chapter
stringlengths 1.97k
1.53M
| path
stringlengths 47
241
|
|---|---|
Learning Objective
• accurately and precisely use reaction mechanism notation and symbols including curved arrows to show the flow of electrons
The Arrow Notation in Mechanisms
Since chemical reactions involve the breaking and making of bonds, a consideration of the movement of bonding (and non-bonding) valence shell electrons is essential to this understanding. It is now common practice to show the movement of electrons with curved arrows, and a sequence of equations depicting the consequences of such electron shifts is termed a mechanism. In general, two kinds of curved arrows are used in drawing mechanisms:
A full head on the arrow indicates the movement or shift of an electron pair:
A partial head (fishhook) on the arrow indicates the shift of a single electron:
The use of these symbols in bond-breaking and bond-making reactions is illustrated below. If a covalent single bond is broken so that one electron of the shared pair remains with each fragment, as in the first example, this bond-breaking is called homolysis. If the bond breaks with both electrons of the shared pair remaining with one fragment, as in the second and third examples, this is called heterolysis.
Bond-Breaking Bond-Making
Other Arrow Symbols
Chemists also use arrow symbols for other purposes, and it is essential to use them correctly.
The Reaction Arrow
The Equilibrium Arrow
The Resonance Arrow
The following equations illustrate the proper use of these symbols:
Reactive Intermediates
The products of bond breaking, shown above, are not stable in the usual sense, and cannot be isolated for prolonged study. Such species are referred to as reactive intermediates, and are believed to be transient intermediates in many reactions. The general structures and names of four such intermediates are given below.
A pair of widely used terms, related to the Lewis acid-base notation, should also be introduced here.
• Electrophile: An electron deficient atom, ion or molecule that has an affinity for an electron pair, and will bond to a base or nucleophile.
• Nucleophile: An atom, ion or molecule that has an electron pair that may be donated in bonding to an electrophile (or Lewis acid).
Using these definitions, it is clear that carbocations ( called carbonium ions in the older literature ) are electrophiles and carbanions are nucleophiles. Carbenes have only a valence shell sextet of electrons and are therefore electron deficient. In this sense they are electrophiles, but the non-bonding electron pair also gives carbenes nucleophilic character. As a rule, the electrophilic character dominates carbene reactivity. Carbon radicals have only seven valence electrons, and may be considered electron deficient; however, they do not in general bond to nucleophilic electron pairs, so their chemistry exhibits unique differences from that of conventional electrophiles. Radical intermediates are often called free radicals.
The importance of electrophile / nucleophile terminology comes from the fact that many organic reactions involve at some stage the bonding of a nucleophile to an electrophile, a process that generally leads to a stable intermediate or product. Reactions of this kind are sometimes called ionic reactions, since ionic reactants or products are often involved. Some common examples of ionic reactions and their mechanisms may be examined below.
The shapes ideally assumed by these intermediates becomes important when considering the stereochemistry of reactions in which they play a role. A simple tetravalent compound like methane, CH4, has a tetrahedral configuration. Carbocations have only three bonds to the charge bearing carbon, so it adopts a planar trigonal configuration. Carbanions are pyramidal in shape ( tetrahedral if the electron pair is viewed as a substituent), but these species invert rapidly at room temperature, passing through a higher energy planar form in which the electron pair occupies a p-orbital. Radicals are intermediate in configuration, the energy difference between pyramidal and planar forms being very small. Since three points determine a plane, the shape of carbenes must be planar; however, the valence electron distribution varies.
Ionic Reactions
The principles and terms introduced in the previous sections can now be summarized and illustrated by the following three examples. Reactions such as these are called ionic or polar reactions, because they often involve charged species and the bonding together of electrophiles and nucleophiles. Ionic reactions normally take place in liquid solutions, where solvent molecules assist the formation of charged intermediates.
The substitution reaction shown on the left can be viewed as taking place in three steps. The first is an acid-base equilibrium, in which HCl protonates the oxygen atom of the alcohol. The resulting conjugate acid then loses water in a second step to give a carbocation intermediate. Finally, this electrophile combines with the chloride anion nucleophile to give the final product.
The addition reaction shown on the left can be viewed as taking place in two steps. The first step can again be considered an acid-base equilibrium, with the pi-electrons of the carbon-carbon double bond functioning as a base. The resulting conjugate acid is a carbocation, and this electrophile combines with the nucleophilic bromide anion.
The elimination reaction shown on the left takes place in one step. The bond breaking and making operations that take place in this step are described by the curved arrows. The initial stage may also be viewed as an acid-base interaction, with hydroxide ion serving as the base and a hydrogen atom component of the alkyl chloride as an acid.
There are many kinds of molecular rearrangements called isomerizations. The examples shown on the left are from an important class called tautomerization or, more specifically, keto-enol tautomerization. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom (colored red here) and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples.
Exercise
1. Add curved arrows to explain the indicated reactivity and classify the reaction as "homolytic cleavage" or "heterolytic cleavage".
2. Add the correct arrow to each expression below using your knowledge of chemistry.
3. Classify the following reactions as substituion, addition, elimination, or tautomerization (an example of isomerization).
Answer
1.
2.
3. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.02%3A_Reaction_Mec.txt |
Learning Objective
• identify nucleophiles and electrophiles in polar reactions
• relate bond polarity to chemical reactivity
Nucleophiles and Electrophiles
The reactants of polar reactions are often called the "nucleophile" and "electrophile". These terms are related to Lewis acid-base notation, so it can be helpful to apply and transfer the knowledge and wisdom gained from this definiation of acid-base chemistry.
• Electrophile (Lewis Acid): An electron deficient atom, ion or molecule that has an affinity for an electron pair, and will bond to a base or nucleophile
• Nucleophile (Lewis Base): An atom, ion or molecule that has an electron pair that may be donated in bonding to an electrophile
Nucleophile
Nucleophilic functional groups are those which have electron-rich atoms able to donate a pair of electrons to form a new covalent bond. In both laboratory and biological organic chemistry, the most relevant nucleophilic atoms are oxygen, nitrogen, and sulfur, and the most common nucleophilic functional groups are water, alcohols, phenols, amines, thiols, and occasionally carboxylates. More specifically in laboratory reactions, halide and azide (N3-) anions are commonly seen acting as nucleophiles.
Enolate ions are the most common carbon nucleophiles in biochemical reactions, while the cyanide ion (CN-) is just one example of a carbon nucleophile commonly used in the laboratory. Hydrocarbons carbons with pi bonds can also be nucleophiles. Reactions with carbon nucleophiles will be dealt with in later chapters In this chapter, we will concentrate on non-carbon nucleophiles.
When thinking about nucleophiles, the first thing to recognize is that, for the most part, the same quality of 'electron-richness' that makes a something nucleophilic also makes it basic: nucleophiles can be bases, and bases can be nucleophiles. It should not be surprising, then, that most of the trends in basicity that we have already discussed also apply to nucleophilicity.
Electrophiles
In the vast majority of polar reactions, the electrophilic atom is a carbon atom bonded to an electronegative atom, usually oxygen, nitrogen, sulfur, or a halogen. The concept of electrophilicity is relatively simple: an electron-poor atom with partial positive charge is an attractive target an electron-rich nucleophile. Electrophiles can be challenging to recognize because their partial positive charge is hidden in polar bonds and/or resonance. Allkyl halides and carbonyl groups are useful electrophiles for synthetic organic chemistry.
Electrophilicity of Alkyl Halides
With respect to electronegativity, halogens are more electronegative than carbons. This results in a carbon-halogen bond that is polarized. As shown in the image below, carbon atom has a partial positive charge, while the halogen has a partial negative charge.
Allkyl halides are useful electrophiles for synthetic organic chemistry. Of the four halogens, fluorine is the most electronegative and iodine the least. That means that the electron pair in the carbon-fluorine bond will be dragged most towards the halogen end. Looking at the methyl halides as simple examples:
The following image shows the relative electronegativity of the halogens. Notice, as we move up the periodic table from iodine to fluorine, electronegativity increases.
The following image shows the relationships between bond length, bond strength, and molecular size. As we progress down the periodic table from fluorine to iodine, molecular size increases. As a result, we also see an increase in bond length. Conversely, as molecular size increases and we get longer bonds, the strength of those bonds decreases.
You might have thought that either of these would be more effective in the case of the carbon-fluorine bond with the quite large amounts of positive and negative charge already present. But that's not so - quite the opposite is true! The thing that governs the reactivity is the strength of the bonds which have to be broken. If is difficult to break a carbon-fluorine bond, but easy to break a carbon-iodine one. The relative electrophilicity of alkyl halides is summarized below.
Electrophilicity of the Carbonyl Group
The carbon atom of the carbonyl group (C=O) is electrophilic because the carbon-oxygen double bond is polar and one of the resonance contributors is ionized with a full positiv echarge on the carbonyl carbon. Oxygen is more electronegative than carbon, so electron density is higher on the oxygen side of the bond and lower on the carbon side. Both of these factors combine to increase the electrophilicity of carbonyl groups. Carbonyl chemistry is studied in greater detail in the second semester of organic chemistry.
Exercise
1. Recognizing organic compounds as nucleophiles or electrophiles is an important first step in recognizing and learning patterns of chemical reactivity. Classify the following compounds as nucleophiles or electrophiles.
a) methoxide (CH3O-)
b) formaldehye (CH2O)
c) bromocyclopentane
d) water
e) sodium cyanide
f) methanamine (CH3NH2)
Answer
a) charged nucleophile
b) electrophile (Carbonyl carbon has partial positive charge.)
c) electrophile (Alkyl halides are always electrophiles - one reason they are an o-chem student's best friend.)
d) neutral nucleophile
e) charge nucleophile (Don't let the cation distract us from the CN-)
f) neutral nucleophile (The lone pair electrons on the nitrogen are nucleophilic in the same way they are Lewis bases (electron donators). | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.03%3A_Polar_Reacti.txt |
Learning Objective
• perform calculations using the equation $ΔGº = –RT \ln K = –2.303 RT \log_{10} K \label{eq2}$ and explain the relationship between equilibrium and free energy
Equilibrium Constant
For the hypothetical chemical reaction:
$aA + bB \rightleftharpoons cC + dD$
the equilibrium constant is defined as:
$K_C = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}$
where the notation [A] signifies the molar concentration of species A. Note that the expression for the equilibrium constant includes only solutes and gases; pure solids and liquids do not appear in the expression. For example, the equilibrium expression for the reaction
$\ce{CaH_2 (s) + 2H_2O (g) \rightleftharpoons Ca(OH)_2 (s) + 2H_2 (g)}$
is the following:
$K_C = \dfrac{[H_2]^2}{[H_2O]^2}$
Observe that the gas-phase species $\ce{H2O}$ and $\ce{H2}$ appear in the expression but the solids $\ce{CaH2}$ and $\ce{Ca(OH)2}$ do not appear.
The equilibrium constant is most readily determined by allowing a reaction to reach equilibrium, measuring the concentrations of the various solution-phase or gas-phase reactants and products, and substituting these values into the Law of Mass Action.
Free Energy
The interaction between enthalpy and entropy changes in chemical reactions is best observed by studying their influence on the equilibrium constants of reversible reactions. To this end a new thermodynamic function called Free Energy (or Gibbs Free Energy), symbol $ΔG$, is defined as shown in the first equation below. Two things should be apparent from this equation. First, in cases where the entropy change is small, $ΔG ≅ ΔH$. Second, the importance of $ΔS$ in determining $ΔG$ increases with increasing temperature.
$ΔGº = ΔHº – TΔSº$
where the temperature is measured in absolute temperature (K).
The free energy function provides improved insight into the thermodynamic driving forces that influence reactions. A negative $ΔGº$ is characteristic of an exergonic reaction, one which is thermodynamically favorable and often spontaneous, as is the melting of ice at 1 ºC. Likewise a positive $ΔGº$ is characteristic of an endergonic reaction, one which requires an input of energy from the surroundings.
For an example of the relationship of free energy to enthalpy consider the decomposition of cyclobutane to ethene, shown in the following equation. The standard state for all the compounds is gaseous.
This reaction is endothermic, but the increase in number of molecules from one (reactants) to two (products) results in a large positive ΔSº.
At 25 ºC (298 K):
ΔGº = 19 kcal/mol – 298(43.6) cal/mole = 19 – 13 kcal/mole = +6 kcal/mole.
Thus, the entropy change opposes the enthalpy change, but is not sufficient to change the sign of the resulting free energy change, which is endergonic. Indeed, cyclobutane is perfectly stable when kept at room temperature.
Because the entropy contribution increases with temperature, this energetically unfavorable transformation can be made favorable by raising the temperature. At 200 ºC (473 K),
\begin{align} ΔGº &= 19\, kcal/mol – 473(43.6)\, cal/mole \[4pt] &= 19 – 20.6\, kcal/mole \[4pt] &= –1.6 kcal/mole.\end{align}
This is now an exergonic reaction, and the thermal cracking of cyclobutane to ethene is known to occur at higher temperatures.
$ΔGº = –RT \ln K = –2.303 RT \log_{10} K \label{eq2}$
where R = 1.987 cal/ K mole T = temperature in K and K = equilibrium constant
Note
Equation \ref{eq2} is important because it demonstrates the fundamental relationship of $ΔGº$ to the equilibrium constant, $K$. Because of the negative logarithmic relationship between these variables, a negative ΔGº generates a K>1, whereas a positive ΔGº generates a K<1. When ΔGº = 0, K = 1. Furthermore, small changes in ΔGº produce large changes in K. A change of 1.4 kcal/mole in ΔGº changes K by approximately a factor of 10. This interrelationship may be explored with the calculator on the right. Entering free energies outside the range -8 to 8 kcal/mole or equilibrium constants outside the range 10-6 to 900,000 will trigger an alert, indicating the large imbalance such numbers imply.
Applications to Organic Reactions
The equation below can also be useful without performing any calculations.
$ΔGº = –RT \ln K = –2.303 RT \log_{10} K \label{eq2}$
Conceptually, this equation helps us compare the energetics of reaction mechanisms to predict the major products. For example, if ΔG° < 0, then K > 1, and products are favored over reactants. If ΔG° > 0, then K < 1, and reactants are favored over products. If ΔG° = 0, then K = 1, and the system is at equilibrium. Recognizing the underlying energetics of equilibrium, the stability of charged reactants and products can be used to predict reaction equilibrium. Reaction conditions can also be adjusted or controlled to shift the equilibrium in the desired direction by a range of experimental methods. A theoretical understanding of the reaction free energy and equilibrium helps us predict and design the optimum reaction conditions for a desired product.
Exercises
1. At 155°C, the equilibrium constant, Keq, for the reaction
$\ce{\sf{CH3CO2H + CH3CH2OH <=> CH3CO2CH2CH3 + H2O}}$
has a value of 4.0. Calculate ΔG° for this reaction at 155°C.
2. Acetylene (C2H2) can be converted into benzene (C6H6) according to the equation:
$\ce{\sf{3H-C#C-H (g) <=> C6H6 (l)}}$
At 25°C, ΔG° for this reaction is −503 kJ and ΔH° is −631 kJ. Determine ΔS° and indicate whether the size of ΔS° agrees with what you would have predicted simply by looking at the chemical equation.
Answer
Δ G ° = − R T ln K eq = − ( 8.314 J ⋅ K − 1 ⋅ mol − 1 ) ( 428 K ) ln ( 4.0 ) = − ( 8.314 J ⋅ K − 1 ⋅ mol − 1 ) ( 428 K ) ( 1.386 ) = − 4.9 × 10 3 J ⋅ mol − 1 = − 4.9 KJ ⋅ mol − 1 Δ G ° = Δ H ° − T Δ S ° Δ S ° = ( Δ H ° − Δ G ° ) T = − 631 kJ − ( − 503 kJ ) 298 K = − 128 kJ 298 K = − 0.430 KJ ⋅ mol − 1 = − 430 J ⋅ mol − 1
The entropy change is negative, as one would expect from looking at the chemical equation, since three moles of reactants yield one mole of product; that is, the system becomes much more “ordered” as it goes from reactants to products. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.04%3A_Describing_a.txt |
Learning Objective
• calculate reaction enthalpies from bond dissociation energies
Introduction
The homolytic bond dissociation energy is the amount of energy needed to break apart one mole of covalently bonded gases into a pair of radicals. The SI units used to describe bond energy are kiloJoules per mole of bonds (kJ/Mol). It indicates how strongly the atoms are bonded to each other. Solvation is the interaction between solvent molecules and the ions or molecules dissolved in that solvent.
Breaking a covalent bond between two partners, A-B, can occur either heterolytically, where the shared pair of electron goes with one partner or another
$A-B \rightarrow A^+ + B:^-$
or
$A-B \rightarrow A:^- + B^+ $
or homolytically, where one electron stays with each partner.
$A-B \rightarrow A^• + B^•$
The products of homolytic cleavage are radicals and the energy that is required to break the bond homolytically is called the Bond Dissociation Energy (BDE) and is a measure of the strength of the bond.
Calculation of the BDE
The BDE for a molecule A-B is calculated as the difference in the enthalpies of formation of the products and reactants for homolysis
$BDE = \Delta_fH(A^•) + \Delta_fH(B^•) - \Delta_fH(A-B)$
Officially, the IUPAC definition of bond dissociation energy refers to the energy change that occurs at 0 K, and the symbol is $D_o$. However, it is commonly referred to as BDE, the bond dissociation energy, and it is generally used, albeit imprecisely, interchangeably with the bond dissociation enthalpy, which generally refers to the enthalpy change at room temperature (298K). Although there are technically differences between BDEs at 0 K and 298 K, those difference are not large and generally do not affect interpretations of chemical processes.
Bond Breakage/Formation
Bond dissociation energy (or enthalpy) is a state function and consequently does not depend on the path by which it occurs. Therefore, the specific mechanism in how a bond breaks or is formed does not affect the BDE. Bond dissociation energies are useful in assessing the energetics of chemical processes. For chemical reactions, combining bond dissociation energies for bonds formed and bonds broken in a chemical reaction using Hess's Law can be used to estimate reaction enthalpies.
Example: Chlorination of Methane
Consider the chlorination of methane
$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$
the overall reaction thermochemistry can be calculated exactly by combining the BDEs for the bonds broken and bonds formed
CH4 → CH3• + H• BDE(CH3-H)
Cl2 → 2Cl• BDE(Cl2)\]
H• + Cl• → HCl -BDE(HCl)
CH3• + Cl• → CH3Cl -BDE(CH3-Cl)
---------------------------------------------------
$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$
$\Delta H = BDE(R-H) + BDE(Cl_2) - BDE(HCl) - BDE(CH_3-Cl)$
Because reaction enthalpy is a state function, it does not matter what reactions are combined to make up the overall process using Hess's Law. However, BDEs are convenient to use because they are readily available.
Alternatively, BDEs can be used to assess individual steps of a mechanism. For example, an important step in free radical chlorination of alkanes is the abstraction of hydrogen from the alkane to form a free radical.
RH + Cl• → R• + HCl
The energy change for this step is equal to the difference in the BDEs in RH and HCl
$\Delta H = BDE(R-H) - BDE(HCl)$
This relationship shows that the hydrogen abstraction step is more favorable when BDE(R-H) is smaller. The difference in energies accounts for the selectivity in the halogenation of hydrocarbons with different types of C-H bonds.
Table 6.8.1: Representative C-H BDEs in Organic Molecules
R-H Do, kJ/mol D298, kJ/mol R-H Do, kJ/mol D298, kJ/mol
CH3-H 432.7±0.1 439.3±0.4 H2C=CH-H 456.7±2.7 463.2±2.9
CH3CH2-H 423.0±1.7 C6H5-H 465.8±1.9 472.4±2.5
(CH3)2CH-H 412.5±1.7 HCCH 551.2±0.1 557.8±0.3
(CH3)3C-H 403.8±1.7
H2C=CHCH2-H 371.5±1.7
HC(O)-H 368.6±0.8 C6H5CH2-H 375.3±2.5
CH3C(O)-H 374.0±1.2
It is important to remember that C-H BDEs refer to the energy it takes to break the bond, and is the difference in energy between the reactants and the products. Therefore, it is not appropriate to interpret BDEs solely in terms of the "stability of the radical products" as is often done.
Analysis of the BDEs shown in the table above shows that there are some systematic trends:
1. BDEs vary with hybridization: Bonds with sp3 hybridized carbons are weakest and bonds with sp hybridized carbons are much stronger. The vinyl and phenyl C-H bonds are similar, reflecting their sp2 hybridization. The correlation with hybridization can be viewed as a reflection of the C-H bond lengths. Longer bonds formed with sp3 orbitals are consequently weaker. Shorter bonds formed with orbitals that have more s-character are similarly stronger.
2. C-H BDEs vary with substitution: Among sp3 hybridized systems, methane has the strongest C-H bond. C-H bonds on primary carbons are stronger than those on secondary carbons, which are stronger than those on tertiary carbons.
Interpretation of C-H BDEs for sp3 Hybridized Carbons
The interpretation of the BDEs in saturated molecules has been subject of recent controversy. As indicated above, the variation in BDEs with substitution has traditionally been interpreted as reflecting the stabilities of the alkyl radicals, with the assessment that more highly substituted radicals are more stable, as with carbocations. Although this is a popular explanation, it fails to account fo the fact the bonds to groups other than H do not show the same types of variation.
R BDE(R-CH3) BDE(R-Cl) BDE(R-Br) BDE(R-OH)
CH3- 377.0±0.4 350.2±0.4 301.7±1.3 385.3±0.4
CH3CH2-
372.4±1.7
354.8±2.1 302.9±2.5 393.3±1.7
(CH3)2CH- 370.7±1.7 356.5±2.1 309.2±2.9 399.6±1.7
(CH3)3C- 366.1±1.7 355.2±2.9 303.8±2.5 400.8±1.7
Therefore, although C-CH3 bonds get weaker with more substitution, the effect is not nearly as large as that observed with C-H bonds. The strengths of C-Cl and C-Br bonds are not affected by substitution, despite the fact that the same radicals are formed as when breaking C-H bonds, and the C-OH bonds in alcohols actually increase with more substitution.
Gronert has proposed that the variation in BDEs is alternately explained as resulting from destabilization of the reactants due to steric repulsion of the substituents, which is released in the nearly planar radicals.1 Considering that BDEs reflect the relative energies of reactants and products, either explanation can account for the trend in BDEs.
Another factor that needs to be considered is the electronegativity. The Pauling definition of electronegativity says that the bond dissociation energy between unequal partners is going to be dependent on the difference in electrongativities, according to the expression
$D_o(A-B) = \dfrac{D_o(A-A) + D_o(B-B)}{2} + (X_A - X_B)^2$
where $X_A$ and $X_B$ are the electronegativities and the bond energies are in eV. Therefore, the variation in BDEs can be interpreted as reflecting variation in the electronegativities of the different types of alkyl fragments.
There is likely some merit in all three interpretations. Since Gronert's original publication of his alternate explanation, there have been many desperate attempts to defend the radical stability explanation.
Exercise
1. Given that ΔH° for the reaction
CH4(g) + 4F2(g) -> CF4(g) + 4HF(g)
is −1936 kJ, use the following data to calculate the average bond energy of the $\ce{{\sf{C-F}} bonds in CF4. Bond Average Bond Energy \(\ce{\sf{C-H}}$ 413 kJ · mol−1
$\ce{\sf{F-F}}$ 155 kJ · mol−1
$\ce{\sf{H-F}}$ 567 kJ · mol−1
2. Calculate ΔH° for the reactions given below.
1. CH3CH2OCH3 + HI -> CH3CH2OH + CH3I
2. CH3Cl + NH3 -> CH3NH2 + HCl
Answer
$1.$CH4(g) + 4 F2(g) CF4(g) + 4HF(g) ΔH°=416kcal
Bonds broken:
4 mol C-H bonds $×\text{}\frac{\left(413\text{\hspace{0.17em}}\text{kJ}\right)}{\left(1\text{\hspace{0.17em}}\text{mol}\right)}=1652\text{\hspace{0.17em}}\text{kJ}$
4 mol F-F bonds $×\text{}\frac{\left(155\text{\hspace{0.17em}}\text{kJ}\right)}{\left(1\text{\hspace{0.17em}}\text{mol}\right)}=620\text{\hspace{0.17em}}\text{kJ}$
Bonds formed:
4 mol C-F bonds $×\text{}\frac{\left(x\text{\hspace{0.17em}}\text{kJ}\right)}{\left(1\text{\hspace{0.17em}}\text{mol}\right)}=4x\text{\hspace{0.17em}}\text{kJ}$
(where x = the average energy of one mole of C-F bonds in CF4, expressed in kJ)
4 mol H-F bonds $×\text{}\frac{\left(567\text{\hspace{0.17em}}\text{kJ}\right)}{\left(1\text{\hspace{0.17em}}\text{mol}\right)}=2268\text{\hspace{0.17em}}\text{kJ}$
$\begin{array}{rl}\Delta H°& =\Delta H°\left(\text{bonds broken}\right)-\Delta H°\left(\text{bonds formed}\right)\ & =\left(1652\text{\hspace{0.17em}}\text{kJ}+620\text{\hspace{0.17em}}\text{kJ}\right)-\left(4x+2268\text{\hspace{0.17em}}\text{kJ}\right)\ & =1652\text{\hspace{0.17em}}\text{kJ}+620\text{\hspace{0.17em}}\text{kJ}-4x-2268\text{\hspace{0.17em}}\text{kJ}\ & =-1936\text{\hspace{0.17em}}\text{kJ}\end{array}$
Thus,
$\begin{array}{rl}4x& =1936\text{\hspace{0.17em}}\text{kJ}-2268\text{\hspace{0.17em}}\text{kJ}+620\text{\hspace{0.17em}}\text{kJ}+1652\text{\hspace{0.17em}}\text{kJ}\ & =1940\text{\hspace{0.17em}}\text{kJ}\end{array}$
and
$\begin{array}{rl}x& =\frac{1940\text{\hspace{0.17em}}\text{kJ}}{4\text{\hspace{0.17em}}\text{mol}}\ & =385\text{\hspace{0.17em}}\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$
The average energy of a C-F bond in CF4 is 385 kJ · mol-1
2. CH3Cl + NH3 -> CH3NH2 + HCl
$\dfrac { \begin{array}{cc} \textrm{Reactant bonds broken} & D \[6pt] \ce{CH3-Cl} & {351 \,\, \textrm{kJ/mol}} \ \ce{NH2-H} & {449 \,\, \textrm{kJ/mol}} \ \end{array} } { \begin{array}{cc} {\phantom{\textrm{Reactant bonds broken}}} & {800 \,\, \textrm{kJ/mol}} \end{array} } \quad \dfrac { \begin{array}{cc} \textrm{Product bonds formed} & D \[6pt] \ce{CH3-NH2} & {335 \,\, \textrm{kJ/mol}} \ \ce{H-Cl} & {432 \,\, \textrm{kJ/mol}} \ \end{array} } { \begin{array}{cc} {\phantom{\textrm{Product bonds formed}}} & {767 \,\, \textrm{kJ/mol}} \end{array} }$
\begin{align*} \Delta H^{\circ} & = D_{\textrm{bonds broken}} + D_{\textrm{bonds formed}}\ & = {800 \,\, \textrm{kJ/mol} - 767 \,\, \textrm{kJ/mol}} \ & = {+33 \,\, \textrm{kJ/mol}} \end{align*}
Further Reading
MasterOrganicChemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.05%3A_Homolytic_Cl.txt |
Learning Objective
• draw Reaction Energy Diagrams from the thermodynamic and kinetic data/information
• use a Reaction Energy Diagram to discuss transition states, Ea, intermediates & rate determining step
• draw the transition state of a reaction
You may recall from general chemistry that it is often convenient to describe chemical reactions with energy diagrams. In an energy diagram, the vertical axis represents the overall energy of the reactants, while the horizontal axis is the ‘reaction coordinate’, tracing from left to right the progress of the reaction from starting compounds to final products. The energy diagram for a typical one-step reaction might look like this:
Despite its apparent simplicity, this energy diagram conveys some very important ideas about the thermodynamics and kinetics of the reaction. Recall that when we talk about the thermodynamics of a reaction, we are concerned with the difference in energy between reactants and products, and whether a reaction is ‘downhill’ (exergonic, energy releasing) or ‘uphill (endergonic, energy absorbing). When we talk about kinetics, on the other hand, we are concerned with the rate of the reaction, regardless of whether it is uphill or downhill thermodynamically.
First, let’s review what this energy diagram tells us about the thermodynamics of the reaction illustrated by the energy diagram above. The energy level of the products is lower than that of the reactants. This tells us that the change in standard Gibbs Free Energy for the reaction (Δrnx) is negative. In other words, the reaction is exergonic, or ‘downhill’. Recall that the Δrnx term encapsulates both Δrnx, the change in enthalpy (heat) and Δrnx , the change in entropy (disorder):
$ΔG˚ = ΔH˚- TΔS˚$
where T is the absolute temperature in Kelvin. For chemical processes where the entropy change is small (~0), the enthalpy change is essentially the same as the change in Gibbs Free Energy. Energy diagrams for these processes will often plot the enthalpy (H) instead of Free Energy for simplicity.
The standard Gibbs Free Energy change for a reaction can be related to the reaction's equilibrium constant ($K_{eq}\_) by a simple equation: $ΔG˚ = -RT \ln K_{eq}$ where: • Keq = [product] / [reactant] at equilibrium • R = 8.314 J×K-1×mol-1 or 1.987 cal× K-1×mol-1 • T = temperature in Kelvin (K) If you do the math, you see that a negative value for Δrnx (an exergonic reaction) corresponds - as it should by intuition - to Keq being greater than 1, an equilibrium constant which favors product formation. In a hypothetical endergonic (energy-absorbing) reaction the products would have a higher energy than reactants and thus Δrnx would be positive and Keq would be less than 1, favoring reactants. Now, let's move to kinetics. Look again at the energy diagram for exergonic reaction: although it is ‘downhill’ overall, it isn’t a straight downhill run. First, an ‘energy barrier’ must be overcome to get to the product side. The height of this energy barrier, you may recall, is called the ‘activation energy’ (ΔG). You may have been taught to use the term “activated complex” rather than “transition state,” as the two are often used interchangeably. Similarly, the activation energy of a reaction is often represented by the symbol Eact or Ea. The activation energy is what determines the kinetics of a reaction: the higher the energy hill, the slower the reaction. At the very top of the energy barrier, the reaction is at its transition state (TS), which is the point at which the bonds are in the process of breaking and forming. The transition state is an ‘activated complex’: a transient and dynamic state that, unlike more stable species, does not have any definable lifetime. It may help to imagine a transition state as being analogous to the exact moment that a baseball is struck by a bat. Transition states are drawn with dotted lines representing bonds that are in the process of breaking or forming, and the drawing is often enclosed by brackets. Here is a picture of a likely transition state for a substitution reaction between hydroxide and chloromethane: $CH_3Cl + HO^- \rightarrow CH_3OH + Cl^-$ This reaction involves a collision between two molecules: for this reason, we say that it has second order kinetics. The rate expression for this type of reaction is: rate = k[reactant 1][reactant 2] . . . which tells us that the rate of the reaction depends on the rate constant k as well as on the concentration of both reactants. The rate constant can be determined experimentally by measuring the rate of the reaction with different starting reactant concentrations. The rate constant depends on the activation energy, of course, but also on temperature: a higher temperature means a higher k and a faster reaction, all else being equal. This should make intuitive sense: when there is more heat energy in the system, more of the reactant molecules are able to get over the energy barrier. Here is one more interesting and useful expression. Consider a simple reaction where the reactants are A and B, and the product is AB (this is referred to as a condensation reaction, because two molecules are coming together, or condensing). If we know the rate constant k for the forward reaction and the rate constant kreverse for the reverse reaction (where AB splits apart into A and B), we can simply take the quotient to find our equilibrium constant \(K_{eq}$:
This too should make some intuitive sense; if the forward rate constant is higher than the reverse rate constant, equilibrium should lie towards products.
Exercise
1. Which reaction is faster, ΔG = + 55 kJ/mol or ΔG = + 75 kJ/mol?
Answer
1. The + 55 kJ/mol reaction is the faster reaction. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.06%3A_Reaction_Ene.txt |
Learning Objective
• describe the structure & relative stabilities of carbocations
Carbocations and their Stability
A carbocation is an ion with a positively-charged carbon atom. A carbocation is very electron-poor, and thus anything which donates electron density to the center of electron poverty will help to stabilize it. Conversely, a carbocation will be destabilized by an electron withdrawing group.
Alkyl groups are electron donating and carbocation-stabilizing because the electrons around the neighboring carbons are drawn towards the nearby positive charge, thus slightly reducing the electron poverty of the positively-charged carbon. What this means is that, in general, more substituted carbocations are more stable: a tert-butyl carbocation, for example, is more stable than an isopropyl carbocation. Primary carbocations are highly unstable and not often observed as reaction intermediates; methyl carbocations are even less stable.
It is not accurate to say, however, that carbocations with higher substitution are always more stable than those with less substitution. Just as electron-donating groups can stabilize a carbocation, electron-withdrawing groups act to destabilize carbocations. Carbonyl groups are electron-withdrawing by inductive effects, due to the polarity of the C=O double bond. It is possible to demonstrate in the laboratory (see section 16.1D) that carbocation A below is more stable than carbocation B, even though A is a primary carbocation and B is secondary.
The difference in stability can be explained by considering the electron-withdrawing inductive effect of the ester carbonyl. Recall that inductive effects - whether electron-withdrawing or donating - are relayed through covalent bonds and that the strength of the effect decreases rapidly as the number of intermediary bonds increases. In other words, the effect decreases with distance. In species B the positive charge is closer to the carbonyl group, thus the destabilizing electron-withdrawing effect is stronger than it is in species A.
Stabilization of a carbocation can also occur through resonance effects, and as we have already discussed in the acid-base chapter, resonance effects as a rule are more powerful than inductive effects. Consider the simple case of a benzylic carbocation:
This carbocation is comparatively stable. In this case, electron donation is a resonance effect. Three additional resonance structures can be drawn for this carbocation in which the positive charge is located on one of three aromatic carbons. The positive charge is not isolated on the benzylic carbon, rather it is delocalized around the aromatic structure: this delocalization of charge results in significant stabilization. As a result, benzylic and allylic carbocations (where the positively charged carbon is conjugated to one or more non-aromatic double bonds) are significantly more stable than even tertiary alkyl carbocations.
Because heteroatoms such as oxygen and nitrogen are more electronegative than carbon, you might expect that they would by definition be electron withdrawing groups that destabilize carbocations. In fact, the opposite is often true: if the oxygen or nitrogen atom is in the correct position, the overall effect is carbocation stabilization. This is due to the fact that although these heteroatoms are electron withdrawing groups by induction, they are electron donating groups by resonance, and it is this resonance effect which is more powerful. (We previously encountered this same idea when considering the relative acidity and basicity of phenols and aromatic amines in section 7.4). Consider the two pairs of carbocation species below:
In the more stable carbocations, the heteroatom acts as an electron donating group by resonance: in effect, the lone pair on the heteroatom is available to delocalize the positive charge. In the less stable carbocations the positively-charged carbon is more than one bond away from the heteroatom, and thus no resonance effects are possible. In fact, in these carbocation species the heteroatoms actually destabilize the positive charge, because they are electron withdrawing by induction.
Finally, vinylic carbocations, in which the positive charge resides on a double-bonded carbon, are very unstable and thus unlikely to form as intermediates in any reaction.
