature of the bromine species in the reaction is complicated, and may even be different under different conditions.
Problem AR2.1.
In the case of uncatalyzed bromination reactions, the addition of salts such as NaBr has no effect on the reaction rate, indicating that the arene reacts directly with Br2 rather than Br+. Explain this line of reasoning.
Problem AR2.2.
Given each of the following electrophiles, provide a mechanism for electrophilic aromatic substitution.
a) NO2+ b) CH3CH2+ c) SO3H+ d) CH3CO+
AR3. Formation of Electrophiles
The mechanism of electrophilic aromatic substitution follows two elementary steps. First, donation of a pair of π electrons to the electrophile results in a loss of aromaticity and formation of a cation. Second, removal of a proton from that cation restores aromaticity.
How does the electrophile form in the first place? The details of that part of the reaction vary from case to case. With the catalysed bromine reaction, the Lewis acid activates the halogen to render it more electrophilic. The activation may even go so far as to form a bromine cation, as suggested earlier. Otherwise, the positive charge on the bromine atom that ligates the Lewis acid can be nullified, indirectly, when the arene donates to the terminal bromine atom.
The appearance of a bromide ion to deprotonate the cation simply results fom the equilibrium of the Lewis acid-base complex.
Problem AR3.1.
Show the mechanism for chlorination of benzene in the presence of ferric chloride.
The reactions of alkyl and acyl halides also involve Lewis acid catalysts; frequently, aluminum chloride (AlCl3) is employed. These two reactions are called Friedel-Crafts reactions after the French and American co-discoverers of the reaction. Typically, Friedel-Crafts reactions are believed to occur through initial formation of cationic electrophiles, which then react with aromatics in the same way as halogen electrophiles.
Because Friedel-Crafts alkylations occur via alkyl cations, the reactions of primary alkyl halides are generally pretty slow. Also, in some cases, multiple products may result via rearrangements. These observations provide additional evidence for the cationic nature of the intermediate.
Problem AR3.2.
Show the mechanism for the Friedel-Crafts alkylation of benzene with 2-chloropropane in the presence of aluminum chloride.
Problem AR3.3.
Why is the Friedel-Crafts reaction of 1-chloropropane so much slower than the reaction of 2-chloropropane? Explain using a mechanism and intermediates.
Problem AR3.4.
Show why Friedel-Crafts alkylation of benzene with 2-chloropentane results in the formation of two different products.
Problem AR3.5.
Show the mechanism for the Friedel-Crafts acylation of benzene with ethanoyl chloride (acetyl chloride) in the presence of aluminum chloride.
Problem AR3.6.
Show why the Friedel-Crafts acylation of benzene with pentanoyl chloride results in only one product, with no rearrangement.
Nitration and sulfonation reactions differ from the other substitutions that we have seen because they do not utilize Lewis acid catalysis. These reactions depend on equilibria that occur in strongly acidic media. When nitric acid is dissolved in sulfuric acid, there is spectroscopic evidence than NO2+ forms, providing an electrophile. Similarly, when sulfuric acid is concentrated by boiling off residual water, sulfur trioxide results. The latter probably forms via SO3H+, the electrophile in sulfonation.
Problem AR3.7.
Provide a mechanism for the formation of NO+ from nitric and sulfuric acid.
Problem AR3.8.
Provide a mechanism for the formation of SO3H+ from sulfuric acid. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Electrophilic_Aromatic_Substitution/AR1._Introduction_to_Electrophilic_Aromatic_Substitution.txt |
Because the benzene acts as a nucleophile in electrophilic aromatic substitution, substituents that make the benzene more electron-rich can accelerate the reaction. Substituents that make the benzene moor electron-poor can retard the reaction. In the mid-twentieth century, physical organic chemists including Christopher Ingold conducted a number of kinetic studies on electrophilic aromatic substitution reactions. In table 1, you can see that some substituents confer a rate of reaction that is much higher than that of benzene (R = H). Phenol, C6H5OH, undergoes nitration a thousand times faster than benzene does. Nitrobenzene, C6H5NO2, undergoes the reaction millions of times more slowly.
Table 1: Rate of nitration in benzene derivatives
R in C6H5R Relative rate
OH 1,000
CH3 25
H 1
CH2Cl 0.71
I 0.18
F 0.15
Cl 0.033
Br 0.030
CO2Et 0.0037
NO2 6 x 10-8
NMe3+ 1.2 x 10-8
These observations are consistent with the role of the aromatic as a nucleophile in this reaction. Substituents that draw electron density away from the aromatic ring slow the reaction down. These groups are called deactivating groups in this reaction. Substituents that readily donate electron desnity to the ring, or that effectively stabilize the cationic intermediate, promote the reaction. These groups are called activating groups in this reaction.
The roles of these groups are related to their electronic interactions with the electrons in the ring. Some groups might be π-donors, providing additional electron density to the benzene ring via conjugation.
Other groups may be π-acceptors, drawing electron density away from the ring via conjugation.
Still others may be σ-acceptors, drawing electron density away from the ring via a simple inductive effect which arises from the electronegativity of a substituent.
In some cases, there may be multiple effects, and the overall influence of the substituents is determined by the balance of the effects. One effect may be stronger in one case than the other, so it wins out in one case and loses in another.
Problem AR4.1.
Explain why a fluorine atom would slow down an electrophilic substitution on an adjacent benzene ring.
Problem AR4.2.
Show, with structures, how the OH group in phenol makes the benzene ring more nucleophilic.
Problem AR4.3.
Show, with structures, how the CO2Et group makes the benzene ring less nucleophilic.
Problem AR4.4.
Show, with structures, how a methyl group stabilizes the cationic intermediate during a nitration reaction.
In general, deactivating groups fall into two classes. Π-acceptors, such as carbonyls, if placed directly adjacent to the aromatic ring, slow down the reaction. Highly electronegative atoms, typically halogens, attached directly to the aromatic ring also slow down the reaction.
• deactivating groups make electrophilic aromatic substitutionslower than in benzene
• π-acceptors are deactivating groups
• halogens are deactivating groups
Activating groups also fall into two categories. Π-donors, typically oxygen or nitrogen atoms, accelerate the reaction. This observation is true even though these atoms are also highly electronegative. Alkyl groups attached to the aromatic ring also accelerate the reaction.
• activating groups make electrophilic aromatic substitution faster than in benzene
• oxygen and nitrogen π-donors are activating groups
• alkyls are activating groups
Note that halogens are also π-donors, but they are less effective in this regard than nitrogen or oxygen. That's because nitrogen and oxygen are similar in size to carbon, and they form effective π-overlap with the adjacent carbon on the benzene ring. In the halogens, electronegativity wins by default, because their π-donating effects are not good enough to make them activators.
Conversely, nitrogen and oxygen are both very electronegative, but their exceptional π-donating ability makes them activators rather than deactivators. Thus, in many cases, there is a subtle balance between activating and directing effects. In some cases, the activating effect is more pronounced, and that is what is observed. In other cases, it is the deactivating effect that wins out.
Alkyl groups behave almost as sigma donors, although that may be a misleading way to think about them.
Instead, their mild activating effect arises from hyperconjugation, in which a pair of C-H bonding electrons can weakly interact with a cationic site, providing a little extra stability to the cation.
Problem AR4.5.
One of the groups in the table, CH2Cl, does not quite fit the general rules. It is very slightly deactivating. Explain why this group acts in this way.
Problem AR4.6.
Predict whether each of the following groups would be activating or deactivating towards electrophilic aromatic substitution.
a) NH2 b) CN c) OCH3 d) SMe2+ e) C(O)CH3
Problem AR4.7.
Explain the trend in activating effects among the different halogens. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Electrophilic_Aromatic_Substitution/AR4._Activation_and_Deactivation.txt |
In addition to exerting an effect on the speed of reaction, substituents on the benzene ring also influence the regiochemistry of the reaction. That is, they control where the new substituent appears in the product.
Remember, there are three different position on the bezene ring where a new substituent can attach, relative to the original substituent. Substitution could actually occur on five positions around the ring, but two pairs are related by symmetry. Isomerism in disubstituted benzenes can be described by numbering the substituents (1,2- etc) or by the relationships ortho-, meta- and para-. There are two positions ortho- to the initial substituent and two positions meta- to it.
Ingold and colleagues investigated the question of regiochemistry in nitration. They reported the following observations:
Table: Substitution patterns during nitration of benzene derivatives
R in C6H5R % o- product % m- product % p- product
CH3 56 3 41
Cl 30 0 70
Br 38 0 62
OH 10 0 90
CHO 19 72 9
CO2Et 28 68 3
CN 17 81 2
NO2 6 94 0
In looking at the table, you might see that there are two groups of substituents. One group reacts to make mixtures of ortho- and para- products. There may be different ratios of ortho- to para- and there may be small amounts of meta-, but don't get bogged down in the details. Focus on the bigger picture. Some groups are "ortho-/para-directors".
The other group reacts to makemostly meta-substituted products. here may be small amounts ofortho- and para- products, but don't worry about that. Focus on the bigger picture. Some groups are "meta-directors". These regiochemical effects are very closely related to the activating and directing effects we have already seen. If we want to understand this data, we need to think about things like π-donation, π-acceptance, inductive effects and cation stability.
AR5.1.
Show resonance structures for the cationic intermediate that results during nitration of toluene (methylbenzene). Explain why a mixture of ortho- and para- substitution results.
AR5.2.
Show resonance structures for the cationic intermediate that results during nitration of chlorobenzene. Explain why a mixture of ortho- and para- substitution results.
AR5.3.
Show resonance structures for the cationic intermediate that results during nitration of acetophenone (C6H5COCH3). Explain why mostly meta- substitution results.
AR5.4.
Show resonance structures for the cationic intermediate that results during nitration of acetanilide (C6H5NH(CO)CH3). Explain why a mixture of ortho- and para- substitution results.
In general, we can divide these substituents into three groups:
• π-acceptors are meta- directors.
• π-donors are ortho-/para- directors.
• alkyls are ortho-/para- directors.