Carbocation Rearrangements
Carbocations typically undergo rearrangement reactions from less stable structures to equally stable or more stable ones with rate constants in excess of 109/sec. This fact complicates synthetic pathways to many compounds, so it is important to look for carbocation rearrangements anytime they are formed. It is possible for either a neighboring hydrogen atom or methyl group to shift to the carbocation to create a more stable intermediate. In the 1,2-hydride shift shown below, the secondary carbocatin rearranges to a more stable tertiary carbocation. The numbers, 1,2- refer to the vicinal location of the rearrangement, not the nomenclature numbers.
In this next example, the methyl group shifts to stabilize the carbocation.
Exercise
1. In which of the structures below is the carbocation expected to be more stable? Explain.
2. Draw a resonance structure of the crystal violet cation in which the positive charge is delocalized to one of the nitrogen atoms.
State which carbocation in each pair below is more stable, or if they are expected to be approximately equal. Explain your reasoning.
Answer
1. In the carbocation on the left, the positive charge is located in a position relative to the nitrogen such that the lone pair of electrons on the nitrogen can be donated to fill the empty orbital. This is not possible for the carbocation species on the right.
2.
3.
a) 1 (tertiary vs. secondary carbocation)
b) equal
c) 1 (tertiary vs. secondary carbocation)
d) 2 (positive charge is further from electron-withdrawing fluorine)
e) 1 (lone pair on nitrogen can donate electrons by resonance)
f) 1 (allylic carbocation – positive charge can be delocalized to a second carbon)
History
The history of carbocations dates back to 1891 when G. Merling[8] reported that he added bromine to tropylidene (cycloheptatriene) and then heated the product to obtain a crystalline, water-soluble material, \(C_7H_7Br\). He did not suggest a structure for it; however, Doering and Knox[9] convincingly showed that it was tropylium (cycloheptatrienylium) bromide. This ion is predicted to be aromatic by Hückel's rule.
In 1902, Norris and Kehrman independently discovered that colorless triphenylmethanol gives deep-yellow solutions in concentrated sulfuric acid. Triphenylmethyl chloride similarly formed orange complexes with aluminium and tin chlorides. In 1902, Adolf von Baeyer recognized the salt-like character of the compounds formed.
He dubbed the relationship between color and salt formation halochromy, of which malachite green is a prime example.
Carbocations are reactive intermediates in many organic reactions. This idea, first proposed by Julius Stieglitz in 1899,[10] was further developed by Hans Meerwein in his 1922 study[11][12] of the Wagner-Meerwein rearrangement. Carbocations were also found to be involved in the \(S_N1\) reaction, the \(E1\0 reaction, and in rearrangement reactions such as the Whitmore 1,2 shift. The chemical establishment was reluctant to accept the notion of a carbocation and for a long time the Journal of the American Chemical Society refused articles that mentioned them.
The first NMR spectrum of a stable carbocation in solution was published by Doering et al.[13] in 1958. It was the heptamethylbenzenium ion, made by treating hexamethylbenzene with methyl chloride and aluminium chloride. The stable 7-norbornadienyl cation was prepared by Story et al. in 1960[14] by reacting norbornadienyl chloride with silver tetrafluoroborate in sulfur dioxide at −80 °C. The NMR spectrum established that it was non-classically bridged (the first stable non-classical ion observed).
In 1962, Olah directly observed the tert-butyl carbocation by nuclear magnetic resonance as a stable species on dissolving tert-butyl fluoride in magic acid. The NMR of the norbornyl cation was first reported by Schleyer et al.[15] and it was shown to undergo proton-scrambling over a barrier by Saunders et al.[16]
References
1. Hansjörg Grützmacher, Christina M. Marchand (1997), "Heteroatom stabilized carbenium ions", Coordination Chemistry Reviews, volume 163, pages 287-344. doi:10.1016/S0010-8545(97)00043-X
2. Gold Book definition carbonium ion HTML
3. George A. Olah (1972), "Stable carbocations. CXVIII. General concept and structure of carbocations based on differentiation of trivalent (classical) carbenium ions from three-center bound penta- of tetracoordinated (nonclassical) carbonium ions. Role of carbocations in electrophilic reactions." Journal of the American Chemical Society, volume 94, issue 3, pages 808–820 doi:10.1021/ja00758a020
4. Organic chemistry 5th Ed. John McMurry ISBN 0-534-37617-7
5. Organic Chemistry, Fourth Edition Paula Yurkanis Bruice ISBN 0-13-140748-1
6. Clayden, Jonathan; Greeves, Nick; Warren, Stuart; Wothers, Peter (2001). Organic Chemistry (1st ed.). Oxford University Press. ISBN 978-0-19-850346-0.
7. Organic Chemistry by Marye Anne Fox and James K. Whitesell ISBN 0-7637-0413-X
8. Chem. Ber. 24, 3108 1891
9. The Cycloheptatrienylium (Tropylium) Ion W. Von E. Doering and L. H. Knox J. Am. Chem. Soc.; 1954; 76(12) pp 3203 - 3206; doi:10.1021/ja01641a027
10. On the Constitution of the Salts of Imido-Ethers and other Carbimide Derivatives; Am. Chem. J. 21, 101; ISSN: 0096-4085
11. H. Meerwein and K. van Emster, Berichte, 1922, 55, 2500.
12. Rzepa, H. S.; Allan, C. S. M. (2010). "Racemization of Isobornyl Chloride via Carbocations: A Nonclassical Look at a Classic Mechanism". Journal of Chemical Education 87 (2): 221. Bibcode:2010JChEd..87..221R. doi:10.1021/ed800058c. edit
13. The 1,1,2,3,4,5,6-heptamethylbenzenonium ion W. von E. Doering and M. Saunders H. G. Boyton, H. W. Earhart, E. F. Wadley and W. R. Edwards G. Laber Tetrahedron Volume 4, Issues 1-2 , 1958, Pages 178-185 doi:10.1016/0040-4020(58)88016-3
14. The 7-norbornadienyl carbonium ion Paul R. Story and Martin Saunders J. Am. Chem. Soc.; 1960; 82(23) pp 6199 - 6199; doi:10.1021/ja01508a058
15. Stable Carbonium Ions. X.1 Direct Nuclear Magnetic Resonance Observation of the 2-Norbornyl Cation Paul von R. Schleyer, William E. Watts, Raymond C. Fort, Melvin B. Comisarow, and George A. Olah J. Am. Chem. Soc.; 1964; 86(24) pp 5679 - 5680; doi:10.1021/ja01078a056
16. Stable Carbonium Ions. XI.1 The Rate of Hydride Shifts in the 2-Norbornyl Cation Martin Saunders, Paul von R. Schleyer, and George A. Olah J. Am. Chem. Soc.; 1964; 86(24) pp 5680 - 5681; doi:10.1021/ja01078a057
17. George A. Olah - Nobel Lecture
18. Nuclear magnetic double resonance studies of the dimethylcyclopropylcarbinyl cation. Measurement of the rotation barrier David S. Kabakoff, , Eli. Namanworth J. Am. Chem. Soc. 1970, 92 (10), pp 3234–3235 doi:10.1021/ja00713a080
19. Stable Carbonium Ions. XVII.1a Cyclopropyl Carbonium Ions and Protonated Cyclopropyl Ketones Charles U. Pittman Jr., George A. Olah J. Am. Chem. Soc., 1965, 87 (22), pp 5123–5132 doi:10.1021/ja00950a026
20. F.A. Carey, R.J. Sundberg Advanced Organic Chemistry Part A 2nd Ed. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.07%3A_Reactive_Int.txt |
Learning Objective
• describe the structure & relative stabilities of free radicals
Radicals
In chemistry, a radical (more precisely, a free radical) is an atom, molecule, or ion that has unpaired valence electrons or an open electron shell, and therefore may be seen as having one or more "dangling" covalent bonds.
With some exceptions, these "dangling" bonds make free radicals highly chemically reactive towards other substances, or even towards themselves: their molecules will often spontaneously dimerize or polymerize if they come in contact with each other. Most radicals are reasonably stable only at very low concentrations in inert media or in a vacuum.
A notable example of a free radical is the hydroxyl radical (HO•), a molecule that is one hydrogen atom short of a water molecule and thus has one bond "dangling" from the oxygen. Two other examples are the carbene molecule (:CH2), which has two dangling bonds; and the superoxide anion (•O−2), the oxygen molecule O2 with one extra electron, which has one dangling bond. In contrast, the hydroxyl anion (HO−), the oxide anion (O2−) and thecarbenium cation (CH+3) are not radicals, since the bonds that may appear to be dangling are in fact resolved by the addition or removal of electrons.
Free radicals may be created in a number of ways, including synthesis with very dilute or rarefied reagents, reactions at very low temperatures, or breakup of larger molecules. The latter can be affected by any process that puts enough energy into the parent molecule, such as ionizing radiation, heat, electrical discharges, electrolysis, and chemical reactions. Indeed, radicals are intermediate stages in many chemical reactions.
Free radicals play an important role in combustion, atmospheric chemistry, polymerization, plasma chemistry, biochemistry, and many other chemical processes. In living organisms, the free radicals superoxide and nitric oxideand their reaction products regulate many processes, such as control of vascular tone and thus blood pressure. They also play a key role in the intermediary metabolism of various biological compounds. Such radicals can even be messengers in a process dubbed redox signaling. A radical may be trapped within a solvent cage or be otherwise bound.
Until late in the 20th century the word "radical" was used in chemistry to indicate any connected group of atoms, such as a methyl group or a carboxyl, whether it was part of a larger molecule or a molecule on its own. The qualifier "free" was then needed to specify the unbound case. Following recent nomenclature revisions, a part of a larger molecule is now called a functional group or substituent, and "radical" now implies "free". However, the old nomenclature may still occur in the literature.
Formation
The formation of radicals may involve breaking of covalent bonds homolytically, a process that requires significant amounts of energy. For example, splitting H2 into 2H· has a ΔH° of +435 kJ/mol, and Cl2 into 2Cl· has a ΔH° of +243 kJ/mol. This is known as the homolytic bond dissociation energy, and is usually abbreviated as the symbol ΔH°. The bond energy between two covalently bonded atoms is affected by the structure of the molecule as a whole, not just the identity of the two atoms. Likewise, radicals requiring more energy to form are less stable than those requiring less energy. Homolytic bond cleavage most often happens between two atoms of similar electronegativity. In organic chemistry this is often the O-O bond in peroxide species or O-N bonds. Sometimes radical formation is spin-forbidden, presenting an additional barrier. However, propagation is a very exothermic reaction. Likewise, although radical ions do exist, most species are electrically neutral. Radicals may also be formed by single electron oxidation or reduction of an atom or molecule. An example is the production of superoxide by the electron transport chain. Early studies of organometallic chemistry, especially tetra-alkyl lead species by F.A. Paneth and K. Hahnfeld in the 1930s supported heterolytic fission of bonds and a radical based mechanism.
Depiction in chemical reactions
In chemical equations, free radicals are frequently denoted by a dot placed immediately to the right of the atomic symbol or molecular formula as follows:
$\mathrm{Cl}_2 \; \xrightarrow{UV} \; {\mathrm{Cl} \cdot} + {\mathrm{Cl} \cdot}$
Chlorine gas can be broken down by ultraviolet light to form atomic chlorine radicals.
Radical reaction mechanisms use single-headed arrows to depict the movement of single electrons:
The homolytic cleavage of the breaking bond is drawn with a 'fish-hook' arrow to distinguish from the usual movement of two electrons depicted by a standard curly arrow. It should be noted that the second electron of the breaking bond also moves to pair up with the attacking radical electron; this is not explicitly indicated in this case.
Relative Stability
Radical alkyl intermediates are stabilized by similar physical processes to carbocations: as a general rule, the more substituted the radical center is, the more stable it is. This directs their reactions. Thus, formation of a tertiary radical (R3C·) is favored over secondary (R2HC·), which is favored over primary (RH2C·). Likewise, radicals next to functional groups such as carbonyl, nitrile, and ether are more stable than tertiary alkyl radicals.
Exercise
1. State which carbon radical (free radical) in each pair below is more stable or if they are expected to have comparable stability. Explain your reasoning.
Answer
1.
a) Cpd 1: Tertiary radicals are more stable than secondary radicals.
b) Cpds 1 and 2 are both secondary so they have comparable stability.
c) Cpd 1: Tertiary radicals are more stable than secondary radicals with similar effects from the Cl atom.
d) Cpd 2: Both compounds are secondary, but positive charge is further from electron-withdrawing chlorine on Cpd 2.
e) Cpd 1: Lone pair on nitrogen can donate electrons by resonance.
f) Cpd 1: Secondary allylic radicals are more stable than tertiary radicals. (Primary allylic radicals are comparable in stability to tertiary radicals.) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.08%3A_Reactive_Int.txt |
Carbanions
A carbanion is an anion in which carbon has an unshared pair of electrons and bears a negative charge usually with three substituents for a total of eight valence electrons.[1] The carbanion exists in a trigonal pyramidal geometry. Formally, a carbanion is the conjugate base of a carbon acid.
$\ce{R_3C-H + B^- \rightarrow R_3C^- + H-B}$
where B stands for the base. A carbanion is one of several reactive intermediates in organic chemistry.
A carbanion is a nucleophile, which stability and reactivity determined by several factors:
1. The inductive effect. Electronegative atoms adjacent to the charge will stabilize the charge;
2. Hybridization of the charge-bearing atom. The greater the s-character of the charge-bearing atom, the more stable the anion;
3. The extent of conjugation of the anion. Resonance effects can stabilize the anion. This is especially true when the anion is stabilized as a result of aromaticity.
A carbanion is a reactive intermediate and is encountered in organic chemistry for instance in the E1cB elimination reaction and in organometallic chemistry in for instance a Grignard reaction or in alkyl lithium chemistry. Stable carbanions do however exist. In 1984 Olmstead presented the lithium crown ether salt of the triphenylmethyl carbanion from triphenylmethane, n-butyllithium and 12-crown-4 at low temperatures:[2]
Adding n-butyllithium to triphenylmethane in THF at low temperatures followed by 12-crown-4 results in a red solution and the salt complex precipitates at −20 °C. The central C-C bond lengths are 145 pm with the phenyl ring propelled at an average angle of 31.2°. This propeller shape is less pronounced with a tetramethylammonium counterion.[3] One tool for the detection of carbanions in solution is proton NMR.[4] A spectrum of cyclopentadiene in DMSO shows four vinylic protons at 6.5 ppm and two methylene bridge protons at 3 ppm whereas the cyclopentadienyl anion has a single resonance at 5.50 ppm.
Carbon acids
Any molecule containing a C-H can lose a proton forming the carbanion. Hence any hydrocarbon containing C-H bonds can be considered an acid with a corresponding pKa value. Methane is certainly not an acid in its classical meaning yet its estimated pKa is 56. Compare this to acetic acid with pKa 4.76. The same factors that determine the stability of the carbanion also determine the order in pKa in carbon acids. These values are determined for the compounds either in water in order to compare them to ordinary acids, indimethyl sulfoxide in which the majority of carbon acids and their anions are soluble or in the gas phase. With DMSO the acidity window for solutes is limited to its own pKa of 35.5.
Table 1. Carbon acid acidities in pKa in DMSO [5]. Reference acids in bold.
name formula structural formula pKa
Methane CH4 ~ 56
Ethane C2H6 ~ 50
Anisole C7H8O ~ 49
Cyclopentane C5H10 ~ 45
Propene C3H6 ~ 44
Benzene C6H6 ~ 43
Toluene C6H5CH3 ~ 43
Dimethyl sulfoxide (CH3)2SO 35.5
Diphenylmethane C13H12 32.3
Aniline C6H5NH2 30.6
Triphenylmethane C19H16 30.6
Xanthene C13H10O 30
Ethanol C2H5OH 29.8
Phenylacetylene C8H6 28.8
Thioxanthene C13H10S 28.6
Acetone C3H6O 26.5
Acetylene C2H2 25
Benzoxazole C7H5NO 24.4
Fluorene C13H10 22.6
Indene C9H8 20.1
Cyclopentadiene C5H6 18
Malononitrile C3H2N2 11.2
Hydrogen cyanide HCN 9.2
Acetylacetone C5H8O2 8.95
Dimedone C8H12O2 5.23
Meldrum's acid C6H8O4 4.97
Acetic acid CH3COOH 4.76
Barbituric acid C4H2O3(NH)2 4.01
Trinitromethane HC(NO2)3 0.17
Fulminic acid HCNO -1.07
Carborane superacid HCHB11Cl11 -9
Note that the anions formed by ionization of acetic acid, ethanol or aniline are not carbanions.
Starting from methane in Table 1, the acidity increases:
• when the anion is aromatic, either because the added electron causes the anion to become aromatic (as in indene and cyclopentadiene), or because the negative charge on carbon can be delocalized over several already-aromatic rings (as in triphenylmethane or the carborane superacid).
• when the carbanion is surrounded by strongly electronegative groups, through the partial neutralisation of the negative charge (as in malononitrile).
• when the carbanion is immediately next to a carbonyl group. The α-protons of carbonyl groups are acidic because the negative charge in the enolate can be partially distributed in the oxygen atom. Meldrum's acid and barbituric acid, historically named acids, are in fact a lactone and a lactam respectively, but their acidic carbon protons make them acidic. The acidity of carbonyl compounds is an important driving force in many organic reactions such as the aldol reaction.
Chiral carbanions
With the molecular geometry for a carbanion described as a trigonal pyramid the question is whether or not carbanions can display chirality, because if the activation barrier for inversion of this geometry is too low any attempt at introducing chirality will end inracemization, similar to the nitrogen inversion. However, solid evidence exists that carbanions can indeed be chiral for example in research carried out with certain organolithium compounds.
The first ever evidence for the existence of chiral organolithium compounds was obtained in 1950. Reaction of chiral 2-iodooctane with sec-butyllithium in petroleum ether at −70 °C followed by reaction with dry ice yielded mostly racemic 2-methylbutyric acid but also an amount of optically active 2-methyloctanoic acid which could only have formed from likewise optical active 2-methylheptyllithium with the carbon atom linked to lithium the carbanion:[6]
On heating the reaction to 0 °C the optical activity is lost. More evidence followed in the 1960s. A reaction of the cis isomer of 2-methylcyclopropyl bromide with sec-butyllithium again followed by carboxylation with dry ice yielded cis-2-methylcyclopropylcarboxylic acid. The formation of the trans isomer would have indicated that the intermediate carbanion was unstable.[7]
In the same manner the reaction of (+)-(S)-l-bromo-l-methyl-2,2-diphenylcyclopropane with n-butyllithium followed by quench with methanol resulted in product with retention of configuration:[8]
Of recent date are chiral methyllithium compounds:[9]
The phosphate 1 contains a chiral group with a hydrogen and a deuterium substituent. The stannyl group is replaced by lithium to intermediate 2 which undergoes a phosphate-phosphorane rearrangement to phosphorane 3 which on reaction with acetic acid givesalcohol 4. Once again in the range of −78 °C to 0 °C the chirality is preserved in this reaction sequence.[10]
History
A carbanionic structure first made an appearance in the reaction mechanism for the benzoin condensation as correctly proposed by Clarke and Lapworth in 1907.[11] In 1904 Schlenk prepared Ph3C-NMe4+ in a quest for pentavalent nitrogen (fromTetramethylammonium chloride and Ph3CNa) [12] and in 1914 he demonstrated how triarylmethyl radicals could be reduced to carbonions by alkali metals.[13] The phrase carbanion was introduced by Wallis and Adams in 1933 as the negatively charged counterpart of the carbonium ion. [14][15]
• Large database of Bordwell pKa values at www.chem.wisc.edu Link
• Large database of Bordwell pKa values at daecr1.harvard.edu Link | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.09%3A_Reactive_Int.txt |
Alkanes (the most basic of all organic compounds) undergo very few reactions. One of these reactions is halogenation, or the substitution of a single hydrogen on the alkane for a single halogen to form a haloalkane. This reaction is very important in organic chemistry because it opens a gateway to further chemical reactions.
Introduction
While the reactions possible with alkanes are few, there are many reactions that involve haloalkanes. In order to better understand the mechanism (a detailed look at the step by step process through which a reaction occurs), we will closely examine the chlorination of methane. When methane (CH4) and chlorine (Cl2) are mixed together in the absence of light at room temperature nothing happens. However, if the conditions are changed, so that either the reaction is taking place at high temperatures (denoted by Δ) or there is ultra violet irradiation, a product is formed, chloromethane (CH3Cl).
Energetics
Why does this reaction occur? Is the reaction favorable? A way to answer these questions is to look at the change in enthalpy ($\Delta{H}$) that occurs when the reaction takes place.
ΔH = (Energy put into reaction) – (Energy given off from reaction)
If more energy is put into a reaction than is given off, the ΔH is positive, the reaction is endothermic and not energetically favorable. If more energy is given off in the reaction than was put in, the ΔH is negative, the reaction is said to be exothermic and is considered favorable. The figure below illustrates the difference between endothermic and exothermic reactions.
ΔH can also be calculated using bond dissociation energies (ΔH°):
$\Delta{H} = \sum \Delta{H^°} \text{ of bonds broken} - \sum \Delta{H^°} \text{ of bonds formed}$
Let’s look at our specific example of the chlorination of methane to determine if it is endothermic or exothermic:
Since, the ΔH for the chlorination of methane is negative, the reaction is exothermic. Energetically this reaction is favorable. In order to better understand this reaction we need to look at the mechanism ( a detailed step by step look at the reaction showing how it occurs) by which the reaction occurs.
Radical Chain Mechanism
The reaction proceeds through the radical chain mechanism. The radical chain mechanism is characterized by three steps: initiation, propagation and termination. Initiation requires an input of energy but after that the reaction is self-sustaining. The first propagation step uses up one of the products from initiation, and the second propagation step makes another one, thus the cycle can continue until indefinitely.
Step 1: Initiation
Initiation breaks the bond between the chlorine molecule (Cl2). For this step to occur energy must be put in, this step is not energetically favorable. After this step, the reaction can occur continuously (as long as reactants provide) without input of more energy. It is important to note that this part of the mechanism cannot occur without some external energy input, through light or heat.
Step 2: Propagation
The next two steps in the mechanism are called propagation steps. In the first propagation step, a chlorine radical combines with a hydrogen on the methane. This gives hydrochloric acid (HCl, the inorganic product of this reaction) and the methyl radical. In the second propagation step more of the chlorine starting material (Cl2) is used, one of the chlorine atoms becomes a radical and the other combines with the methyl radical.
The first propagation step is endothermic, meaning it takes in heat (requires 2 kcal/mol) and is not energetically favorable. In contrast the second propagation step is exothermic, releasing 27 kcal/mol. Once the reaction is initiated, the exothermic energy released from the second propagation step provides the activation energy for the first propagation step creating a cyclic chain reaction following Le Chatelier's principle until termination.
Step 3: Termination
In the termination steps, all the remaining radicals combine (in all possible manners) to form more product (CH3Cl), more reactant (Cl2) and even combinations of the two methyl radicals to form a side product of ethane (CH3CH3).
Limitations of the Chlorination
The chlorination of methane or any other alkane does not necessarily stop after one chlorination. It may actually be very hard to get a monosubstituted chloromethane. Instead di-, tri- and even tetra-chloromethanes are formed. One way to avoid this problem is to use a much higher concentration of methane or other alkane in comparison to chloride. This reduces the chance of a chlorine radical running into a chloromethane and starting the mechanism over again to form a dichloromethane. Through this method of controlling product ratios one is able to have a relative amount of control over the product.
Exercises
1. Compounds other than chlorine and methane can react via free-radical halogenation. Write out the complete mechanism for the monobromination of ethane.
2. Explain how the energetically unfavorable first propagation step can continue to occur without the input of energy from an external source.
3. Which step of the radical chain mechanism requires outside energy? What can be used as this energy?
4. Use the table provided below to calculate the change in enthalpy for the monobromination of ethane.
Compound Bond Dissociation Energy (kcal/mol)
CH3CH2-H 101
CH3CH2-Br 70
H-Br 87
Br2 46
Solutions
1.
2. The exothermic energy released from the second propagation step provides the activation energy for the first propagation step creating a cyclic chain reaction following Le Chatelier's principle until termination.
3. The initiation step requires energy from heat o lit. For maximum photoefficiency, the wavelength of light is correlated with bond being homolytically cleaved.
4.
References
1. Matyjaszewski, Krzysztof, Wojciech Jakubowski, Ke Min, Wei Tang, Jinyu Huang, Wade A. Braunecker, and Nicolay V. Tsarevsky. "Diminishing Catalyst Concentration in Atom Transfer Radical Polymerization with Reducing Agents." Science 72 (1930): 379-90.
2. Phillips, Francis C. "# Researches upon the Chemical Properties of Gases." Researches upon the Chemical Properties of Gases 17 (1893): 149-236.
Contributors and Attributions
• Kristen Kelley and Britt Farquharson | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.10%3A_The_Free-Rad.txt |
Comparing Reactivity
Given the knowledge that a particular reaction will proceed at a suitable rate, a host of practical considerations are necessary for satisfactory operation. These considerations include interference by possible side reactions that give products other than those desired, the ease of separation of the desired product from the reaction mixture, and costs of materials, apparatus, and labor. We shall consider these problems in connection with the important synthetic reactions discussed in this book.
The chlorination of saturated hydrocarbons can be induced by light, but also can be carried out at temperatures of about $300^\text{o}$ in the dark. Under such circumstances the mechanism is similar to that of light-induced chlorination, except that the chlorine atoms are formed by thermal dissociation of chlorine molecules. Solid carbon surfaces catalyze thermal chlorination, possibly by aiding in the cleavage of the chlorine molecules.
Direct monohalogenation of saturated hydrocarbons works satisfactorily only with chlorine and bromine. For the general reaction
the calculated $\Delta H^\text{0}$ value is negative and very large for fluorine, negative and moderate for chlorine and bromine, and positive for iodine (see Table 4-7). With fluorine, the reaction evolves so much heat that it may be difficult to control, and products from cleavage of carbon-carbon as well as of carbon-hydrogen bonds may be obtained. The only successful, direct fluorination procedure for hydrocarbons involves diffusion of minute amounts of fluorine mixed with helium into liquid or solid hydrocarbons at low temperatures, typically $-78^\text{o}$ (Dry Ice temperature). As fluorination proceeds, the concentration of fluorine can be increased. The process is best suited for preparation of completely fluorinated compounds, and it has been possible to obtain in this way amounts of $\left( CF_3 \right)_4C$ and $\left( CF_3 \right)_3 C-C \left( CF_3 \right)_3$ from 2,2-dimethylpropane and 2,2,3,3-tetramethylbutane corresponding to $10$-$15\%$ yields based on the fluorine used.
Bromine generally is much less reactive toward hydrocarbons than chlorine is, both at high temperatures and with activation by light. Nonetheless, it usually is possible to brominate saturated hydrocarbons successfully. Iodine is unreactive.
Table: Calculated Heat of Reaction for Halogenation fo Hydrocarbons
Halogen (X) $\Delta H^o$ (kcal/mole)a
F -116
Cl -27
Br -10
I 13
aCalculated from the bond energies of Table 4-3.
The chlorination of methane does not have to stop with the formation of chloromethane (methyl chloride). It is usual when chlorinating methane to obtain some of the higher chlorination products: dichloromethane (methylene chloride), trichloromethane (chloroform), and tetrachloromethane (carbon tetrachloride):
In practice, one can control the degree of substitution to a considerable extent by controlling the methane-chlorine ratio. For example, for monochlorination to predominate, a high methane-chlorine ratio is necessary such that the chlorine atoms react with $CH_4$ and not with $CH_3Cl$.
Selectivity in Alkane Halogenation
For propane and higher hydrocarbons for which more than one monosubstitution product is generally possible, difficult separation problems bay arise when a particular product is desired. For example, the chlorination of 2-methylbutane $3$ at $300^\text{o}$ gives all four possible monosubstitution products. On a purely statistical basis, we may expect the ratio of products from 2-methylbutane to correlate with the number of available hydrogens at the various positions of substitution in the ratio 6:1:2:3 ($50\%$:$8\%$:$17\%$:$25\%$:). However, as can be seen from the strengths of bonds between hydrogen and primary, secondary, and tertiary carbons are not the same and we would expect the weaker $C-H$ bonds to preferentially react with $Cl \cdot$. As such, the proportion of the tertiary halide is about three times that expected on a statistical basis which is in accord with our expectation that the tertiary $C-H$ bond of 2-methylbutane should be the weakest of the $C-H$ bonds.
Bromine atoms are far more selective than chlorine atoms. This is not unexpected because is endothermic, whereas corresponding reactions with a chlorine atoms usually are exothermic (data from Table 4-6). Bromine removes only those hydrogens that are relatively weakly bonded to a carbon atom. As predicted, attack of $Br \cdot$ on 2-methylbutane leads mostly to 2-bromo-2-methylbutane, some secondary bromide, and essentially no primary bromides:
When the structure of the alkane is symmetrical, then the fast reactivity of chlorination can be used for efficiency. When the structure of the alkane can produce a range of monohalogenated products, then the selectivity of bromination can be used to produce the most stable product in the greatest percentage.
Exercises
1. Specify the optimum halogenation conditions (Cl2/heat or Br2/heat) to produce the indicated major product.
Solution
1.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.11%3A_Reactivity_a.txt |
Objectives
• No objectives have been identified for this section
Key Terms
Make certain that you can define, and use in context, the key term below.
• enzyme
Study Notes
This section is a brief (but perhaps interesting) overview of some of the key differences between reactions performed in the lab and those in living systems. At this point, do not concern yourself with memorizing large biological molecules and reactions.
The active site
A critical element in the three-dimensional structure of any enzyme is the presence of an ‘active site’, which is a pocket, usually located in the interior of the protein, that serves as a docking point for the enzyme’s substrate(s) (‘substrate’ is the term that biochemists use for a reactant molecule in an enzyme-catalyzed reaction). It is inside the active site pocket that enzymatic catalysis occurs. Shown below is an image of the glycolytic enzyme fructose-1,6-bisphosphate aldolase, with its substrate bound inside the active site pocket.
When the substrate binds to the active site, a large number of noncovalent interactions form with the amino acid residues that line the active site. The shape of the active site, and the enzyme-substrate interactions that form as a result of substrate binding, are specific to the substrate-enzyme pair: the active site has evolved to 'fit' one particular substrate and to catalyze one particular reaction. Other molecules do not fit in this active site nearly so well as fructose 1,6-bisphosphate.
Here are two close-up views of the same active site pocket, showing some of the specific hydrogen-bonding interactions between the substrate and active site amino acids. The first image below is a three-dimensional rendering directly from the crystal structure data. The substrate is shown in 'space-filling' style, while the active site amino acids are shown in the 'ball and stick' style. Hydrogens are not shown. The color scheme is grey for carbon, red for oxygen, blue for nitrogen, and orange for phosphorus.
Below is a two-dimensional picture of the substrate (colored red) surrounded by hydrogen-bonding active site amino acids. Notice that both main chain and side chain groups contribute to hydrogen bonding: in this figure, main chain H-bonding groups are colored blue, and side chain H-bonding groups are colored green.
Looking at the last three images should give you some appreciation for the specific manner in which a substrate fits inside its active site.
Transition state stabilization
One of the most important ways that an enzyme catalyzes any given reaction is through entropy reduction: by bringing order to a disordered situation (remember that entropy is a component of Gibbs Free Energy, and thus a component of the activation energy). Let’s turn again to our previous example (from section 6.1C) of a biochemical nucleophilic substitution reaction, the methylation of adenosine in DNA. The reaction is shown below with non-reactive sections of the molecules depicted by variously shaped 'bubbles' for the sake of simplicity.
In order for this reaction to occur, the two substrates (reactants) must come into contact in precisely the right way. If they are both floating around free in solution, the likelihood of this occurring is very small – the entropy of the system is simply too high. In other words, this reaction takes place very slowly without the help of a catalyst.
Here’s where the enzyme’s active site pocket comes into play. It is lined with various functional groups from the amino acid main and side chains, and has a very specific three-dimensional architecture that has evolved to bind to both of the substrates. If the SAM molecule, for example, diffuses into the active site, it can replace its (favorable) interactions with the surrounding water molecules with (even more favorable) new interactions with the functional groups lining the active site.
In a sense, SAM is moving from one solvent (water) to another 'solvent' (the active site), where many new energetically favorable interactions are possible. Remember: these new contacts between SAM and the active site groups are highly specific to SAM and SAM alone – no other molecule can ‘fit’ so well in this precise active site environment, and thus no other molecule will be likely to give up its contacts to water and bind to the active site.
The second substrate also has a specific spot reserved in the active site. (Because in this case the second substrate is a small segment of a long DNA molecule, the DNA-binding region of the active site is more of a 'groove' than a 'pocket').
So now we have both substrates bound in the active site. But they are not just bound in any random orientation – they are specifically positioned relative to one another so that the nucleophilic nitrogen is held very close to the electrophilic carbon, with a free path of attack. What used to be a very disordered situation – two reactants diffusing freely in solution – is now a very highly ordered situation, with everything set up for the reaction to proceed. This is what is meant by entropy reduction: the entropic component of the energy barrier has been lowered.
Looking a bit deeper, though, it is not really the noncovalent interaction between enzyme and substrate that are responsible for catalysis. Remember: all catalysts, enzymes included, accelerate reactions by lowering the energy of the transition state. With this in mind, it should make sense that the primary job of an enzyme is to maximize favorable interactions with the transition state, not with the starting substrates. This does not imply that enzyme-substrate interactions are not strong, rather that enzyme-TS interactions are far stronger, often by several orders of magnitude. Think about it this way: if an enzyme were to bind to (and stabilize) its substrate(s) more tightly than it bound to (and stabilized) the transition state, it would actually slow down the reaction, because it would be increasing the energy difference between starting state and transition state. The enzyme has evolved to maximize favorable noncovalent interactions to the transition state: in our example, this is the state in which the nucleophilic nitrogen is already beginning to attack the electrophilic carbon, and the carbon-sulfur bond has already begun to break.
In many enzymatic reactions, certain active site amino acid residues contribute to catalysis by increasing the reactivity of the substrates. Often, the catalytic role is that of acid and/or base. In our DNA methylation example, the nucleophilic nitrogen is deprotonated by a nearby aspartate side chain as it begins its nucleophilic attack on the methyl group of SAM. We will study nucleophilicity in greater detail in chapter 8, but it should make intuitive sense that deprotonating the amine increases the electron density of the nitrogen, making it more nucleophilic. Notice also in the figure below that the main chain carbonyl of an active site proline forms a hydrogen bond with the amine, which also has the effect of increasing the nitrogen's electron density and thus its nucleophilicity (Nucleic Acids Res. 2000, 28, 3950).
How does our picture of enzyme catalysis apply to multi-step reaction mechanisms? Although the two-step nucleophilic substitution reaction between tert-butyl chloride and hydroxide (section 6.1C) is not a biologically relevant process, let’s pretend just for the sake of illustration that there is a hypothetical enzyme that catalyzes this reaction.
The same basic principles apply here: the enzyme binds best to the transition state. But therein lies the problem: there are two transition states! To which TS does the enzyme maximize its contacts?
Recall that the first step – the loss of the chloride leaving group to form the carbocation intermediate – is the slower, rate-limiting step. It is this step that our hypothetical enzyme needs to accelerate if it wants to accelerate the overall reaction, and it is thus the energy of TS1 that needs to be lowered.
By Hammond’s postulate, we also know that the intermediate I is a close approximation of TS1. So the enzyme, by stabilizing the intermediate, will also stabilize TS1 (as well as TS2) and thereby accelerate the reaction.
If you read scientific papers about enzyme mechanisms, you will often see researchers discussing how an enzyme stabilizes a reaction intermediate. By virtue of Hammond's postulate, they are, at the same time, talking about how the enzyme lowers the energy of the transition state.
An additional note: although we have in this section been referring to SAM as a 'substrate' of the DNA methylation reaction, it is also often referred to as a coenzyme, or cofactor. These terms are used to describe small (relative to protein and DNA) biological organic molecules that bind specifically in the active site of an enzyme and help the enzyme to do its job. In the case of SAM, the job is methyl group donation. In addition to SAM, we will see many other examples of coenzymes in the coming chapters, a number of which - like ATP (adenosine triphosphate), coenzyme A, thiamine, and flavin - you have probably heard of before. The full structures of some common coenzymes are shown in table 6 in the tables section.
5.13: Additional
Kinetics and the Rate Equation
5-1 For a chemical reaction, what is the rate equation used to correlate?
5-2 Write the rate equation that describes the rate of the following reaction.
aA + bB → cC + dD
5-3 What is the overall order of the following reaction with multiple reactants?
Rate = k [A]1[B]1/2
Halogenation of Alkanes
5-4 For the following compounds, give all possible monochlorinated derivatives.
5-5 For the following compounds, identify the major product of free-radical bromination.
5-6 Explain why radical bromination is significantly more selective than radical chlorination.
5.14: Solutions t
Kinetics and the Rate Equation
5-1 The rate equation is an experimentally derived equation that explains the relationship between the concentration of reactants and the rate of the reaction.
5-2 Rate = k [A]m[B]n
5-3 Overall order = 1.5
Halogenation of Alkanes
5-4
5-5
5-6 Radical bromination is more selective because of its slightly higher activation energy required to break a C-H bond during the propagation steps (when the bromine radical abstracts a proton from the substrate). Though the difference in activation energy is not huge (Cl = ~1 kcal/mol and Br = ~3 kcal/mol), it leads to a significant difference in selectivity. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.12%3A_A_Comparison.txt |
Learning Objectives
After reading this chapter and completing the exercises and homework, a student can be able to:
• recognize and classify molecules as chiral or achiral and identify planes of symmetry - refer to section 6.1
• draw, interpret, and convert between perspective formulae and Fischer projections for chiral compounds - refer to section 6.2
• name chiral compounds using (R) & (S) nomenclature - refer to section 6.3
• recognize and classify diastereomers and meso compounds - refer to section 6.4 and 6.5 respectively
• explain how physical properties differ for different types of stereoisomers - refer to section ?????
• distinguish and discern the structural and chemical relationships between isomeric compounds - refer to section 6.6
• define and explain the lack of optical activity of racemic mixtures - refer to section 6.7
• determine the percent composition of an enantiomeric mixture from polarimetry data and the for specific rotation formula - refer to section 6.7
• explain how to resolve (separate) a pair of enantiomers - refer to section 6.8
• interpret the stereoisomerism of compounds with three or more chiral centers - refer to section 6.9
• compare and contrast absolute configuration with relative configuration - refer to section 6.10
• interpret the stereoisomerism of compounds with nitrogen, phosphorus, or sulfur as chiral centers - refer to section 6.11
• recognize and explain biochemical applications of chirality - refer to section 6.12
• describe Jean Baptiste Biot and Louis Pasteur's contributions to the understanding of optical isomers - refer to section 6.13
• 6.1: Chirality
Chiral carbons are tetrahedral carbons bonded to four unique groups. At first glance, many carbons may look alike, but upon closer inspection, we can discern their differences.
• 6.2: Fischer Projections to communicate Chirality
Converting between perspective formula structures (wedges and dashes) and Fischer projections can be useful when evaluating stereochemistry, especially for carbohydrate chemistry.
• 6.3: Absolute Configuration and the (R) and (S) System
The absolute configuration of chiral centers as R or S is determined by applying the Cahn-Ingold-Prelog rules.
• 6.4: Diastereomers - more than one chiral center
Diastereomers are stereoisomers with two or more chiral centers that are not enantiomers. Diastereomers have different physical properties (melting points, boiling points, and densities). Depending on the reaction mechanism, diastereomers can produce different stereochemical products.
• 6.5: Meso Compounds
A meso compound is an achiral compound that has two or more chiral centers. Molecular symmetry allows the mirror images to super-impose so that they are not enantiomers.
• 6.6: Isomerism Summary Diagram
A simple diagram is helpful in distinguishing between the different types of isomers that are possible.
• 6.7: Optical Activity and Racemic Mixtures
Optical activity is one of the few ways to distinguish between enantiomers. A racemic mixture is a 50:50 mixture of two enantiomers. Racemic mixtures were an interesting experimental discovery because two optically active samples were combined to create an optically INACTIVE sample.
• 6.8: Resolution (Separation) of Enantiomers
The most commonly used procedure for separating enantiomers is to convert them to a mixture of diastereomeric salts that can be separated based on their differences in their physical properties. After separation, the isolated D or the L enantiomer can be recovered.
• 6.9: Stereochemistry of Molecules with Three or More Asymmetric Carbons
As the number of chiral carbons increases, the number of stereoisomers also increases. This sections shows a short cut for compounds with three or more stereocenters.
• 6.10: Absolute and Relative Configuration - the distinction
The absolute configuration at a chiral center in a molecule is a time-independent and unambiguous symbolic description of the spatial arrangement of ligands (atoms) around it. The relative configuration is the experimentally determined relationship between two enantiomers even though we may not know the absolute configuration.
• 6.11: Chirality at Nitrogen, Phosphorus, and Sulfur
Chirality can also occur with central atoms other than carbon.
• 6.12: Biochemistry of Enantiomers
Biological activity and chirality are strongly correlated. The section explores a few examples.
• 6.13: The Discovery of Enantiomers
The initial work carried out by Biot and Pasteur contributed to the concepts of chirality.
• 6.14: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
• 6.15: Solutions to Additional Exercises
This section has the solutions to the additional exercises from the previous section.
06: Stereochemistry at Tetrahedral Centers
Learning Objective
• recognize and classify molecules as chiral or achiral and identify planes of symmetry
Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of their most interesting type of isomer is the mirror-image stereoisomers, a non-superimposable set of two molecules that are mirror image of one another. The existence of these molecules are determined by concept known as chirality.
Introduction
Organic compounds, molecules created around a chain of carbon atom (more commonly known as carbon backbone), play an essential role in the chemistry of life. These molecules derive their importance from the energy they carry, mainly in a form of potential energy between atomic molecules. Since such potential force can be widely affected due to changes in atomic placement, it is important to understand the concept of an isomer, a molecule sharing same atomic make up as another but differing in structural arrangements. This article will be devoted to a specific isomers called stereoisomers and its property of chirality (Figure 1).
The concepts of steroisomerism and chirality command great deal of importance in modern organic chemistry, as these ideas helps to understand the physical and theoretical reasons behind the formation and structures of numerous organic molecules, the main reason behind the energy embedded in these essential chemicals. In contrast to more well-known constitutional isomerism, which develops isotopic compounds simply by different atomic connectivity, stereoisomerism generally maintains equal atomic connections and orders of building blocks as well as having same numbers of atoms and types of elements.
What, then, makes stereoisomers so unique? To answer this question, the learner must be able to think and imagine in not just two-dimensional images, but also three-dimensional space. This is due to the fact that stereoisomers are isomers because their atoms are different from others in terms of spatial arrangement.
Spatial Arrangement
First and foremost, one must understand the concept of spatial arrangement in order to understand stereoisomerism and chirality. Spatial arrangement of atoms concern how different atomic particles and molecules are situated about in the space around the organic compound, namely its carbon chain. In this sense, spatial arrangement of an organic molecule are different another if an atom is shifted in any three-dimensional direction by even one degree. This opens up a very broad possibility of different molecules, each with their unique placement of atoms in three-dimensional space .
Stereoisomers
Stereoisomers are, as mentioned above, contain different types of isomers within itself, each with distinct characteristics that further separate each other as different chemical entities having different properties. Type called entaniomer are the previously-mentioned mirror-image stereoisomers, and will be explained in detail in this article. Another type, diastereomer, has different properties and will be introduced afterwards.
The Many Synonyms of the Chiral Carbon
Be aware - all of the following terms can be used to describe a chiral carbon.
chiral carbon = asymmetric carbon = optically active carbon = stereo carbon
Enantiomers
This type of stereoisomer is the essential mirror-image, non-superimposable type of stereoisomer introduced in the beginning of the article. Figure 3 provides a perfect example; note that the gray plane in the middle demotes the mirror plane.
Note that even if one were to flip over the left molecule over to the right, the atomic spatial arrangement will not be equal. This is equivalent to the left hand - right hand relationship, and is aptly referred to as 'handedness' in molecules. This can be somewhat counter-intuitive, so this article recommends the reader try the 'hand' example. Place both palm facing up, and hands next to each other. Now flip either side over to the other. One hand should be showing the back of the hand, while the other one is showing the palm. They are not same and non-superimposable.
This is where the concept of chirality comes in as one of the most essential and defining idea of stereoisomerism.
Chirality
Chirality essentially means 'mirror-image, non-superimposable molecules', and to say that a molecule is chiral is to say that its mirror image (it must have one) is not the same as it self. Whether a molecule is chiral or achiral depends upon a certain set of overlapping conditions. Figure 4 shows an example of two molecules, chiral and achiral, respectively. Notice the distinct characteristic of the achiral molecule: it possesses two atoms of same element. In theory and reality, if one were to create a plane that runs through the other two atoms, they will be able to create what is known as bisecting plane: The images on either side of the plan is the same as the other (Figure 4).
In this case, the molecule is considered 'achiral'. In other words, to distinguish chiral molecule from an achiral molecule, one must search for the existence of the bisecting plane in a molecule. All chiral molecules are deprive of bisecting plane, whether simple or complex.
As a universal rule, no molecule with different surrounding atoms are achiral. Chirality is a simple but essential idea to support the concept of stereoisomerism, being used to explain one type of its kind. The chemical properties of the chiral molecule differs from its mirror image, and in this lies the significance of chilarity in relation to modern organic chemistry.
Exercise 1
Identify the following as either a constitutional isomer or stereoisomer. If stereoisomer, determine if it is an enantiomer or diastereomer. Explain the reason behind the answer. Also mark chirality for each molecule.
a) b) c)
Solutions
a) achiral
b) chiral
c) chiral
Exercise 2
Identify the chiral centers in each of the following:
Solutions
Contributors and Attributions
• Dan Chong
• Jonathan Mooney (McGill University)
6.02: Fischer Projections to communicate Chirality
Learning Objective
• draw, interpret, and convert between perspective formulae and Fischer projections for chiral compounds
Perspective Formulas and Fischer Projections
So far, we have communicated the stereochemical orentation of compounds using the wedges and dashes of perspective formulas. For example, the perspective formula for (R)-Lactic acid is shown below.
A Fischer projection is a convention used to depict stereochemistry in two dimensions. The horizontal bonds are seen as wedges and the vertical bonds are seen as dashed lines as shown below in the example below for glyceraldehyde.
Converting between Perspective and Fischer Formulas
It can be useful to convert between perspective formulas and Fischer projections. Below is one approach using (R)-lactic acid as an example..
Step 1: Hold the molecule so that
a) the chiral center is on the plane of the paper,
b) two bonds are coming out of the plane of the paper and are on a horizontal plane,
c) the two remaining bonds are going into the plane of the paper and are on a vertical plane
Step 2: Push the two bonds coming out of the plane of the paper onto the plane of the paper.
Step 3: Pull the two bonds going into the plane of the paper onto the plane of the paper.
Step 4: Omit the chiral atom symbol for convenience. This is the Fischer Projection of (R)-Lactic acid.
The stereochemical formula for (R)-lactic acid can be drawn using either method. To build this skill, we begin by drawing the structures and converting them step wise. Models can also be helpful. Eventually, we will be able to mentally conversion between these two structures.
Exercise 1
1. Convert each compound into the alternate sterochemical structure (Perspective Fischer).
Answer
See also D,L-convention. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.01%3A_Chirality.txt |
Learning Objective
• name chiral compounds using (R) & (S) nomenclature
USE YOUR MODELING KIT: Models assist in visualizing the structure. When using a model, make sure the lowest priority is pointing away from you. Then determine the direction from the highest priority substituent to the lowest: clockwise (R) or counterclockwise (S).
IF YOU DO NOT HAVE A MODELING KIT: remember that the dashes mean the bond is going into the screen and the wedges means that bond is coming out of the screen. If the lowest priority bond is not pointing to the back, mentally rotate it so that it is. However, it is very useful when learning organic chemistry to use models.
If you have a modeling kit use it as you read through this section and work the practice problems.
Introduction and the Cahn-Ingold-Prelog rules of Priority
To name the enantiomers of a compound unambiguously, their names must include the "handedness" of the molecule. The letters "R" and "S" are determined by applying the Cahn-Ingold-Prelog (CIP) rules. The optical activity (+/-) can also be communicated in the name, but must be empirically derived. There are also biochemical conventions for carbohydrates (sugars) and amino acids (the building blocks of proteins).
The method of unambiguously assigning the handedness of molecules was originated by three chemists: R.S. Cahn, C. Ingold, and V. Prelog and, as such, is also often called the Cahn-Ingold-Prelog rules. In addition to the CIP system, there are two ways of experimentally determining the absolute configuration of an enantiomer:
1. X-ray diffraction analysis. Note that there is no correlation between the sign of rotation and the structure of a particular enantiomer.
2. Chemical correlation with a molecule whose structure has already been determined via X-ray diffraction.
However, for non-laboratory purposes, it is beneficial to focus on the R/S system. The sign of optical rotation, although different for the two enantiomers of a chiral molecule,at the same temperature, cannot be used to establish the absolute configuration of an enantiomer; this is because the sign of optical rotation for a particular enantiomer may change when the temperature changes.
The Cahn-Ingold-Prelog rules of priority are based on the atomic numbers of the atoms of interest. For chirality, the atoms of interest are the atoms bonded to the chiral carbon.
1. The atom with higher atomic number has higher priority (I > Br > Cl > S > P > F > O > N > C > H).
2. When comparing isotopes, the atom with the higher mass number has higher priority [18O > 16O or 15N > 14N or 13C > 12C or T (3H) > D (2H) > H].
3. When there is a tie in (2) above, establish relative priority by proceeding to the next atom(s) along the chain until the first difference is observed.
Multiple bonds are treated as if each bond of the multiple bond is bonded to a unique atom. For example, the ethenyl group (CH2=CH) has higher priority than the ethyl group (CH3CH2). The ethenyl carbon priority is "two" bonds to carbon atoms and one bond to a hydrogen atom compared with the ethyl carbon that has only one bond to a carbon atom and two bonds to two hydrogen atoms. Similarly, the carbon-carbon triple bond of acetylene would give it higher CIP priority than the ethenyl group as summarized below.
Stereocenters are labeled R or S
The "right hand" and "left hand" nomenclature is used to name the enantiomers of a chiral compound. The stereocenters are labeled as R or S.
Consider the diagram above on the left: a curved arrow is drawn counter-clockwise (c-cw) from the highest priority substituent (1) to the lowest priority substituent (4) in the S-configuration ("Sinister" → Latin= "left"). The counterclockwise direction can be recognized by the movement left when leaving the 12 o' clock position. Now consider the diagram above on the right where a curved arrow is drawn clockwise (cw) from the highest priority substituent (1) to the lowest priority substituent (4) in the R configuration ("Rectus" → Latin= "right"). The R or S is then added as a prefix, in parenthesis, to the name of the enantiomer of interest. A locator number is required if there is more than one chiral center. Otherwise, the person reading the name is expected to recognize the chiral center.
Example 1
The two chiral compounds below are drawn to emphasize the chiral carbon with the full chemical name below each structure.
Absolute Configurations of Perspective Formulas
Chemists need a convenient way to distinguish one stereoisomer from another. The Cahn-Ingold-Prelog system is a set of rules that allows us to unambiguously define the stereochemical configuration of any stereocenter, using the designations 'R ’ (from the Latin rectus, meaning right-handed) or ' S ’ (from the Latin sinister, meaning left-handed).
The rules for this system of stereochemical nomenclature are, on the surface, fairly simple.
Rules for assigning an R/S designation to a chiral center
1: Assign priorities to the four substituents, with #1 being the highest priority and #4 the lowest. Priorities are based on the atomic number.
2: Trace a circle from #1 to #2 to #3.
3: Determine the orientation of the #4 priority group. If it is oriented into the plane of the page (away from you), go to step 4a. If it is oriented out of the plane of the page (toward you) go to step 4b.
4a: (#4 group pointing away from you): a clockwise circle in part 2 corresponds to the R configuration, while a counterclockwise circle corresponds to the S configuration.
4b: (#4 group pointing toward you): a clockwise circle in part 2 corresponds to the S configuration, while a counterclockwise circle corresponds to the R configuration.
We’ll use the 3-carbon sugar glyceraldehyde as our first example. The first thing that we must do is to assign a priority to each of the four substituents bound to the chiral center. We first look at the atoms that are directly bonded to the chiral center: these are H, O (in the hydroxyl), C (in the aldehyde), and C (in the CH2OH group).
Assigning R/S configuration to glyceraldehyde:
Two priorities are easy: hydrogen, with an atomic number of 1, is the lowest (#4) priority, and the hydroxyl oxygen, with atomic number 8, is priority #1. Carbon has an atomic number of 6. Which of the two ‘C’ groups is priority #2, the aldehyde or the CH2OH? To determine this, we move one more bond away from the chiral center: for the aldehyde we have a double bond to an oxygen, while on the CH2OH group we have a single bond to an oxygen. If the atom is the same, double bonds have a higher priority than single bonds. Therefore, the aldehyde group is assigned #2 priority and the CH2OH group the #3 priority.
With our priorities assigned, we look next at the #4 priority group (the hydrogen) and see that it is pointed back away from us, into the plane of the page - thus step 4a from the procedure above applies. Then, we trace a circle defined by the #1, #2, and #3 priority groups, in increasing order. The circle is clockwise, which by step 4a tells us that this carbon has the ‘R’ configuration, and that this molecule is (R)-glyceraldehyde. Its enantiomer, by definition, must be (S)-glyceraldehyde.
Next, let's look at one of the enantiomers of lactic acid and determine the configuration of the chiral center. Clearly, H is the #4 substituent and OH is #1. Owing to its three bonds to oxygen, the carbon on the acid group takes priority #2, and the methyl group takes #3. The #4 group, hydrogen, happens to be drawn pointing toward us (out of the plane of the page) in this figure, so we use step 4b: The circle traced from #1 to #2 to #3 is clockwise, which means that the chiral center has the S configuration.
The drug thalidomide is an interesting - but tragic - case study in the importance of stereochemistry in drug design. First manufactured by a German drug company and prescribed widely in Europe and Australia in the late 1950's as a sedative and remedy for morning sickness in pregnant women, thalidomide was soon implicated as the cause of devastating birth defects in babies born to women who had taken it. Thalidomide contains a chiral center, and thus exists in two enantiomeric forms. It was marketed as a racemic mixture: in other words, a 50:50 mixture of both enantiomers.
Let’s try to determine the stereochemical configuration of the enantiomer on the left. Of the four bonds to the chiral center, the #4 priority is hydrogen. The nitrogen group is #1, the carbonyl side of the ring is #2, and the –CH2 side of the ring is #3.
The hydrogen is shown pointing away from us, and the prioritized substituents trace a clockwise circle: this is the R enantiomer of thalidomide. The other enantiomer, of course, must have the S configuration.
Although scientists are still unsure today how thalidomide works, experimental evidence suggests that it was actually the R enantiomer that had the desired medical effects, while the S enantiomer caused the birth defects. Even with this knowledge, however, pure (R)-thalidomide is not safe, because enzymes in the body rapidly convert between the two enantiomers - we will see how that happens in chapter 12.
As a historical note, thalidomide was never approved for use in the United States. This was thanks in large part to the efforts of Dr. Frances Kelsey, a Food and Drug officer who, at peril to her career, blocked its approval due to her concerns about the lack of adequate safety studies, particularly with regard to the drug's ability to enter the bloodstream of a developing fetus. Unfortunately, though, at that time clinical trials for new drugs involved widespread and unregulated distribution to doctors and their patients across the country, so families in the U.S. were not spared from the damage caused.
Very recently a close derivative of thalidomide has become legal to prescribe again in the United States, with strict safety measures enforced, for the treatment of a form of blood cancer called multiple myeloma. In Brazil, thalidomide is used in the treatment of leprosy - but despite safety measures, children are still being born with thalidomide-related defects.
Exercise 1.: Determine the stereochemical configurations of the chiral centers in the biomolecules shown below.
Exercise 2.: Should the (R) enantiomer of malate have a solid or dashed wedge for the C-O bond in the figure below?
Exercise 3.: Using solid or dashed wedges to show stereochemistry, draw the (R) enantiomer of ibuprofen and the (S) enantiomer of 2-methylerythritol-4-phosphate (structures are shown earlier in this chapter without stereochemistry).
Solutions to exercises
Absolute Configurations of Fischer Projections
To determine the absolute configuration of a chiral center in a Fisher projection, use the following two-step procedure.
Step 1
Assign priority numbers to the four ligands (groups) bonded to the chiral center using the CIP priority system.
Step 2 - vertical option
If the lowest priority ligand is on a Vertical bond, then it is pointing away from the viewer.
Trace the three highest-priority ligands starting at the highest-priority ligand (① → ② → ) in the direction that will give a Very correct answer.
In the compound below, the movement is clockwise indicating an R-configuration. The complete IUPAC name for this compound is (R)-butan-2-ol.
Step 2 - horizontal option
If the lowest-priority ligand is on a Horizontal bond, then it is pointing toward the viewer.
Trace the three highest-priority ligands starting at the highest-priority ligand (① → ② → ) in the direction that will give a Horribly wrong answer. Note in the table below that the configurations are reversed from the first example.
In the compound below, the movement is clockwise (R) which is Horribly wrong, so the actual configuration is S. The complete IUPAC name for this compound is (S)-butan-2-ol.
Manipulating Fischer Projections with NO Change to Configuration
A Fischer projection restricts a three-dimensional molecule into two dimensions. Consequently, there are limitations as to the operations that can be performed on a Fischer projection without changing the absolute configuration at chiral centers. The operations that do not change the absolute configuration at a chiral center in a Fischer projections can be summarized as two rules.
Rule 1: Rotation of the Fischer projection by 180º in either direction without lifting it off the plane of the paper does not change the absolute configuration at the chiral center.
Rule 2: Rotation of three ligands on the chiral center in either direction, keeping the remaining ligand in place, does not change the absolute configuration at the chiral center.
Manipulating Fischer Projections with Change to Configuration
The operations that do change the absolute configuration at a chiral center in a Fischer projection can be summarized as two rules.
Rule 1: Rotation of the Fischer projection by 90º in either direction changes the absolute configuration at the chiral center.
Rule 2: Interchanging any two ligands on the chiral center changes the absolute configuration at the chiral center.
The above rules assume that the Fischer projection under consideration contains only one chiral center. However, with care, they can be applied to Fischer projections containing any number of chiral centers.
Exercise 1
Classify the following compounds as R or S?
Solution
1. S: I > Br > F > H. The lowest priority substituent, H, is already going towards the back. It turns left going from I to Br to F, so it's a S.
2. R: Br > Cl > CH3 > H. You have to switch the H and Br in order to place the H, the lowest priority, in the back. Then, going from Br to Cl, CH3 is turning to the right, giving you a R.
3. Neither R or S: This molecule is achiral. Only chiral molecules can be named R or S.
4. R: OH > CN > CH2NH2 > H. The H, the lowest priority, has to be switched to the back. Then, going from OH to CN to CH2NH2, you are turning right, giving you a R.
5. (5) S: \(\ce{-COOH}\) > \(\ce{-CH_2OH}\) > \(\ce{C#CH}\) > \(\ce{H}\). Then, going from \(\ce{-COOH}\) to \(\ce{-CH_2OH}\) to \(\ce{-C#CH}\) you are turning left, giving you a S configuration.
Exercises
6. Orient the following so that the least priority (4) atom is paced behind, then assign stereochemistry (R or S).
7. Draw (R)-2-bromobutan-2-ol.
8. Assign R/S to the following molecule.
Solutions
6.
A = S; B = R
7.
8. The stereo center is R. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.03%3A_Absolute_Configuration_and_the_%28R%29_and_%28S%29_Sy.txt |
Learning Objective
• recognize and classify diastereomers
Diastereomers are stereoisomers with two or more chiral centers that are not enantiomers. Diastereomers have different physical properties (melting points, boiling points, and densities). Depending on the reaction mechanism, diastereomers can produce different stereochemical products.
Introduction
So far, we have been analyzing compounds with a single chiral center. Next, we turn our attention to those which have multiple chiral centers. We'll start with some stereoisomeric four-carbon sugars with two chiral centers.
To avoid confusion, we will simply refer to the different stereoisomers by capital letters.
Look first at compound A below. Both chiral centers in have the R configuration (you should confirm this for yourself!). The mirror image of Compound A is compound B, which has the S configuration at both chiral centers. If we were to pick up compound A, flip it over and put it next to compound B, we would see that they are not superimposable (again, confirm this for yourself with your models!). A and B are nonsuperimposable mirror images: in other words, enantiomers.
Now, look at compound C, in which the configuration is S at chiral center 1 and R at chiral center 2. Compounds A and C are stereoisomers: they have the same molecular formula and the same bond connectivity, but a different arrangement of atoms in space (recall that this is the definition of the term 'stereoisomer). However, they are not mirror images of each other (confirm this with your models!), and so they are not enantiomers. By definition, they are diastereomers of each other.
Notice that compounds C and B also have a diastereomeric relationship, by the same definition.
So, compounds A and B are a pair of enantiomers, and compound C is a diastereomer of both of them. Does compound C have its own enantiomer? Compound D is the mirror image of compound C, and the two are not superimposable. Therefore, C and D are a pair of enantiomers. Compound D is also a diastereomer of compounds A and B.
This can also seem very confusing at first, but there some simple shortcuts to analyzing stereoisomers:
Stereoisomer shortcuts
If all of the chiral centers are of opposite R/S configuration between two stereoisomers, they are enantiomers.
If at least one, but not all of the chiral centers are opposite between two stereoisomers, they are diastereomers.
(Note: these shortcuts to not take into account the possibility of additional stereoisomers due to alkene groups: we will come to that later)
Here's another way of looking at the four stereoisomers, where one chiral center is associated with red and the other blue. Pairs of enantiomers are stacked together.
We know, using the shortcut above, that the enantiomer of RR must be SS - both chiral centers are different. We also know that RS and SR are diastereomers of RR, because in each case one - but not both - chiral centers are different.
Diastereomers vs. Enantiomers in Wine Chemistry
Tartaric acid, C4H6O6, is an organic compound that can be found in grape, bananas, and in wine. The structures of tartaric acid itself is really interesting. Naturally, it is in the form of (R,R) stereocenters. Artificially, it can be in the meso form (R,S), which is achiral. R,R tartaric acid is enantiomer to is mirror image which is S,S tartaric acid and diasteromers to meso-tartaric acid (Figure 5.6.2).
(R,R) and (S,S) tartaric acid have similar physical properties and reactivity. However, meso-tartaric acid have different physical properties and reactivity. For example, melting point of (R,R) & (S,S) tartaric is about 170 degree Celsius, and melting point of meso-tartaric acid is about 145 degree Celsius.
Diastereomers vs. Enantiomers in Sugar Chemistry
D-erythrose is a common four-carbon sugar.
A note on sugar nomenclature: biochemists use a special system to refer to the stereochemistry of sugar molecules, employing names of historical origin in addition to the designators 'D' and 'L'. You will learn about this system if you take a biochemistry class. We will use the D/L designations here to refer to different sugars, but we won't worry about learning the system.
As you can see, D-erythrose is a chiral molecule: C2 and C3 are stereocenters, both of which have the R configuration. In addition, you should make a model to convince yourself that it is impossible to find a plane of symmetry through the molecule, regardless of the conformation. Does D-erythrose have an enantiomer? Of course it does – if it is a chiral molecule, it must. The enantiomer of erythrose is its mirror image, and is named L-erythrose (once again, you should use models to convince yourself that these mirror images of erythrose are not superimposable).
Notice that both chiral centers in L-erythrose both have the S configuration.
Note
In a pair of enantiomers, all of the chiral centers are of the opposite configuration.
What happens if we draw a stereoisomer of erythrose in which the configuration is S at C2 and R at C3? This stereoisomer, which is a sugar called D-threose, is not a mirror image of erythrose. D-threose is a diastereomer of both D-erythrose and L-erythrose.
The definition of diastereomers is simple: if two molecules are stereoisomers (same molecular formula, same connectivity, different arrangement of atoms in space) but are not enantiomers, then they are diastereomers by default. In practical terms, this means that at least one - but not all - of the chiral centers are opposite in a pair of diastereomers. By definition, two molecules that are diastereomers are not mirror images of each other.
L-threose, the enantiomer of D-threose, has the R configuration at C2 and the S configuration at C3. L-threose is a diastereomer of both erythrose enantiomers.
Erythronolide B, a precursor to the 'macrocyclic' antibiotic erythromycin, has 10 stereocenters. It’s enantiomer is that molecule in which all 10 stereocenters are inverted.
In total, there are 210 = 1024 stereoisomers in the erythronolide B family: 1022 of these are diastereomers of the structure above, one is the enantiomer of the structure above, and the last is the structure above.
We know that enantiomers have identical physical properties and equal but opposite degrees of specific rotation. Diastereomers, in theory at least, have different physical properties – we stipulate ‘in theory’ because sometimes the physical properties of two or more diastereomers are so similar that it is very difficult to separate them. In addition, the specific rotations of diastereomers are unrelated – they could be the same sign or opposite signs, and similar in magnitude or very dissimilar.
Exercises
1. Draw the structures of L-galactose (the enantiomer of D-galactose) and two more diastereomers of D-glucose (one should be an epimer).
2. Determine the stereochemistry of the following molecule:
Answer
1.
2. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.04%3A_Diastereomers_-_more_than_one_chiral_center.txt |
Learning Objective
• recognize and classify meso compounds
A meso compound is an achiral compound that has chiral centers. It is superimposed on its mirror image and is optically inactive although it contains two or more stereocenters.
Introduction
In general, a meso compound should contain two or more identical substituted stereocenters. Also, it has an internal symmetry plane that divides the compound in half. These two halves reflect each other by the internal mirror. The stereochemistry of stereocenters should "cancel out". What it means here is that when we have an internal plane that splits the compound into two symmetrical sides, the stereochemistry of both left and right side should be opposite to each other, and therefore, result in optically inactive. Cyclic compounds may also be meso.
Identification
If A is a meso compound, it should have two or more stereocenters, an internal plane, and the stereochemistry should be R and S.
1. Look for an internal plane, or internal mirror, that lies in between the compound.
2. The stereochemistry (e.g. R or S) is very crucial in determining whether it is a meso compound or not. As mentioned above, a meso compound is optically inactive, so their stereochemistry should cancel out. For instance, R cancels S out in a meso compound with two stereocenters.
trans-1,2-dichloro-1,2-ethanediol
(meso)-2,3-dibromobutane
Tips: An interesting thing about single bonds or sp3-orbitals is that we can rotate the substituted groups that attached to a stereocenter around to recognize the internal plane. As the molecule is rotated, its stereochemistry does not change. For example:
Another case is when we rotate the whole molecule by 180 degree. Both molecules below are still meso.
Remember the internal plane here is depicted on two dimensions. However, in reality, it is three dimensions, so be aware of it when we identify the internal mirror.
Example \(1\):
1 has a plane of symmetry (the horizontal plane going through the red broken line) and, therefore, is achiral; 1 has chiral centers. Thus, 1 is a meso compound.
Example \(2\):
This molecules has a plane of symmetry (the vertical plane going through the red broken line perpendicular to the plane of the ring) and, therefore, is achiral, but has has two chiral centers. Thus, its is a meso compound.
Exercise \(1\)
Which of the following are meso-compounds:
1. C – 2,3-dibromobutane
2. D – 2,3-dibromopentane
Answer
Compounds A and C are meso.
Other Examples of meso compounds
Meso compounds can exist in many different forms such as pentane, butane, heptane, and even cyclobutane. They do not necessarily have to be two stereocenters, but can have more.
The chiral centers in the preceding examples have all been different. In the case of 2,3-dihydroxybutanedioic acid, known as tartaric acid, the two chiral centers have the same four substituents and are equivalent. As a result, two of the four possible stereoisomers of this compound are identical due to a plane of symmetry, so there are only three stereoisomeric tartaric acids. Two of these stereoisomers are enantiomers and the third is an achiral diastereomer, called a meso compound. Meso compounds are achiral (optically inactive) diastereomers of chiral stereoisomers. Investigations of isomeric tartaric acid salts, carried out by Louis Pasteur in the mid 19th century, were instrumental in elucidating some of the subtleties of stereochemistry. Some physical properties of the isomers of tartaric acid are given in the following table.
(+)-tartaric acid: [α]D = +13º m.p. 172 ºC
(–)-tartaric acid: [α]D = –13º m.p. 172 ºC
meso-tartaric acid: [α]D = 0º m.p. 140 ºC
Fischer projection formulas provide a helpful view of the configurational relationships within the structures of these isomers. In the following illustration a mirror line is drawn between formulas that have a mirror-image relationship. In demonstrating the identity of the two meso-compound formulas, remember that a Fischer projection formula may be rotated 180º in the plane.
Optical Activity Analysis
When the optical activity of a meso compound is attempted to be determined with a polarimeter, the indicator will not show (+) or (-). It simply means there is no certain direction of rotation of the polarized light, neither levorotatory (-) and dexorotatory (+).
Achiral Diastereomers (meso-Compounds)
Exercise \(1\)
Beside meso, there are also other types of molecules: enantiomer, diastereomer, and identical. Determine if the following molecules are meso.
Answer
A C, D, E are meso compounds.
6.06: Isomerism Summary Diagram
Learning Objective
• distinguish and discern the structural and chemical relationships between isomeric compounds
The various types of isomers have been introduced and explored over several chapters. It can be helpful to review, compare, and contrast all of the forms of isomerism to build our skills of discernment. A brief review of each type of isomerism follows the summary diagram. See the respective chapter for a complete explanation.
Conformational Isomers
The rotation of C–C single bonds both carbon chains creates conformers (the same compound shown in different rotations). Consequently, many different arrangements of the atoms are possible, each corresponding to different degrees of rotation. Differences in three-dimensional structure resulting from rotation about a σ bond are called differences in conformation, and each different arrangement is called a conformational isomer (or conformer). While complete rotation of C-C single bonds is not possible in rings. The freedom of bond movement does allow the rings to assume different conformations, such as the chair and boat for 6-membered rings.
Structural (Constitutional) Isomers
Unlike conformational isomers, structural isomers differ in connectivity, as illustrated below for 1-propanol and 2-propanol. Although these two alcohols have the same molecular formula C3H8O, the position of the –OH group differs creating a unique compounds with differences in their physical and chemical properties.
Consider, for example, the following five structures represented by the formula C5H12. In the conversion of one structural isomer to another, at least one bond must be broken and reformed at a different position in the molecule.