Note that, once again, we may have two competing effects in one substituent, such as a halogen. In halogens, although the net effect may be to slow the reaction down, that weak π-donation is still enough to tilt the balance of products in favour of ortho- and para- substitution.
AR5.5.
Fill in the major organic products of the following reactions.
AR5.6.
Fill in the starting materials and reagents needed to obtain the major product shown via electrophilic aromatic substitution.
AR5.7.
Given two different substituents on a benzene, there can sometimes be a conflict in predicting which substitution pattern will result. Generally, the group with the stronger activating effect wins out. Predict the major products of the following reactions.
AR6. Solutions to Selected Problems
Problem AR2.1.
In the case of uncatalyzed bromination reactions, there is clear evidence that the Br-Br bond-breaking step does not start the reaction off. If that were the first step, there would presumably be an equilibrium between Br2 and Br+/Br- ions. That equilibrium would be shifted back toward Br2 if bromide salts were added. In that case, the amount of bromine cation would be suppressed and the reaction would slow down. No such salt effects are observed, however. That evidence suggests that, in the uncatalyzed reaction, the aromatic reacts directly with Br2.
Problem AR2.2.
In each case, a base must remove the proton from the cationic intermediate. An anion that would be present in solution has been chosen for this role.
a)
b)
c)
d)
Problem AR3.3.
The primary cation formed is very unstable. As a result, there is a high barrier to cation formation.
Problem AR3.6.
The cation that results is stabilized via π-donation from oxygen.
Problem AR4.5.
This is a substituted alkyl group. An alkyl group should be moderately activating, but the presence of a halogen exerts an inductive electron-withdrawing effect. The cation-stabilizing effect of the alkyl substituent is completely counteracted by the halogen.
Problem AR4.6.
a) activating b) deactivating c) activating d) deactivating e) deactivating
Problem AR5.1.
The tertiary cations that result during ortho- and meta- substitution offer extra stability, leading to preferential formation of these cations.
Problem AR5.2.
The π-donation that occurs in the cations arising from ortho- and meta- substitution results in extra stability, leading to preferential formation of these cations.
Problem AR5.3.
The cation directly adjacent to the carbonyl is destabilized by the electron withdrawing effect of the ketone. By default, the other intermediate is preferentially formed.
Problem AR5.4.
The π-donation that occurs in the cations arising from ortho- and meta- substitution results in extra stability, leading to preferential formation of these cations.
Problem AR5.7.
In cases leading to mixtures of ortho and para products, only one product was chosen, based on minimal steric interactions. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Electrophilic_Aromatic_Substitution/AR5._Directing_Effects.txt |
• LS1. Introduction to Ligand Substitution
• LS2. Mechanism of Ligand Substitution
• LS3. Kinetics of Associative Mechanism
• LS4. Kinetics of Dissociative Mechanism
• LS5. Activation Parameters
• LS6. Some Reasons for Differing Mechanisms
• LS7. The Trans Effect
• LS8. Solutions to Selected Problems
Ligand Substitution in Coordination Complexes
Ligand substitution refers to the replacement of one ligand in a coordination complex with another ligand.
Figure LS1.1. Substitution of one ligand for another in a coordination complex.
Remember, a ligand in coordination chemistry is just a Lewis base that binds to a metal atom or ion. It does so by donating a lone pair (or other pair of electrons). Generallly, this donation is reversible. The donor can always take its electrons back. Typically, there may be some balance between the metal's need for more electrons and the donor's attraction for its own electrons; donor atoms are frequently more electronegative than the metal.
Figure LS1.2. An example of ligand substitution. THF replaces a carbon monoxide in this molybdenum complex.
Even though the reaction is pretty simple, it can occur in different ways.
That is, the elementary steps involved in the reaction can occur in different orders. The elementary reactions are the individual bond-making or bond-breaking events that lead to an overall change. Sometimes the order of steps is referred to as the mechanism or the mechanistic pathway.
• The mechanism is the order of elementary reaction steps.
• Elementary reaction steps are individual bond-making and breaking steps.
You may have seen reaction mechanisms before. For example, carbonyl addition chemistry can involve lengthy mechanisms, in which a number of proton transfers and other bond-making and bond-breaking steps must occur to get from one state to another. Because ligand substitution is simpler than that, it is a good place to study mechanism in a little more depth, without getting overwhelmed by the details.
The sequence of steps in the mechanism influences how different factors will impact the reaction. For example, changing concentrations of different components in a reaction mixture can affect the time it takes for a reaction to finish.
• The mechanism can have a dramatic impact on the outcome of the reaction under different circumstances.
These kinds of considerations have a dramatic impact on industrial processes such as pharmaceutical production. In that setting, chemical engineers need to make decisions about how much of each reactant must be admitted to a reaction mixture and how long they should be allowed to react together. If they allow the reaction to proceed for too, long, there may be “side-reactions” that start to occur, interfering with the quality of the product, and they will waste valuable time in the production pipeline. If they don’t allow it to react long enough, the reaction may not finish, and the product will be contaminated with leftover starting materials.
In this chapter, we will look at how this simple reaction can occur in different ways. We will see some different methods that are used to tell which way the reaction occurs (i.e. evidence of what is really happening). We will also look at some different factors that may influence whether the reaction is likely to occur one way or the other (i.e. reasons it is happening that way, or reasons we expect it will happen that way).
Problem LS1.1.
Some kind of substitution occurs in each of the following reactions: an atom or group replaces another. In each case, identify what is being replaced, and what replaces it.
LS2. Mechanism of Ligand Sub
There are two basic steps in ligand substitution: association and dissociation. Association, in this case, refers to the binding of a ligand to the metal. The ligand donates an electron pair to the metal and the two molecules come together to form a new bond. Dissociation, in this case, refers to the release of a ligand from a metal. The metal-ligand bond breaks and the ligand leaves with its electron pair.
Two mechanistic possibilities seem pretty obvious. Either the new ligand binds first and then the old one leaves, or the old ligand leaves first and then the new one binds.
• Associative mechanism is association first. The new ligand binds and then the old one leaves.
• Dissociative mechanism is dissociation first. The old ligand leaves and then the new one binds.
Knowing the mechanism is important because the mechanism has an impact on what factors affect the reaction. For example, if the reaction is associative, adding lots more new ligand may speed up the reaction, because then it becomes more likely that the new ligand will find the metal complex and bind with it. However, if the old ligand is supposed to leave before the new ligand arrives, then it doesn’t matter how much new ligand is around. It has to wait for the old ligand to leave before it can bind, anyway, so adding a lot more new ligand won’t speed things up.
Problem LS2.1.
1. Which kind of step costs more energy: bond-making or bond-breaking?
2. What would be the rate-determining step in the associative mechanism?
3. What would be the rate-determining step in the dissociative mechanism?
4. What would be the rate law for the associative mechanism?
5. What would be the rate law for the dissociative mechanism?
Problem LS2.2.
Draw a mechanism, with arrows, for the following substitution. Assume an associative mechanism.
Problem LS2.3.
Draw a mechanism, with arrows, for the following substitution. Assume a dissociative mechanism.
Problem LS2.4.
Draw a mechanism, with arrows, for the following isomerization.
Problem LS2.5.
The ability to substitute for a ligand depends partly on its ability to leave. Rank the following ligands, from the easiest to replace to the hardest to replace:
CO Cl- PPh NH3 NO3- H2O
Problem LS2.6.
Draw curved arrows for the following steps. Classify each step as associative or dissociative. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Ligand_Substitution_in_Coordination_Complexes/LS1._Intro_to_Ligand_Substitution.txt |
We can measure the rate of an associative reaction and make changes in the reaction conditions to see how the rate is affected. For example, we could easily change the concentrations of the two reactants. All we have to do is change the amount of reactant we dissolve in the solution.
If we did that, we would find a linear relationship between each concentration and rate. If we double the concentration of new ligand, the rate of reaction doubles. If we triple it, the rate triples. Also, if we double the amount of metal complex, the rate doubles and so on. We can write the following expression, called the rate law, to describe this relationship:
$\text {Rate Law: Rate} = {-d[MLn]\over dt } = k[MLn][L']$
This type of reaction is sometimes called a second order reaction. That term just refers to the mathematical form of the rate law, which depends on concentration times concentration, or concentration squared. The "order" of the reaction is the number of concentrations multiplied together in the rate law.
Why does the associative mechanism depend on concentrations in this specific way?
This is a case of two molecules coming together. If both compounds are dissolved in solution, they must “swim around” or travel through the solution until they bump into each other and react. The more concentrated the solution is, or the more crowded it is with molecules, the more likely are the reactants to bump into each other. If we double the amount of new ligand in solution, an encounter between ligand and complex becomes twice as likely. If we double the amount of metal complex in solution, an encounter also becomes twice as likely.
Figure LS3.1. The effect of concentration on collision probability. In the first beaker, there is a chance that a black molecule and white molecule will meet and react together. The chance of a meeting is much higher in both the second beaker, where there are lots more black molecules, and in the third beaker, where there are many more white molecules.
Problem LS3.1.
Given the associative rate law above, what would happen to the reaction rate for an associative substitution in the following cases?
1. the concentration of ligand is doubled, and the concentration of metal complex is doubled
2. the concentration of ligand is tripled, and the concentration of metal complex is doubled
3. the concentration of ligand is tripled, and the concentration of metal complex is tripled
4. the concentration of ligand is halved, and the concentration of metal complex is doubled
Problem LS3.2.
Plot graphs of initial rate vs. concentration to show what you would see in associative substitution.
1. The concentration of new ligand, [X], is held constant at 0.1 mol/L and the concentration of metal complex is changed from 0.5 mol/L to 1 mol/L and then to 1.5 mol/L.
2. The concentration of metal complex, [MLn], is held constant at 0.1 mol/L and the concentration of ligand is changed from 0.5 mol/L to 1 mol/L and then to 1.5 mol/L.
Problem LS3.3.
In the previous problem, the experiment was run in a particular way for particular reasons.
1. Why was one concentration held constant while the other one was changed? Why not change both?
2. Why does the graph report "initial rate" -- just the rate at the very beginning of the reaction?
Problem LS3.4.
Given the following sets of initial rate data (rates measured at the beginning of a reaction), determine whether each case represents an associative substitution.
a)
b)
c)
d)
Problem LS3.5.
What information can be gained from the slopes of lines in Problem LS3.4.? | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Ligand_Substitution_in_Coordination_Complexes/LS3._Kinetics_of_Assoc._Mechanism.txt |
What about a dissociative reaction? Does it also depend on concentration? We could also change concentrations of the two reactants, the new ligand and the metal complex, in a dissociative reaction. If we did that, we would find a linear relationship between the concentration of metal complex and rate. If we doubled the amount of metal complex, the rate would double and so on.
However, changes in concentration of the new ligand would have little effect on the rate. If we double the concentration of new ligand, the rate wouldn't change.
We can write the following expression, called the rate law, to describe this relationship:
$\text {Rate Law: Rate} = -{d[MLn] \over dt} = k[MLn]$
This type of reaction is sometimes called a first order reaction. That means the rate law depends on only one concentration term.
Why does the dissociative mechanism depend on concentrations in this specific way?
This is a case of one molecule losing a ligand. Once it does so, a second ligand can replace the one that left. However, losing a ligand may be harder to do than gaining a new one. To lose a ligand, a bond must be broken, which costs energy. To gain a new ligand, a bond is made, releasing energy. That first step is harder to do, so it takes longer. It is a bottleneck that slows the reaction down. It is called the rate-determining step.
• The rate-determining step is the slow step of the reaction.
• The rate-determining step controls the rate of the overall reaction; everything else has to wait for that step to happen.
• Once the rate-determining step has occurred, everything else follows very quickly.
Figure LS5.1. A bottleneck in a reaction. If the reaction must wait for the white molecule to dissociate, it doesn't matter how many black molecules there are; the reaction still goes just as slow. However, the more white molecules there are, the more frequently they will be able to react with the black molecules as dissociation occurs.
No collision is necessary for the metal complex to lose a ligand. Instead, a bond in the metal complex has to break. That just takes time and energy. As a result, concentrations matter very little.
We should think a little more about energy requirements, available energy and reaction rate. It takes a certain amount of energy to break a bond. Over any given period of time, a specific amount of energy is available in the surroundings to use. That energy is not available uniformly. Some molecules will get more energy from their surroundings and others will get less. There will be a statistical distribution, like a bell curve, of energy available in different molecules. That means bond-breaking events are governed by statistics.
Figure LS5.2. The relationship between temperature, energy available, and energy barrier. The black line represents energy needed to start the reaction, also called the energy barrier or the activation barrier. The blue curve is the distribution of available energy in a group of molecules at a cooler temperature. The yellow curve is for a group of molecules that is a little warmer, and the red curve even warmer.
In figure LS5.2, most of the molecules at the low temperature (blue) do not have enough energy to begin the reaction. A small portion do, and so the reaction will proceed, but very slowly. In the yellow curve, there is more energy available, and so a large fraction of molecules have the energy necessary to begin the reaction. In the red curve, the vast majority have sufficient energy to react. Thus, one of the factors governing how quickly a reaction will happen is the energy needed, or activation barrier. A second factor is the energy available, as indicated by the temperature.
Of course, even if there is enough energy for the reaction, the reaction might not occur yet. Energy is necessary but not sufficient to start a reaction. There are also statistical factors in terms of whether a molecule has its energy allotted into the right places, or in some cases, whether two molecules that need to react together are oriented properly.
Suppose at a given temperature it takes a specific amount of time for half the molecules to gain enough energy so that they can undergo the reaction. That amount of time is called the half life of the reaction. After one half life, half the molecules have reacted and half remain. After a second half life, half the remaining molecules (another quarter, for three quarters of the original material in all) have also reacted, and a quarter still remain. After a third half life, half the remaining ones (another eighth, making it seven eighths reacted in total) will have reacted, leaving an eighth of the original material behind.
• Exponential decay is based on a statistical distribution of energy availability.
• The concept of half life is related to exponential decay.
• It takes a fixed period of time for a half of the metal complex obtain enough energy to dissociate.
Thus, the time it takes for the reaction to happen does not really depend on the concentration of anything.
However, the change in concentration over time -- the quantity that we can usually measure most easily -- depends on the original concentration, and for that reason the concentration of the metal complex appears in the rate law.
Figure LS4.3. The reactions in the top row and bottom row are proceeding with the same half-life as we move from left to right. However, the top row starts out more concentrated than the bottom row. As a result, the concentrations in the top row are changing more quickly than in the bottom row.
Suppose the half life for a particular case of ligand substitution is one second. After a half life, a 1 M solution becomes 0.5 M, so the rate of change in concentration per time is 0.5M/s. But after the same half life, a 0.5 M solution becomes 0.25 M, so the change in concentration is 0.25 M/s.
Problem LS4.1.
If a first order reaction has a half-life of 120 seconds, how much of the original material is left after
a) four minutes? b) six minutes? c) eight minutes? d) ten minutes?
Problem LS4.2.
Given the dissociative rate law above, what would happen to the reaction rate for an associative substitution in the following cases?
1. the concentration of ligand is doubled, and the concentration of metal complex is doubled
2. the concentration of ligand is tripled, and the concentration of metal complex is doubled
3. the concentration of ligand is tripled, and the concentration of metal complex is tripled
4. the concentration of ligand is halved, and the concentration of metal complex is doubled
Problem LS4.3.
Plot graphs of initial rate vs concentration to show what you would see in dissociative substitution.
1. The concentration of metal complex, [MLn], is held constant at 0.1 mol/L and the concentration of ligand is changed from 0.5 mol/L to 1 mol/L and then to 1.5 mol/L.
2. The concentration of new ligand, [X], is held constant at 0.1 mol/L and the concentration of metal complex is changed from 0.5 mol/L to 1 mol/L and then to 1.5 mol/L.
Problem LS4.4.
Given the following sets of initial rate data, determine whether each case represents a dissociative substitution.
a)
b)
c)
Problem LS4.5.
In the following data, the concentration of the metal complex and the incoming ligand were held constant, but more of the departing ligand was added to the solution.
1. Explain what the data says about rate dependence on this concentration.
2. Explain this rate dependence in terms of the reaction.
Problem LS4.6.
In certain solvents, such as THF, acetonitrile and pyridine, the rate law for substitution often appears to be Rate = k1[MLn] + k2[MLn][X], in which X is the incoming ligand and MLn is the metal complex.
1. What do these solvents have in common?
2. What is a possible explanation for this rate law?
3. This rate law has been shown to be consistent with an entirely associative mechanism. How is that possible? | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Ligand_Substitution_in_Coordination_Complexes/LS4._Kinetics_of_Diss._Mechanism.txt |
LS5. Activation Parameters
The rate law shows how the rate of a reaction depends on concentrations of different species in solution. The proportionality constant, k, is called the rate constant. It contains other information about the energetic requirements of the reaction.
All reactions must overcome activation barriers in order to occur. The activation barrier is the sum of the energy that must be expended to get the reaction going. An activation barrier is often thought of, cartoonishly, as a hill the molecule has to climb over during the reaction. Once, there, it can just slide down the other side of the hill to become products. At the top of the hill, the molecule exists in what is called the "transition state". At the transition state, the structure is somewhere between its original form and the structure of the products.
Figure LS5.1. The activation barrier for a ligand dissociation step.
The type of diagram shown in figure LS6.1 is sometimes called a "reaction progress diagram". It shows energy changes in the system as a reaction proceeds. One or more activation barriers may occur along the reaction pathways, as various elementary steps occur in the reaction. In the above case, it is easy to imagine the source of the energy barrier, because some energy must be expended to break the bond to ligand C.
However, after that barrier is passed, energy is lowered again. This can happen for several reasons. Once C has separated from the metal complex, it is free to vibrate, tumble, roll and zip around all on its own. That means it can put its energy into any of those modes, independently of the metal complex. As a result, the entropy of the system increases. That lowers the overall "free energy" of the system. In addition, there may be some relief of crowding as the molecule changes from a four-coordinate complex to a three-coordinate complex, so strain energy is also lowered.
Problem LS5.1.
Make drawings depicting the relationship between reaction progress and energy for the following cases:
1. a new ligand binds to a four-coordinate complex, forming a five coordinate complex.
2. a two-step process in which a new ligand binds to a four-coordinate complex, forming a five coordinate complex, and then an old ligand dissociates to form a new, four-coordinate complex.
The rate constant gives direct insight into what is happening at the transition state, because it gives us the energy difference between the reactants and the transition state. Based on that information, we get some ideas of what is happening on the way to the transition state.
The rate constant can be broken down into pieces. Mathematically, it is often expressed as
$k = \left ( RT \over Nh \right ) e^{- \Delta G \ddagger \over RT}$
In which R = the ideal gas constant, T = temperature, N = Avogadro's number, h = Planck's constant and Δ G = the free energy of activation.
The ideal gas constant, Planck's constant and Avogadro's number are all typical constants used in modeling the behaviour of molecules or large groups of molecules. The free energy of activation is essentially the energy requirement to get a molecule (or a mole of them) to undergo the reaction.
Note that k depends on just two variables:
• Δ G or the energy required for the reaction
• T or the temperature of the surroundings, which is an index of the available energy
The ratio of activation free energy to temperature compares the energy needs to the energy available. The more energy available compared to the energy needed, the lower this ratio becomes. As a result, the exponential part of the function becomes larger (since the power has a minus sign). That makes the rate constant bigger, and the reaction becomes faster.
The activation free energy is constant for a given reaction. It can be broken down in turn to:
$\Delta G^{\ddagger} = \Delta H^{\ddagger} - T \Delta S^{\ddagger}$
in which Δ H = activation enthalpy and Δ S = activation entropy.
The activation enthalpy is the energy required for the reaction. The activation entropy deals with how the energy within the molecule must be redistributed for the reaction to occur. These two parameters can be useful in understanding events leading to the transition state.