Structures (a) and (d) above represent the same compound, n-pentane. Structures (b) and (c) represent the same compound, 2-methylbutane. No bonds need to be broken and reformed to convert between (a) and (d) or between (b) and (c). The molecules are simply rotated 180° about a vertical axis. Structure (e) is named 2,2-dimethylpropane. There are only three structural isomers possible with the chemical formula C5H12: n-pentane, 2-methylbutane, and 2,2-dimethylpropane. Structural isomers have distinct physical and chemical properties.
Stereoisomers
Enantiomers are pairs of compounds that are non-superimposable images. When there are two or more chiral centers in a compounds, the diatereomers can exist. Diastereomers are stereoisomers that are NOT enantiomers. Enatiomers share all physical properties except for their interaction with plane polarized light. Diastereomers have different physical properties (melting points and boiling points and densities).
Exercise
1. What kind of isomers are the following pairs? Note: It can be difficult to answer this question directly from the names. It can be helpful to draw the structures.
1. (R)-5-chlorohexene and 6-chlorohexene
2. (2R,3R)-dibromohexane and (2R,3S)-dibromohexane
Answer
1.
a. Structural Isomers
b. Diastereomers | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.05%3A_Meso_Compounds.txt |
Learning Objective
• define and explain the lack of optical activity of racemic mixtures
• determine the percent composition of an enantiomeric mixture from polarimetry data and the for specific rotation formula
Racemic Mixtures (Racimates)
A racemic mixture is a 50:50 mixture of two enantiomers. Racemic mixtures were an interesting experimental discovery because two optically active samples can be combined in a 1:1 ratio to create an optically INACTIVE sample. Polarimetry is used to measure optical activity. The history and theoretical foundation are discussed below.
Identifying and distinguishing enantiomers is inherently difficult, since their physical and chemical properties are largely identical. Fortunately, a nearly two hundred year old discovery by the French physicist Jean-Baptiste Biot has made this task much easier. This discovery disclosed that the right- and left-handed enantiomers of a chiral compound perturb plane-polarized light in opposite ways. This perturbation is unique to chiral molecules, and has been termed optical activity.
Polarimetry
Plane-polarized light is created by passing ordinary light through a polarizing device, which may be as simple as a lens taken from polarizing sun-glasses. Such devices transmit selectively only that component of a light beam having electrical and magnetic field vectors oscillating in a single plane. The plane of polarization can be determined by an instrument called a polarimeter, shown in the diagram below.
Monochromatic (single wavelength) light, is polarized by a fixed polarizer next to the light source. A sample cell holder is located in line with the light beam, followed by a movable polarizer (the analyzer) and an eyepiece through which the light intensity can be observed. In modern instruments an electronic light detector takes the place of the human eye. In the absence of a sample, the light intensity at the detector is at a maximum when the second (movable) polarizer is set parallel to the first polarizer (α = 0º). If the analyzer is turned 90º to the plane of initial polarization, all the light will be blocked from reaching the detector.
Chemists use polarimeters to investigate the influence of compounds (in the sample cell) on plane polarized light. Samples composed only of achiral molecules (e.g. water or hexane), have no effect on the polarized light beam. However, if a single enantiomer is examined (all sample molecules being right-handed, or all being left-handed), the plane of polarization is rotated in either a clockwise (positive) or counter-clockwise (negative) direction, and the analyzer must be turned an appropriate matching angle, α, if full light intensity is to reach the detector. In the above illustration, the sample has rotated the polarization plane clockwise by +90º, and the analyzer has been turned this amount to permit maximum light transmission.
The observed rotations ($\alpha$) of enantiomers are opposite in direction. One enantiomer will rotate polarized light in a clockwise direction, termed dextrorotatory or (+), and its mirror-image partner in a counter-clockwise manner, termed levorotatory or (–). The prefixes dextro and levo come from the Latin dexter, meaning right, and laevus, for left, and are abbreviated d and l respectively. If equal quantities of each enantiomer are examined , using the same sample cell, then the magnitude of the rotations will be the same, with one being positive and the other negative. To be absolutely certain whether an observed rotation is positive or negative it is often necessary to make a second measurement using a different amount or concentration of the sample. In the above illustration, for example, α might be –90º or +270º rather than +90º. If the sample concentration is reduced by 10%, then the positive rotation would change to +81º (or +243º) while the negative rotation would change to –81º, and the correct α would be identified unambiguously.
Since it is not always possible to obtain or use samples of exactly the same size, the observed rotation is usually corrected to compensate for variations in sample quantity and cell length. Thus it is common practice to convert the observed rotation, α, to a specific rotation, by the following formula:
$[\alpha]_D = \dfrac{\alpha}{l c} \tag{5.3.1}$
where
• $[\alpha]_D$ is the specific rotation
• $l$ is the cell length in dm
• $c$ is the concentration in g/ml
• $D$ designates that the light used is the 589 line from a sodium lamp
Compounds that rotate the plane of polarized light are termed optically active. Each enantiomer of a stereoisomeric pair is optically active and has an equal but opposite-in-sign specific rotation. Specific rotations are useful in that they are experimentally determined constants that characterize and identify pure enantiomers. For example, the lactic acid and carvone enantiomers discussed earlier have the following specific rotations.
Carvone from caraway: [α]D = +62.5º this isomer may be referred to as (+)-carvone or d-carvone
Carvone from spearmint: [α]D = –62.5º this isomer may be referred to as (–)-carvone or l-carvone
Lactic acid from muscle tissue: [α]D = +2.5º this isomer may be referred to as (+)-lactic acid or d-lactic acid
Lactic acid from sour milk: [α]D = –2.5º this isomer may be referred to as (–)-lactic acid or l-lactic acid
A 50:50 mixture of enantiomers has no observable optical activity. Such mixtures are called racemates or racemic modifications, and are designated (±). When chiral compounds are created from achiral compounds, the products are racemic unless a single enantiomer of a chiral co-reactant or catalyst is involved in the reaction. The addition of HBr to either cis- or trans-2-butene is an example of racemic product formation (the chiral center is colored red in the following equation).
CH3CH=CHCH3 + HBr (±) CH3CH2CHBrCH3
Chiral organic compounds isolated from living organisms are usually optically active, indicating that one of the enantiomers predominates (often it is the only isomer present). This is a result of the action of chiral catalysts we call enzymes, and reflects the inherently chiral nature of life itself. Chiral synthetic compounds, on the other hand, are commonly racemates, unless they have been prepared from enantiomerically pure starting materials.
There are two ways in which the condition of a chiral substance may be changed:
1. A racemate may be separated into its component enantiomers. This process is called resolution.
2. A pure enantiomer may be transformed into its racemate. This process is called racemization.
Enantiomeric Excess
The "optical purity" is a comparison of the optical rotation of a pure sample of unknown stereochemistry versus the optical rotation of a sample of pure enantiomer. It is expressed as a percentage. If the sample only rotates plane-polarized light half as much as expected, the optical purity is 50%.
Because R and S enantiomers have equal but opposite optical activity, it naturally follows that a 50:50 racemic mixture of two enantiomers will have no observable optical activity. If we know the specific rotation for a chiral molecule, however, we can easily calculate the ratio of enantiomers present in a mixture of two enantiomers, based on its measured optical activity. When a mixture contains more of one enantiomer than the other, chemists often use the concept of enantiomeric excess (ee) to quantify the difference. Enantiomeric excess can be expressed as:
For example, a mixture containing 60% R enantiomer (and 40% S enantiomer) has a 20% enantiomeric excess of R: ((60-50) x 100) / 50 = 20 %.
Example
The specific rotation of (S)-carvone is (+)61°, measured 'neat' (pure liquid sample, no solvent). The optical rotation of a neat sample of a mixture of R and S carvone is measured at (-)23°. Which enantiomer is in excess, and what is its ee? What are the percentages of (R)- and (S)-carvone in the sample?
Solution
The observed rotation of the mixture is levorotary (negative, counter-clockwise), and the specific rotation of the pure S enantiomer is given as dextrorotary (positive, clockwise), meaning that the pure R enantiomer must be levorotary, and the mixture must contain more of the R enantiomer than of the S enantiomer.
Rotation (R/S Mix) = [Fraction(S) × Rotation (S)] + [Fraction(R) × Rotation (R)]
Let Fraction (S) = x, therefore Fraction (R) = 1 – x
Rotation (R/S Mix) = x[Rotation (S)] + (1 – x)[Rotation (R)]
–23 = x(+61) + (1 – x)(–61)
Solve for x: x = 0.3114 and (1 – x) = 0.6885
Therefore the percentages of (R)- and (S)-carvone in the sample are 68.9% and 31.1%, respectively.
ee = [(% more abundant enantiomer – 50) × 100]/50
= [68.9 – 50) × 100]/50 = 37.8%
Chiral molecules are often labeled according to the sign of their specific rotation, as in (S)-(+)-carvone and (R)-(-)-carvone, or (±)-carvone for the racemic mixture. However, there is no relationship whatsoever between a molecule's R/S designation and the sign of its specific rotation. Without performing a polarimetry experiment or looking in the literature, we would have no idea that (-)-carvone has the R configuration and (+)-carvone has the S configuration.
Separation of Chiral Compounds
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. For example, if a racemic mixture of a chiral alcohol is reacted with a enantiomerically pure carboxylic acid, the result is a mixture of diastereomers: in this case, because the pure (R) entantiomer of the acid was used, the product is a mixture of (R-R) and (R-S) diastereomeric esters, which can, in theory, be separated by their different physical properties. Subsequent hydrolysis of each separated ester will yield the 'resolved' (enantiomerically pure) alcohols. The used in this technique are known as '
Exercise 1
A sample with a concentration of 0.3 g/mL was placed in a cell with a length of 5 cm. The resulting rotation at the sodium D line was +1.52°. What is the [α]D?
Solution
5 cm = 0.5 dm
[α]D = α/(c x l) = +1.52/(0.3 x 0.5) = +10.1° | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.07%3A_Optical_Activity_and_Racemic_Mixtures.txt |
Learning Objective
• explain how to resolve (separate) a pair of enantiomers
Introduction and Overview
A racemic mixture is a 50:50 mixture of two enantiomers. Because they are mirror images, each enantiomer rotates plane-polarized light in an equal but opposite direction and is optically inactive. If the enantiomers are separated, the mixture is said to have been resolved. A common experiment in the laboratory component of introductory organic chemistry involves the resolution of a racemic mixture.
The dramatic biochemical consequences of chirality are illustrated by the use, in the 1950s, of the drug Thalidomide, a sedative given to pregnant women to relieve morning sickness. It was later realized that while the (+)‑form of the molecule, was a safe and effective sedative, the (−)‑form was an active teratogen. The drug caused numerous birth abnormalities when taken in the early stages of pregnancy because it contained a mixture of the two forms.
Chiral Resolution
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. For example, if a racemic mixture of a chiral alcohol is reacted with a enantiomerically pure carboxylic acid, the result is a mixture of diastereomers: in this case, because the pure (R) entantiomer of the acid was used, the product is a mixture of (R-R) and (R-S) diastereomeric esters, which can, in theory, be separated by their different physical properties. Subsequent hydrolysis of each separated ester will yield the 'resolved' (enantiomerically pure) alcohols. The used in this technique are known as 'Moscher's esters', after Harry Stone Moscher, a chemist who pioneered the method at Stanford University.
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. Reversing the first reaction then leads to the separated enantiomers plus the recovered reagent.
Many kinds of chemical and physical reactions, including salt formation, may be used to achieve the diastereomeric intermediates needed for separation. Figure 5.8.1 illustrates this general principle by showing how a nut having a right-handed thread (R) could serve as a "reagent" to discriminate and separate a mixture of right- and left-handed bolts of identical size and weight. Only the two right-handed partners can interact to give a fully-threaded intermediate, so separation is fairly simple. The resolving moiety, i.e. the nut, is then removed, leaving the bolts separated into their right and left-handed forms. Chemical reactions of enantiomers are normally not so dramatically different, but a practical distinction is nevertheless possible.
Because the physical properties of enantiomers are identical, they seldom can be separated by simple physical methods, such as fractional crystallization or distillation. It is only under the influence of another chiral substance that enantiomers behave differently, and almost all methods of resolution of enantiomers are based upon this fact. We include here a discussion of the primary methods of resolution
Chiral Amines as Resolving Agents and Resolution of Racemic Acids
The most commonly used procedure for separating enantiomers is to convert them to a mixture of diastereomers that will have different physical properties: melting point, boiling point, solubility, and so on (Section 5-5). For example, if you have a racemic or D,L mixture of enantiomers of an acid and convert this to a salt with a chiral base having the D configuration, the salt will be a mixture of two diastereomers, (D acid . D base) and (L acid . D base). These diastereomeric salts are not identical and they are not mirror images. Therefore they will differ to some degree in their physical properties, and a separation by physical methods, such as crystallization, may be possible. If the diastereomeric salts can be completely separated, the acid regenerated from each salt will be either exclusively the D or the L enantiomer:
Resolution of chiral acids through the formation of diastereomeric salts requires adequate supplies of suitable chiral bases. Brucine, strychnine, and quinine frequently are used for this purpose because they are readily available, naturally occurring chiral bases. Simpler amines of synthetic origin, such as 2-amino- 1 -butanol, amphetamine, and 1 -phenylethanamine, also can be used, but first they must be resolved themselves.
Resolution of Racemic Bases
Chiral acids, such as (+)-tartaric acid, (-)-malic acid, (-)-mandelic acid, and (+)-camphor- 10-sulfonic acid, are used for the resolution of a racemic base.
The principle is the same as for the resolution of a racemic acid with a chiral base, and the choice of acid will depend both on the ease of separation of the diastereomeric salts and, of course, on the availability of the acid for the scale of the resolution involved. Resolution methods of this kind can be tedious, because numerous recrystallizations in different solvents may be necessary to progressively enrich the crystals in the less-soluble diastereomer. To determine when the resolution is complete, the mixture of diastereomers is recrystallized until there is no further change in the measured optical rotation of the crystals. At this stage it is hoped that the crystalline salt is a pure diastereomer from which one pure enantiomer can be recovered. The optical rotation of this
enantiomer will be a maximum value if it is "optically" pure because any amount of the other enantiomer could only reduce the magnitude of the measured rotation $\alpha$.
Resolution of Racemic Alcohols
To resolve a racemic alcohol, a chiral acid can be used to convert the alcohol to a mixture of diastereomeric esters. This is not as generally useful as might be thought because esters tend to be liquids unless they are very high-molecularweight compounds. If the diastereomeric esters are not crystalline, they must be separated by some other method than fractional crystallization (for instance, by chromatography methods, Section 9-2). Two chiral acids that are useful resolving agents for alcohols are:
The most common method of resolving an alcohol is to convert it to a half-ester of a dicarboxylic acid, such as butanedioic (succinic) or 1,2-benzenedicarboxylic (phthalic) acid, with the corresponding anhydride. The resulting half-ester has a free carboxyl function and may then be resolvable with a chiral base, usually brucine:
Other Methods of Resolution
Qne of the major goals in the field of organic chemistry is the development of reagents with the property of "chiral recognition" such that they can effect a clean separation of enantiomers in one operation without destroying either of the enantiomers. We have not achieved that ideal yet, but it may not be far in the future. Chromatographic methods (Section 9-2), whereby the stationary phase is a chiral reagent that adsorbs one enantiomer more strongly than the other, have been used to resolve racemic compounds, but such resolutions seldom have led to both pure enantiomers on a preparative scale. Other methods, called kinetic resolutions, are excellent when applicable. The procedure takes advantage of differences in reaction rates of enantiomers with chiral reagents. One enantiomer may react more rapidly, thereby leaving an excess of the other enantiomer behind. For example, racemic tartaric acid
can be resolved with the aid of certain penicillin molds that consume the dextrorotatory enantiomer faster than the levorotatory enantiomer. As a result, almost pure (-)-tartaric acid can be recovered from the mixture:
(±)-tartaric acid + mold $\rightarrow$ (-)-tartaric acid + more mold
A disadvantage of resolutions of this type is that the more reactive enantiomer usually is not recoverable from the reaction mixture.
The crystallization procedure employed by Pasteur for his classical resolution of (±)-tartaric acid (Section 5-1C) has been successful only in a very few cases. This procedure depends on the formation of individual crystals of each enantiomer. Thus if the crystallization of sodium ammonium tartrate is carried out below 27", the usual racemate salt does not form; a mixture of crystals of the (+) and (-) salts forms instead. The two different kinds of crystals, which are related as an object to its mirror image, can be separated manually with the aid of a microscope and subsequently may be converted to the tartaric acid enantiomers by strong acid. A variation on this method of resolution is the seeding of a saturated solution of a racemic mixture with crystals of one pure enantiomer in the hope of causing crystallization of just that one enantiomer, thereby leaving the other in solution. Unfortunately, very
few practical resolutions have been achieved in this way.
Even when a successful resolution is achieved, some significant problems remain. For instance, the resolution itself does not provide information on the actual configuration of the (+) or (-) enantiomer. This must be determined by other means (see Section 19-5). Also, it is not possible to tell the enantiomeric purity (optical purity) of the resolved enantiomers without additional information. This point is discussed further in the next section.
Exercise
1. Indicate the reagents you would use to resolve the following compounds. Show the reactions involved and specify the physical method you believe would be the best to separate the diastereomers.
1. 1 -phenyl-2-propanamine
2. 2,3-pentadienedioic acid
3. 1 -phenylethanol
Solutions:
1.
a. React 1-phenyl-2-propanamine racemic mixture with a chiral acid such as (+)-tartaric acid (R, R).
Reaction will produce a mixture of diastereomeric salts (i.e. R, R, R and S, R, R).
Separate diastereomers through crystallization.
Treat salt with strong base (e.g. KOH) to recover the pure enantiomeric amine.
b. React 2,3-pentadienedioic acid mixture with a chiral base such as (R)‑1‑phenylethylamine.
Reaction will produce a mixture of diastereomeric salts.
Separate diastereomers through crystallization.
Treat salt with strong acid (e.g. HCl) to recover the pure enantiomer acid.
c. React 1-phenylethanol mixture with 1,2-benzenedicarboxylic anhydride.
Reaction will produce a mixture of diastereomeric salts.
Separate diastereomers through crystallization.
Then alkaline hydrolysis treatment to recover the pure enantiomeric alcohol.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.08%3A_Resolution_%28Separation%29_of_Enantiomers.txt |
Learning Objective
• interpret the stereoisomerism of compounds with three or more chiral centers
Possible Number of Stereoisomers
In general, a structure with n stereocenters will have 2n different stereoisomers. (We are not considering, for the time being, the stereochemistry of double bonds – that will come later). For example, let's consider the glucose molecule in its open-chain form (recall that many sugar molecules can exist in either an open-chain or a cyclic form). There are two enantiomers of glucose, called D-glucose and L-glucose. The D-enantiomer is the common sugar that our bodies use for energy. It has n = 4 stereocenters, so therefore there are 2n = 24 = 16 possible stereoisomers (including D-glucose itself).
In L-glucose, all of the stereocenters are inverted relative to D-glucose. That leaves 14 diastereomers of D-glucose: these are molecules in which at least one, but not all, of the stereocenters are inverted relative to D-glucose. One of these 14 diastereomers, a sugar called D-galactose, is shown above: in D-galactose, one of four stereocenters is inverted relative to D-glucose. Diastereomers which differ in only one stereocenter (out of two or more) are called epimers. D-glucose and D-galactose can therefore be refered to as epimers as well as diastereomers.
The epimer term is useful because in biochemical pathways, compounds with multiple chiral centers are isomerized at one specific center by enzymes known as epimerases. Two examples of epimerase-catalyzed reactions are below.
Now, let's extend our analysis to a sugar molecule with three chiral centers. Going through all the possible combinations, we come up with eight total stereoisomers - four pairs of enantiomers.
Let's draw the RRR stereoisomer. Being careful to draw the wedge bonds correctly so that they match the RRR configurations, we get:
Now, using the above drawing as our model, drawing any other stereoisomer is easy. If we want to draw the enantiomer of RRR, we don't need to try to visualize the mirror image, we just start with the RRR structure and invert the configuration at every chiral center to get SSS.
Try making models of RRR and SSS and confirm that they are in fact nonsuperimposable mirror images of each other.
There are six diastereomers of RRR. To draw one of them, we just invert the configuration of at least one, but not all three, of the chiral centers. Let's invert the configuration at chiral center 1 and 2, but leave chiral center 3 unchanged. This gives us the SSR configuration.
One more definition at this point: diastereomers which differ at only a single chiral center are called epimers. For example, RRR and SRR are epimers:
The RRR and SSR stereoisomers shown earlier are diastereomers but not epimers because they differ at two of the three chiral centers.
Example \(1\)
1. Draw the structure of the enantiomer of the SRS stereoisomer of the sugar used in the previous example.
2. List (using the XXX format, not drawing the structures) all of the epimers of SRS.
3. List all of the stereoisomers that are diastereomers, but not epimers, of SRS.
Solutions to exercises
Solution
Add text here.
Example \(2\)
The sugar below is one of the stereoisomers that we have been discussing.
The only problem is, it is drawn with the carbon backbone in a different orientation from what we have seen. Determine the configuration at each chiral center to determine which stereoisomer it is.
Exercise \(3\)
Draw the enantiomer of the xylulose-5-phosphate structure in the previous figure.
Exercise \(4\)
The structure of the amino acid D-threonine, drawn without stereochemistry, is shown below. D-threonine has the (S) configuration at both of its chiral centers. Draw D-threonine, it's enantiomer, and its two diastereomers.
Answer
Solutions to exercises
Comparing Stereoisomerism with Structural Isomerism
D-glucose and D-fructose are not stereoisomers, because they have different bonding connectivity: glucose has an aldehyde group, while fructose has a ketone. The two sugars do, however, have the same molecular formula, so by definition they are constitutional isomers.
D-glucose and D-ribose are not isomers of any kind, because they have different molecular formulas.
Exercise 5: Identify the relationship between each pair of structures. Your choices are: not isomers, constitutional isomers, diastereomers but not epimers, epimers, enantiomers, or same molecule
Exercise 6: Identify the relationship between each pair of structures. Hint - figure out the configuration of each chiral center.
Solutions to exercises
Kahn Academy video tutorial on stereoisomeric relationships
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.09%3A_Stereochemistry_of_Molecules_with_Three_or_More_Asymm.txt |
Learning Objective
• compare and contrast absolute configuration with relative configuration
Absolute Configuration
The absolute configuration at a chiral center in a molecule is a time-independent and unambiguous symbolic description of the spatial arrangement of ligands (groups) bonded to the chiral center.
The chiral centers in 1 and 2 bear the same ligands: a,b,d, and e. However, 1 and 2 are not superimposable on each other, meaning that the arrangement of ligands around the chiral center in 1 and in 2 is different. 1 and 2 are mirror images of each other, meaning that the arrangement of ligands around the chiral center in 1 is the exact opposite of that in 2. Chiral centers in 1and 2 are said to have opposite absolute configurations.
According to R,S convention, if the absolute configuration at the chiral center in 1 is R, that at the chiral center in 2 is S or vice versa.
Relative Configuration
The relative configuration is the experimentally determined relationship between two enantiomers even though we may not know the absolute configuration. The sign of rotation of plane-polarized light by an enantiomer is not easily related to its configuration. This is true even for substances with very similar structures. Thus, given lactic acid, $\ce{CH_3CHOHCO_2H}$, with a specific rotation $+3.82^\text{o}$, and methyl lactate, $\ce{CH_3CHOHCO_2CH_3}$, with a specific rotation $-8.25^\text{o}$, we cannot tell from the rotation alone whether the acid and ester have the same or a different arrangement of groups about the chiral center. Their relative configurations have to be obtained by other means.
If we convert $\left( + \right)$-lactic acid into its methyl ester, we can be reasonably certain that the ester will be related in configuration to the acid, because esterification should not affect the configuration about the chiral carbon atom. It happens that the methyl ester so obtained is levorotatory, so we know that $\left( + \right)$-lactic acid and $\left( - \right)$-methyl lactate have the same relative configuration at the asymmetric carbon, even if they possess opposite signs of optical rotation. However, we still do not know the absolute configuration; that is, we are unable to tell which of the two possible configurations of lactic acid, $2a$ or $2b$, corresponds to the dextro or $\left( + \right)$-acid and which to the levo or $\left( - \right)$-acid:
Until 1956, the absolute configuration of no optically active compound was known. Instead, configurations were assigned relative to a standard, glyceraldehyde, which originally was chosen by E. Fischer (around 1885) for the purpose of correlating the configuration of carbohydrates. Fischer arbitrarily assigned the configuration $3a$ to dextrorotatory glyceraldehyde, which was known as $D$-$\left( + \right)$-glyceraldehyde. The levorotatory enantiomer, $3b$, is designated as $L$-$\left( - \right)$-glyceraldehyde. (If you are unsure of the terminology $D$ and $L$, or of the rules for writing Fischer projection formulas, review Sections 5-3C and 5-4.)
The configurations of many compounds besides sugars now have been related to glyceraldehyde, including $\alpha$-amino acids, terpenes, steroids, and other biochemically important substances. Compounds whose configurations are related to $D$-$\left( + \right)$-glyceraldehyde are said to belong to the $D$ series, and those related to $L$-$\left( - \right)$-glyceraldehyde belong to the $L$ series.
At the time the choice of absolute configuration for glyceraldehyde was made, there was no way of knowing whether the configuration of $\left( + \right)$-glyceraldehyde was in reality $3a$ or $3b$. However, the choice had a $50\%$ chance of being correct, and we now know that $3a$, the $D$ configuration, is in fact the correct configuration of $\left( + \right)$-glyceraldehyde. This was established through use of a special x-ray crystallographic technique, which permitted determination of the absolute disposition of the atoms in space of sodium rubidium $\left( + \right)$-tartrate. The configuration of $\left( + \right)$-tartaric acid (Section 5-5) previously had been shown by chemical means to be opposite to that of $\left( + \right)$-glyceraldehyde. Consequently the absolute configuration of any compound now is known once it has been correlated directly or indirectly with glyceraldehyde. For example,
Chemical transformation showing how the configuration of natural $\left( + \right)$-alanine has been related to $L$-$\left( + \right)$-lactic acid and hence to $L$-$\left( - \right)$-glyceraldehyde. The transformations shown involve two $S_\text{N}2$ reactions, each of which is stereospecific and inverts the configuration (Section 8-5). Reduction of the azide group leaves the configuration unchanged.
When there are several chiral carbons in a molecule, the configuration at one center usually is related directly or indirectly to glyceraldehyde, and the configurations at the other centers are determined relative to the first. Thus in the aldehyde form of the important sugar, $\left( + \right)$-glucose, there are four chiral centers, and so there are $2^4 = 16$ possible stereoisorners. The projection formula of the isomer that corresponds to the aldehyde form of natural glucose is $4$. By convention for sugars, the configuration of the highest-numbered chiral carbon is referred to glyceraldehyde to determine the overall configuration of the molecule. For glucose, this atom is $\ce{C5}$, next to the $\ce{CH_2OH}$ group, and has the hydroxyl group on the right. Therefore, naturally occurring glucose, which has a $\left( + \right)$ rotation, belongs to the $D$ series and is properly called $D$-$\left( + \right)$-glucose:
However, the configurations of $\alpha$-amino acids possessing more than one chiral carbon are determined by the lowest-numbered chiral carbon, which is the carbon alpha to the carboxyl group. Thus, even though the natural $\alpha$-amino acid, threonine, has exactly the same kind of arrangement of substituents as the natural sugar, threose, threonine by the amino-acid convention belongs to the $L$-series, whereas threose by the sugar convention belongs to the $D$-series:
A serious ambiguity arises for compounds such as the active tartaric acids. If the amino-acid convention is used, $\left( + \right)$-tartaric acid falls in the $D$ series; by the sugar convention, it has the $L$ configuration. One way out of this dilemma is to use the subscripts $s$ and $g$ to denote the amino-acid or carbohydrate conventions, respectively. Then the absolute configuration of $\left( + \right)$-tartaric acid can be designated as either $D_s$-$\left( + \right)$-tartaric acid of $L_g$-$\left( + \right)$-tartaric acid.
Contributors and Attributions
• John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
6.11: Chirality at Nitrogen Phosphorus and Sulfur
Learning Objective
• interpret the stereoisomerism of compounds with nitrogen, phosphorus, or sulfur as chiral centers
Stereogenic Nitrogen
Single-bonded nitrogen is pyramidal in shape, with the non-bonding electron pair pointing to the unoccupied corner of a tetrahedral region. Since the nitrogen in these compounds is bonded to three different groups, its configuration is chiral. The non-identical mirror-image configurations are illustrated in the following diagram (the remainder of the molecule is represented by R, and the electron pair is colored yellow). If these configurations were stable, there would be four additional stereoisomers of ephedrine and pseudoephedrine. However, pyramidal nitrogen is normally not configurationally stable. It rapidly inverts its configuration (equilibrium arrows) by passing through a planar, sp2-hybridized transition state, leading to a mixture of interconverting R and S configurations. If the nitrogen atom were the only chiral center in the molecule, a 50:50 (racemic) mixture of R and S configurations would exist at equilibrium. If other chiral centers are present, as in the ephedrin isomers, a mixture of diastereomers will result. The take-home message is that nitrogen does not contribute to isolable stereoisomers.
Asymmetric quaternary ammonium groups are also chiral. Amines, however, are not chiral, because they rapidly invert, or turn ‘inside out’, at room temperature.
The phosphorus center of phosphate ion and organic phosphate esters, for example, is tetrahedral, and thus is potentially a stereocenter.
We will see in chapter 10 how researchers, in order to investigate the stereochemistry of reactions at the phosphate center, incorporated sulfur and/or 17O and 18O isotopes of oxygen (the ‘normal’ isotope is 16O) to create chiral phosphate groups. Phosphate triesters are chiral if the three substituent groups are different. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.10%3A_Absolute_and_Relative_Configuration_-_the_distinction.txt |
Learning Objective
• recognize and explain biochemical applications of chirality
Some Chiral Organic Biomolecules
There are a number of important biomolecules that could occur as enantiomers, including amino acids and sugars. In most cases, only one enantiomer occurs (although some fungi, for example, are able to produce mirror-image forms of these compounds). We will look later at some of these biomolecules, but first we will look at a compound that occurs naturally in both enantiomeric forms.
Carvone is a secondary metabolite. That means it is a naturally-occurring compound that is not directly connected to the very basic functions of a cell, such as self-replication or the production of energy. The role of secondary metabolites in nature is often difficult to determine. However, these compounds often play roles in self-defense, acting as deterrents against competitor species in a sort of small-scale chemical warfare scenario. They are also frequently used in communications; this role has been studied most extensively among insects, which use lots of compounds to send information to each other.
The two naturally-occurring enantiomers of carvone.
Carvone is produced in two enantiomeric forms. One of these forms, called (-)-carvone, is found in mint leaves, and it is a principal contributor to the distinctive odor of mint. The other form, (+)-carvone, is found in caraway seeds. This form has a very different smell, and is typically used to flavor rye bread and other Eastern European foods.
Note that (+)-carvone is the same thing as (S)-carvone. The (+) designation is based on its positive optical rotation value, which is experimentally measured. The (S) designation is determined by the Cahn-Ingold-Prelog rules for designating stereochemistry, which deal with looking at the groups attached to a chiral center and assigning priority based on atomic number. However, carvone's chiral center actually has three carbons attached to it; they all have the same atomic number. We need a new rule to break the tie.
• If two substituent groups have the same atomic number, go one bond further to the next atom.
• If there is a difference among the second tier of atoms, stop.
• The group in which you have encountered a higher atomic number gets the highest priority.
• If there is not a clear difference, proceed one additional bond to the next set of atoms, and so on, until you find a difference.
In carvone, this decision tree works as follows:
• The chiral center is connected to a H, a C, a C and a C.
• The H is lowest priority.
• One C eventually leads to a C=O. However, at the second bond from the chiral center, this C is connected to a C and two H's.
• A second C is also part of the six-membered ring, but the C=O is farther away in this direction. At the second bond from the chiral center, this C is connected to a C and two H's, just like the first one.
• The third C is part of a little three-carbon group attached to the six-membered ring. At the second bond from the chiral center, it is connected to only one H and has two bonds to another C (this is counted as two bonds to C and one to H).
• Those first two carbon groups are identical so far.
• However, the third group is different; it has an extra bond to C, whereas the others have an extra bond to H. C has a higher atomic number than H, so this group has higher priority.
• The second-highest priority is the branch that reaches the oxygen at the third bond from the chiral center.
Comparing atoms step-by-step to assign configuration.
How different, exactly, are these two compounds, (+)- and (-)-carvone? Are they completely different isomers, with different physical properties? In most ways, the answer is no. These two compounds have the same appearance (colorless oil), the same boiling point (230 °C), the same refractive index (1.499) and specific gravity (0.965). However, they have optical rotations that are almost exactly opposite values.
• Two enantiomers have the same physical properties.
• Enantiomers have opposite optical rotations.
Clearly they have different biological properties; since they have slightly different odors, they must fit into slightly different nasal receptors, signaling to the brain whether the person next to you is chewing a stick of gum or a piece of rye bread. This different shape complimentarity is not surprising, just as it isn't surprising that a left hand only fits into a left handed baseball glove and not into a right handed one.
Thalidomide.
There are other reasons that we might concern ourselves with an understanding of enantiomers, apart from dietary and olfactory preferences. Perhaps the most dramatic example of the importance of enantiomers can be found in the case of thalidomide. Thalidomide was a drug commonly prescribed during the 1950's and 1960's in order to alleviate nausea and other symptoms of morning sickness. In fact, only one enantiomer of thalidomide had any therapeutic effect in this regard. The other enantiomer, apart from being therapeutically useless in this application, was subsequently found to be a teratogen, meaning it produces pronounced birth defects. This was obviously not a good thing to prescribe to pregnant women. Workers in the pharmaceutical industry are now much more aware of these kinds of consequences, although of course not all problems with drugs go undetected even through the extensive clinical trials required in the United States. Since the era of thalidomide, however, a tremendous amount of research in the field of synthetic organic chemistry has been devoted to methods of producing only one enantiomer of a useful compound and not the other. This effort probably represents the single biggest aim of synthetic organic chemistry through the last quarter century.
• Enantiomers may have very different biological properties.
• Obtaining enantiomerically pure compounds is very important in medicine and the pharmaceutical industry.
Exercises
1. Draw the two enantiomeric forms of 2-butanol, CH3CH(OH)CH2CH3. Label their configurations.
2. Sometimes, compounds have many chiral centers in them. For the following compounds, identify four chiral centers in each, mark them with asterisks, and identify each center as R or S configuration.
The following is the structure of dysinosin A, a potent thrombin inhibitor that consequently prevents blood clotting.
Ginkgolide B (below) is a secondary metabolite of the ginkgo tree, extracts of which are used in Chinese medicine.
Sanglifehrin A, shown below, is produced by a bacteria that may be found in the soil of coffee plantations in Malawi. It is also a promising candidate for the treatment of organ transplant patients owing to its potent immuno-suppressant activity.
Solution
1.
2.
Contributors and Attributions
• Dr. Dietmar Kennepohl FCIC (Professor of Chemistry, Athabasca University)
• Prof. Steven Farmer (Sonoma State University)
• John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.12%3A_Biochemistry_of_Enantiomers.txt |
Learning Objective
• describe Jean Baptiste Biot and Louis Pasteur's contributions to the understanding of optical isomers
Because enantiomers have identical physical and chemical properties in achiral environments, separation of the stereoisomeric components of a racemic mixture or racemate is normally not possible by the conventional techniques of distillation and crystallization. In some cases, however, the crystal habits of solid enantiomers and racemates permit the chemist (acting as a chiral resolving agent) to discriminate enantiomeric components of a mixture. As background for the following example, it is recommended that the section on crystal properties be reviewed.
Tartaric acid, its potassium salt known in antiquity as "tartar", has served as the locus of several landmark events in the history of stereochemistry. In 1832 the French chemist Jean Baptiste Biot observed that tartaric acid obtained from tartar was optically active, rotating the plane of polarized light clockwise (dextrorotatory). An optically inactive, higher melting, form of tartaric acid, called racemic acid was also known. A little more than a decade later, young Louis Pasteur conducted a careful study of the crystalline forms assumed by various salts of these acids. He noticed that under certain conditions, the sodium ammonium mixed salt of the racemic acid formed a mixture of enantiomorphic hemihedral crystals; a drawing of such a pair is shown on the right. Pasteur reasoned that the dissymmetry of the crystals might reflect the optical activity and dissymmetry of its component molecules. After picking the different crystals apart with a tweezer, he found that one group yielded the known dextrorotatory tartaric acid measured by Biot; the second led to a previously unknown levorotatory tartaric acid, having the same melting point as the dextrorotatory acid. Today we recognize that Pasteur had achieved the first resolution of a racemic mixture, and laid the foundation of what we now call stereochemistry.
Optical activity was first observed by the French physicist Jean-Baptiste Biot. He concluded that the change in direction of plane-polarized light when it passed through certain substances was actually a rotation of light, and that it had a molecular basis. His work was supported by the experimentation of Louis Pasteur. Pasteur observed the existence of two crystals that were mirror images in tartaric acid, an acid found in wine. Through meticulous experimentation, he found that one set of molecules rotated polarized light clockwise while the other rotated light counterclockwise to the same extent. He also observed that a mixture of both, a racemic mixture (or racemic modification), did not rotate light because the optical activity of one molecule canceled the effects of the other molecule. Pasteur was the first to show the existence of chiral molecules.
6.14: Additional Exercises
6-1 For the following compounds, star (*) each chiral center, if any.
6-2 For the following compounds, identify the R or S configuration of each chiral carbon atom.
6-3 Draw out the following molecules, including stereocenters.
a) (2R,4S,6R)-2-bromo-6-chloro-4-methylheptane
b) (4R)-4-bromopent-1-ene
c) (1R,2R,3S)-1-fluoro-2,3-dimethylcyclohexane
d) (3S)-3-methylcyclopent-1-ene
(R) and (S) Nomenclature of Asymmetric Carbon Atoms
6-4 For the following compounds, assign R or S configurations for each stereocenter.
6-5 For the following compounds, assign R or S configurations for each stereocenter.
6-6 Identify each molecule as either (R)- or (S)-Limonene.
Chiral Compounds Without Asymmetric Atoms
6-7 Explain why the following compound is optically active.
6-8 Does the following compound contain a chiral center? Is it a chiral molecule?
6-9 Why is this biaryl compound shown below considered chiral, despite having no chiral center?
Fischer Projections and Diastereomers
6-10 For the following Fischer projections, identify the configuration (R or S) of all chiral centers (some atoms may not be chiral centers).
6-11 For the following pairs of compounds, identify whether they are enantiomers, diastereomers, or the same compound.
6-12 For the following pairs of compounds, identify whether they are enantiomers, diastereomers, or the same compound.
Meso Compounds
6-13 For the following compounds, identify whether they are meso or not meso.
6-14 Are meso compounds optically active? Explain your answer.
6-15 Is the following compound meso or not meso?
6.15: Solutions to Additional Exercises
6-1
6-2
6-3
6-4
6-5
6-6
Chiral Compounds Without Asymmetric Atoms
6-7 Though the molecule does not contain a chiral carbon, it is chiral as it is non-superimposable on its mirror image due to its twisted nature (the twist comes from the structure of the double bonds needing to be at 90° angles to each other, preventing the molecule from being planar). This allows it to be optically active.
6-8 The molecule does not contain a chiral center; however, it is a chiral molecule as it is non-superimposable on its mirror image.
6-9 In the case of this biaryl molecule, the large bulky substituents, located at the ortho positions relative to the sigma bond in the middle, experience enough steric interference with each other to create a large energy barrier to free rotation around the C-C sigma bond. Thus, the molecule cannot freely rotate to its other conformations and is non-superimposable on its mirror image.
6-10
6-11
a) Enantiomers
b) Diastereomers
c) Diastereomers
d) Same compound
6-12
a) Enantiomers
b) Diastereomers
c) Diastereomers
Meso Compounds
6-13
a) Meso
b) Not meso
c) Meso
d) Meso
e) Meso
f) Not meso
6-14 Meso compounds are not optically active as they can be superimposed on their mirror images, making them achiral (which are not optically active). The stereocenters on one half of the molecule will rotate light one direction, while the other half of the molecule will rotate light the opposite direction, giving a net rotation of zero and making the molecule optically inactive.
6-15 The compound is meso, since the opposing stereocenters have opposite absolute configurations and if the molecule is rotated about the sigma bond in the middle, you can see that the two halves of the compound are mirror images of each other. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.13%3A_The_Discovery_of_Enantiomers.txt |
Learning Objectives
After reading the chapter and completing the exercises and homework, a student can be able to:
• classify alkyl halides - refer to section 7.1
• predict relative boiling points and solubility of alkyl halides - refer to section 7.1
• discuss the common uses of alkyl halides - refer to section 7.2
• specify the reagents for the most efficient synthesis of alkyl halides using free-radical halogenation of alkanes (Chapter 5) or allylic halogenation of alkenes with NBS - refer to section 7.3
• apply the alpha and beta labels to alkyl halides for substitution and elimination reactions - refer to section 7.4
• determine the rate law & predict the mechanism based on its rate equation or reaction data for SN1, SN2, E1 & E2 reactions - refer to sections 7.5, 7.6, 7.8, 7.13, and 7.15
• use Zaitsev’s rule to predict major and minor products of elimination reactions including halocyclohexanes - refer to sections 7.14, 7.15, and 7.16
• predict the products and specify the reagents for SN1, SN2, E1 and E2 reactions with stereochemistry - refer to sections 7.6, 7.7, 7.9, 7.14, 7.15, 7.19
• propose mechanisms for SN1, SN2, E1 and E2 reactions - refer to sections 7.5, 7.6, 7.7, 7.8, 7.9, 7.13, 7.14, 7.15, 7.19
• draw, interpret, and apply Reaction Energy Diagrams for SN1, SN2, E1 and E2 reactions - refer to sections 7.5, 7.6, 7.7, 7.8, 7.9, 7.13, 7.14, 7.15, 7.19
• predict carbocation rearrangements in 1st order reactions - refer to section 7.10
• explain and apply Hammond's Postulate to substitution reactions - refer to section 7.11
• explain how the kinetic isotope effect (KIE) can be used to elucidate reaction mechanisms - refer to section 7.17
• distinguish 1st or 2nd order substitution and elimination reactions - refer to sections 7.12 and 7.18
• discuss the importance of leaving groups in biological substitution reactions - refer to section 7.20
• discuss enzymatic elimination reactions of histidine - refer to section 7.21
• 7.1: Alkyl Halides - Structure and Physical Properties
Alkyl halides are classified based upon the structure of the carbon atom bonded to the halogen. Common names and physical properties are discussed.
• 7.2: Common Uses of Alkyl Halides
Halogen containing organic compounds are relatively rare in terrestrial plants and animals, with the thyroid hormones T3 and T4 as notable exceptions. Alkyl halides are excellent electrophiles and quickly become an o-chem student's best friend for synthetic pathways.
• 7.3: Preparation of Alkyl Halides
Alkyl halides can be readily synthesized from alkanes, alkenes, and alcohols. This section expands the ways we can brominate tetrahedral carbons to the allylic position of alkenes.
• 7.4: Reactions of Alkyl Halides- Substitution and Elimination
The two major reaction pathways for alkyl halides (substitution and elimination) are introduced.
• 7.5: The Sₙ2 Reaction
The SN2 mechanism is described mechanistically and kinetically as a one-step (concerted) reaction between two reactants (bimolecular) that inverts the configuration of the carbon at the reactive site. The terms nucleophile, electrophile, and leaving group are explained by application to SN2 reactions.
• 7.6: Characteristics of the Sₙ2 Reaction
In order of decreasing importance, the factors impacting SN2 reaction pathways are the structure of the alkyl halide, the strength of the nucleophile, the stability of the leaving group, and the type of solvent.
• 7.7: Stereochemistry of the SN2 Reaction
The SN2 reaction is stereospecific. A stereospecific reaction is one in which different stereoisomers react to give different stereoisomers of the product.
• 7.8: The Sₙ1 Reaction
In the SN1 reaction, the solvent helps pull apart the halogen and carbon to form a halide and carbocation. A nucleophile can now form a bond with the carbocation to create a new product. The mechanism is explained with stereochemistry and reaction kinetics.
• 7.9: Characteristics of the Sₙ1 Reaction
The formation and stability of the carbocation intermediate strongly influence the SN1 mechanism. The structure of the alkyl halide, the stability of the leaving group, and the type of solvent influence the reaction pathway. Since the nucleophile is not involved in the rate determining step, the strength of the nucleophile has low importance.
• 7.10: Rearrangements of the Carbocation and Sₙ1 Reactions
Carbocation rearrangements are structural reorganizational shifts within an ion to a more stable state (lower energy).
• 7.11: The Hammond Postulate and Transition States
The Hammond postulate states that a transition state resembles the structure of the nearest stable species and helps explain the product distribution differences observed between exergonic and endergonic reactions.
• 7.12: Comparison of SN1 and SN2 Reactions
In comparing the SN1 and SN2 mechanisms, the structure of the alkyl halide (electrophile), the strength of the nucleophile, and the reaction solvent are the primary considerations. The leaving group will have a similar effect for both reactions, so it is not of interest when comparing the mechanistic pathways.
• 7.13: Characteristics of the E2 Reaction
E2, bimolecular elimination, was proposed in the 1920s by British chemist Christopher Kelk Ingold. In E2 reactions, a beta-hydrogen and the leaving group are eliminated from an alkyl halide in reaction with a strong base to form an alkene.
• 7.14: Zaitsev's Rule
Zaitsev's Rule can be used to predict the regiochemistry of elimination reactions. Regiochemistry describes the orientation of reactions about carbon-carbon double bonds (C=C).
• 7.15: Characteristics of the E1 Reaction
The unimolecular E1 mechanism is a first order elimination reaction in which carbocation formation and stability are the primary factors for determining reaction pathway(s) and product(s).
• 7.16: E2 Regiochemistry and Cyclohexane Conformations
Cyclohexyl halides provides the perfect opportunity to learn and understand the regiochemistry of the E2 reaction and why Zaitsev's Rule does not always apply. The anti-coplanar orientation of the E2 mechanism can also be see with certain carbon chain diastereomers.
• 7.17: The E2 Reaction and the Deuterium Isotope Effect
The bimolecular transition state of the E2 reaction illustrates the effects of bond strength on reaction rates when studying the kinetic isotope effect of deuterium.
• 7.18: Comparison of E1 and E2 Reactions
The strength of the base is the primary consideration when distinguishing between the E1 and E2 pathways. The reaction solvent is a secondary consideration.
• 7.19: Comparing Substitution and Elimination Reactions
Chemical reactivity patterns can help us determine the most favorable pathway among the closely related SN1, SN2, E1, and E2 mechanisms.
• 7.20: Biological Substitution Reactions
A few examples of biochemical SN1 and SN2 mechanisms are introduced with an emphasis on the effects of the leaving group.
• 7.21: Biological Elimination Reactions
A few examples of biochemical E1 and E2 mechanisms are introduced.
• 7.22: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
• 7.23: Solutions to Additional Exercises
This section has the solutions to the additional exercises from the previous section.
07: Alkyl Halides- Nucleophilic Substitution and Elimination
Learning Objective
• classify alkyl halides
• predict relative boiling points and solubility of alkyl halides
Introduction
Alkyl halides are also known as haloalkanes. Alkyl halides are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). We will only look at compounds containing one halogen atom like th compounds below.
Alkyl halides fall into different classes depending on how the halogen atom is positioned on the chain of carbon atoms. Alkyl halides can be classified as primary, secondary, or tertiary. The chemical reactivity of alkyl halides is frequently discussed using alkyl halide classifications to help discern patterns and trends. Because the neutral bonding pattern for halogens is one bond and three lone pairs, the carbon and halogen always share a single bond. Alkyl halide classification is determined by the bonding pattern of the carbon atom bonded to the halogen as shown in the diagram below.
Primary alkyl halides
In a primary (1°) haloalkane, the carbon bonded to the halogen atom is only attached to one other alkyl group. Some examples of primary alkyl halides include thecompounds below.
Notice that it doesn't matter how complicated the attached alkyl group is. In each case there is only one linkage to an alkyl group from the CH2 group holding the halogen. There is an exception to this: CH3Br and the other methyl halides are often counted as primary alkyl halides even though there are no alkyl groups attached to the carbon with the halogen on it.
Secondary alkyl halides
In a secondary (2°) haloalkane, the carbon bonded with the halogen atom is joined directly to two other alkyl groups that can be the same or different. Some examples of secondary alkyl halides include thecompounds below.
Tertiary alkyl halides
In a tertiary (3°) halogenoalkane, the carbon atom holding the halogen is attached directly to three alkyl groups, which may be any combination of same or different. Some examples of tertiary alkyl halides include thecompounds below.
Common Names
Many organic compounds are closely related to the alkanes and this similarity is incorporated into many common names. The reactions of alkanes with halogens produce halogenated hydrocarbons, compounds in which one or more hydrogen atoms of a hydrocarbon have been replaced by halogen atoms: The replacement of only one hydrogen atom gives an alkyl halide (or haloalkane). The common names of alkyl halides consist of two parts: the name of the alkyl group plus the stem of the name of the halogen, with the ending -ide.
Examples
Give the common and IUPAC names for each compound.
1. CH3CH2CH2Br
2. (CH3)2CHCl
3. Give the IUPAC name for each compound.
a)
b)
SolutionS
1. The alkyl group (CH3CH2CH2–) is a propyl group, and the halogen is bromine (Br). The common name is therefore propyl bromide. For the IUPAC name, the prefix for bromine (bromo) is combined with the name for a three-carbon chain (propane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-bromopropane.
2. The alkyl group [(CH3)2CH–] has three carbon atoms, with a chlorine (Cl) atom attached to the middle carbon atom. The alkyl group is therefore isopropyl, and the common name of the compound is isopropyl chloride. For the IUPAC name, the Cl atom (prefix chloro-) attached to the middle (second) carbon atom of a propane chain results in 2-chloropropane.
3. a) The parent alkane has five carbon atoms in the longest continuous chain; it is pentane. A bromo (Br) group is attached to the second carbon atom of the chain. The IUPAC name is 2-bromopentane.
b) The parent alkane is hexane. Methyl (CH3) and bromo (Br) groups are attached to the second and fourth carbon atoms, respectively. Listing the substituents in alphabetical order gives the name 4-bromo-2-methylhexane.
Exercise
1. Give common and IUPAC names for each compound.
1. CH3CH2I
2. CH3CH2CH2CH2F
2. Give the IUPAC name for each compound.
a)
b)
Answer
1. a) ethyl iodide and iodoethane, respectively; Note the IUPAC name does not need a locator number because there is only one possible structure with two carbons and one iodine.
b) butyl fluoride and 1-fluorobutane
2. a) 2-chloro-2-methylbutane
b) 1-bromo-2-chloro-4-methylpentane
Halogens and the Character of the Carbon-Halogen Bond
With respect to electronegativity, halogens are more electronegative than carbons. This results in a carbon-halogen bond that is polarized. As shown in the image below, carbon atom has a partial positive charge, while the halogen has a partial negative charge.
The following image shows the relationship between the halogens and electronegativity. Notice, as we move up the periodic table from iodine to fluorine, electronegativity increases.
The following image shows the relationships between bond length, bond strength, and molecular size. As we progress down the periodic table from fluorine to iodine, molecular size increases. As a result, we also see an increase in bond length. Conversely, as molecular size increases and we get longer bonds, the strength of those bonds decreases.
Haloalkanes Have Higher Boiling Points than Alkanes
When comparing alkanes and haloalkanes, we will see that haloalkanes have higher boiling points than alkanes containing the same number of carbons. London dispersion forces are the first of two types of forces that contribute to this physical property. You might recall from general chemistry that London dispersion forces increase with molecular surface area. In comparing haloalkanes with alkanes, haloalkanes exhibit an increase in surface area due to the substitution of a halogen for hydrogen. The incease in surface area leads to an increase in London dispersion forces, which then results in a higher boiling point.
Dipole-dipole interaction is the second type of force that contributes to a higher boiling point. As you may recall, this type of interaction is a coulombic attraction between the partial positive and partial negative charges that exist between carbon-halogen bonds on separate haloalkane molecules. Similar to London dispersion forces, dipole-dipole interactions establish a higher boiling point for haloalkanes in comparison to alkanes with the same number of carbons.
The table below illustrates how boiling points are affected by some of these properties. Notice that the boiling point increases when hydrogen is replaced by a halogen, a consequence of the increase in molecular size, as well as an increase in both London dispersion forces and dipole-dipole attractions. The boiling point also increases as a result of increasing the size of the halogen, as well as increasing the size of the carbon chain.
Solubility
Solubility in water
Alkyl halides have little to no solubility in water in spite of the polar carbon-halogen bond. The attraction between the alkyl halide molecules is stronger than the attraction between the alkyl halide and water. Alkyl halides have little to no solubility in water, but be aware of densities. Polyhalogenated alkanes such as dichloromethane can have densities greater than water.
Solubility in organic solvents
Alkyl halides are soluble in most organic solvents. The London Dispersion forces play a dominant role in solubility.
Exercises
Exercise
3. Classify (primary, secondary, tertiary, vicinal, or geminal) and give the IUPAC name for the following organohalides:
4. Classify (primary, secondary, tertiary, vicinal, or geminal) and draw the bond-line structures of the following compounds:
a) 2-Chloro-3,3-dimethylpentane
b) 1,1-Dichloro-4-isopropylcyclohexane
c) 3-bromo-3-ethylhexane
5. Arrange the following alkyl halides in order of decreasing boiling point.
6. Predict the solvent with great alkyl halide solubility.
a) water or hexane
b) water or 1-octanol
c) water or benzene
d) water or acetone
Solutions
3.
a) secondary; 5-ethyl-4-iodo-3methyl-octane
b) primary; 1-bromo-2,3,4-trimethyl-pentane
c) vicinal dihalide; 4-bromo-5-chloro-2-methyl-heptane
4. (A) secondary; (B) geminal dichloride (C) tertiary
5. A > B > C
6.
a) hexane
b) benzene
c) 1-octanol
d) acetone
Alkyl halides have little to no solubility in water, but be aware of densities. Polyhalogenated alkanes can have densities greater than water. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.01%3A_Alkyl_Halides_-_Structure_and_Physi.txt |
Learning Objective
• Discuss the common uses of alkyl halides
Halogen containing organic compounds are relatively rare in terrestrial plants and animals. The thyroid hormones T3 and T4 are exceptions; as is fluoroacetate, the toxic agent in the South African shrub Dichapetalum cymosum, known as "gifblaar". However, the halogen rich environment of the ocean has produced many interesting natural products incorporating large amounts of halogen. Some examples are shown below.
The ocean is the largest known source for atmospheric methyl bromide and methyl iodide. Furthermore, the ocean is also estimated to supply 10-20% of atmospheric methyl chloride, with other significant contributions coming from biomass burning, salt marshes and wood-rotting fungi. Many subsequent chemical and biological processes produce poly-halogenated methanes.
Synthetic organic halogen compounds are readily available by direct halogenation of hydrocarbons and by addition reactions to alkenes and alkynes. Many of these have proven useful as intermediates in traditional synthetic processes. Some halogen compounds, shown in the box. have been used as pesticides, but their persistence in the environment, once applied, has led to restrictions, including banning, of their use in developed countries. Because DDT is a cheap and effective mosquito control agent, underdeveloped countries in Africa and Latin America have experienced a dramatic increase in malaria deaths following its removal, and arguments are made for returning it to limited use. 2,4,5-T and 2,4-D are common herbicides that are sold by most garden stores. Other organic halogen compounds that have been implicated in environmental damage include the polychloro- and polybromo-biphenyls (PCBs and PBBs), used as heat transfer fluids and fire retardants; and freons (e.g. CCl2F2 and other chlorofluorocarbons) used as refrigeration gases and fire extinguishing agents.
Alkyl halides provide nice examples for learning about two very important organic reaction mechanism types: nucleophilic substitution and beta-elimination. In learning about these mechanisms in the context of alkyl halide reactivity, we will also learn some very fundamental ideas about three main players in many organic reactions: nucleophiles, electrophiles, and leaving groups. We'll start with an overview of the substitution and elimination reactions which alkyl halides undergo.
7.03: Preparation of Alkyl Halides
Learning Objective
• specify the reagents for the most efficient synthesis of alkyl halides using free-radical halogenation of alkanes (Chapter 5) or allylic halogenation of alkenes with NBS
Free radical halogenation of alkanes
Free radical halogenation of alkanes is the substitution of a single hydrogen on the alkane for a single halogen to form a haloalkane. Light is required to initiate the radical formation and is a good example of a photochemical reaction. The simplest example is shown below for methane reacting with chlorine in the presence of light to form chloromethane and hydrogen chloride gas.
\[\ce{CH4 + Cl2 + energy → CH3Cl + HCl}\]
Free radical halogenation of alkanes has been thoroughly explained in chapter 5. The structure of the alkane is evaluated to choose between the high reactivity of chlorine (Cl2) and the high selectivity of bromine (Br2).
Allylic Bromination
When halogens are in the presence of unsaturated molecules such as alkenes, the expected reaction is addition to the double bond carbons resulting in a vicinal dihalide (halogens on adjacent carbons). The reaction is studied in a later chapter. To avoid halogen reactions at the alkene the halogen concentration is kept low enough that a substitution reaction occurs at the allylic position rather than addition at the double bond. The product is an allylic halide (halogen on carbon next to double bond carbons), which is acquired through a radical chain mechanism.
Why Substitution of Allylic Hydrogens?
As the table below shows, the dissociation energy for the allylic C-H bond is lower than the dissociation energies for the C-H bonds at the vinylic and alkylic positions. This is because the radical formed when the allylic hydrogen is removed is resonance-stabilized. Hence, given that the halogen concentration is low, substitution at the allylic position is favored over competing reactions. However, when the halogen concentration is high, addition at the double bond is favored because a polar reaction out competes the radical chain reaction.
Radical Allylic Bromination using NBS and light
Preparation of Bromine (low concentration)
NBS (N-bromosuccinimide) is the most commonly used reagent to produce low concentrations of bromine. When suspended in tetrachloride (CCl4), NBS reacts with trace amounts of HBr to produce a low enough concentration of bromine to facilitate the allylic bromination reaction.
Allylic Bromination Mechanism
Step 1: Initiation
Once the pre-initiation step involving NBS produces small quantities of Br2, the bromine molecules are homolytically cleaved by light to produce bromine radicals.
Step 2: Propagation
One bromine radical produced by homolytic cleavage in the initiation step removes an allylic hydrogen of the alkene molecule. A radical intermediate is generated, which is stabilized by resonance. The stability provided by delocalization of the radical in the alkene intermediate is the reason that substitution at the allylic position is favored over competing reactions such as addition at the double bond.
The intermediate radical then reacts with a Br2 molecule to generate the allylic bromide product and regenerate the bromine radical, which continues the radical chain mechanism. If the alkene reactant is asymmetric, two distinct product isomers are formed.
Step 3: Termination
The radical chain mechanism of allylic bromination can be terminated by any of the possible steps shown below.
Radical Allylic Chlorination
Like bromination, chlorination at the allylic position of an alkene is achieved when low concentrations of Cl2 are present. The reaction is run at high temperatures to achieve the desired results.
Industrial Uses
Allylic chlorination has important practical applications in industry. Since chlorine is inexpensive, allylic chlorinations of alkenes have been used in the industrial production of valuable products. For example, 3-chloropropene, which is necessary for the synthesis of products such as epoxy resin, is acquired through radical allylic chlorination (shown below).
Exercises
Exercises
1. Predict the two products of the allylic chlorination reaction of 1-heptene.
2. What conditions are required for allylic halogenation to occur? Why does this reaction outcompete other possible reactions such as addition when these conditions are met?
3. Predict the product of the allylic bromination reaction of 2-phenylheptane. (Hint: How are benzylic hydrogens similar to allylic hydrogens?)
4. The reactant 5-methyl-1-hexene generates the products 3-bromo-5-methyl-1-hexene and 1-bromo-5-methyl-2-hexene. What reagents were used in this reaction?
5. Predict the products of the following reactions:
Solutions
1. 3-chloro-1-heptene and 1-chloro-2-heptene
2. A low concentration of halide radical is sufficient for reaction at the allylic carbon without creating a reactivty environment for the pi bond of the alkene.
3. 2-bromo-2-phenylheptane
4. NBS with light
5. The product (A) is a 1° halogen which is more stable product even though the (B) had a better transition state with a 2° radical.
6.
7.04: Reactions of Alkyl Halides- Substit
Learning Objective
• apply the alpha and beta labels to alkyl halides for substitution and elimination reactions - refer to section 7.4
Alkyl Halide Structure and Reaction Language
The carbon bonded to a halide is called the alpha-carbon. The carbons bonded to the alpha-carbon are called beta-carbons. Carbon atoms further removed from the alpha carbon are named by continuing the Greek alphabet (alpha, beta, gamma, delta, etc). In discussing the reactions of alkyl halides, it can be effective to use the alpha- and beta- labels. The structure for 2-bromopropane is used below to illustrate the application of these terms.
The Reactions - Nucleophilic Substitution and Elimination
Alkyl halides can undergo two major types of reactions - substitution and/or elimination. The substitution reaction is called a Nucleophilic Substitution reaction because the electrophilic alkyl halide forms a new bond with the nucleophile which substitutes for (replaces) the halogen at the alpha-carbon. Because carbon can only form four bonds, the halogen must leave and is called the "Leaving Group". Alkyl halides are excellent electrophiles because halogens share a polar bond with carbon, are polarizable, and form relatively stable leaving groups as halide anions. In the example below, 2-bromopropane is converted into propan-2-ol in a substitution reaction.
Allkyl halides can also undergo elimination reactions in the presence of strong bases. The elimination of a beta-hydrogen (hydrogen on a carbon vicinal to the alkyl halide carbon) and the halide produces a carbon-carbon double bond to form an alkene. In the example below, 2-bromopropane has undergone an elimination reaction to give an alkene - propene.
What decides whether you get substitution or elimination?
In the examples above, the reagents were the same for both substitution and 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. The product distribution depends on a number of factors. These factors will be explored in the remaining sections of this chapter. Depending on the structure of the alkyl halide, reagent type, reaction conditions, some reactions will only undergo only one pathway - substitution or elimination. While other alkyl halides will always produce a mixture of substitution and elimination products like the example above. The goal of efficient multiple-step synthetic pathways is to maximize the formation of a single product during each step. The reaction conditions explored in this chapter will be useful for future reactions we will study and learn.
Exercise
1. Classify the following reactions as "Substitutions" or "Eliminations".
Answer
1. a) substitution
b) elimination
c) elimination
d) substitution | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.02%3A_Common_Uses_of_Alkyl_Halides.txt |
Learning Objectives
• determine the rate law & predict the mechanism based on its rate equation or reaction data for SN2 reactions
• propose mechanisms for SN2 reactions
• draw and interpret Reaction Energy Diagrams for SN2 reactions
Introduction
In many ways, the proton transfer process of a Brønsted-Lowry acid-base reaction can be thought of as simply a special kind of nucleophilic substitution reaction, one in which the electrophile is a hydrogen rather than a carbon.
In both reaction types, we are looking at very similar players: an electron-rich species (the nucleophile/base) reacts with an electron-poor species (the electrophile/proton), driving off the leaving group/conjugate base.
In the next few sections, we are going to be discussing some general aspects of nucleophilic substitution reactions, and in doing so it will simplify things greatly if we can use some abbreviations and generalizations before we dive into real examples.
What is a nucleophile (Nu)?
Instead of showing a specific nucleophile like hydroxide, we will simply refer to the nucleophilic reactant as 'Nu'. Nucleophilic functional groups are those which have electron-rich atoms able to donate a pair of electrons to form a new covalent bond. Nucleophiles can be negatively charged and some that are neutral with lone pair electrons. In both laboratory and biological organic chemistry, the most relevant nucleophilic atoms are oxygen, nitrogen, and sulfur, and the most common nucleophilic functional groups are water, alcohols, phenols, amines, thiols, and occasionally carboxylates.
More specifically in laboratory reactions, halide and azide (N3-) anions are commonly seen acting as nucleophiles.
When thinking about nucleophiles, the first thing to recognize is that, for the most part, the same quality of 'electron-richness' that makes a something nucleophilic also makes it basic: nucleophiles can be bases, and bases can be nucleophiles. It should not be surprising, then, that most of the trends in basicity that we have already discussed also apply to nucleophilicity.
Some confusion in distinguishing basicity (base strength) and nucleophilicity (nucleophile strength) is inevitable. Since basicity is a less troublesome concept; it is convenient to start with it. Basicity refers to the ability of a base to accept a proton. Basicity may be related to the pKa of the corresponding conjugate acid, as shown below. The strongest bases have the weakest conjugate acids and vice versa. The range of basicities included in the following table is remarkable, covering over fifty powers of ten!
In an acid-base equilibrium the weakest acid and the weakest base will predominate (they will necessarily be on the same side of the equilibrium). Learning the pKa values for common compounds provides a useful foundation on which to build an understanding of acid-base factors in reaction mechanisms.
Base I (–) Cl (–) H2O CH3CO2(–) RS(–) CN(–) RO(–) NH2(–) CH3(–)
Conj. Acid HI HCl H3O(+) CH3CO2H RSH HCN ROH NH3 CH4
pKa -9 -7 -1.7 4.8 8 9.1 16 33 48
Nucleophilicity is a more complex property. It commonly refers to the rate of substitution reactions at the halogen-bearing carbon atom of a reference alkyl halide, such as CH3-Br. Thus the nucleophilicity of the Nu:(–) reactant in the following substitution reaction varies as shown in the chart below:
Nucleophilicity: CH3CO2 (–) < Cl(–) < Br(–) < N3(–) < CH3O(–) < CN(–) < I(–) < CH3S(–)
What is a Leaving Group (X or LG)?
In a similar fashion, we will call the leaving group 'X' for halogens as is customary. For other reactions, it will be more accurate to abbreviate the leaving group as "LG". The context of the reaction will dictate the abbreviation. Leaving groups are sometimes negatively charged, sometimes neutral, and sometimes positively charged. Therefore, in this general picture we will not include a charge designation on the 'X' or 'LG' species. In referring to the comparison between acid-base chemistry and substitution reactions, the stability of the leaving group is evaluated the same way we evaluate the stability of conjugate bases.
When comparing the reactivity of electrophiles that vary only in their leaving groups, then leaving group stability plays a dominant role. The electrophile with the more stable leaving group will be favored. The lower the electron density of the leaving group, the more stable it is. Neutral leaving groups are favoring over charged leaving groups. When comparing charged leaving groups, apply the concepts used to determine the relative stability of conjugate bases:
1) identity or identities of the atom(s) holding the charge
2) delocalization of the charge via resonance
3) inductive effects
4) orbital hybridization
What is an Electrophile (E)?
An electrophile accepts electrons analogous to a Lewis acid. Electrophiles (E) are sometimes protonated and sometimes neutral. Electrophiles can also be called "Substrates". Since nucleophiles, leaving groups, and electrons may be charged or neutral, we will not include charges on 'Nu' or 'X' (or 'LG') or 'E'.
We will generalize the three other groups bonded on the electrophilic alpha-carbon as R1, R2, and R3: these symbols could represent hydrogens as well as alkyl groups. Finally, in order to keep figures from becoming too crowded, we will use in most cases the line structure convention in which the central, electrophilic carbon is not drawn out as a 'C'.
Here, then, is the generalized picture of a concerted (single-step) nucleophilic substitution reaction:
To recognize neutral electrophiles, we will need to identify polarity and/or resonance with compounds to create a partial positive charge to attract the nucleophile. The electrophilicity of alkyl halides comes from the polar carbon-halogen bond.
The common halogens being fluorine, chlorine, bromine and iodine. With the exception of iodine, these halogens have electronegativities significantly greater than carbon. Consequently, this functional group is polarized so that the carbon is electrophilic and the halogen is nucleophilic, as shown in the drawing on the right. Two characteristics other than electronegativity also have an important influence on the chemical behavior of these compounds. The first of these is covalent bond strength. The strongest of the carbon-halogen covalent bonds is that to fluorine. Remarkably, this is the strongest common single bond to carbon, being roughly 30 kcal/mole stronger than a carbon-carbon bond and about 15 kcal/mole stronger than a carbon-hydrogen bond. Because of this, alkyl fluorides and fluorocarbons in general are chemically and thermodynamically quite stable, and do not share any of the reactivity patterns shown by the other alkyl halides. The carbon-chlorine covalent bond is slightly weaker than a carbon-carbon bond, and the bonds to the other halogens are weaker still, the bond to iodine being about 33% weaker. The second factor to be considered is the relative stability of the corresponding halide anions, which is likely the form in which these electronegative atoms will be replaced. This stability may be estimated from the relative acidities of the H-X acids, assuming that the strongest acid releases the most stable conjugate base (halide anion). With the exception of HF (pKa = 3.2), all the hydrohalic acids are very strong, small differences being in the direction HCl < HBr < HI.
Exercise
1. Since everything is relative in chemistry, one reaction's nucleophile can be another reaction's leaving group. Some functional groups can only react as a nuclephile or electrophile, while other functional groups can react as either a nuclephile or electrophile depending on the reaction conditions. Classify the following compounds as nucleophiles, electrophiles, or leaving groups. More than one answer may be possible.
a) bromoethane
b) hydroxide
c) water
d) chlorocyclohexane
e) ethanol
f) bromide
Answer
a) electrophile (Alkyl halides are always electrophiles - one reason they are an o-chem student's best friend.)
b) strong nucleophile
c) weak nucleophile and good leaving group
d) electrophile (Alkyl halides are always electrophiles - one reason they are an o-chem student's best friend.)
e) weak nucleophile, a poor electrophile without clever chemistry (stay tuned for future chapters), good leaving group
f) good nucleophile and a good leaving group
The SN2 mechanism
There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution, SN2 and SN1. The SN2 reaction takes place in a single step with bond-forming and bond-breaking occurring simultaneously. (In all figures in this section, 'X' indicates a halogen substituent).
This is called an 'SN2' mechanism. In the term SN2, S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that it is a bimolecular reaction: the overall rate depends on a step in which two separate molecules (the nucleophile and the electrophile) collide. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
If you look carefully at the progress of the SN2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must react with the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the electron rich, leaving group blocks the way with electrostatic repulsion and steric hindrance.
The result of this backside penetration is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
What this means is that SN2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product. Below is an animation illustrating the principles we have just learned, showing the SN2 reaction between hydroxide ion and methyl iodide. Notice how backside attack by the hydroxide nucleophile results in inversion at the tetrahedral carbon electrophile.