For example, in ligand substitution, an associative pathway is marked by low enthalpy of activation but a negative entropy of activation. The low enthalpy of activation results because bonds don't need to be broken before the transition state, so it doesn't cost much to get there. That's favourable and makes the reaction easier. However, a decrease in entropy means that energy must be partitioned into fewer states. That's not favourable and makes the reaction harder. The reason the energy must be redistributed this way is that two molecules (the metal complex and the new ligand) are coming together to make one bigger molecule. They can no longer move independently of each other, and all of their combined energy must be reapportioned together, with a more limited range of vibrational, rotational and translational states to use for that purpose.
• Associative pathway: more bond making than bond breaking; lower enthalpy needs
• Associative pathway: two molecules must be aligned and come together; fewer degrees of freedom for energy distribution; decrease in entropy
On the other hand, the dissociative pathway is marked by a higher enthalpy of activation but a positive entropy of activation. The higher enthalpy of activation results because a bond must be broken in the rate determining step. That's not favourable. However, the molecule breaks into two molecules in the rate determining step. these two molecules have more degrees of freedom in which to partition their energy than they did as one molecule. That's favourable.
• Dissociative pathway: more bond breaking in rate determining step, higher enthalpy needs
• Dissociative pathway: one molecule converts to two molecules in rate determining step, greater degrees of freedom in two independently moving molecules, entropy increases
Thus, looking at the activation parameters can reveal a lot about what is going on in the transition state.
Problem LS5.2.
What factor(s) other than entropy might raise the free energy of the transition state going into an associative step between a metal complex and an incoming ligand? (What factor might make the first, associative step slower than the second, dissociative step?)
Problem LS5.3.
Other mechanisms for ligand substitution are also possible. The following case is referred to as an associative interchange (IA).
a) Describe in words what happens in an associative interchange.
b) Predict the rate law for the reaction.
c) Qualitatively predict the activation entropy and enthalpy, compared with
i) an associative mechanism and
ii) a dissociative mechanism.
Problem LS5.4.
For the following mechanism:
a) Describe in words what is happening.
b) Predict the rate determining step.
c) Predict the rate law for the reaction.
d) Qualitatively predict the activation entropy and enthalpy, compared with
i) both an associative mechanism and
ii) a dissociative mechanism.
e) Suggest some ligands that may be able to make this mechanism occur. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Ligand_Substitution_in_Coordination_Complexes/LS5._Activation_Parameters.txt |
LS6. Some Reasons for Differing Mechanisms
We usually look for physical reasons why a given compound might undergo a reaction via one mechanism and not another. That ability adds to our understanding of chemistry. If we can take information and give it predictive value, then we may be able to make educated decisions about what is probably happening with new reactions.
Why might a reaction undergo a dissociative reaction rather than an associative one? What factors might prevent an associative pathway?
One reason may be that there is not enough room. In an associative step, an additional ligand comes in and binds to the metal. If it is already crowded, that may be difficult.
Figure LS6.1. The role of steric crowding in ligand substitution. In one of these cases, the associative mechanism is less favoured because of crowding that will occur in the transition state.
• Steric crowding may lead to a dissociative, rather than associative, mechanism.
Another reason has to do with electronics. Maybe the compound cannot easily accept an additional bonding pair. That may be the case if the compound already has eighteen electrons.
Figure LS6.2. The role of electron count in ligand substitution. In one of these cases, the associative mechanism is less favoured because the metal is already electronically saturated.
• Electronic saturation may lead to a dissociative, rather than associative, mechanism.
However, if there is less crowding, and more electrons can be accommodated, an associative pathway may result.
Problem LS6.1.
i) Draw structures for the following reactions. Pay attention to geometry.
ii) Predict whether each of the substitutions would occur through associative or dissociative mechanisms.
a) AuCl3py + Na N3 → Na+ [AuCl3N3]- + py
b) Rh(C2H4)2(acac) + C2D4 → Rh(C2D4)2(acac) + C2H4
c) [Co(NH3)5Cl]2+ + H2O → [Co(NH3)5(OH2)]2+ + Cl-
d) trans-(Et3P)2PtCl2 + -SCN → trans-(Et3P)2PtCl(SCN) + Cl-
e) Rh(NH3)4Cl2+ + -CN → Rh(NH3)4Cl(CN)+ + Cl-
f) trans-[Cr(en)2Br2]+ + -SCN → cis- and trans-[Cr(en)2Br(SCN)]+ + Br-
Problem LS6.2.
Maurice Brookhart of UNC, Chapel Hill, makes organometallic compounds in order to study fundamental questions about reactivity. In this case, he has reported making a new compound capable of "C-H activation", a reaction in which unreactive C-H bonds can be forced to break. This process holds the future promise of converting coal and natural gas into important commodities currently obtained from petroleum.
1. Draw, with curved arrows, a mechanism for the ligand substitution in the synthesis of this C-H activating complex.
2. Explain your reasons for your choice of reaction mechanism.
Problem LS6.3.
Bob Grubbs of Cal Tech was awarded the Nobel Prize in chemistry for his development of catalysts for olefin metathesis. Olefin metathesis is important both in the reforming of petroleum and in the synthesis of important commodities such as pharmaceuticals. In the following study, he replaced chlorides on a "Grubbs Generation I catalyst" to study the effect on the olefin metathesis reaction.
1. Draw, with curved arrows, a mechanism for the ligand substitution in this complex.
2. Explain your reasons for your choice of reaction mechanism.
3. What factor(s) do you think Grubbs hoped to study by making this substitution in the catalyst?
Problem LS6.4.
Sometimes, kinetic studies can give insight into a reaction if controlled changes in the reaction produce measurable results.
1. Draw, with curved arrows, a mechanism for the ligand substitution in this reaction.
2. Explain your reasons for your choice of reaction mechanism.
3. Explain the following kinetic data.
Problem LS6.5.
Several different structures were proposed for Ni(cysteine)22-. Kinetic studies of substitution in this complex showed the rate was dependent in the concentration of both the metal complex and the incoming ligand. Which structure do you think is correct? Why? | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Ligand_Substitution_in_Coordination_Complexes/LS6._Reasons_for_Diff._Mechanisms.txt |
LS7. The Trans Effect
Occasionally in ligand substitution there is a situation in which there are two identical ligands that could be replaced, but two different products would result depending on which ligand left. This situation often happens in square planar complexes, for example. Replacement of one ligand would lead to a cis product. Replacement of the other one would lead to a trans product.
An important example of this issue is in the synthesis of cis-platin, an antitumour medication frequently used to treat ovarian and testicular cancer.
Cis-platin could be made from treatment of tetraammineplatinum(II) with chloride salts. The chloride ion could replace two of the ammonia ligands.
But that doesn't work. That synthesis results in the formation of trans-platin, a compound that has all of the nasty side effects of the cis isomer but with none of the therapeautic benefit.
If instead you were to start with tetrachloroplatinate salts and treat them with ammonia, you could replace two of the chloride ligands. That works really well, and it provides cis-platin, not trans-platin.
Problem LS7.1.
What do you think is the mechanism of substitution of the two reactions above? Why?
This reaction, if run under these conditions, is clearly not always under thermodynamic control. Two different products result, depending on how the reaction is done. One of those isomers is probably more stable than the other; if thermodynamics were in charge, it would form the same thing both times.
Instead, there may be an element of kinetic control for at least one of the pathways. A given product might be made, not because it is more stable, but simply because it forms more quickly than the other one.
Take another look at those two reactions. One of the things that they have in common is that the ligand that gets replaced is trans to a chloride. It is not trans to an ammonia. Maybe the other ligands in the complex can influence how quickly one ligand can leave.
Specifically, the "trans effect" is the role of trans-ligands in influencing substitution rates in square planar complexes.
The following kinetic data were obtained for substitutions on square planar platinum complexes, in the reaction:
trans-(PEt3)2PtLCl + py → trans-(PEt3)2PtLpy+ + Cl-
L kobs (s-1) T, °C
PMe3 0.20 0
H- 0.047 0
PEt3 0.041 0
CH3- 6.0 x 10-4 25
C6H5- 1.2 x 10-4 25
Cl- 3.5 x 10-6 25
Ref: Cooper Langford & Harry Gray, Ligand Substitution Processes, W.A. Benjamin, NY, 1965, p. 25.
Problem LS7.2.
Draw structures for each of the complexes listed in the table.
Problem LS7.3.
Provide a mechanism with arrows for the reaction studied in the table.
Clearly that trans ligand has a dramatic effect on how quickly the chloride can be substituted in the above study. Additional studies like this one have led to some general trends. Below, the ligands on the left have strong trans-effects. Ligands trans to them are substituted very quickly. The ligands on the right have very modest trans effects. Ligands trans to them are substituted only slowly.
Problem LS7.4.
Look for empirical trends in the series of ligands above. Without trying to explain exactly why, find chemically relevant factors that may be responsible for these reactivity trends.
In general, explanations of the trans effect have focused on two separate types of ligands. These are strong sigma donors and strong pi acceptors.
Strong sigma donors donate electrons very effectively to the metal via a sigma bond. Because the ligand trans to this donor would be bonding via donation to the same metal p orbital, there is a competition. The metal p orbital bonds more favourably with the strong sigma donor, and the ligand trans to it is left with a weaker bond.
The strong sigma donor gets good overlap with the metal orbital and the resulting interaction goes down low in energy.
The weak sigma donor gets poorer overlap with the metal orbital and only weak stabilization of the donor electrons.
Place these two choices together, and the metal orbital will engage in a strong bonding interaction with the strong σ-donor. Doing so lowers the electronic energy significantly. It won't interact very much with the weak σ-donor, because doing that won't result in as much lowering of electronic energy. The result is a strong bond one one side of the metal and a weak bond on the other. That weak bond will break easily and that ligand will be replaced easily.
Problem LS7.5.
Which of the ligands in the trans effect series are probably strong σ-donors? Why?
Strong pi acceptors exert their trans effect in a different way. They are thought to stabilize a particular geometry of the five-coordinate intermediate in substitution of square planar complexes. We haven't worried too much about the geometry of that intermediate, but it is probably trigonal bipyramidal. It would have three ligands in an equatorial plane and two more directly opposite each other, in the axial positions.
Essentially, the incoming ligand pushes two of the ligands down from the square plane to form this trigonal bipyramid. When it comes time for a ligand to leave, it is probably going to be one of these ligands that is already on the move. They are already on a trajectory out of the square plane, anyway.