Exercise
2. Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
Solution
2.
SN2 Reactions Occur at sp3 Carbons with a Leaving Group
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons bonded to a leaving group. SN2 reactions cannot occur where the leaving group is attached to an sp2-hybridized carbon:
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module).
For future reference when discerning between substitution and elimination reactions, evaluating the structure of the electrophile can eliminate possible products. If the electrophilic carbon has no beta-hydrogens, then only substitution reactions can occur and elimination reactions are not possible (of course carbocation rearrangements may need to be considered). The first four halides shown on the left below do not give elimination reactions on treatment with base, because they have no β-hydrogens. The two halides on the right do not normally undergo such reactions because the potential elimination products have highly strained double or triple bonds. It is also worth noting that sp2 hybridized C–X compounds, such as the three on the right, do not normally undergo nucleophilic substitution reactions, unless other functional groups perturb the double bond(s).
Exercise
3. Predict which alkyl halides can undergo a SN2 reaction.
a) C6H5Br
b) CH3CH2CH2Br
c) CH2CHBr
d) CH3CH2CH2CHBrCH3
Solutions
3.
a) No, aryl halide.
b) Yes, primary alkyl halide
c) No, vinyl halide
d) Yes, secondary alkyl halide
SN2 Reaction Kinetics
In the term SN2, the S stands for substitution, the N stands for nucleophilic, and the number two stands for bimolecular, meaning there are two molecules involved in the rate determining step. The rate of bimolecular nucleophilic substitution reactions depends on the concentration of both the haloalkane and the nucleophile. To understand how the rate depends on the concentrations of both the haloalkane and the nucleophile, let us look at the following example. The hydroxide ion is the nucleophile and methyl iodide is the haloalkane.
If we were to double the concentration of either the haloalkane or the nucleophile, we can see that the rate of the reaction would proceed twice as fast as the initial rate.
If we were to double the concentration of both the haloalkane and the nucleophile, we can see that the rate of the reaction would proceed four times as fast as the initial rate.
The bimolecular nucleophilic substitution reaction follows second-order kinetics; that is, the rate of the reaction depends on the concentration of two first-order reactants. In the case of bimolecular nucleophilic substitution, these two reactants are the haloalkane and the nucleophile. For further clarification on reaction kinetics, the following links may facilitate your understanding of rate laws, rate constants, and second-order kinetics
Exercise
4. The reaction below follows the SN2 mechanism.
a) Write the rate law for this reaction.
b) Determine the value of the rate coefficient, k, if the initial concentrations are 0.01 M CH3Cl, 0.01 M NaOH, and the initial reaction rate is 6 x 10-10 M/s.
c) Calculate the initial reaction rate if the initial reactant concentrations are changed to 0.02 M CH3Cl and 0.0005 M NaOH.
Solutions
4.
a) rate = k [CH3Cl] [OH-]
b) substitute the data into the rate expression above and apply algebra to solve for k
k = 6 x 10-6 Lmol-1s-1
c) Using the rate law above, substitute the value for k from the previous question along with the new concentrations to determine the new initial rate.
rate = 6 x 10-10 M/s | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.05%3A_The_S2_Reaction.txt |
Learning Objective
• determine the rate law & predict the mechanism based on its rate equation or reaction data for SN2 reactions
• predict the products and specify the reagents for SN2 reactions with stereochemistry
• propose mechanisms for SN2 reactions
• draw and interpret Reaction Energy Diagrams for SN2 reactions
Introduction
To understand why some combinations of alkyl halides and nucleophiles give a substitution reaction, whereas other combinations give elimination, and still others give no observable reaction, we must investigate systematically the way in which changes in reaction variables perturb the course of the reaction. The following general equation summarizes the factors that will be important in such an investigation where X represents the leaving group (a halide for this chapter).
In order of decreasing importance, the factors impacting SN2 reaction pathways are
1) structure of the alkyl halide
2) strength of the nucleophile
3) stability of the leaving group
4) type of solvent.
The bimolecular transition state of the SN2 pathway means that sterics are a primary consideration. The orbitals of the nucleophile must be able to penetrate through the reaction solution and create orbital overlap with the orbitals of the electrophilic carbon. The sterics of this mechanism can be determined by applying the bonding theories for individual compounds and ions to the interaction of the nucleophile and electrophile. The strength of the nucleophile will also influence the reaction along with the stability of the leaving group. Solvents can have a subtle yet measurable effect on SN2 pathway. Solvation may be defined as the interaction between molecules of solvent and particles of solute. The result of solvation is to stabilize (i.e., lower the energy of) the solute particles. Solvents with lone pairs of electrons are good at solvating cations. Protic (i.e., hydroxylic) solvents are able to solvate anions through hydrogen bonding. As water has two lone pairs of electrons and is also protic, it is good at solvating both anions and cations. The role of the solvent is often misunderstood and consequently given way too much importance. Do not drown in the solvent. Solvation effects are less significant than the structure of the alkyl halide, the reactivity of the nucleophile, and the stability of the leaving group.
The following variables and observables can be used to study the SN2 mechanism.
Variables
R change α-carbon from 1º to 2º to 3º
if the α-carbon is a
chiral center, set as (R) or (S)
X change from Cl to Br to I (F is relatively unreactive)
Nu: change from anion to neutral; change basicity; change
polarizability
Solvent polar vs. non-polar; protic vs. non-protic
Observables
Products substitution, elimination, no reaction.
Stereospecificity if the α-carbon is a chiral center what happens to its configuration?
Reaction Rate measure as a function of reactant concentration.
When several reaction variables may be changed, it is important to isolate the effects of each during the course of study. In other words: only one variable should be changed at a time, the others being held as constant as possible. For example, we can examine the effect of changing the halogen substituent from Cl to Br to I, using ethyl as a common R–group, cyanide anion as a common nucleophile, and ethanol as a common solvent. We would find a common substitution product, C2H5–CN, in all cases, but the speed or rate of the reaction would increase in the order: Cl < Br < I. This reactivity order reflects both the strength of the C–X bond, and the stability of X(–) as a leaving group, and leads to the general conclusion that alkyl iodides are the most reactive members of this functional class.
Bimolecular Nucleophilic Substitution Reactions Are Concerted
Bimolecular nucleophilic substitution (SN2) reactions are concerted, meaning they are a one step process. The bond-making between the nucleophile and the electrophilic carbon occurs at the same time as the bond-breaking between the electophilic carbon and the halogen.
The potential energy diagram for an SN2 reaction is shown below. The one-step mechanism means that only a single transition state is formed. A transition state, unlike a reaction intermediate, is a very short-lived species that cannot be isolated or directly observed.
Alkyl halide (Substrate) Structure and SN2 Reaction Rates
Now that we have discussed the effects that the leaving group, nucleophile, and solvent have on biomolecular nucleophilic substitution (SN2) reactions, it's time to turn our attention to how the substrate affects the reaction. Although the substrate, in the case of nucleophilic substitution of haloalkanes, is considered to be the entire molecule circled below, we will be paying particular attention to the alkyl portion of the substrate. In other words, we are most interested in the electrophilic center that bears the leaving group.
The SN2 transition state is very crowded with a total of five groups around the electrophilic center, the nucleophile, the leaving group, and three substituents.
If each of the three substituents in this transition state were small hydrogen atoms, as illustrated in the first example below, there would be little steric repulsion between the incoming nucleophile and the electrophilic center, thereby increasing the ease at which the nucleophilic substitution reaction can occur. Remember, for the SN2 reaction to occur, the nucleophile must be able to overlap orbitals with the electrophilic carbon center, resulting in the expulsion of the leaving group. If one of the hydrogens, however, were replaced with an R group, such as a methyl or ethyl group, there would be an increase in steric repulsion with the incoming nucleophile. If two of the hydrogens were replaced by R groups, there would be an even greater increase in steric repulsion with the incoming nucleophile.
How does steric hindrance affect the rate at which an SN2 reaction will occur? As each hydrogen is replaced by an R group, the rate of reaction is significantly diminished. This is because the addition of one or two R groups shields the backside of the electrophilic carbon impeding nucleophilic penetration.
The diagram below illustrates this concept, showing that electrophilic carbons attached to three hydrogen atoms results in faster nucleophilic substitution reactions, in comparison to primary and secondary haloalkanes, which result in nucleophilic substitution reactions that occur at slower or much slower rates, respectively. Notice that a tertiary haloalkane, that which has three R groups attached, does not undergo nucleophilic substitution reactions at all. The addition of a third R group to this molecule creates a carbon that is entirely blocked.
Substitutes on Neighboring Carbons Slow Nucleophilic Substitution Reactions
Previously we learned that adding R groups to the electrophilic carbon results in nucleophilic substitution reactions that occur at a slower rate. What if R groups are added to neighboring carbons? It turns out that the addition of substitutes on neighboring carbons will slow nucleophilic substitution reactions as well.
In the example below, 2-methyl-1-bromopropane differs from 1-bromopropane in that it has a methyl group attached to the carbon that neighbors the electrophilic carbon. The addition of this methyl group results in a significant decrease in the rate of a nucleophilic substitution reaction.
If R groups were added to carbons farther away from the electrophilic carbon, we would still see a decrease in the reaction rate. However, branching at carbons farther away from the electrophilic carbon would have a much smaller effect.
Nucleophilicity
There are predictable periodic trends in nucleophilicity. Moving horizontally across the second row of the table, the trend in nucleophilicity parallels the trend in basicity:
The reasoning behind the horizontal nucleophilicity trend is the same as the reasoning behind the basicity trend: more electronegative elements hold their electrons more tightly, and are less able to donate them to form a new bond. This horizontal trend also tells us that amines are more nucleophilic than alcohols, although both groups commonly act as nucleophiles in both laboratory and biochemical reactions.
Recall that the basicity of atoms decreases as we move vertically down a column on the periodic table: thiolate ions are less basic than alkoxide ions, for example, and bromide ion is less basic than chloride ion, which in turn is less basic than fluoride ion. Recall also that this trend can be explained by considering the increasing size of the 'electron cloud' around the larger ions: the electron density inherent in the negative charge is spread around a larger area, which tends to increase stability (and thus reduce basicity).
The vertical periodic trend for nucleophilicity is somewhat more complicated because the solvent can influence the nucleophilicity trend in either direction. Let's take the simple example of the SN2 reaction below:
. . .where Nu- is one of the halide ions: fluoride, chloride, bromide, or iodide, and the leaving group I* is a radioactive isotope of iodine (which allows us to distinguish the leaving group from the nucleophile when both are iodide). If this reaction is occurring in a protic solvent (that is, a solvent that has a hydrogen bonded to an oxygen or nitrogen - water, methanol and ethanol are the most important examples), then the reaction will go fastest when iodide is the nucleophile, and slowest when fluoride is the nucleophile, reflecting the relative strength of the nucleophile.
Relative nucleophilicity in a protic solvent
This of course, is opposite that of the vertical periodic trend for basicity, where iodide is the least basic. What is going on here? Shouldn't the stronger base, with its more reactive unbonded valence electrons, also be the stronger nucleophile?
As mentioned above, it all has to do with the solvent. Remember, we are talking now about the reaction occurring in a protic solvent like ethanol. Protic solvent molecules form very strong ion-dipole interactions with the negatively-charged nucleophile, essentially creating a 'solvent cage' around the nucleophile:
In order for the nucleophile to react with the electrophile, it must break free, at least in part, from its solvent cage. The lone pair electrons on the larger, less basic iodide ion interact less tightly with the protons on the protic solvent molecules resulting in weaker solvation - thus the iodide nucleophile is better able to break free from its solvent cage compared the smaller, more basic fluoride ion, whose lone pair electrons are bound more tightly to the protons of the solvent cage.
The picture changes if we switch to a polar aprotic solvent, such as acetone, in which there is a molecular dipole but no hydrogens bound to oxygen or nitrogen. Now, fluoride is the best nucleophile, and iodide the weakest.
Relative nucleophilicity in a polar aprotic solvent
The reason for the reversal is that, with an aprotic solvent, the ion-dipole interactions between solvent and nucleophile are much weaker: the positive end of the solvent's dipole is hidden in the interior of the molecule, and thus it is shielded from the negative charge of the nucleophile.
A weaker solvent-nucleophile interaction means a weaker solvent cage for the nucleophile to break through, so the solvent effect is much less important, and the more basic fluoride ion is also the better nucleophile.
Why not use a completely nonpolar solvent, such as hexane, for this reaction, so that the solvent cage is eliminated completely? The answer to this is simple - the nucleophile needs to be in solution in order to react at an appreciable rate with the electrophile, and a solvent such as hexane will not solvate an a charged (or highly polar) nucleophile at all. That is why chemists use polar aprotic solvents for nucleophilic substitution reactions in the laboratory: they are polar enough to solvate the nucleophile, but not so polar as to lock it away in an impenetrable solvent cage. In addition to acetone, three other commonly used polar aprotic solvents are acetonitrile, dimethylformamide (DMF), and dimethyl sulfoxide (DMSO).
In biological chemistry, where the solvent is protic (water), the most important implication of the periodic trends in nucleophilicity is that thiols are more powerful nucleophiles than alcohols. The thiol group in a cysteine amino acid, for example, is a powerful nucleophile and often acts as a nucleophile in enzymatic reactions, and of course negatively-charged thiolates (RS-) are even more nucleophilic. This is not to say that the hydroxyl groups on serine, threonine, and tyrosine do not also act as nucleophiles - they do.
Resonance effects on nucleophilicity
Resonance effects also come into play when comparing the inherent nucleophilicity of different molecules. The reasoning involved is the same as that which we used to understand resonance effects on basicity. If the electron lone pair on a heteroatom is delocalized by resonance, it is inherently less reactive - meaning less nucleophilic, and also less basic. An alkoxide ion, for example, is more nucleophilic and more basic than a carboxylate group, even though in both cases the nucleophilic atom is a negatively charged oxygen. In the alkoxide, the negative charge is localized on a single oxygen, while in the carboxylate the charge is delocalized over two oxygen atoms by resonance.
The nitrogen atom on an amide is less nucleophilic than the nitrogen of an amine, due to the resonance stabilization of the nitrogen lone pair provided by the amide carbonyl group.
The leaving group
The more stable the leaving group, the lower the transition state energy, the lower the activation energy, the faster the reaction rate. Evaluating leaving group stability is analogous to determining relative acidity by evaluating conjugate base stability. The considerations are the same: identity of the atom(s) and relative position on the periodic table, resonance delocalization, and electronegativity. Orbital hybridization is rarely relevant.
As Size Increases, Basicity Decreases, Leaving Group Stability Increases:In general, if we move from the top of the periodic table to the bottom of the periodic table as shown in the diagram below, the size of an atom will increase. As size increases, basicity will decrease, meaning a species will be less likely to act as a base; that is, the species will be less likely to share its electrons.
When evaluating halogens as leaving groups, the same trend is significant. Fluoride has the highest electron density and is considered the worst leaving group to the point of no reactivity. As move down the column, the leaving groups have lower electron density and greater stability with iodide considered an excellent leaving group.
Resonance Decreases Basicity and Increases Leaving Group Stability:The formation of a resonance stabilized structure delocalizes the electrons over two or more atoms lowering the electron density of the leaving group and increases its stability. For halides as leaving groups there are no applications for this consideration, so we will look briefly at carbonyl chemistry to illustrate this effect. When comparing the hydrolysis rates of anhydrides and esters, anhydrides react spontaneously with water and undergo hydrolysis to form a resonance stabilized carboxylate ion. Whereas, ester hydrolysis is a much slower reaction and requires a catalyst to overcome the alkoxides as poor leaving groups. The details of these two reactions will be studied in greater detail later in this text.
As Electronegativity Increases, Basicity Decreases and Leaving Group Stability Increases: In general, if we move from the left of the periodic table to the right of the periodic table as shown in the diagram below, electronegativity increases. As electronegativity increases, basicity will decrease, meaning a species will be less likely to act as base; that is, the species will be less likely to share its electrons.
The following diagram illustrates this concept, showing -CH3 to be the worst leaving group and F- to be the best leaving group. This particular example should only be used to facilitate your understanding of this concept. In real reaction mechanisms, these groups are not good leaving groups at all. For example, fluoride is such a poor leaving group that SN2 reactions of fluoroalkanes are rarely observed.
Leaving Groups Across a Period
Solvent Effects on an SN2 reaction
The rate of an SN2 reaction is significantly influenced by the solvent in which the reaction takes place. The use of protic solvents (those, such as water or alcohols, with hydrogen-bond donating capability) decreases the power of the nucleophile through strong solvation. WE can view the nucleophile as being locked in a solvent cage through the strong hydrogen-bond interactions between solvent protons and the reactive lone pairs on the nucleophile. A less powerful nucleophile in turn means a slower SN2 reaction.
SN2 reactions are faster in polar, aprotic solvents: those that lack hydrogen-bond donating capability. Below are several polar aprotic solvents that are commonly used in the laboratory:
These aprotic solvents are polar but, because they do not form hydrogen bonds with the anionic nucleophile, there is a relatively weak interaction between the aprotic solvent and the nucleophile. By using an aprotic solvent we can raise the reactivity of the nucleophile. This can sometimes have dramatic effects on the rate at which a nucleophilic substitution reaction can occur. For example, if we consider the reaction between bromoethane and potassium iodide, the reaction occurs 500 times faster in acetone than in methanol.
Example
In each pair (A and B) below, which electrophile would be expected to react more rapidly in an SN2 reaction with the thiol group of cysteine as the common nucleophile?
Explanations to explain differences in chemical reactivity need to discuss structural and/or electrostatic differences between the reactants
a) Cpd B b/c it has a more stable leaving group.
The larger atomic size of S relative to O means the sulfide (CH3S-) will have a lower electron density than the alkoxide (CH3O-).
b) Cpd A b/c it has a more stable leaving group.
The neutral leaving group, (CH3)2S, is more stable than the charged sulfie leaving group (CH3S-).
c) Cpd B b/c the leaving group is resonance stabilized delocalizing the negative charge over two oxygen atoms. d) Cpd B b/c the leaving group has inductive electron withdrawal stabilization from the three fluorine atoms in addition to the resonance stabilzation.
Exercise
1. What product(s) do you expect from the reaction of 1-bromopentane with each of the following reagents in an SN2 reaction?
a) KI
b) NaOH
c) CH3C≡C-Li
d) NH3
2. Which in the following pairs is a better nuceophile?
a) (CH3CH2)2N- or (CH3CH2)2NH
b) (CH3CH2)3N or (CH3CH2)3B
c) H2O or H2S
3. Order the following in increasing reactivity for an SN2 reaction.
CH3CH2Br CH3CH2OTos (CH3CH2)3CCl (CH3CH2)2CHCl
4. Solvents benzene, ether, chloroform are non-polar and not strongly polar solvents. What effects do these solvents have on an SN2 reaction?
Answer
1. (a) - (d)
2.
a) (CH3CH2)2N- as there is a charge present on the nitrogen.
b) (CH3CH2)3N because a lone pair of electrons is present.
c) H2O as oxygen is more electronegative.
3.
4. They will decrease the reactivity of the reaction. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.06%3A_Characteristics_of_the_S2_Reaction.txt |
Learning Objective
• predict the products and specify the reagents for SN2reactions with stereochemistry
• propose mechanisms for SN2 reactions
• draw and interpret Reaction Energy Diagrams for SN2reactions
SN2 Reactions Are Stereospecific
The SN2 reaction is stereospecific like other concerted reactions.. A stereospecific reaction is one in which different stereoisomers react to give different stereoisomers of the product. The nuclephile and electrophile must be correctly oriented for orbital overlap to occur and trigger chemical reactivity. Experimental observations show that all SN2 reactions proceed with inversion of configuration; that is, the nucleophile will always penetrate from the backside in SN2 reactions. To think about why this might be true, remember that the nucleophile has a lone pair of electrons to be shared with the electrophilic carbon center and the leaving group is going to take a lone pair of electrons with it upon leaving. Because like charges repel each other, the nucleophile will always proceed by a backside displacement mechanism.
• Frontside Orientation: In a frontside orientation, the nucleophile approaches the electrophilic center on the same side as the leaving group. With frontside orientation, the stereochemistry of the product remains the same; that is, we have retention of configuration.
• Backside Orientation: In a backside orientation, the nucleophile approachss the electrophilic center on the side that is opposite to the leaving group. With backside orientation, the stereochemistry of the product does not stay the same. There is inversion of configuration.
For example, if the substrate is an R enantiomer, a frontside nucleophilic orientation results in retention of configuration, and the formation of the R enantiomer. A backside nucleophilic orientation results in inversion of configuration, and the formation of the S enantiomer.
Conversely, if the substrate is an S enantiomer, a frontside nucleophilic orientation results in retention of configuration, and the formation of the S enantiomer. A backside nucleophilic orientation results in inversion of configuration, and the formation of the R enantiomer.
Empirically, SN2 reactions that begin with the R enantiomer as the substrate will form the S enantiomer as the product. Those that begin with the S enantiomer as the substrate will form the R enantiomer as the product. This concept of retention and inversion of configuration can also be applied to substrates that can exist as geometric isomers (cis and trans). If the cis configuration is the substrate, the resulting product will be trans. Conversely, if the trans configuration is the substrate, the resulting product will be cis.
Exercise
1. Predict the product of a nucleophilic substitution of (S)-2-bromopentane reacting with CH3CO2-, Show stereochemistry.
Answer
1.
Contributors and Attributions
• Racheal Curtis (UCD)
7.08: The S1 Reaction
Learning Objective
• determine the rate law & predict the mechanism based on its rate equation or reaction data for SN1 reactions
• predict the products and specify the reagents for SN1 reactions with stereochemistry
• propose mechanisms for SN1 reactions
• draw and interpret Reaction Energy Diagrams for SN1 reactions
The SN1 mechanism with Stereochemistry
A second model for a nucleophilic substitution reaction is called the 'dissociative' or 'SN1' mechanism. In many cases, the nucleophile is the solvent, so this mechanism can also be called "solvolysis".
Step1: In the SN1 mechanism, the carbocation forms when the C-X bond breaks first, before the nucleophile approaches
Th carbocation has a central carbon with only three bonds and bears a formal charge of +1. Recall that a carbocation should be pictured as sp2 hybridized, with trigonal planar geometry. Perpendicular to the plane formed by the three sp2 hybrid orbitals is an empty, unhybridized p orbital.
Step 2: The nucleophile reacts with the empty, 'electron hungry' p orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry. Because of this trigonal planar geometry, the nucleophile can approach the carbocation from either lobe of the empty p orbital (aka either side of the carbocation). This means that about half the time the product has the same stereochemical configuration as the starting material (retention of configuration), and about half the time the stereochemistry has been inverted. In other words, racemization of the product occurs during SN1 reactions if the electrophilic carbon is chiral. If the intermediate from a chiral alkyl halide survives long enough to encounter a random environment, the products are expected to be racemic (a 50:50 mixture of enantiomers). On the other hand, if the departing halide anion temporarily blocks the front side, or if a nucleophile is oriented selectively at one or the other face, then the substitution might occur with predominant inversion or even retention of configuration.
As an example, the tertiary alkyl bromide below, (S)-3-bromo-3-methylhexane, would be expected to form a racemic mix of R- and S-3-methyl-3-hexanol after an SN1 reaction with water as the nucleophile.
Exercise
1. Draw the structure of the intermediate in the two-step nucleophilic substitution reaction (SN1) above.
Solution
1.
The SN1 Reaction Energy Diagram
The SN1 reaction is an example of a two-step reaction with a reaction intermediate. Evaluating reactive intermediates is a very important skill in the study of organic reaction mechanisms. Many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. In the SN1 reaction, the carbocation species is a reaction intermediate. A potential energy diagram for an SN1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states.
Exercise
2. Draw structures representing transition state 1 (TS1) and transition state 2 (TS2) in the reaction above. Use the solid/dash wedge convention to show three dimensions.
Solution
2.
SN1 Reaction Kinetics
In the first step of an SN1 mechanism, two charged species are formed from a neutral molecule. This step is much the slower of the two steps, and is therefore rate-determining. In the reaction energy diagram, the activation energy for the first step is higher than that for the second step indicating that the SN1 reaction has first order kinetics because the rate determining step involves one molecule splitting apart, not two molecules colliding. It is important to remember that first order refers to the rate law expression where the generic term substrate is used to describe the alkyl halide.
rate = k [substrate]
Because an SN1 reaction is first order overall the concentration of the nucleophile does not affect the rate. The implication is that the nucleophile does not participate in the rate limiting step or any prior steps, which suggests that the first step is the rate limiting step. Since the nucleophile is not involved in the rate-limiting first step, the nature of the nucleophile does not affect the rate of an SN1 reaction.
Exercise
3. Consider two nucleophilic substitutions that occur uncatalyzed in solution. Assume that reaction A is SN2, and reaction B is SN1. Predict, in each case, what would happen to the rate of the reaction if the concentration of the nucleophile were doubled, while all other conditions remained constant.
4. Give the products of the following SN1 reaction. Show stereochemistry.
Solution
3. For Reaction A, the rate law is rate = k[CH3I][CH3S-]. Therefore, if the concentration of the nucleophile, CH3S-, is doubled and the concentration of the alkyl halide remains the same, then the reaction rate will double.
For Reaction B, the rate law is rate = k[CH3)3Br]. Therefore, if the concentration of the nucleophile, CH3SH, is doubled and the concentration of the alkyl halide remains the same, then reaction rate stays the same.
4. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.07%3A_Stereochemistry_of_the_SN2_Reaction.txt |
Learning Objective
• determine the rate law & predict the mechanism based on its rate equation or reaction data for SN1 reactions
• predict the products and specify the reagents for SN1 reactions with stereochemistry
• propose mechanisms for SN1 reactions
• draw and interpret Reaction Energy Diagrams for SN1 reactions
In order of decreasing importance, the factors impacting SN1 reaction pathways are
1. structure of the alkyl halide
2. stability of the leaving group
3. type of solvent.
The unimolecular transition state of the SN1 pathway means that structure of the alkyl halide and stability of the leaving group are the primary considerations. Alkyl halides that can ionize to form stable carbocations are more reactive via the SN1 mechanism. Because carbocation stability is the primary energetic consideration, stabilization of the carbocation via solvation is also an important consideration.
Alkyl Halide Structure and Carbocation Stability
The first order kinetics of SN1 reactions suggest a two-step mechanism in which the rate-determining step consists of carbocation formation from the ionization of the alkyl halide as shown in the diagram below. In this mechanism, the carbocation is a high-energy intermediate the bonds immediately to nearby nucleophiles. The only reactant that is undergoing change in the first (rate-determining) step is the alkyl halide, so we expect such reactions would be unimolecular and follow a first-order rate equation. Hence the name SN1 is applied to this mechanism.
The Hammond postulate suggests that the activation energy of the rate-determining first step will be inversely proportional to the stability of the carbocation intermediate: the more stable the carbocation, the lower the activation energy, the faster the reactivity. Therefore, carbocation stability is a primary consideration in SN1 reactions. Carbocations can be stabilized by delocalizing the charge via resonance and through inductive electron donation of alkyl groups. Carbocations can also stabilize by rearrangement via 1,2-hydride or 1,2-methyl shifts. Carbocation rearrangements are explained in a subsequent section of this chapter.
Benzyl Carbocation
The relative stability of carbocations is summarized below.
Carbocation Stability CH3(+) < CH3CH2(+) < (CH3)2CH(+) CH2=CH-CH2(+) < C6H5CH2(+) (CH3)3C(+)
Consequently, we expect that 3º-alkyl halides will be more reactive than their 2º and 1º-counterparts in reactions that follow an SN1 mechanism. This is opposite to the reactivity order observed for the SN2 mechanism. Allylic and benzylic halides are exceptionally reactive by either mechanism. This trend is summarized in the diagram below.
Effects of Leaving Group
An SN1 reaction speeds up with a good leaving group. This is because the leaving group is involved in the rate-determining step. A good leaving group wants to leave so it breaks the C-Leaving Group bond faster. Once the bond breaks, the carbocation is formed and the faster the carbocation is formed, the faster the nucleophile can come in and the faster the reaction will be completed.
A good leaving group is a weak base because weak bases can hold the charge. They're happy to leave with both electrons and in order for the leaving group to leave, it needs to be able to accept electrons. Strong bases, on the other hand, donate electrons which is why they can't be good leaving groups. As you go from left to right on the periodic table, electron donating ability decreases and thus ability to be a good leaving group increases. Halides are an example of a good leaving group whos leaving-group ability increases as you go down the column.
The two reactions below is the same reaction done with two different leaving groups. One is significantly faster than the other. This is because the better leaving group leaves faster and thus the reaction can proceed faster.
Methyl Sulfate Ion Mesylate Ion Triflate Ion Tosylate Ion
CH3SO42- CH3SO32- CF3SO32- CH3C6H4SO32-
Solvent Effects on the SN1 Reaction
To facilitate the charge separation of the ionization reaction in the first step, a good ionizing solvent is needed. Two solvent characteristics will be particularly important - the polarity and the solvating power. The dielectric constant, ε, measures polarity of solvent molecules and their ability to orient themselves between ions to attenuate (reduce) the electrostatic force one ion exerts on the other. The higher the dielectric constant the more polar the substance and in the case of SN1 reactions, the faster the rate. A dielectric constant below 15 is usually considered non-polar. Solvents having high dielectric constants, such as water (ε=81), formic acid (ε=58), dimethyl sulfoxide (ε=45) & acetonitrile (ε=39) are generally considered better ionizing solvents than are some common organic solvents such as ethanol (ε=25), acetone (ε=21), methylene chloride (ε=9) & ether (ε=4). Below is the same reaction conducted in two different solvents. The relative reaction rate in water (ε=81) is 150,000 times faster than in methanol (ε=33).
Solvation refers to the solvent's ability to stabilize ions by encasing them in a sheath of weakly bonded solvent molecules. Anions are solvated by partial positive charges of hydrogen-bonding solvents. Cations are often best solvated by the nucleophilic sites on a solvent molecule (e.g. oxygen & nitrogen atoms). The interaction of the carbocations with these nucleophilic solvents may be strong enough to form covalent bonds to carbon, thus converting the intermediate to a substitution product and creating the reaction name "solvolysis". When solvolysis occurs with water, the actions are called "hydrolysis reactions" as shown in the reaction below.
Polar Protic and Polar Aprotic Solvents
Protic solvents contain polarized hydrogen. Whereas, aprotic solvents do NOT contain polarized hydrogen. For SN2 reactions, solvation of the nucleophile by polar protic solvents slows the reaction rate. However, in SN1 reaction the nucleophile is not a part of the rate-determining step so this concern is not relevant. In fact, polar protic solvents actually speed up the rate of SN1 reactions because the polar solvent helps stabilize the transition state and carbocation intermediate. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction. Polar aprotic solvents have a dipole moment, but their hydrogen is not highly polarized. Polar aprotic solvents are not used in SN1 reactions because some of them can react with the carbocation intermediate and give an unwanted side-product. Rather, polar protic solvents are preferred for unimolecular substitution reactions.
Effects of Nucleophile
The strength of the nucleophile does not affect the reaction rate of SN1 because the nucleophile is not involved in the rate-determining step. Since nucleophiles only participate in the fast second step, their relative molar concentrations rather than their nucleophilicities should be the primary product-determining factor. If a nucleophilic solvent such as water is used, its high concentration will assure that alcohols are the major product. However, if you have more than one nucleophile competing to bond to the carbocation, the strengths and concentrations of those nucleophiles affects the distribution of products. For example, if t-butylchloride reacts with a mixture of water and formic acid where the water and formic acid are competing nucleophiles, two different products are formed: (CH3)3COH and (CH3)3COCOH. The relative yields of these products depends on the concentrations and relative reactivities of the nucleophiles. With a higher electron density, water is considered the stronger nucleophile and the tertiary alcohol will be the major product if there are equal concentrations of competing nucleophiles.
Exercises
1. Rank the following by increasing reactivity in an SN1 reaction.
2. 3-bromo-1-pentene and 1-bromo-2-pentene undergo SN1 reaction at almost the same rate, but one is a secondary halide while the other is a primary halide. Explain why this is.
3. Label the following reactions as most likely occuring by an SN1 or SN2 mechanism. Suggest why.
Answers
1. Consider the stability of the intermediate, the carbocation.
A < D < B < C (most reactive)
2. They have the same intermediates when you look at the resonance forms.
3. A – SN1 *poor leaving group, protic solvent, secondary cation intermediate
B – SN2 *good leaving group, polar solvent, primary position.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.09%3A_Characteristics_of_the_S1_Reaction.txt |
Learning Objective
• predict carbocation rearrangements in 1st order reactions
Whenever reactants like alkyl halides form carbocations, the carbocations are subject to a phenomenon known as carbocation rearrangement. A carbocation is highly reactive and holds the positive charge on carbon with a sextet rather than an octet. There are two types of rearrangements: hydride shift and alkyl shift. Rearrangements occur to create more stable carbocations. Reviewing carbocation stability from chapter 5 is helpful in identifying carbocations that can undergo rearrangement.
Once rearranged, the molecules can also undergo further unimolecular substitution (SN1) or unimolecular elimination (E1). Nucleophilic reactions often produce two products, 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). Similarly, we will see in subsequent sections of this chapter that for the unimolecular elimination reaction, a more substituted alkene can form through carbocation rearrangements ("stay tuned for coming attractions").
Hydride Shift
The hydride shift can also be called the 1,2-Hydride Shift because rearrangements primarily occur between adjacent carbon atoms. The 1,2 are communicating that the carbons are vicinal (adjacent). These numbers have nothing to do with the nomenclature of the reactant. We can see the phenomenon of hydride shift in solvolysis (SN1) reactions like the example below.
As shown in the following mechanism, the polarized carbon-chlorine bonds is heterolytically broken to produce a chloride ion and carbocation. The secondary carbocation undergoes a 1,2 hydride shift to produce the more stable tertiary carbocation. The oxygen of a water molecule acts as the nucleophile and reacts with the carbocation to form a protonated alcohol. The intermediate is deprotonated to form the final product, an alcohol. The mechanism for hydride shift occurs in multiple steps that includes various intermediates and transition states.
Exercise
1. Draw the bond-line structure for the major solvolysis product of each reaction.
Answer
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 molecule.
Exercise
2. Draw the bond-line structure for the major solvolysis product of each reaction.
Answer
Alkyl Halide Classification and Carbocation Rearrangements
Reactions of tertiary carbocations react much faster than that of secondary carbocations and will form the major product almost exclusively. Alkyl shifts from a secondary carbocation to tertiary carbocation in SN1 reactions occur by independent steps. When the alkyl halide is primary, then slight variations and differences between the two reaction mechanisms. 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.
Exercise
3. Draw the bond-line structure for the major solvolysis product of each reaction.
Answer
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 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.10%3A_Rearrangements_of_the_Carbocation_a.txt |
Learning Objective
• explain and apply Hammond's Postulate to substitution reactions
Now, back to transition states. Chemists are often very interested in trying to learn about what the transition state for a given reaction looks like, but addressing this question requires an indirect approach because the transition state itself cannot be observed. In order to gain some insight into what a particular transition state looks like, chemists often invoke the Hammond postulate, which states that a transition state resembles the structure of the nearest stable species. For an exergonic reaction, therefore, the transition state resembles the reactants more than it does the products.