A strong pi acceptor like CO exerts its trans effect by making sure it, along with the ligand opposite it, gets into that equatorial plane. It does that by a stabilizing delocalization that happens when the π-acceptor is in the electron-rich equatorial plane. In that position, it can draw electron density via π-donation from two different donors. If it were in an axial position, it could still delocalize electrons this way, but it would draw most effectively from just one donor rather than two.
So which of those two ligands is going to keep moving and leave the complex? It certainly won't be the one that is exerting a stabilizing, delocalizing effect on the complex via its strong bonding interactions. It will be the unlucky trans ligand that got dragged along with it.
Problem LS7.6.
Which of the ligands in the trans-effect series are probably strong π-acceptors? Show why.
Problem LS7.7.
Draw an orbital picture showing the -delocalization described in the trigonal bipyramidal intermediate. Label the orbitals, assuming the z axis is along the axis of the trigonal bipyramid (i.e. the equatorial plane is the xy plane). | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Ligand_Substitution_in_Coordination_Complexes/LS7._The_Trans_Effect.txt |
Problem LS1.1.
1. a bromine atom replaces a hydrogen atom
2. an acetate group (ethanoyl ester) replaces a bromine atom
3. a bromine atom replaces a chlorine atom (or ions)
4. a methoxy group replaces a chlorine atom
5. a methylamino group replaces an oxygen atom
Problem LS2.1.
1. Energy is released when bonds are formed. Energy must be added to break bonds. In general, bond-breaking costs more energy than bond-making.
2. On the basis of question (a), we would assume that the second, bond-breaking step is the rate determining step in association mechanisms.
3. The first, dissociative step would be the rate-determining step, on the basis of question (a).
4. The rate law would include steps prior t the rate determining step. Rate = k[MLn][X] if MLn is the complex and X is the new ligand.
5. Rate = k[MLn].
Problem LS2.5.
This problem is answered through consideration of the spectrochemical series.
from easiest to hardest to replace: NO3- Cl- H2O NH3 PPh3 CO
Problem LS3.1.
Associative Rate Law: Rate = [MLn][X], if MLn is the complex and X is the new ligand.
1. Rate will quadruple: Rate = (2 x [MLn]0) x (2 x [X]0) = 4 x [MLn]0[X]0, if [X]0 and [MLn]0 are the original concentrations.
2. Rate will sextuple: Rate = (3 x [MLn]0) x (2 x [X]0) = 6 x [MLn]0[X]0.
3. Rate will nonuple: Rate = (3 x [MLn]0) x (3 x [X]0) = 9 x [MLn]0[X]0.
4. Rate will stay the same: Rate = (0.5 x [MLn]0) x (2 x [X]0) = 1 x [MLn]0[X]0.
Problem LS3.3.
1. Changing both concentrations at once would leave some doubt about whether one concentration had affected the rate, or the other concentration, or both. In practice, one concentration is usually held constant while the other is kept in excess and varied.
2. Rate changes over time because the concentrations of reactants change as they are consumed. By reporting only the initial rate (usually meaning less than 5% or 10% complete, but possibly even less than that if a lot of data can be gathered very quickly), the concentrations are still about what you started with. That means you can report a rate that corresponds to a given concentration with confidence.
Problem LS3.4.
1. The rate would change with the ligand concentration if associative. This rate is constant over a range of ligand concentrations, so the reaction is not associative.
2. The rate increases linearly with ligand concentration. This reaction proceeds via an associative mechanism.
3. The rate changes over the concentration range, but it decreases. This is the opposite of what should happen. This reaction does not follow a simple associative pathway.
4. The rate increases linearly with ligand concentration. This reaction proceeds via an associative mechanism.
Problem LS3.5.
Because Rate = k[MLn][L] and [MLn] is held constant while [L] is varied, then the slope of the line is k [MLn]. Since you would know the value of [MLn], you could obtain the rate constant from the quantity (slope/ [MLn]).
Problem LS4.1.
1. 4 minutes = 240 seconds = 2 x 120 second = 2 half lives. Material left = 50% x 50% = 0.5 x 0.5 = 0.25 = 25% left
2. 6 minutes = 360 seconds = 3 x 120 second = 3 half lives. Material left = 0.5 x 0.5 x 0.5 = 0.125 = 12.5% left
3. 8 minutes = 480 seconds = 4 x 120 second = 4 half lives. Material left = 0.5 x 0.5 x 0.5 x 0.5 = 0.0625 = 6.25% left
4. 10 minutes = 600 seconds = 5 x 120 second = 5 half lives. Material left = 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 0.03125 = 3.125% left
Problem LS4.2.
Dissociative Rate Law: Rate = [MLn], if MLn is the complex. There is no dependence on [X], if X is the new ligand.
1. Rate will double: Rate = 2 x [MLn]0, if [MLn]0 is the original concentration.
2. Rate will triple: Rate = 3 x [MLn]0.
3. Rate will triple: Rate = 3 x [MLn]0.
4. Rate will be halved: Rate = 0.5 x [MLn]0.
Problem LS4.4.
1. The rate increases with both concentration of metal complex and incoming ligand. This looks like an associative mechanism.
2. The rate depends on concentration of the metal complex, but not the incoming ligand. This looks like a dissociative mechanism.
3. The rate depends on the concentration of incoming ligand, but not the metal complex. Whatever is going on here, it isn't a simple dissociative mechanism.
Problem LS4.5.
1. The rate of the reaction is depressed when the concentration of the departing ligand is increased.
2. This dependence could indicate an equilibrium in the dissociative step. The more departing ligand is added, the more the equilibrium is pushed back towards the original metal complex. With less dissociated metal complex around, the entering ligand cannot form the new complex as quickly.
Problem LS4.6.
1. These solvents all have lone pairs. They could be Lewis bases or nucleophiles.
2. It looks like two competing mechanisms. On term suggests a dissociative mechanism, whereas the other term suggests a dissociative mechanism. They could be happening in competition with each other.
3. On the other hand, it could be that there is one mechanism with two different nucleophiles. If the incoming ligand is the nucleophile, the term on the right shows up in the rate law. If the solvent is the nucleophile, forming a third complex, the term on the left shows up in the rate law. That's because we would typically change the amount of metal complex and the amount of ligand that we add to the solution in order to determine the rate law, but we wouldn't normally be able to change the concentration of the solvent, so it would be a constant. (How could you confirm this explanation in an experiment?)
Problem LS5.2.
The metal centre is becoming more crowded as the new ligand arrives, so an increase in energy owing to steric hindrance may also play a role in the transition state energetics.
Problem LS5.3.
a) The new ligand, B, is arriving at the same time as the old ligand, A, is departing. We might also describe it as new ligand B pushing old ligand A out of the complex.
b) Rate = k[ML5A][B], which looks like an associative rate law.
c) This is a thought-provoking question without a definite answer. Associative mechanisms typically have lower activation enthalpy than dissociative mechanisms, because there has also been some bond-making prior to the bond-breaking in the rate determining step. The associative interchange would be a little more like the associative mechanism than dissociative. The mix of bond-making and bond-breaking at the transition state would make the enthalpy of activation relatively low.
Associative mechanisms have negative activation entropies, whereas dissociative mechanisms have positive activation entropies. The associative interchange could be in between the two, given that the elementary step would be close to entropically neutral overall. What happens at the transition state is a little harder to imagine, but it might reflect the small changes in entropy through the course of the reaction, producing a small entropy of activation. On the other hand, if the incoming ligand is forced to adopt some specific approach as it comes into the molecule (to stay out of the way of the departing ligand, for example) then that restriction could show up as a small negative activation entropy.
Problem LS5.4.
a) The lone pair donation from one ligand appears to push another ligand out.
b) The first step is probably rate determining, because of the bond breaking involved.
c) If the first step is rate determining, Rate = k [ML5A].
d) Another question without a very clear answer. Compared with an associative mechanism, the activation entropy is probably much more positive, because additional degrees of freedom are being gained as the molecule heads over the activation barrier and one of the ligands separates to be on its own. However, the activation entropy may be less positive than in a regular dissociation, because in this case the breaking of one bond has to be coordinated with the formation of another.
The enthalpy of activation has both a bond-making and bond-breaking component, a little like in an associative mechanism. However, the amount of bond making here is probably less important, because pi bonds are typically not as strong as sigma bonds. The activation enthalpy is probably higher than an associative pathway but not as high as a dissociative one.
e) The donor ligand must have a lone pair. Oxygen donors would be good candidates, because even if one lone pair is already donating in a sigma bond, an additional lone pair may be available for pi donation. The same thing is true for halogen donors. It would also be true for anionic nitrogen donors but not for neutral nitrogen donors, because a neutral nitrogen has only one lone pair.
Problem LS7.1.
This is probably an associative mechanism because of the square planar geometry.
Problem LS7.4.
There is an electronegativity trend: the less electronegative, the greater the trans effect (see the halogens, as well as the series O,N,C and also the orders within several pairings: S,P; O,S and N,P).
Alternatively, some of the above could be described by a polarizability trend: more polarizable atom, greater trans effect (for example, the halogens).
Most of the ligands containing π-bonds have strong trans influence (but not all).
Most of the π-donors have a weaker trans influence. However, these ligand cover a very broad range in this series.
Problem LS7.5.
The strongest σ-donors are typically those with more polarizable donor atoms (such as S, P, I) as well as those with less electronegative donor ions such as C- and H-. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Ligand_Substitution_in_Coordination_Complexes/LS8._Solutions_to_Selected_Problems.txt |
NS1. Introduction to Aliphatic Nucleophilic Substitution
Aliphatic nucleophilic substitution is a mouthful, but each piece tells you something important about this kind of reaction.
In substitution reactions, one piece of a molecule is replaced by another. For example, ligands can be replaced in transition metal complexes. Oxygen atoms in organic carbonyl compounds can be replaced by nitrogen atoms or sulfur atoms, in a particular variation of carbonyl addition reactions.
These reactions all involve the addition of a nucleophile to an electrophilic atom or ion. They are all nucleophilic substitution reactions.
Aliphatic systems involve chains of saturated hydrocarbons, in which carbons are attached to each other only through single bonds. Aliphatic nucleophilic substitution is the substitution of a nucleophile at a tetrahedral or sp3 carbon.
Aliphatic nucleophilic substitutions do not play a glamourous, central role in the world of chemistry. They don't happen in every important process, the way carbonyl additions and carboxyloid substitutions appear to in biochemistry. Instead, they are ubiquitous little reactions that play important, small roles in all kinds of places.
For example, polyethylene gloycol (PEG) is a commonly used polymer in lots of biomedical applications. PEG frequently has hydroxyl groups at each end of the polymer. Capping the ends of the polymer through reaction with another group can lead to very different physical properties.
For another example, many biochemical processes require prenylation of proteins. That would involve a nucleophilic substitution in which a sulfur in a cysteine residue adds to a tetrahedral carbon in a prenyl group, replacing a phosphate group.
In order to be an electrophile, that tetrahedral carbon should have at least some partial positive charge on it. In the simplest cases, this electrophilic carbon is attached to a halogen: chlorine, bromine or iodine. These compounds are called alkyl halides (or alkyl chlorides, alkyl bromides and alkyl iodides).
Problem NS1.1.
Draw structures of the following alkyl halides.
a) 2-bromopentane b) 2-methyl-2-chlorobutane c) benzyl iodide d) allyl chloride
Lots of things can be nucleophiles in these reactions. Sometimes, the nucleophile is a neutral compound with a lone pair, such as ammonia or water (or, by extension, an amine or an alcohol).
Problem NS1.2.
Sometimes, addition of a mild base is helpful in reactions of neutral nucleophiles. Show, with mechanistic arrows, how sodium carbonate (K2CO3) would play a role in the reaction.
The third row analogs of these nucleophiles, in which the nucleophlic atom is a phosphorus or a sulfur, are also good nucleophiles in these reactions.
Sometimes, the nucleophile is an anion. Cyanide anion is a good nucleophile, as are the structurally similar acetylides.
Enols, enolates and enamines are also very good nucleophiles in this type of reaction.
Semi-anionic nucleophiles such as Grignard (or organomagnesium) reagents and alkyl lithium reagents can sometimes act as nucleophiles in this reactions, but they are not very reliable. Complications often lead to other reactions instead. Gilman (or organocopper) reagents, in which a carbon atom is attached to a copper atom, can usually react with alkyl halides. However, they probably act via a different mechanism from the ones described in this chapter.
NS10. Leaving Group Formation
NS10. Leaving Group Formation
Aliphatic Nucelophilic Substitutions can be useful reactions. A minor drawback is the low natural occurrence of alkyl halides. Alcohols are much more common.
We could do nucleophilic substitutions on alcohols. The trouble is, oxygens are less polarizable than halides. A hydroxide ion is less stable, and harder to form than a halide ion. They don't make very good leaving groups, comparatively.
One way around that problem would be to protonate the oxygen. Attached to the carbon, it is a cation. Once it leaves, it becomes a neutral. The issue of ion stability is sidestepped. The trouble is, plunking a compound into concentrated acid is not always a reliable way to get things done.
For that reason, it is pretty helpful to be able to turn alcohols into alkyl halides, or otherwise turn hydroxyls into stable, anionic leaving groups.
One of the most common synthetic methods of converting alcohols into good candidates for nucleophilic substitution is to convert the hydroxyl into a halide through the use of a phosphorus reagent. Phosphorus tribromide is frequently used to make alkyl bromides from alcohols.
This reaction itself involves a sequence of nucleophilic substitution reactions. In the first, the oxygen atom in a hydroxyl group acts as the nucleophile and replaces a bromide on phosphorus. In the second, the displaced bromide ion rebounds to displace the oxygen atom from the tetrahedral carbon. This mechanism is aided by the strength of the strong phosphorus-oxygen bond that is formed. The phosphite that forms is a very good leaving group.
Another common method is to turn the hydroxyl into a sulfonate ester, such as a mesylate or tosylate. Again, the oxygen atom acts as a nucleophile, displacing a halide from the sulfur in a sulfonyl chloride. This is very similar to the bromination with phosphorus tribromide, but the sulfonate ester waits, poised to be displaced by a nucleophile. In fact, tosylates are generally even better leaving groups than halides.
Biologically, something very similar to both of these processes sometimes happens. The alcohol unit is converted into a phosphate. The alcohol can be phosphorylated by a molecule of ATP. Again, the phosphate portion of the molecule is a very good leaving group.
Problem NS10.1.
Provide a mechanism for the reaction of phosphorus tribromide with 2-pentanol, based on the description provided.
Problem NS10.2.
Provide a mechanism for the reaction of mesyl chloride with 2-pentanol, based on the description provided.
Problem NS10.3.
Sometimes in formation of sulfonate esters, halogenation occurs by accident, forming an alkyl chloride. Show how that might happen by continuing on from the mechanism of formation of a mesylate ester from mesyl chloride and 2-pentanol.
Problem NS10.4.
One way of preventing side reactions during synthesis of sulfonate esters, like the one in the previous question, is to perform a two-phase reaction. For example, the reaction might be performed in a mixture of water and dichloromethane, with a little added sodium hydroxide to act as a base. Show how this approach would limit the chlorination reaction.
Problem NS10.5.
Provide reagents (or series of reagents) to accomplish the following transformations.
Problem NS10.6.
Mitsunobu addition is a one-step method of replacing an alcohol's OH group with a nucleophile. In the following case, the original oxygen atom has been completely replaced. You may be able to guess where it goes based on other reactions you have seen here.
Propose a mechanism for this reaction, but don't worry about what DEAD does ( that is a more advanced topic).
NS11. NA to Strained Rings
Addition to Strained Rings: Epoxides
Oxygen is a very common element in all kinds of compounds, whether they are biological molecules, minerals from the earth or petrochemicals. Exploiting oxygen's electronegativity and giving it a little help to become a leaving group is a common way to make connections and build new molecules in nature, the laboratory or the production facility.
Sometimes oxygen does not need much help to become a leaving group. Epoxides, or oxiranes, are three-membered ring ethers. They are good electrophiles, and a C-O bond breaks easily when a nucleophile donates electrons to the carbon.
Problem NS11.1.
Explain why the C-O bond in an epoxide breaks easily.
Problem NS11.2.
Use a potential energy diagram to show why epoxides are susceptible to react with nucleophiles, whereas other ethers are not.
Epoxides are very useful in the synthesis of important molecules. The Nu-C-C-O motif that is formed in nucleophilic addition to an epoxide is very valuable. Whereas other nucleophilic additions simply replace a halide or leaving group with a nucleophile, exchanging one reactive site with another, addition to an epoxide makes a product that has gone from having one reatcive site to two reactive sites. That can open the door to lots of useful strategies when trying to make a valuable commodity.
Problem NS11.3.
Show how you could carry out the following transformation. More than one step is involved.
Problem NS11.4.
One of the most widespread uses of epoxides is in making polymers. The polyethylene glycol produced in polymerization of an epoxide is frequently used in biomedical applications. Provide a mechanism with arrows for the following polymerization of ethylene oxide, in the presence of:
1. an acid catalyst
2. a basic catalyst.
Problem NS11.5.
Tetrahydrofuran can also be polymerized, forming polytetramethylene glycol.
1. Compare the rate of polymerization of THF with that of ethylene oxide.
2. Polymerization of THF generally requires an acid catalyst, rather than a basic one. Why? | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Nucleophilic_Substitution_at_Tetrahedral_Carbon/NS1._Introduction_to_ANS.txt |
NS12. Elimination
Sometimes, elimination reactions occur instead of aliphatic nucleophilic substitutions. In an elimination reaction, instead of connecting to the electrophilic carbon, the nucleophile takes a proton from the next carbon away from it. The halide or other leaving group is still displaced. A double bond forms between the two carbons.
Thus, there are actually more than two competing mechanisms occurring at once here. In addition to unimolecular and bimolecular substitution, a reaction involving deprotonation is also possible.
Problem NS12.1.
Draw a mechanism for the elimination reaction above. Assume the reaction is bimolecular.
The mechanism of an elimination reaction is almost exactly the same as an aliphatic nucleophilic substitution, except that the nucelophile misses its mark. It hits a proton instead of a carbon and acts as a base instead of a nucleophile. This process can happen at the same time as the leaving group's departure or it can happen afterwards. These mechanisms are called E1 and E2.
Problem NS12.2.
Draw another mechanism for the elimination reaction above, but this time, suppose the reaction is unimolecular.
Problem NS12.3.
Given the mechanism in NS12.2., other products would also be expected.
1. What are they?
2. What does their absence suggest about the mechanism of the reaction?
Why might a reaction undergo elimination rather than substitution? The most important reason concerns the nature of the nucleophile. The more basic the nucleophile, the more likely it will induce elimination.
• Basic nucleophiles lead to elimination.
Very strong bases include carbon and nitrogen anions and semi-anions. Examples include butyllithium and sodium amide. Very strong bases are highly likely to engage in elimination, rather than substitution.
Strong bases include non-stabilized oxygen anions. Examples include sodium hydroxide as well as alkoxides such as potassium tert-butoxide or sodium ethoxide. Strong bases favour elimination, too. Nevertheless, they can sometimes undergo either elimination or substitution, depending on other factors (see below).
Weak bases include cyanide, stabilized oxygen anions such as carboxylates and aryloxides, fluoride ion and neutral amines. Weak bases are much more likely to undergo substitution than elimination.
Very weak bases include heavy halides such as chloride, bromide or iodide, as well as phosphorus and sulfur nucleophiles. Very weak bases undergo elimination only rarely.
Problem NS12.4.
Typically, strong bases and very strong bases are more likely to react via the E2 mechanism; they react so quickly that the deprotonation step triggers C-LGp ionization, rather than the other way around. However, E1 mechanisms also occur with these bases, especially at low concentrations. Explain why.
Problem NS12.5.
Why is it that an anion such as cyanide is a weak base, whereas CH3Li is a strong base? Give two reasons.