If we consider a hypothetical exergonic reaction between compounds A and B to form AB, the distance between A and B would be relatively large at the transition state, resembling the starting state where A and B are two isolated species. In the hypothetical endergonic reaction between C and D to form CD, however, the bond formation process would be much further along at the TS point, resembling the product.
The Hammond Postulate is a very simplistic idea, which relies on an assumption that potential energy surfaces are parabolic. Although such an assumption is not rigorously true, it is fairly reliable and allows chemists to make energetic arguments about transition states by employing arguments about the stability of a related species. Since the formation of a reactive intermediate is very reliably endergonic, arguments about the stability of reactive intermediates can serve as proxy arguments about transition state stability.
The Hammond Postulate and the SN1 Reaction
The Hammond postulate suggests that the activation energy of the rate-determining first step will be inversely proportional to the stability of the carbocation intermediate. The stability of carbocations is shown qualitatively below:
Carbocation Stability
\[\ce{CH3(+) < CH3CH2(+) < (CH3)2CH(+) ≈ CH2=CH-CH2(+) < C6H5CH2(+) ≈ (CH3)3C(+)}\]
Consequently, we expect that 3º-alkyl halides will be more reactive than their 2º and 1º-counterparts in reactions that follow an SN1 mechanism. This is opposite to the reactivity order observed for the SN2 mechanism. Allylic and benzylic halides are exceptionally reactive by either mechanism.
7.12: Comparison of SN1 and SN2 Reactions
Learning Objective
• distinguish 1st or 2nd order substitution reactions
Predicting SN1 vs. SN2 mechanisms
When considering whether a nucleophilic substitution is likely to occur via an SN1 or SN2 mechanism, we really need to consider three factors:
1) The electrophile: when the leaving group is attached to a methyl group or a primary carbon, an SN2 mechanism is favored (here the electrophile is unhindered by surrounded groups, and any carbocation intermediate would be high-energy and thus unlikely). When the leaving group is attached to a tertiary, allylic, or benzylic carbon, a carbocation intermediate will be relatively stable and thus an SN1 mechanism is favored. These patterns of reactivity of summarized below.
Alkyl Halide Structure Possible Substitution Reactions
methyl and primary SN2 only
secondary SN2 and SN1
tertiary SN1 only
primary and secondary benzylic and allylic SN2 and SN1
tertiary benzylic and allylic SN1 only
vinyl and aryl NO reaction
2) The nucleophile: powerful nucleophiles, especially those with negative charges, favor the SN2 mechanism. Weaker nucleophiles such as water or alcohols favor the SN1 mechanism.
3) The solvent: Polar aprotic solvents favor the SN2 mechanism by enhancing the reactivity of the nucleophile. Polar protic solvents favor the SN1 mechanism by stabilizing the transition state and carbocation intermediate. SN1 reactions are called solvolysis reactions when the solvent is the nucleophile.
These patterns of reactivity are summarized in the table below.
Comparison between SN2 and SN1 Reactions
Reaction Parameter SN2 SN1
alkyl halide structure methyl > primary > secondary >>>> tertiary tertiary > secodary >>>> primary > methyl
nucleophile high concentration of a strong nucleophile poor nucleophile (often the solvent)
mechanism 1-step 2-stp
rate limiting step bimolecular transition state carbocation formation
rate law rate = k[R-X][Nu] rate = k[R-X]
stereochemisty inversion of configuration mixed configuration
solvent polar aprotic polar protic
For example, the reaction below has a tertiary alkyl bromide as the electrophile, a weak nucleophile, and a polar protic solvent (we’ll assume that methanol is the solvent). Thus we’d confidently predict an SN1 reaction mechanism. Because substitution occurs at a chiral carbon, we can also predict that the reaction will proceed with racemization.
In the reaction below, on the other hand, the electrophile is a secondary alkyl bromide – with these, both SN1 and SN2 mechanisms are possible, depending on the nucleophile and the solvent. In this example, the nucleophile (a thiolate anion) is strong, and a polar protic solvent is used – so the SN2 mechanism is heavily favored. The reaction is expected to proceed with inversion of configuration.
Exercise
1. Determine whether each substitution reaction shown below is likely to proceed by an SN1 or SN2 mechanism and explain your reasoning.
Answer
a) SN2 b/c primary alkyl halide with a strong nucleophile in a polar aprotic solvent.
b) SN1 b/c tertiary alkyl halide with a weak nucleophile that is also the solvent (solvolysis).
c) SN2 b/c secondary alkyl halides favor this mechanism when reacted with a strong nucleophile (and weak base) in a polar aprotic solvent. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.11%3A_The_Hammond_Postulate_and_Transitio.txt |
Learning Objective
• determine the rate law & predict the mechanism based on its rate equation or reaction data for E2 reactions
• predict the products and specify the reagents for E2 reactions with stereochemistry
• propose mechanisms for E2 reactions
• draw and interpret Reaction Energy Diagrams for E2 reactions
In order of decreasing importance, the factors impacting E2 reaction pathways are
1) structure of the alkyl halide
2) strength of the base
3) stability of the leaving group
4) type of solvent.
The bimolecular transition state of the E2 pathway means that orientation of the base and leaving group are a primary consideration. Both the base and leaving group are electron rich and electrostatically repel each other forcing an anti-coplanar orientation between the base and leaving group. The structure of the alkyl halide must assume the orientation for an anti-coplar transition state. The strength of the base will also influence the reaction along with the stability of the leaving group. Solvents play a very minor role in E2 pathway.
Introduction
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.
To get a clearer picture of the interplay of these factors involved in a a reaction between a nucleophile/base and an alkyl halide, consider the reaction of a 2º-alkyl halide, isopropyl bromide, with two different nucleophiles. In one pathway, a methanethiolate nucleophile substitutes for bromine in an SN2 reaction. In the other (bottom) pathway, methoxide ion acts as a base (rather than as a nucleophile) in an elimination reaction. As we will soon see, the mechanism of this reaction is single-step, and is referred to as the E2 mechanism.
General Reaction
Below is a mechanistic diagram of an elimination reaction by the E2 pathway:
.
In this reaction, ethoxide (CH3CH2O-) represents the base and Br representents 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 anticoplanar (antiperiplanar) transition states. Staggered conformation is usually the major product because of its lower energy confirmation.
An E2 reaction has certain requirements to proceed:
• A strong base is necessary especially necessary for primary alkyl halides. Secondary and tertirary primary halides will procede 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, a strically hindered base will result in the least substituted product.
E2 Reaction Coordinate
In the reaction energy diagram below, the base is represented as Ba-. The bimolecular transition state determines the overall reaction rate. It is important to note the anti-coplanar orientation of the base and the leaving group. Both the base and leaving group are electron rich and electrostatically repel each other forcing an anti-coplanar orientation between the base and leaving group.
anti-coplanar transition state
The Leaving Group Effect in E2 Reactions
As Size Increases, The Electron Density Decrease, The Ability of the Leaving Group to Leave Increases: Here we revisit the effect size has on basicity. If we move down the periodic table, size increases. With an increase in size, basicity decreases, and the ability of the leaving group to leave increases. The relationship among the following halogens, unlike the previous example, is true to what we will see in upcoming reaction mechanisms.
Exercise
1. Ignoring the alkene stereochemistry show the elimination product(s) of the following compounds.
Answer
1.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
• Layne A. Morsch (University of Illinois Springfield)
7.14: Zaitsev's Rule
Learning Objective
• use Zaitsev’s rule to predict major and minor products of elimination reactions
Zaisev's Rule and Regioselectivity
The prefix "regio" indicates the interaction of reactants during bond making and/or bond breaking occurs preferentially by one orientation. Because the beta-carbons of an alkyl halide may not be equivalent, there can be more than one possible elimination product. Zaitsev's Rule can be used to predict the regiochemistry of elimination reactions.
Zaitsev’s or Saytzev’s (anglicized spelling) rule is an empirical rule used to predict regioselectivity of beta-elimination reactions occurring via the E1 or E2 mechanisms. It states that in a regioselective E1 or E2 reaction the major product is the more stable alkene with the more highly substituted double bond as shown in the example below.
If two or more structurally distinct groups of beta-hydrogens are present in a given reactant, then several constitutionally isomeric alkenes may be formed by an E2 elimination. This situation is illustrated by the 2-bromobutane and 2-bromo-2,3-dimethylbutane elimination examples given below.
By using the strongly basic hydroxide nucleophile, we direct these reactions toward elimination. In both cases there are two different sets of beta-hydrogens available to the elimination reaction (these are colored red and magenta and the alpha carbon is blue). If the rate of each possible elimination was the same, we might expect the amounts of the isomeric elimination products to reflect the number of hydrogens that could participate in that reaction. For example, since there are three 1º-hydrogens (red) and two 2º-hydrogens (magenta) on beta-carbons in 2-bromobutane, statistics would suggest a 3:2 ratio of 1-butene and 2-butene in the products. This is not observed, and the latter predominates by 4:1. This departure from statistical expectation is even more pronounced in the second example, where there are six 1º-beta-hydrogens compared with one 3º-hydrogen. These results point to a strong regioselectivity favoring the more highly substituted product double bond, an empirical statement generally called the Zaitsev Rule.
Exercise
1. Ignoring the alkene stereochemistry show the elimination product(s) of the following compounds:
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.13%3A__Characteristics_of_the_E2_Reaction.txt |
Learning Objective
• determine the rate law & predict the mechanism based on its rate equation or reaction data for E1 reactions
• predict the products and specify the reagents for E1 reactions with stereochemistry
• propose mechanisms for E1 reactions
• draw and interpret Reaction Energy Diagrams for E1 reactions
General Reaction
Unimolecular Elimination (E1) is a reaction in which loss of the leaving group followed by removal of he beta-hydrogen 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 as the rate determining (slow) step, hence the name unimolecular. . Alkyl halides that can ionize to form stable carbocations are more reactive via the E1 mechanism. Because carbocation stability is the primary energetic consideration, stabilization of the carbocation via solvation is also an important consideration. Because carbocations are highly reactive, the strength of the base is not important and weak bases can be used. Since SN1 and E1 reactions behave similarly, they often 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. Heating the reaction favors elimination over substitution.
In order of decreasing importance, the factors impacting E1 reaction pathways are
1) structure of the alkyl halide
2) stability of the carbocation
3) type of solvent
4) strength of the base.
Mechanism for Alkyl Halides
As can be seen in the E1 mechanism below, 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 to form 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 and leaving group salt.
Once again, we see the two steps of the E1 mechanism.
1. 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).
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.
Regiochemistry & Stereochemistry of the E1 Reaction
The E1 reaction is regiospecific because it follows Zaitsev's rule that states 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, the E1 reaction is not stereospecific. Thus, a hydrogen is not required to be anti-coplanar to the leaving group because the leaving group is gone. In the mechanism below, we can see two possible pathways for the reaction. 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.
Exercises
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.
Answer
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 simaltaneously).
Outside Sources
1. McMurry, J., Simanek, E. Fundamentals of Organic Chemistry, 6th edition. Cengage Learning, 2007.
Contributors
• Satish Balasubramanian | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.15%3A_Characteristics_of_the_E1_Reaction.txt |
Learning Objective
• use Zaitsev’s rule to predict major and minor products of elimination reactions including halocyclohexanes
Cyclohexane Conformation & Anti-coplanar Orientation
The concerted mechanism of the E2 reaction requires that the base and leaving group are orientated anti-coplanar to each other. When the beta-hydrogen and leaving group (halide for this chapter) are located on a 6-membered ring, then Zaitsev's Rule may not be followed. The beta-hydrogen and leaving group must both be in the axial position for the E2 reaction to occur. Consequently, E2 reactions of certain cycloalkyl halides show unusual rates and regioselectivity that are not explained by the principles thus far discussed. For example, trans-2-methyl-1-chlorocyclohexane reacts with alcoholic KOH at a much slower rate than does its cis-isomer. Furthermore, the product from elimination of the trans-isomer is 3-methylcyclohexene (not predicted by the Zaitsev rule), whereas the cis-isomer gives the predicted 1-methylcyclohexene as the chief product. These differences are described by the first two equations in the following diagram.
Unlike open chain structures, cyclic compounds generally restrict the spatial orientation of ring substituents to relatively few arrangements. Consequently, reactions conducted on such substrates often provide us with information about the preferred orientation of reactant species in the transition state. Stereoisomers are particularly suitable in this respect, so the results shown here contain important information about the E2 transition state.
The most sensible interpretation of the elimination reactions of 2- and 4-substituted halocyclohexanes is that this reaction prefers an anti orientation of the halogen and the beta-hydrogen which is attacked by the base. These anti orientations are colored in red in the above equations. The compounds used here all have six-membered rings, so the anti orientation of groups requires that they assume a diaxial conformation. The observed differences in rate are the result of a steric preference for equatorial orientation of large substituents, which reduces the effective concentration of conformers having an axial halogen. In the case of the 1-bromo-4-tert-butylcyclohexane isomers, the tert-butyl group is so large that it will always assume an equatorial orientation, leaving the bromine to be axial in the cis-isomer and equatorial in the trans. Because of symmetry, the two axial beta-hydrogens in the cis-isomer react equally with base, resulting in rapid elimination to the same alkene (actually a racemic mixture). This reflects the fixed anti orientation of these hydrogens to the chlorine atom. To assume a conformation having an axial bromine the trans-isomer must tolerate serious crowding distortions. Such conformers are therefore present in extremely low concentration, and the rate of elimination is very slow. Indeed, substitution by hydroxide anion predominates.
A similar analysis of the 1-chloro-2-methylcyclohexane isomers explains both the rate and regioselectivity differences. Both the chlorine and methyl groups may assume an equatorial orientation in a chair conformation of the trans-isomer, as shown in the top equation. The axial chlorine needed for the E2 elimination is present only in the less stable alternative chair conformer, but this structure has only one axial beta-hydrogen (colored red), and the resulting elimination gives 3-methylcyclohexene. In the cis-isomer the smaller chlorine atom assumes an axial position in the more stable chair conformation, and here there are two axial beta hydrogens. The more stable 1-methylcyclohexene is therefore the predominant product, and the overall rate of elimination is relatively fast.
An orbital drawing of the anti-transition state is shown on the right. Note that the base attacks the alkyl halide from the side opposite the halogen, just as in the SN2 mechanism. In this drawing the α and β carbon atoms are undergoing a rehybridization from sp3 to sp2 and the developing π-bond is drawn as dashed light blue lines. The symbol R represents an alkyl group or hydrogen. Since both the base and the alkyl halide are present in this transition state, the reaction is bimolecular and should exhibit second order kinetics. We should note in passing that a syn-transition state would also provide good orbital overlap for elimination, and in some cases where an anti-orientation is prohibited by structural constraints syn-elimination has been observed.
Alkyl Halide Chains
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.
Exercise
1. Which of the following compounds will react faster in an E2 reaction; trans-1-bromo-2-isopropylcyclohexane or cis-1-bromo-2-isopropylcyclohexane?
Answer
1. The cis isomer will react faster than the trans. The cis isomer has two possible perpendicular hydrogen in which it can eliminate from. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.16%3A_E2_Regiochemistry_and_Cyclohexane_C.txt |
Learning Objective
• explain how the kinetic isotope effect (KIE) can be used to elucidate reaction mechanisms
Kinetic Isotope Effects
Kinetic Isotope Effects (KIEs) are used to determine reaction mechanisms by determining rate limiting steps and transition states and are commonly measured using NMR to detect isotope location or GC/MS to detect mass changes. In a KIE experiment an atom is replaced by its isotope and the change in rate of the reaction is observed. A very common isotope substitution is when hydrogen is replaced by deuterium. This is known as a deuterium effect and is expressed by the ratio kH/kD (as explained above). Normal KIEs for the deuterium effect are around 1 to 7 or 8. Large effects are seen because the percentage mass change between hydrogen and deuterium is great. Heavy atom isotope effects involve the substitution of carbon, oxygen, nitrogen, sulfur, and bromine, with effects that are much smaller and are usually between 1.02 and 1.10. The difference in KIE magnitude is directly related to the percentage change in mass. Large effects are seen when hydrogen is replaced with deuterium because the percentage mass change is very large (mass is being doubled) while smaller percent mass changes are present when an atom like sulfur is replaced with its isotope (increased by two mass units).
Primary KIEs
Primary kinetic isotope effects are rate changes due to isotopic substitution at a site of bond breaking in the rate determining step of a reaction.
Example
Consider the bromination of acetone: kinetic studies have been performed that show the rate of this reaction is independent of the concentration of bromine. To determine the rate determining step and mechanism of this reaction the substitution of a deuterium for a hydrogen can be made.
When hydrogen was replaced with deuterium in this reaction a $k_H \over k_D$ of 7 was found. Therefore the rate determining step is the tautomerization of acetone and involves the breaking of a C-H bond. Since the breaking of a C-H bond is involved, a substantial isotope effect is expected.
7.18: Comparison of E1 and E2 Reactions
Learning Objective
• distinguish 1st or 2nd order elimination reactions
Elimination reactions of alkyl halides can occur via the bimolecular E2 mechanism or unimolecular E1 mechanism as shown in the diagram below.
Comparing E1 and E2 mechanisms
When considering whether an elimination reaction is likely to occur via an E1 or E2 mechanism, we really need to consider three factors:
1) The base: strong bases favor the E2 mechanism, whereas, E1 mechanisms only require a weak base.
2) The solvent: good ionizing xolvents (polar protic) favor the E1 mechanism by stabilizing the carbocation intermediate.
3) The alkyl halide: primary alkyl halides have the only structure useful in distinguishing between the E2 and E1 pathways. Since primary carbocations do not form, only the E2 mechanism is possible.
Reaction Parameter E2 E1
alkyl halide structure tertiary > secondary > primary tertiary > secondary >>>> primary
nucleophile high concentration of a strong base weak base
mechanism 1-step 2-step
rate limiting step anti-coplanar bimolecular transition state carbocation formation
rate law rate = k[R-X][Base] rate = k[R-X]
stereochemisty retained configuration mixed configuration
solvent not important polar protic
Exercises
1. Predict the dominant elimination mechanism (E1 or E2) for each reaction below. Explain your reasoning.
2. Which one of the following groups of compounds would eliminate \(\ce{HCl}\) most readily on reaction with potassium hydroxide? Explain your reasoning, draw the bond-line structure and give the IUPAC name of the product.
a) \(\ce{(CH_3)_3CCl}\) \(\ce{CH_3CH_2CH_2CH_2Cl}\) \(\ce{CH_3CH(Cl)CH_2CH_3}\)
b) \(\ce{(CH_3)_3CCH_2Cl}\) \(\ce{(CH_3)_2CHCH_2Cl}\)
c)
3. Specify the reaction conditions to favor the indicated elimination mechanism.
Answer
1.
2.
3. a) strong base, such as hydroxide, an alkoxide, or equivalent
b) water or alcohol or equivalent weak base with heat
c) strong base, such as hydroxide, an alkoxide, or equivalent
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.17%3A_The_E2_Reaction_and_the_Deuterium_I.txt |
Learning Objective
• predict the products and specify the reagents for SN1, SN2, E1 and E2 reactions with stereochemistry
• propose mechanisms for SN1, SN2, E1 and E2 reactions
• draw, interpret, and apply Reaction Energy Diagrams for SN1, SN2, E1 and E2 reactions
Summary of Reaction Patterns
Having discussed the many factors that influence nucleophilic substitution and elimination reactions of alkyl halides, we must now consider the practical problem of predicting the most likely outcome when a given alkyl halide is reacted with a given nucleophile. As we noted earlier, several variables must be considered, the most important being the structure of the alkyl group and the nature of the nucleophilic reactant.In general, in order for an SN1 or E1 reaction to occur, the relevant carbocation intermediate must be relatively stable. Strong nucleophile favor substitution, and strong bases, especially strong hindered bases (such as tert-butoxide) favor elimination.
The nature of the halogen substituent on the alkyl halide is usually not very significant if it is Cl, Br or I. In cases where both SN2 and E2 reactions compete, chlorides generally give more elimination than do iodides, since the greater electronegativity of chlorine increases the acidity of beta-hydrogens. Indeed, although alkyl fluorides are relatively unreactive, when reactions with basic nucleophiles are forced, elimination occurs (note the high electronegativity of fluorine).
The following table summarizes the expected outcome of alkyl halide reactions with nucleophiles. It is assumed that the alkyl halides have one or more beta-hydrogens, making elimination possible; and that low dielectric solvents (e.g. acetone, ethanol, tetrahydrofuran & ethyl acetate) are used. When a high dielectric solvent would significantly influence the reaction this is noted in red. Note that halogens bonded to sp2 or sp hybridized carbon atoms do not normally undergo substitution or elimination reactions with nucleophilic reagents.
Nucleophile
Anionic Nucleophiles
( Weak Bases: I, Br, SCN, N3,
CH3CO2 , RS, CN etc. )
pKa's from -9 to 10 (left to right)
Anionic Nucleophiles
( Strong Bases: HO, RO )
pKa's > 15
Neutral Nucleophiles
( H2O, ROH, RSH, R3N )
pKa's ranging from -2 to 11
Alkyl Group
Primary
RCH2
Rapid SN2 substitution. The rate may be reduced by substitution of β-carbons, as in the case of neopentyl. Rapid SN2 substitution. E2 elimination may also occur. e.g.
ClCH2CH2Cl + KOH ——> CH2=CHCl
SN2 substitution. (N ≈ S >>O)
Secondary
R2CH–
SN2 substitution and / or E2 elimination (depending on the basicity of the nucleophile). Bases weaker than acetate (pKa = 4.8) give less elimination. The rate of substitution may be reduced by branching at the β-carbons, and this will increase elimination. E2 elimination will dominate. SN2 substitution. (N ≈ S >>O)
In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be formed slowly.
Tertiary
R3C–
E2 elimination will dominate with most nucleophiles (even if they are weak bases). No SN2 substitution due to steric hindrance. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be expected. E2 elimination will dominate. No SN2 substitution will occur. In high dielectric ionizing solvents SN1 and E1 products may be formed. E2 elimination with nitrogen nucleophiles (they are bases). No SN2 substitution. In high dielectric ionizing solvents SN1 and E1 products may be formed.
Allyl
H2C=CHCH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides. E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Benzyl
C6H5CH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides (note there are no β hydrogens). E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Experimental Observations
Experimental observations are reported for the following reactions. These reactions include a range of alkyl halide structures in a variety of reaction conditions to illustrate the reaction patterns summarized above. In describing these, it is useful to designate the halogen-bearing carbon as alpha and the carbon atom(s) adjacent to it as beta, as noted in the first four equations shown below. Replacement or substitution of the halogen on the α-carbon (colored maroon) by a nucleophilic reagent is a commonly observed reaction, as shown in equations 1, 2, 5, 6 & 7 below. Also, since the electrophilic character introduced by the halogen extends to the β-carbons, and since nucleophiles are also bases, the possibility of base induced H-X elimination must also be considered, as illustrated by equation 3. Finally, there are some combinations of alkyl halides and nucleophiles that fail to show any reaction over a 24 hour period, such as the example in equation 4. For consistency, alkyl bromides have been used in these examples. Similar reactions occur when alkyl chlorides or iodides are used, but the speed of the reactions and the exact distribution of products will change.
In order to understand why some combinations of alkyl halides and nucleophiles give a substitution reaction, whereas other combinations give elimination, and still others give no observable reaction, we must investigate systematically the way in which changes in reaction variables perturb the course of the reaction. The following general equation summarizes the factors that will be important in such an investigation.
Friendly Reminder: One conclusion, relating the structure of the R-group to possible products, should be immediately obvious. If R- has no beta-hydrogens an elimination reaction is not possible, unless a structural rearrangement occurs first. The first four halides shown on the left below do not give elimination reactions on treatment with base, because they have no β-hydrogens. The two halides on the right do not normally undergo such reactions because the potential elimination products have highly strained double or triple bonds. It is also worth noting that sp2 hybridized C–X compounds, such as the three on the right, do not normally undergo nucleophilic substitution reactions, unless other functional groups perturb the double bond(s).
Exercise
1. Identify the dominant reaction mechanism (SN1, SN2, E1, or E2) and predict the major product for the following reactions.
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.19%3A_Comparing_Substitution_and_Eliminat.txt |
Objective
• discuss the importance of leaving groups in biological substitution reactions
Leaving Groups in Biochemical Reactions
In biological reactions, we do not often see halides serving as leaving groups (in fact, outside of some marine organisms, halogens are fairly unusual in biological molecules). More common leaving groups in biochemical reactions are phosphates, water, alcohols, and thiols. In many cases, the leaving group is protonated by an acidic group on the enzyme as bond-breaking occurs. For example, hydroxide ion itself seldom acts as a leaving group – it is simply too high in energy (too basic). Rather, the hydroxide oxygen is generally protonated by an enzymatic acid before or during the bond-breaking event, resulting in a (very stable) water leaving group.
More often, however, the hydroxyl group of an alcohol is first converted enzymatically to a phosphate ester in order to create a better leaving group. This phosphate ester can take the form of a simple monophosphate (arrow 1 in the figure below), a diphosphate (arrow 2), or a nucleotide monophosphate (arrow 3).
Due to resonance delocalization of the developing negative charge, phosphates are excellent leaving groups.
Here’s a specific example (from DNA nucleotide biosynthesis):
Here, the OH group on ribofuranose is converted to a diphosphate, a much better leaving group. Ammonia is the nucleophile in the second step of this SN1-like reaction.
What is important for now is that in each case, an alcohol has been converted into a much better leaving group, and is now primed for a nucleophilic substitution reaction.
SAM Methyltransferases
Some of the most important examples of SN2 reactions in biochemistry are those catalyzed by S-adenosyl methionine (SAM) – dependent methyltransferase enzymes. We have already seen, in chapter 6 and again in chapter 8, how a methyl group is transferred in an SN2 reaction from SAM to the amine group on the nucleotide base adenosine:
Another SAM-dependent methylation reaction is catalyzed by an enzyme called catechol-O-methyltransferase. The substrate here is epinephrine, also known as adrenaline.
Notice that in this example, the attacking nucleophile is an alcohol rather than an amine (that’s why the enzyme is called an O-methyltransferase). In both cases, though, a basic amino acid side chain is positioned in the active site in just the right place to deprotonate the nucleophilic group as it attacks, increasing its nucleophilicity. The electrophile in both reactions is a methyl carbon, so there is little steric hindrance to slow down the nucleophilic attack. The methyl carbon is electrophilic because it is bonded to a positively-charged sulfur, which is a powerful electron withdrawing group. The positive charge on the sulfur also makes it an excellent leaving group, as the resulting product will be a neutral and very stable sulfide. All in all, in both reactions we have a reasonably good nucleophile, an electron-poor, unhindered electrophile, and an excellent leaving group.
Because the electrophilic carbon in these reactions is a methyl carbon, a stepwise SN1-like mechanism is extremely unlikely: a methyl carbocation is very high in energy and thus is not a reasonable intermediate to propose. We can confidently predict that this reaction is SN2. Does this SN2 reaction occur, as expected, with inversion of stereochemistry? Of course, the electrophilic methyl carbon in these reactions is achiral, so inversion is not apparent. To demonstrate inversion, the following experiment has been carried out with catechol-O-methyltransferase:
Here, the methyl group of SAM was made to be chiral by incorporating hydrogen isotopes tritium (3H, T) and deuterium (2H, D). The researchers determined that the reaction occurred with inversion of configuration, as expected for an SN2 displacement (J. Biol. Chem. 1980, 255, 9124).
Example
Contributors | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.20%3A_Biological_Substitution_Reactions.txt |
Learning Objective
• discuss enzymatic elimination reactions of histidine
Enzymatic E1 and E2 reactions
While most biochemical b-elimination reactions are of the E1cb type, some enzymatic E2 and E1 reactions are known. Like the enzymatic SN2 and SN1 substitution mechanisms discussed in chapters 8 and 9, the E2 and E1 models represent two possible mechanistic extremes, and actual enzymatic elimination reactions may fall somewhere in between. In an E1/E2 hybrid elimination, for example, Cβ-X bond cleavage may be quite advanced (but not complete) before proton abstraction takes place - this would lead to the build-up of transient partial positive charge on Cβ, but a discreet carbocation intermediate would not form. The extent to which partial positive charge builds up determines whether we refer to the mechanism as 'E1-like' or 'E2-like'.
A reaction in the histidine biosynthetic pathway provides a good example of a biological E1-like elimination step (we're looking specifically here at the first, enol-forming step in the reaction below - the second step is simply a tautomerization from the enol to the ketone product.
Notice in this mechanism that an E1cb elimination is not possible - there is no electron-withdrawing group (like a carbonyl) to stabilize the carbanion intermediate that would form if the proton were abstracted first. There is, however, an electron-donating group (the lone pair on a nitrogen) that can stabilize a positively-charged intermediate that forms when the water leaves.
Another good example of a biological E1-like reaction is the elimination of phosphate in the formation of 5-enolpyruvylshikimate-3-phosphate (EPSP), an intermediate in the synthesis of aromatic amino acids.
Experimental evidence indicates that significant positive charge probably builds up on Cβ of the starting compound, implying that C-O bond cleavage is advanced before proton abstraction occurs (notice the parallels to the Cope elimination in the previous section):
The very next step in the aromatic acid biosynthesis pathway is also an elimination, this time a 1,6-conjugated elimination rather than a simple beta-elimination.
An E1-like mechanism (as illustrated above) has been proposed for this step, but other evidence suggests that a free-radical mechanism may be involved.
While most E1 and E2 reactions involve proton abstraction, eliminations can also incorporate a decarboxylation step.
Isopentenyl diphosphate, the 'building block' for all isoprenoid compounds, is formed from a decarboxylation-elimination reaction.
Phenylpyruvate, a precursor in the biosynthesis of phenylalanine, results from a conjugated 1,6 decarboxylation-elimination.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
7.22: Additional Exercises
SN2
7-1 Predict which compound in each pair would undergo the SN2 reaction faster.
a) or
b) or
c) or
d) or
7-2 Predict the products of these nucleophilic substitution reactions, including stereochemistry when appropriate.
a)
b)
c)
d)
7-3 Show how each compound might be synthesized using SN2 reaction.
a)
b)
c)
7-4 Show 2 ways to synthesize the ether below using SN2 reaction
7-5 Of the two ways of synthesizing the compound in previous question (7-4), which one would be the most efficient? Why? Show the mechanism of the reaction as part of your explanation.
7-6 Arrange the compounds below in increasing order of nucleophilicity.
H2O; NH2-; CH3O-; CH3CH2O-
SN1
7-7 List the following carbocations in order of increasing stability.
7-8 Give the solvolysis product expected when the compound is heated in methanol.
a)
b)
c)
d)
7-9 Predict with compound in each pair will undergo an SN1 reaction more quickly.
a) or
b) or
c) or
d) or
7-10 Show how the following carbocations would rearrange to become more stable. Draw the mechanism of the rearrangement.
a)
b)
c)
d)
SN2 vs SN1
7-11 Predict whether each compound below would be more likely to undergo a SN2 or SN1 reaction.
a)
b)
c)
d)
7-12 Predict the product of the following reactions.
a)
b)
c)
d)
7-13 Show how each compound may be synthesized using nucleophilic substitution reactions.
a)
b)
c)
d)
e)
f)
g)
E2 vs E1
7-14 Predict the major products of the following reactions.
a)
b)
c)
7-15 Draw the expected major product when each of the following compounds is treated with hydroxide to give an E2 reaction.
a)
b)
c)
7-16 Predict all the elimination products of the following reactions and label the major product.
a)
b)
c)
d)
Substitution vs Elimination
7-17 Identify the function of the following reagents. The reagents will be a strong/weak nucleophile and/or a strong/weak base.
a) Cl-
b) NaH
c) t-BuO-
d) OH-
e) H2O
f) HS-
g) MeOH
7-18 Identify which mechanism the following reactions would undergo.
a)
b)
c)
d)
e)
7-19 Identify all the products of the following reactions and specify the major product.
a)
b)
c)
d)
e)
7-20 The following reaction yields five different products. Give the mechanisms for how each is formed.
7.23: Solutions to Additional Exercises
SN2
7-1
a)
b)
c)
d)
7-2
a)
b)
c) No reaction
d)
7-3
a)
b)
c)
7-4
First method:
Second method:
7-5 The second method is more efficient since the alkyl halide is not sterically hindered
7-6 H2O < NH2- < CH3CH2O-< CH3O-
7-7
7-8
a)
b)
c) No reaction
d)
7-9
a)
b)
c)
d)
7-10
a)
b)
c)
d)
7-11
a) SN2
b) SN2
c) SN1
d) SN1
7-12
a)
b)
c)
d)
7-13
a)
b)
c)
d)
e)
f)
g)
7-14
a)
b)
c)
7-15
a)
b)
c)
7-16
a)
b)
c)
d)
Substitution vs Elimination
7-17
a) Cl- ; strong nucleophile
b) NaH ; strong base
c) t-BuO- ; strong base
d) OH- ; strong nucleophile ; strong base
e) H2O ; weak nucleophile ; weak base
f) HS- ; strong nucleophile
g) MeOH ; weak nucleophile ; weak base
7-18
a) E2, SN1
b) SN2, E2
c) SN2
d) SN1, E1
e) E2, SN1
7-19
a)
b)
c)
d)
e)
7-20 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.21%3A_Biological_Elimination_Reactions.txt |
Learning Objectives:
After reading the chapter and completing the exercises and homework, a student can be able to:
• describe the electronic structure of alkenes using Molecular Orbital (MO) Theory and Orbital Hybridization - refer to section 8.1
• memorize the common names for vinylic and allylic groups including isoprene and styrene refer to section 8.2
• predict the relative physical properties of alkenes - refer to section 8.2
• recognize and classify the stereochemistry of alkenes using the cis/trans and E/Z systems - refer to section 8.3
• calculate the Degrees of Unsaturation (DU) and apply it to alkene structure - refer to section 8.4
• give the IUPAC names for alkenes given their structure & vice versa including E/Z isomers - refer to section 8.5 and chapter 3
• use heats of hydrogenation to compare the stabilities of alkenes - refer to section 8.6
• interpret and draw reaction energy diagrams for dehydrohalogenation of R-X’s and alcohol dehydration reactions - refer to sections 8.7 and 8.8 respectively and chapter 7
• propose mechanisms for a dehydrohalogenation or dehydration reactions - refer to sections 8.7 and 8.8 respectively and chapter 7
• predict the products and specify the reagents for alkene synthesis from dehydrohalogenation of R-X’s and alcohol dehydration reactions - refer to sections 8.7 and 8.8 respectively
• predict and explain the stereochemistry of E2 eliminations to form alkenes, especially from cyclohexanes - refer to sections 8.7 and 8.8 and chapter 7
• discuss the uses and sources of alkenes including catalytic cracking - refer to section 8.9
• 8.1: Alkene Structure
Alkenes are a class of hydrocarbons (i.e., containing only carbon and hydrogen). They are unsaturated compounds with at least one carbon-to-carbon double bond.
• 8.2: Physical Properties and Important Common Names
Alkenes are non-polar hydrocarbons with physical properties similar to alkanes. At room temperature, alkenes exist in all three phases, solid, liquids, and gases. The stereochemistry of the geometric isomers (cis/trans) can influence the physical properties.
• 8.3: The Alkene Double Bond and Stereoisomerism
The two lobes of the pi bond in the alkenes prevent rotation and are responsible for their rigid nature. The lack of rotation creates the potential for geometric isomers (cis/trans).
• 8.4: Degrees of Unsaturation
Calculating the degrees of unsaturation (DU) can provide useful information about the chemical structure from the molecular formula. The DU indicates the presence of rings and π bonds, but cannot distinguish between them.