Another factor is sterics. The more crowded the electyrophile, the more likely the nucleophile will encounter a proton on its way to the electrophilic carbon.
As a nucleophile approaches tert-butyl bromide, coming from the side opposite the bromine in order to undergo nucleophilic substitution, it is pretty likely to collide with a proton on its way to the electrophilic carbon. The same thing has a good chance of happening with iso-propyl bromide. However, it is much less likely to happen with bromoethane. Finally, bromomethane doesn't even have a beta-hydrogen, so the chance of elimination in that case is zero.
• Crowding leads to elimination.
Note that the crowding could involve either the structure of the base or the structure of the electrophile. A large, bulky base may be more likely to deprotonate than find its way in to the electrophilic carbon atom.
Problem NS12.6.
Given the following pairs of nucleophiles, which one is more likely to undergo elimination?
Problem NS12.7.
Although acetylides (such as sodium acetylide, Na CCH) are actually more basic than alkoxides (such as sodium isopropoxide, Na OCH(CH3)2), acetylides frequently undergo substitution rather than elimination. Propose a reason for this difference.
A third factor is temperature. An elimination reaction involves the cleavage of two bonds, whereas a substitution reaction requires only one bond to break. Thus, an elimination reaction is more energy-intensive, and it is more likely to occur at higher temperatures, when more energy is available.
• Higher temperatures lead to elimination.
Problem NS12.8.
An additional factor in the energy dependence of eliminations and substitutions is entropy.
1. Use simple rules about to determine which products are favoured by entropy: Elimination or substitution?
2. Given the relationship ΔG = ΔH - T ΔS, which thermodynamic factor dominates free energy change at high temperature?
3. Therefore, which product is favoured at high temperature?
NS13. Regiochem. in Elimination
NS13. Regiochemistry in Elimination
Sometimes, an elimination reaction could lead to formation of a double bond in more than one place. If the halide is on one carbon and there are protons that could be removed on either side, then taking one proton or the other might lead to two different products. This reaction could have different regiochemical outcomes, meaning it could happen at two different places in the molecule.
What factors might influence which product forms? We might think about product stability, in case there are corresponding differences in barriers leading to those products. We already know about stereochemical effects in alkene stability, but what about other effects?
It is well-established that alkene stability is influenced by degree of substitution of the double bond. The greater the number of carbons attached to the double bond, the more stable it is.
That effect is related to hyperconjugation. Specifically, it's an interaction between bonding orbitals and antibonding orbitals on neighbouring carbons. The interaction allows the bonding electrons to drop a little lower in energy through delocalization. The interaction also pushes the antibonding orbitals a little higher in energy, but since they have no electrons, they don't contribute to the real energy of the molecule. Overall, the molecules goes down in energy.
• In most cases, the most-substituted alkene results from elimination reactions.
However, alkene stability isn't the only factor that plays a role in elimination. Steric hindrance can play a role, too.
In a case in which there are two different hydrogens from which to select, the one leading to the more-substituted double bond is sometimes a little bit crowded. That leaves the base with fewer viable pathways to approach the proton.
On the other hand, the removal of a proton leading to the less stable alkene is often less crowded, allowing the base to approach much more easily from a number of angles.
• In some cases, especially with very bulky bases, the least substituted alkene forms, even though it is less stable. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Nucleophilic_Substitution_at_Tetrahedral_Carbon/NS12._Elimination_Reactions.txt |
NS14. Stereochemistry in Elimination
Sometimes, elimination reactions may lead to multiple stereoisomers; that is, they could lead to either the cis or the trans isomer, or in more complicated structures, either the Z or the E isomer.
Of course, if there were some inherent stability difference between these isomers, that could be a factor that plays a role in influencing the outcome. Elimination reactions aren't generally reversible, so products are not directly determined by alkene isomer stabilities. Nevertheless, sometimes the barrier leading to a more stable product is a little lower than the barrier leading to a less stable product.
We do know that in simple cis vs. trans cases, the trans isomer is generally lower in energy because of fewer steric interactions between the substituents on the double bond. In the absence of other information, we could take that as a starting point. Let's see whether elimination reactions generally lead to trans isomers.
It turns out that sometimes this is true: eliminations often lead to the more stable product. Sometimes it isn't true, though. The answer depends on the mechanism.
• E1 eliminations generally lead to the more stable stereochemistry.
• E2 eliminations may or may not lead to the more stable stereochemistry.
Instead, in an E2 reaction, stereochemistry of the double bond -- that is, whether the E or Z isomer results -- is dictated by the stereochemistry of the starting material, if it is diastereomeric. In other words, if the carbon with the hydrogen and the carbon with the halogen are both chiral, then one diastereomer will lead to one product, and the other diastereomer will lead to the other product.
The following reactions of potassium ethoxide with dibromostilbene (1,2-dibromo-1,2-diphenylethane) both occurred via an E2 mechanism. Two different diastereomers were used. Two different stereoisomers (E vs. Z) resulted.
Problem NS14.1.
Provide the stereochemical configurations of the following compounds from the above reactions:
1. 1,2-dibromo-1,2-diphenylethane (upper example)
2. 1-bromo-1,2-diphenylethene (upper example)
3. 1,2-dibromo-1,2-diphenylethane (lower example)
4. 1-bromo-1,2-diphenylethene (lower example)
In E2 eliminations, the spatial relationship between the proton and leaving group determines the product stereochemistry. That's because pi bond formation happens at the same time that the halide leaves and at the same time that the base removes the proton. All of these events have to be coordinated together. The central, tricky event is the pi bond formation. The leaving group can leave in any direction, and the base can approach from many directions, but unless the pi bond is ready to form, nothing else happens.
Let's slow the reaction down and imagine it takes place in slightly different stages.
As the leaving group leaves, it takes its electrons with it. It begins to leave a positive charge behind. That positive charge will be centered on the carbon from which the halide is departing. That carbocation, if it fully formed, would have only three neighbours to bond with. It would be trigonal planar. It would have an unoccupied, non-bonding p orbital.
As the base takes the proton, the hydrogen leaves behind the electrons from the C-H bond that held it in place. These electrons stay behind on the carbon atom. They are left in a non-bonding carbon valence orbital, a p orbital or something quite like it.
Now we have a filled p orbital next to an empty p orbital. They overlap to form a pi bond.
Of course, in an E2 reaction, things don't happen in stages. Everything happens at once. That means that, as the base removes the proton, the pi bond must already start forming. Because a pi bond requires parallel alignment of two p orbitals, and the p orbitals are forming from the C-H and C-LGp bonds, then those bonds must line up in order for the elimination to occur.
So let's look again at that dibromostilbene example.
In the first case, we need to spin the molecule so that we can see how the H on one carbon and the Br on the other are aligned and ready to eliminate via an E2 reaction. The substituents coming towards us in the reactant will still be coming towards us in the product. The substituents pointing away from us in the reactant will still be pointing away from us in the product.
So the relationships between the substituents on the nascent double bond are determined by their relationship once the reactant is aligned for the E2 reaction.
In the second case, we can spin the molecule but quickly realize the C-H and C-Br bonds are not lined up in this conformer. We need a bond rotation. Once we have made a conformational change, the C-H and C-Br bonds line up. It doesn't matter if this conformer is not favoured; if there is going to be any E2 reaction at all, this is the conformer it will have to go through.
Again, the relationships between on the new double bond are determined by their relationship once the reactant is aligned for the E2 reaction.
Problem NS14.2.
Sometimes it is easier to see the relationships between substituents by using a Newman projection. Draw Newman projections showing how the two isomers above proceed to different products in an E2 reaction.
Problem NS14.3.
Predict the product of each of the following E2 reactions. Note that the compounds differ in the incorporation of a 2H isotope (deuterium, or D) in place of a regular 1H isotope (protium, or H).
Problem NS14.4.
Conformational analysis of cyclohexanes requires the use of diamond lattice projections ("chairs"). Show why elimination can't proceed through the E2 mechanism in the following compound.
Problem NS14.5.
Predict the E2 elimination products from the following compounds.
In contrast to E2 reactions, E1 reactions do not occur in one step. That means there is time for reorganization in the intermediate. Once the leaving group leaves, the cation can sort itself into the most stable conformer. When the proton is taken, generally the most stable stereoisomer results because it comes from the most stable conformer of the cation. Any steric interactions in the alkene would also have occurred in the cation, so this interaction would have been sorted out at that point.
Thus, in the case of the dibromostilbenes examined before, E1 elimination would result in the same product in either case.
• E1 reactions can, in principle, lead to either stereochemistry of alkene.
• Free rotation around bonds in the carbocation intermediate allows the cation to adopt either conformer prior to elimination.
• However, steric interactions will lead to a preponderance of one conformer.
• The more stable conformer will lead to the more stable alkene. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Nucleophilic_Substitution_at_Tetrahedral_Carbon/NS14._Stereochem_in_Elimination.txt |
NS13. Factors Influencing the Elimination Mechanism
The factors that influence whether an elimination reaction proceeds through an E1 or E2 reaction are almost exactly the same as the factors that influence the SN1/SN2 pathway. Cation stability, solvents and basicity play prominent roles.
Problem NS13.1.
By analogy with substitution reaction, in which elimination mechanism does cation stability play a strong role: E1 or E2?
Problem NS13.2.
Draw an example of an alkyl halide that is likely to undergo an E1 elimination.
Problem NS13.3.
By analogy with substitution reactions, what mechanism would be promoted by protic solvents: E1 or E2?
Problem NS13.4.
Basicity refers to the strength of the base. Which mechanism is more likely to occur with strong bases: E1 or E2?
NS16. Solutions for Problems
NS16. Solutions to selected problems
Problem NS2.1.
The electronegativity of carbon (2.55 on Pauling scale) is less than that of fluorine (3.98), chlorine (3.16), bromine (2.96) or iodine (2.66).
1. On that basis, the carbon attached to a halogen is electrophilic because it has a partial positive charge resulting from the polar carbon-halogen bond.
2. We would expect an alkyl fluoride to be the most electrophilic of these compounds, based on electronegativity.
3. Assuming the energy required for breaking the carbon-halogen bond plays a major role in the activation barrier (not guaranteed), we would expect the activation barrier to be lowest with the alkyl iodide, then the alkyl bromide, then the alkyl chloride and finally the alkyl fluoride. This prediction contrasts with what we might expect based on electronegativity.
4. The stability of alkyl fluorides towards this reactions suggests that there is, in fact, a prominent role played by bond strengths, at least in that case. The carbon-fluoride bond is strong enough to hinder nucleophilic substitution in this compound.
Problem NS2.2.
1. In mechanism B, the dissociative one, we would expect a higher activation enthalpy. The first step, which appears to be rate determining, is a bond-breaking step, which will cost energy. In mechanism C, the bond-breaking is compensated by some bond-making; overall, this probably costs less energy.
2. In mechanism B, the dissociative case, we expect a more positive entropy of activation. As the bond to the halide begins to break, the halide and carbocation fragments begin to move independently of each other, gaining degrees of freedom and increasing in entropy. In mechanism C, the incoming nucleophile appears to coordinate its motion with that of the departing halide; as a result, there are fewer degrees of freedom in this case.
Problem NS2.3.
1. Charged intermediates are present in the dissociative mechanism (B).
2. It seems like a more polar solvent would favour both mechanisms, because both involve the interaction of an anionic nucleophile with an electrophile and loss of an anionic leaving group. However, the dissociative case (B) involves a build-up of charge in the intermediate. It is possible that a more epolar solvent could reduce the barrier to that buildup of charge separation, accelerating this mechanism.
Problem NS2.4.
1. The rate-determining step is probably the bond-breaking one (the first one).
2. Because the nucleophile has not yet participated at that point, Rate = k [R-X], if R-X = the alkyl halide.
3. There is only one step; it is the rate-determining step, by default.
4. Rate = k [R-X][Nu].
Problem NS6.2.
Keep in mind that there are other factors that can influence the reaction pathway; what we have here are just the most likely mechanisms.
a) SN2 b) Both pathways are very possible c) Both pathways are very possible d) SN1
e) SN1 f) SN2 g) SN1 h) SN1 i) SN1 j) SN1
Problem NS7.1.
1. ethanol, isopropanol, trifluoroacetic acid
2. hexane, toluene
3. THF, acetonitrile, DMF, dichloromethane, ether, DMSO, triethylamine, pyridine
4. DMSO > DMF > ACN > pyridine > DCM > THF > ether > TEA, based on dielectric constants. In general, the ones with multiple bonds between two different atoms are the most polar.
5. pyridine and triethylamine. The lone pair on the nitrogen atom is basic toward protons. The trend in basicity is triethylamine > pyridine >> acetonitrile; as the percent s character in the lone pair increases, the electrons are lower in energy and less available for donation.
Problem NS10.6.
DEAD acts as an oxidizing agent to convert the phosphorus product to a stable side-product, triphenylphosphine oxide, Ph3P=O. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Nucleophilic_Substitution_at_Tetrahedral_Carbon/NS15._Elim._Mechanism_Factors.txt |
Aliphatic nucleophilic substitution clearly involves the donation of a lone pair from the nucleophile to the tetrahedral, electrophilic carbon bonded to a halogen. We might expect this carbon to be electrophilic because of the halogen attached to it. For that reason, it attracts to nucleophile. However, the mechanism of the reaction might happen in a couple of different ways.
Problem NS2.1.
Compare the electronegativity of carbon to that of fluorine, chlorine, bromine and iodine.
1. On this basis alone, explain why the carbon attached to the halogen would be electrophilic.
2. Which compound should be most electrophilic based on electronegativity: fluoromethane, chloromethane, bromomethane or iodomethane?
3. Use the following bond strengths to estimate the qualitative trend in activation barriers for nucleophilic substitution in the four compounds in part (b): C-F 115 kcal/mol; C-Cl 84 kcal/mol; C-Br 72 kcal/mol; C-I 58 kcal/mol.
4. Fluorocarbons are quite stable towards aliphatic nucleophilic substitution; in general, they do not undergo this reaction. Explain why.
In considering possible mechanisms for this reaction, we ought to think about overall bond-making and bond-breaking steps. In the addition of sodium cyanide to alkyl chloride to make an alkyl nitrile, there is one bond-making step (the C-C bond) and one bond-breaking step (the C-Cl bond). The simplest reaction mechanism would involve some combination of these steps.
Two possibilities immediately present themselves:
Mechanism A
The C-C bond forms and then the C-Cl bond breaks.
Mechanism B
The C-Cl bond breaks and then the C-C bond forms.
However, some familiarity with bonding in the second row of the periodic table may suggest to you that mechanism A is not very likely. That mechanism would require forming five bonds to carbon before the C-Cl bond eventually breaks. We can safely ignore this possibility.
Instead, there may be a third possibility to consider.
Mechanism C
The C-Cl bond breaks and the C-C bond forms at the same time.
Mechanism C is a concerted mechanism; two bond-making and -breaking events happen at once. However, no octet rules are violated.
Reaction progress diagrams for these two reactions would look like the illustrations below.
Mechanism B, ionization and then addition of nucleophile:
Mechanism C, direct displacement of leaving group by nucelophile:
Problem NS2.2.
Compare mechanism B and C in terms of your expectations of the following parameters:
1. Activation enthalpy.
2. Activation entropy.
There isn't necessarily a reason to believe that mechanism B is the correct mechanism and mechanism C is the wrong one, or vice versa. Either one may be possible. You may need to do some work in order to figure out which one really happens. Some experiments may help to highlight what is going on.
Problem NS2.3.
If charged intermediates are suspected along a reaction pathway, insight can sometimes be gained by running a reaction in a more polar solvent and comparing its rate to that of the reaction in a less polar solvent.
1. Are charged intermediates present, either in mechanism B or C?
2. Explain how each of these mechanisms might behave in a more polar solvent.
Problem NS2.4.
Sometimes, a distinction between two possible mechanisms can be gained by comparing rate laws expected from each mechanism.
1. What do you think is the likely rate-determing step in mechanism B?
2. What do you expect will be the rate law for mechanism B?
3. What do you think is the likely rate-determing step in mechanism C?
4. What do you expect will be the rate law for mechanism C? | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Nucleophilic_Substitution_at_Tetrahedral_Carbon/NS2._Possible_Mechanisms_of_NS.txt |
Aliphatic Nucleophilic Substitution: Rate Laws
In aliphatic nucleophilic substitution, a nucleophile (abbreviated Nu) replaces a halogen "leaving group" (abbreviated LGp) from a tetrahedral carbon. Aliphatic nucleophilic substitution may take place through two different mechanisms:
• C-LGp bond breaking, followed by C-Nu bond formation.
or
• C-Nu bond formation at the same time as C-LGp bond breaking.
A look at the reaction progress diagrams for these two reactions illustrates some big differences. We will look at cyanide anion, a nucleophile, substituting for chloride in 2-chloropropane.
In the first case, some energy must be added in order to break the carbon-chlorine bond. The chlorine forms an anion, leaving a cation on the carbon. This ion pair is an intermediate along the reaction pathway. The cyanide ion then connects with this cation to form the nitrile product. Thus, there are two elementary steps in this mechanism.
Most likely, the first step is the rate-determining step. Breaking bonds costs energy, whereas making bonds releases energy. It is hard to imagine that there could be a significant barrier to the second step; the anion and cation should come together almost automatically.
The rate law for this stepwise mechanism is:
\[ Rate = k[^iPrCl] \]
that is, the rate depends on the first elementary step, but not on the second one. The second step happens pretty much automatically as soon as the first one has finally gotten around to happening.
In the second case, the nucleophile displaces the chloride directly in one step. There is only one elementary step in this reaction, and it requires both compounds to come together at once.
The rate law for this concerted mechanism is:
\[ Rate = k[^iPrCl][^-CN]\]
These two rate laws are very different, and offer an additional way for us to tell how this reaction is taking place. In principle, if we try the reaction with different concentrations of cyanide (but keep the 2-chloropropane concentration constant), we can see whether that has an effect on how quickly the product appears. If it has the predictable effect, maybe the reaction happens in one step. If not, maybe it is a two-step reaction.
Because the rate laws for these two mechanisms are so different, there has arisen a catchy shorthand for describing these reactions based on their rate laws, coined by C.K. Ingold. The rate of the stepwise reaction depends only on one concentration and is referred to as a "unimolecular reaction"; Ingold's shorthand for this kind of nucleophilic substitution was "SN1".
The rate of the concerted reaction depends on two different concentrations and is referred to as a bimolecular reaction; Ingold's shorthand for this reaction was "SN2".
Problem NS3.1.
Suppose you run this reaction with three different concentrations of cyanide: 0.1 mol/L, 0.2 mol/L and 0.3 mol/L. You keep the 2-chloropropane concentration constant at 0.05 mol/L.
1. The reaction turns out to be proceeding via a SN1 mechanism. Plot a graph of rate vs.[-CN].
2. The reaction turns out to be proceeding via a SN2 mechanism. Plot a graph of rate vs.[-CN].
Now you switch things up and run this reaction with three different concentrations of 2-chloropropane: 0.1 mol/L, 0.2 mol/L and 0.3 mol/L. You keep the 2-chloropropane concentration constant at 0.05 mol/L.
1. The reaction turns out to be proceeding via a SN1 mechanism. Plot a graph of rate vs.[iPrCl].
2. The reaction turns out to be proceeding via a SN2 mechanism. Plot a graph of rate vs.[iPrCl].
Why would the mechanism proceed in one way and not the other? Molecular choices between pathways like this are often described on the basis of "steric and electronic effects"; in other words, it's either something to do with charge or something to do with crowdedness. We will see soon how these effects can influence the course of the reaction, and how the mechanism can itself have consequences in the formation of different products.
Problem NS3.2.
How might crowdedness or steric effects influence the pathway taken by the reaction between cyanide and 2-chloropropane?
Problem NS3.3.
How might charge stability influence the pathway taken by the reaction between cyanide and 2-chloropropane? | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Nucleophilic_Substitution_at_Tetrahedral_Carbon/NS3._Rate_Laws_in_NS.txt |
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