• 8.5: The E/Z System (when cis/trans does not work)
Some alkenes cannot be unambiguously named using the cis/trans system. The Cahn-Ingold-Prelog (CIP) rules were used to develop the E/Z system for naming the stereoisomers of alkenes.
• 8.6: Stability of Alkenes
The energy released during alkene hydrogenation is called the heat of hydrogenation and indicates the relative stability of the double bond in the molecule.
• 8.7: Alkene Synthesis by Elimination of Alkyl Halides
The alkyl halide elimination reactions (E1 and E2) to synthesize alkenes are briefly reviewed. Refer to chapter 7 sections 13 through 18 for a complete explanation.
• 8.8: Alkene Synthesis by Dehydration of Alcohols
The dehydration reaction of alcohols to generate alkene proceeds by heating the alcohols in the presence of a strong acid, such as sulfuric or phosphoric acid, at high temperatures.
• 8.9: Uses and Sources of Alkenes
Among the most important and most abundant organic chemicals produced worldwide are the two simple alkenes, ethylene and propylene. Thermal cracking is briefly explained.
• 8.10: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
• 8.11: Solutions to Additional Exercises
This section has the solutions to the additional exercises from the previous section.
08: Structure and Synthesis of Alkenes
Learning Objective
• describe the electronic structure of alkenes using Molecular Orbital (MO) Theory and Orbital Hybridization
Alkenes are a class of hydrocarbons (i.e., containing only carbon and hydrogen). They are unsaturated compounds with at least one carbon-to-carbon double bond. The double bond makes alkenes more reactive than alkanes. Olefin is another term used to describe alkenes. The alkene group can also be called a vinyl group and the carbons sharing the double bond can be called vinyl carbons.
Alkenes
Nomenclature of alkenes is covered in chapter 3. Condensed structural formulas for the first eight alkenes are listed in Table $1$ along with some relevant physical propoerties. Thus, CH2=CH2 stands for
The double bond is shared by the two carbon atoms and does not involve the hydrogen atoms, although the condensed formula does not make this point obvious. Note that the molecular formula for ethene is C2H4, whereas that for ethane is C2H6.
Structure of Ethene - the simplest alkene
Ethene is not a very complicated molecule. It is made up of four 1s1 hydrogen atoms and two 2s2 2$p_x$1 2$p_y$2 carbon atoms. These carbon atoms already have four electrons, but they each want to get four more so that they have a full eight in the valence shell. Having eight valence electrons around carbon gives the atom itself the same electron configuration as neon, a noble gas. Carbon wants to have the same configuration as Neon because when it has eight valence electrons carbon is at its most stable, lowest energy state, it has all of the electrons that it wants, so it is no longer reactive.
This forms a total of three bonds to each carbon atom, giving them an $sp^2$ hybridization. Since the carbon atom is forming three sigma bonds instead of the four that it can, it only needs to hybridize three of its outer orbitals, instead of four. It does this by using the $2s$ electron and two of the $2p$ electrons, leaving the other unchanged. This new orbital is called an $sp^2$ hybrid because that's exactly what it is, it is made from one s orbital and two p orbitals.
When atoms are an $sp^2$ hybrid they have a trigonal planar structure. These structures are very similar to a 'peace' sign, there is a central atom with three atoms around it, all on one plane. Trigonal planar molecules have an ideal bond angle of 120° on each side.
The H-C-H bond angle is 117°, which is very close to the ideal 120° of a carbon with $sp^2$ hybridization. The other two angles (H-C=C) are both 121.5°.
Rigidity in Ethene
There is rigidity in the ethene molecule due to the double-bonded carbons. A double bond consists of one sigma bond formed by overlap of sp2 hybrid orbitals and one pi bond formed by overlap of parallel 2 p orbitals. In ethene there is no free rotation about the carbon-carbon sigma bond because these two carbons also share a $\pi$ bond. A $\pi$ bond is only formed when there is adequate overlap between both top and bottom p-orbitals. Free rotation the p-orbitals cause them to be 90° from each other breaking the $\pi$ bond because there would be no overlap. Since the $\pi$ bond is essential to the structure of ethene it must not break, so there can be not free rotation about the carbon-carbon sigma bond. The two carbon atoms of a double bond and the four atoms attached to them lie in a plane, with bond angles of approximately 120° as shownn in the figure below
(a) The σ-bonded framework is formed by the overlap of two sets of singly occupied carbon sp2 hybrid orbitals and four singly occupied hydrogen 1s orbitals to form electron-pair bonds. This uses 10 of the 12 valence electrons to form a total of five σ bonds (four C–H bonds and one C–C bond).
(b) One singly occupied unhybridized 2pz orbital remains on each carbon atom to form a carbon–carbon π bond. (Note: by convention, in planar molecules the axis perpendicular to the molecular plane is the z-axis.)
The first two alkenes in Table $1$, ethene and propene, are most often called by their common names—ethylene and propylene, respectively (Figure $1$). Ethylene is a major commercial chemical. The US chemical industry produces about 25 billion kilograms of ethylene annually, more than any other synthetic organic chemical. More than half of this ethylene goes into the manufacture of polyethylene, one of the most familiar plastics. Propylene is also an important industrial chemical. It is converted to plastics, isopropyl alcohol, and a variety of other products.
Exercise
1. Although there is only one alkene with the formula C2H4 (ethene) and only one with the formula C3H6 (propene), there are several alkenes with the formula C4H8. Draw all of the possible bond line structures for alkenes with the formula C4H8 including all possible structural and stereoisomers.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.01%3A_Alkene_Structure.txt |
Learning Objectives
• memorize the common names for vinylic and allylic groups including isoprene and styrene
• predict the relative physical properties of alkenes
Common names
The carbon atoms sharing the double bond can be referred to as the "vinyl carbons". The carbon atoms adjacent to the vinyl carbon atoms are called "allylic carbons". These carbon atoms have unique reactivity because of the potential for interaction with the pi bond.
Overall, common names remove the -ane suffix and add -ylene. There are a couple of unique ones like ethenyl's common name is vinyl and 2-propenyl's common name is allyl that need to be memorized.
• vinyl substituent H2C=CH-
• allyl substituent H2C=CH-CH2-
• allene molecule H2C=C=CH2
• isoprene is shown below
Physical Properties of Selected Alkenes
Some representative alkenes—their names, structures, and physical properties—are given in the table below.
Physical Properties of Some Selected Alkenes
IUPAC Name Molecular Formula Condensed Structural Formula Melting Point (°C) Boiling Point (°C)
ethene C2H4 CH2=CH2 –169 –104
propene C3H6 CH2=CHCH3 –185 –47
1-butene C4H8 CH2=CHCH2CH3 –185 –6
1-pentene C5H10 CH2=CH(CH2)2CH3 –138 30
1-hexene C6H12 CH2=CH(CH2)3CH3 –140 63
1-heptene C7H14 CH2=CH(CH2)4CH3 –119 94
1-octene C8H16 CH2=CH(CH2)5CH3 –102 121
Polarity and Physical Properties
Alkenes are non-polar hydrocarbons. The dominant intermolecular forces shared by alkenes are the London dispersion forces. These interactions are weak and temporary, so they are easily disrupted.
Physical States: The physical states reflect the weak attractive forces between molecules. Ethene, propene, and butene exist as colorless gases. Alkenes with 5 to 14 carbons are liquids, and alkenes with 15 carbons or more are solids.
Density: Alkenes are less dense than water with most densities in the range of 0.6 to 0.7 g/mL. Alkenes float on top of water.
Solubility: Alkenes are virtually insoluble in water, but dissolve in organic solvents. The reasons for this are exactly the same as for the alkanes.
Boiling Points: The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Boiling points of alkenes depend on more molecular mass (chain length). The more intermolecular mass is added, the higher the boiling point. Intermolecular forces of alkenes gets stronger with increase in the size of the molecules. In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains.
Compound Boiling points (oC)
Ethene -104
Propene -47
Trans-2-Butene 0.9
Cis-2-butene 3.7
Trans 1,2-dichlorobutene 155
Cis 1,2-dichlorobutene 152
1-Pentene 30
Trans-2-Pentene 36
Cis-2-Pentene 37
1-Heptene 115
3-Octene 122
3-Nonene 147
5-Decene 170
Melting Points: Melting points of alkenes depends on the packaging of the molecules so the stereochemistry of the carbon-carbon double bond has a strong influence on the relative melting points. Alkenes have similar melting points to that of alkanes, however, in cis isomers molecules are package in a U-bending shape, therefore, will display a lower melting points to that of the trans isomers. This effect is notable when comparing the melting points of fats and oils. The differences in the melting points is strongly influenced by the long hydrocarbon tails. Oils have a greater number of cis double bonds and exist as liquids at room temperature. Whereas, fats are primarily saturated and exist as solids at room temperature.
Compound Melting Points (0C)
Ethene -169
Propene -185
Butene -138
1-Pentene -165
Trans-2-Pentene -135
Cis-2-Pentene -180
1-Heptene -119
3-Octene -101.9
3-Nonene -81.4
5-Decene -66.3
Exercise
1. Draw the bond-line structures for the following compounds in order of increasing boiling point: 2-methyl-2-pentene; 2-hexene; isoprene; 2-heptene.
2. Which phase will contain the most 3-octene?
a) water or hexane
b) water or benzene
c) methanol or 1-octanol
Answer
1. relative boiling points
2. a) hexane (Hydrocarbons are hydrophobic and lipophilic.)
b) benzene (Hydrocarbons are hydrophobic and lipophilic.)
c) 1-octanol (Hydrocarbons seek the solvent with the most carbons and fewest polar groups.)
Contributors
• Trung Nguyen
• Jim Clark (Chemguide.co.uk)
• Layne A. Morsch (University of Illinois Springfield) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.02%3A_Physical_Properties_and_Important_Common_Names.txt |
Learning Objective
• recognize and classify the stereochemistry of alkenes using the cis/trans system
Steroisomerism in Alkenes
There is free rotation about the carbon-to-carbon single bonds (C–C) in alkanes. For example, in 1,2-dichloroethane, there is free rotation about the C–C bond. The two models shown represent exactly the same molecule; they are not isomers. You can draw structural formulas that look different, but if you bear in mind the possibility of this free rotation about single bonds, you should recognize that these two structures represent the same molecule:
In contrast, the structure of alkenes requires that the carbon atoms of a double bond and the two atoms bonded to each carbon atom all lie in a single plane, and that each doubly bonded carbon atom lies in the center of a triangle. This part of the molecule’s structure is rigid; rotation about doubly bonded carbon atoms is not possible without rupturing the bond. In 1,2-dichloroethene, restricted rotation about the double bond means that the relative positions of substituent groups above or below the double bond become significant. This leads to a special kind of isomerism. The isomer in which the two chlorine (Cl) atoms lie on the same side of the molecule is called the cis isomer (Latin cis, meaning “on this side”) and is named cis-1,2-dichloroethene. The isomer with the two Cl atoms on opposite sides of the molecule is the trans isomer (Latin trans, meaning “across”) and is named trans-1,2-dichloroethene. These two compounds are cis-trans isomers (or geometric isomers), compounds that have different configurations (groups permanently in different places in space) because of the presence of a rigid structure in their molecule.
The diagram below summarizes the differences between alkanes and alkenes with respect to rotation where the carbons are red, hydrogens are off-white, and the chlorines are green.
In 1,2-dichloroethane (a), free rotation about the C–C bond allows the two structures to be interconverted by a twist of one end relative to the other. In 1,2-dichloroethene (b), restricted rotation about the double bond means that the relative positions of substituent groups above or below the double bond are significant.
Geometric Isomers have Different Physical Properties
Consider the alkene with the condensed structural formula CH3CH=CHCH3. We could name it 2-butene, but there are actually two such compounds; the double bond results in cis-trans isomerism. Cis-2-butene has both methyl groups on the same side of the molecule. Trans-2-butene has the methyl groups on opposite sides of the molecule as shown in the diagram below.
Cis-2-butene and trans-2-butene are unique compounds with slightly different physical properties as shown below. For stereospecific reactions, these compounds produce different stereoisomeric products under the same reaction conditions. This phenonenom of reactivity will be explored more closely in the next chapter on alkene reactivity.
Example
Which compounds can exist as cis-trans (geometric) isomers? Draw them.
1. CHCl=CHBr
2. CH2=CBrCH3
3. (CH3)2C=CHCH2CH3
4. CH3CH=CHCH2CH3
Solutions Explained
All four structures have a double bond and thus meet rule 1 for cis-trans isomerism.
1. This compound meets rule 2; it has two nonidentical groups on each carbon atom (H and Cl on one and H and Br on the other). It exists as both cis and trans isomers:
2. This compound has two hydrogen atoms on one of its doubly bonded carbon atoms; it fails rule 2 and does not exist as cis and trans isomers.
3. This compound has two methyl (CH3) groups on one of its doubly bonded carbon atoms. It fails rule 2 and does not exist as cis and trans isomers.
4. This compound meets rule 2; it has two nonidentical groups on each carbon atom and exists as both cis and trans isomers:
Exercise
Exercise
1. If the compound below can exist as cis-trans isomers, then draw both bond-line structures.
1. CH2=CHCH2CH2CH3
2. CH3CH=CHCH2CH3
3. CH3CH2CH=CHCH2CH3
4. CH2C(CH3)CH2CH3
5. CH3C(CH3)CHCH3
2. Write out the condensed structure for ethene.
3. Draw the Kekulé (Lewis) structure for ethene.
4. Draw the bond-line structure for ethene.
5. Why is it that the carbons in ethene cannot freely rotate around the carbon-carbon double bond?
Answer
1.
2. CH2CH2 The carbon-carbon double bond is not shown, but can be recognized by knowing the neutral bonding patterns of carbon.
3.
4." - " Note how easy it would be to misinterpret this small dash, therefore, the structure for ethene is typically shown with a condensed or Kekule structure.
5. The carbons cannot freely rotate about the carbon-carbon double bond because rotating p-orbitals would have to pass through a 90° point where there would no longer be any overlap, so the $\pi$ bond would have to break for there to be free rotation. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.03%3A_The_Alkene_Double_Bond_and_Stereoisomerism.txt |
Learning Objectives
• calculate the Degrees of Unsaturation (DU) and apply it to alkene structure
Saturated and Unsaturated Molecules
In the lab, saturation may be thought of as the point when a solution cannot dissolve anymore of a substance added to it. In terms of degrees of unsaturation, a molecule only containing single bonds with no rings is considered saturated.
CH3CH2CH3 1-methyoxypentane
Unlike saturated molecules, unsaturated molecules contain double bond(s), triple bond(s) and/or ring(s).
CH3CH=CHCH3 3-chloro-5-octyne
There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (NMR) and infrared radiation (IR) are the primary ways of determining molecular structures, these techniques require expensive instrumentation and are not always readily available. Fortunately, calculating the degrees of unsaturation provides useful information about the structure. The degree of unsaturation indicates the total number of pi bonds and rings within a molecule which makes it easier for one to figure out the molecular structure.
DU = Degrees of Unsaturation = (number of pi bonds) + (number of rings)
Alkenes (R2C=CR2) and alkynes (R–C≡C–R) are called unsaturated hydrocarbons because they have fewer hydrogen atoms than does an alkane with the same number of carbon atoms, as is indicated in the following general formulas:
Calculating The Degree of Unsaturation (DU)
If the molecular formula is given, plug in the numbers into this formula:
$DoU= \dfrac{2C+2+N-X-H}{2} \tag{7.2.1}$
• $C$ is the number of carbons
• $N$ is the number of nitrogens
• $X$ is the number of halogens (F, Cl, Br, I)
• $H$ is the number of hydrogens
The molecular formula of a hydrocarbon provides information about the possible structural types it may represent. A saturated molecule contains only single bonds and no rings. Another way of interpreting this is that a saturated molecule has the maximum number of hydrogen atoms possible to be an acyclic alkane. Thus, the number of hydrogens can be represented by 2C+2, which is the general molecular representation of an alkane. As an example, for the molecular formula C3H4 the number of actual hydrogens needed for the compound to be saturated is 8
[2C+2=(2x3)+2=8.]
The compound needs 4 more hydrogens in order to be fully saturated (expected number of hydrogens-observed number of hydrogens=8-4=4). Degrees of unsaturation is equal to 2, or half the number of hydrogens the molecule needs to be classified as saturated. Hence, the DoB formula divides by 2. The formula subtracts the number of X's because a halogen (X) replaces a hydrogen in a compound. For instance, in chloroethane, C2H5Cl, there is one less hydrogen compared to ethane, C2H6. For example, consider compounds having the formula C5H8. The formula of the five-carbon alkane pentane is C5H12 so the difference in hydrogen content is 4. This difference suggests such compounds may have a triple bond, two double bonds, a ring plus a double bond, or two rings. Some examples are shown here, and there are at least fourteen others!
For a compound to be saturated, there is one more hydrogen in a molecule when nitrogen is present. Therefore, we add the number of nitrogens (N). This can be seen with C3H9N compared to C3H8. Oxygen and sulfur are not included in the formula because saturation is unaffected by these elements. As seen in alcohols, the same number of hydrogens in ethanol, C2H5OH, matches the number of hydrogens in ethane, C2H6.
The following chart illustrates the possible combinations of the number of double bond(s), triple bond(s), and/or ring(s) for a given degree of unsaturation. Each row corresponds to a different combination.
• One degree of unsaturation is equivalent to 1 ring or 1 double bond (1 $\pi$ bond).
• Two degrees of unsaturation is equivalent to 2 double bonds, 1 ring and 1 double bond, 2 rings, or 1 triple bond (2 $\pi$ bonds).
When the DU is 4 or greater, the presence of benzene rings is very likely.
DU
Possible combinations of rings/ bonds
# of rings
# of double bonds
# of triple bonds
1
1
0
0
0
1
0
2
2
0
0
0
2
0
0
0
1
1
1
0
Remember, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings+1 double bond, 1 ring+2 double bonds, 1 ring+1 triple bond, 1 double bond+1 triple bond, or 3 double bonds.
Example: Benzene
What is the Degree of Unsaturation for Benzene?
Solution
The molecular formula for benzene is C6H6. Thus,
DU= 4, where C=6, N=0,X=0, and H=6. 1 DoB can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds.
However, when given the molecular formula C6H6, benzene is only one of many possible structures (isomers). The following structures all have DU of 4 and have the same molecular formula as benzene. However, these compounds are very rare, unlike benzene. We will learn more about the reasons for benzen's high stability when we studey aromaticity in later chapters.
Exercises
1. Are the following molecules saturated or unsaturated:
1. (b.) (c.) (d.) C10H6N4
2. Using the molecules from (1) above, give the degrees of unsaturation for each.
3. Calculate the degrees of unsaturation, classify the compound as saturated or unsaturated, and list all the ring/pi bond combination possible for the following molecular formulas: (a.) C9H20 (b.) C7H8 (c.) C5H7Cl (d.) C9H9NO4
4. Calculate degrees of unsaturation (DoU) for the following, and propose a structure for each.
a) C5H8
b) C4H4
5. Calculate the degree of unsaturation (DoU) for the following molecules
a) C5H5N
b) C5H5NO2
c) C5H5Br
6. The following molecule is caffeine (C8H10N4O2), determine the degrees of unsaturation (DoU).
Answer
1.
(a.) unsaturated (Even though rings only contain single bonds, rings are considered unsaturated.)
(b.) unsaturated
(c.) saturated
(d.) unsaturated
2. If the molecular structure is given, the easiest way to solve is to count the number of double bonds, triple bonds and/or rings. However, you can also determine the molecular formula and solve for the degrees of unsaturation by using the formula.
(a.) 2
(b.) 2 (one double bond and the double bond from the carbonyl)
(c.) 0
(d.) 10
3.
(a.) DU =0 ; saturated (Remember-a saturated molecule only contains single bonds)
(b.) DU = 4; unsaturated The molecule can contain any of these combinations of rings and pi bonds that add up to 4, such as (i) 4 double bonds (ii) 4 rings (iii) 2 double bonds+2 rings (iv) 1 double bond+3 rings (v) 3 double bonds+1 ring (vi) 1 triple bond+2 rings (vii) 2 triple bonds (viii) 1 triple bond+1 double bond+1 ring (ix) 1 triple bond+2 double bonds
(c.) DU = 2; unsaturated (i) 1 triple bond (ii) 1 ring+1 double bond (iii) 2 rings (iv) 2 double bonds
(d.) DU = 6; (i) 3 triple bonds (ii) 2 triple bonds+2 double bonds (iii) 2 triple bonds+1 double bond+1 ring (iv)... (As you can see, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or ring. Thus, the formula may give numerous possible structures for a given molecular formula.)
4.
5. a) 4 b) 4 c) 3
6. DU = 6 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.04%3A_Degrees_of_Unsaturation.txt |
Learning Objective
• recognize and classify the stereochemistry of alkenes using the cis/trans and E/Z systems
• give the IUPAC names for alkenes given their structure & vice versa including E/Z isomers
E/Z Nomenclature in Alkenes
The traditional system for naming the geometric isomers of an alkene, in which the same groups are arranged differently, is to name them as cis or trans. However, it is easy to find examples where the cis-trans system is not easily applied. IUPAC has a more complete system for naming alkene isomers. The R-S system for chirality is based on a set of "priority rules", which allow you to rank any groups. The rigorous IUPAC system for naming alkene isomers, called the E-Z system, is based on the same priority rules.
The priority rules are often called the Cahn-Ingold-Prelog (CIP) rules, after the chemists who developed the system.
The general strategy of the E-Z system is to analyze the two groups at each end of the double bond. For each vinyl carbon, rank the two groups using the CIP priority rules. Determine whether the higher priority group at one end of the double bond and the higher priority group at the other end of the double bond are on the same side (Z, from German zusammen = together or "on Zee Zame Zide") or on opposite sides (E, from German entgegen = opposite) of the double bond.
Example \(1\)
The figure below shows the two isomers of 2-butene. You should recognize them as cis and trans. Let's analyze them to see whether they are E or Z. Start with the left hand structure (the cis isomer). On C2 (the left end of the double bond), the two atoms attached to the double bond are C and H. By the CIP priority rules, C is higher priority than H (higher atomic number). Now look at C3 (the right end of the double bond). Similarly, the atoms are C and H, with C being higher priority. We see that the higher priority group is "down" at C2 and "down" at C3. Since the two priority groups are both on the same side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is (Z)-2-butene.
Now look at the right hand structure (the trans isomer). In this case, the priority group is "down" on the left end of the double bond and "up" on the right end of the double bond. Since the two priority groups are on opposite sides of the double bond, they are entgegen = opposite. Therefore, this is (E)-2-butene.
E/Z will work -- even when cis/trans fails
In simple cases, such as 2-butene, Z corresponds to cis and E to trans. However, that is not a rule. This section and the following one illustrate some idiosyncrasies that happen when you try to compare the two systems. The real advantage of the E-Z system is that it will always work. In contrast, the cis-trans system breaks down with many ambiguous cases.
Example \(1\)
The following figure shows two isomers of an alkene with four different groups on the double bond, 1-bromo-2-chloro-2-fluoro-1-iodoethene.
It should be apparent that the two structures shown are distinct chemicals. However, it is impossible to name them as cis or trans. On the other hand, the E-Z system works fine... Consider the left hand structure. On C1 (the left end of the double bond), the two atoms attached to the double bond are Br and I. By the CIP priority rules, I is higher priority than Br (higher atomic number). Now look at C2. The atoms are Cl and F, with Cl being higher priority. We see that the higher priority group is "down" at C1 and "down" at C2. Since the two priority groups are both on the same side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is the (Z) isomer. Similarly, the right hand structure is (E).
E/Z will work, but may not agree with cis/trans
There are also molecules for which the E/Z system will not agree with the cis/trans system. Let's use 2-bromo-2-butene to explore this option. Is this compound cis or trans? This molecule is clearly cis. The two methyl groups are on the same side. More rigorously, the "parent chain" is cis.
Is this compound E or Z? There is a methyl at each end of the double bond. On the left, the methyl is the high priority group because the other group is -H. On the right, the methyl is the low priority group because the other group is -Br. That is, the high priority groups are -CH3 (left) and -Br (right). Thus the two priority groups are on opposite sides = entgegen = E.
This example should convince you that cis and Z are not synonyms. Cis/trans and E,Z are determined by distinct criteria. There may seem to be a simple correspondence, but it is not a rule. Be sure to determine cis,trans or E,Z separately, as needed.
Multiple double bonds
If the compound contains more than one double bond, then each one is analyzed and declared to be E or Z.
Example \(3\)
The configuration at the left hand double bond is E; at the right hand double bond it is Z. Thus this compound is (1E,4Z)-1,5-dichloro-1,4-hexadiene.
The double-bond rule in determining priorities
Example \(4\)
Consider the compound below
This is 1-chloro-2-ethyl-1,3-butadiene -- ignoring, for the moment, the geometric isomerism. There is no geometric isomerism at the second double bond, at 3-4, because it has 2 H at its far end.
What about the first double bond, at 1-2? On the left hand end, there is H and Cl; Cl is higher priority (by atomic number). On the right hand end, there is -CH2-CH3 (an ethyl group) and -CH=CH2 (a vinyl or ethenyl group). Both of these groups have C as the first atom, so we have a tie so far and must look further. What is attached to this first C? For the ethyl group, the first C is attached to C, H, and H. For the ethenyl group, the first C is attached to a C twice, so we count it twice; therefore that C is attached to C, C, H. CCH is higher than CHH; therefore, the ethenyl group is higher priority. Since the priority groups, Cl and ethenyl, are on the same side of the double bond, this is the Z-isomer; the compound is (Z)-1-chloro-2-ethyl-1,3-butadiene.
The "first point of difference" rule
Which is higher priority, by the CIP rules: a C with an O and 2 H attached to it or a C with three C? The first C has one atom of high priority but also two atoms of low priority. How do these "balance out"? Answering this requires a clear understanding of how the ranking is done. The simple answer is that the first point of difference is what matters; the O wins.
To illustrate this, consider the molecule at the left. Is the double bond here E or Z? At the left end of the double bond, Br > H. But the right end of the double bond requires a careful analysis.
At the right hand end, the first atom attached to the double bond is a C at each position. A tie, so we look at what is attached to this first C. For the upper C, it is CCC (since the triple bond counts three times). For the lower C, it is OHH -- listed in order from high priority atom to low. OHH is higher priority than CCC, because of the first atom in the list. That is, the O of the lower group beats the C of the upper group. In other words, the O is the highest priority atom of any in this comparison; thus the O "wins".
Therefore, the high priority groups are "up" on the left end (the -Br) and "down" on the right end (the -CH2-O-CH3). This means that the isomer shown is opposite = entgegen = E. And what is the name? The "name" feature of ChemSketch says it is (2E)-2-(1-bromoethylidene)pent-3-ynyl methyl ether.
Example \(1\)
The configuration about double bonds is undoubtedly best specified by the cis-trans notation when there is no ambiguity involved. Unfortunately, many compounds cannot be described adequately by the cis-trans system. Consider, for example, configurational isomers of 1 -fluoro- 1 -chloro-2- bromo-2-iodo-ethene, 9 and 10. There is no obvious way in which the cis-trans system can
be used:
A system that is easy to use and which is based on the sequence rules already described for the R,S system works as follows:
1. An order of precedence is established for the two atoms or groups attached to each end of the double bond according to the sequence rules of Section 19-6. When these rules are applied to 1-fluoro- 1-chloro-2-bromo-2- iodoethene, the priority sequence is:
• at carbon atom 1, C1 > F
• at carbon atom 2, I > Br
1. Examination of the two configurations shows that the two priority groups- one on each end- are either on the same side of the double bond or on opposite sides:
The Z isomer is designated as the isomer in which the top priority groups are on the same side (Z is taken from the German word zusammen- together). The E isomer has these groups on opposite sides (E, German for entgegen across). Two further examples show how the nomenclature is used:
Exercises
1. Order the following in increasing priority.
A) –H, –Cl, –OH
B) –CH3, CH2OH, CH2CH3
C) –C≡CH, –CH=CH2, –CH=O
2. Label the following as E or Z conformations.
Answer
1. A) –H < OH < Cl (highest priority)
B) –CH3 < CH2CH3 < CH2OH (highest priority)
C) –CH=CH2 < C≡CH < CH=O (highest priority)
2. A) Z B) Z C) E
Contributors and Attributions
>Robert Bruner (http://bbruner.org)
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.05%3A_The_E_Z_System_%28when_cis_trans_does_not_work%29.txt |
Learning Objective
• use heats of hydrogenation to compare the stabilities of alkenes
Heats of Hydrogenation
The stability of an alkene can be determined by measuring the amount of energy associated with the hydrogenation of the molecule. Since the double bond is breaking in this reaction, the energy released in hydrogenation is proportional to the energy in the double bond of the molecule. This is a useful tool because heats of hydrogenation can be measured very accurately. The \(\Delta H^o\) is usually around -30 kcal/mol for alkenes. Stability is simply a measure of energy. Lower energy molecules are more stable than higher energy molecules. More substituted alkenes are more stable than less substituted ones due to hyperconjugation. They have a lower heat of hydrogenation. The following illustrates stability of alkenes with various substituents:
In disubstituted alkenes, trans isomers are more stable than cis isomers due to steric hindrance. Also, internal alkenes are more stable than terminal ones. See the following isomers of butene:
In cycloalkenes smaller than cyclooctene, the cis isomers are more stable than the trans as a result of ring strain.
Exercises
1. When looking at their heats of hydrogenation, is the cis or the trans isomer generally more stable?
2. Arrange the following alkenes in order of increasing stability: 2,3-dimethyl-2-butene; trans-2-hexene; 2-methyl-2-pentene; cis-2-hexene
3. Which is the more stable alkene in each pair?
Answer
1. Trans alkenes are more stable as demonstrated by the lower heats of hydrogenation when compared to their cis-isomers.
2. (least substituted and cis) cis-2-hexene < trans-2-hexene < 2-methyl-2-pentene < 2,3-dimethyl-2-butene (most substituted)
3. A) 2 b/c trans with same substitution at C=C B) 1 b/c the C=C is more substituted
8.07: Alkene Synthesis by Elimination of Alkyl Halides
Learning Objective
• interpret and draw reaction energy diagrams for dehydrohalogenation of R-X’s
• propose mechanisms for a dehydrohalogenation reactions -
• predict the products and specify the reagents for alkene synthesis from dehydrohalogenation of R-X’s
• predict and explain the stereochemistry of E2 eliminations to form alkenes, especially from cyclohexanes
Alkene Synthesis by Elimination of Alkyl Halides is discussed in detail in chapter 7 sections 13 - 18. The major learning objectives are summarized briefly in this section.
Alkene Synthesis by Elimination of Alkyl Halides
When considering whether an elimination reaction is likely to occur via an E1 or E2 mechanism, we really need to consider three factors:
1. 1) The base: strong bases favor the E2 mechanism, whereas, E1 mechanisms only require a weak base.
2. 2) The solvent: good ionizing xolvents (polar protic) favor the E1 mechanism by stabilizing the carbocation intermediate.
3. 3) The alkyl halide: primary alkyl halides have the only structure useful in distinguishing between the E2 and E1 pathways. Since primary carbocations do not form, only the E2 mechanism is possible.
Reaction Parameter E2 E1
alkyl halide structure tertiary > secondary > primary tertiary > secondary >>>> primary
nucleophile high concentration of a strong base weak base
mechanism 1-step 2-step
rate limiting step anti-coplanar bimolecular transition state carbocation formation
rate law rate = k[R-X][Base] rate = k[R-X]
stereochemisty retained configuration mixed configuration
solvent not important polar protic | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.06%3A_Stability_of_Alkenes.txt |
Learning Objective
• interpret and draw reaction energy diagrams for alcohol dehydration reactions
• propose mechanisms for dehydration reactions
• predict the products and specify the reagents for alkene synthesis from alcohol dehydration reactions
• predict and explain the stereochemistry of E2 eliminations to form alkenes, especially from cyclohexanes
Dehydration of Alcohols to Yield Alkenes
One way to synthesize alkenes is by dehydration of alcohols. Alcohols undergo E1 or E2 mechanisms to lose water and form a double bond. This mechanism is analogous to the alkyl halide mechanism. The only difference is that hydroxide is a very poor leaving group so an extra step is required. The hydroxyl group is protonated so that water is now the leaving group. Therefore, the dehydration reaction of alcohols to generate alkene proceeds by heating the alcohols in the presence of a strong acid, such as sulfuric or phosphoric acid, at high temperatures.
The required reaction temperature range decreases with increasing substitution of the hydroxy-containing carbon:
• 1° alcohols: 170° - 180°C
• 2° alcohols: 100°– 140 °C
• 3° alcohols: 25°– 80°C
If the reaction is not sufficiently heated, the alcohols do not dehydrate to form alkenes, but react with one another to form ethers (e.g., the Williamson Ether Synthesis).
Mechanism for the Dehydration of Alcohols
Different types of alcohols may dehydrate through a slightly different mechanism pathway. However, the general idea behind each dehydration reaction is that the –OH group in the alcohol donates two electrons to H+ from the acid reagent to form an alkyloxonium ion. This ion acts as a very good leaving group for either the E1 or E2 mechanism. The deprotonated acid (the conjugate base) then reacts with one of the beta-hydrogens to form a double bond.
Primary Alcohols and the E2 Mechanism
Primary alcohols undergo bimolecular elimination (E2 mechanism) while secondary and tertiary alcohols undergo unimolecular elimination (E1 mechanism). The relative reactivity of alcohols in dehydration reaction is ranked as the following
Methanol < primary < secondary < tertiary
Primary alcohol dehydrates through the E2 mechanism. Oxygen donates two electrons to a proton from sulfuric acid H2SO4, forming an alkyloxonium ion. The resulting conjugate base (HSO4) approaches in an anti-coplanar orientation relative to the leaving group and reacts with one adjacent hydrogen while the alkyloxonium ion simultaneously leaves in a concerted process, making a double bond.
Secondary and Tertiary Alcohols and the E1 Mechanism
Secondary and tertiary alcohols dehydrate through the E1 mechanism. Similarly to the reaction above, secondary and tertiary –OH protonate to form alkyloxonium ions. However, the ion leaves first and forms a carbocation as the reaction intermediate. The water molecule (which is a stronger base than the HSO4- ion) then abstracts a proton from an adjacent carbon, forming a double bond.
For example, in the mechanism below that the alkene formed depends on which proton is abstracted: the red arrows show formation of the more substituted 2-butene, while the blue arrows show formation of the less substituted 1-butene. Recall the general rule that more substituted alkenes are more stable than less substituted alkenes, and trans alkenes are more stable than cis alkenes. Thereore, the trans diastereomer of the 2-butene product is most abundant.
The dehydration mechanism for a tertiary alcohol is analogous to that shown above for a secondary alcohol. The E2 elimination of 3º-alcohols under relatively non-acidic conditions may be accomplished by treatment with phosphorous oxychloride (POCl3) in pyridine. This procedure is also effective with hindered 2º-alcohols, but for unhindered and 1º-alcohols an SN2 chloride ion substitution of the chlorophosphate intermediate competes with elimination. Examples of these and related reactions are given in the following figure. The first equation shows the dehydration of a 3º-alcohol. The predominance of the non-Zaitsev product (less substituted double bond) is presumed due to steric hindrance of the methylene group hydrogen atoms, which interferes with the approach of base at that site. The second example shows two elimination procedures applied to the same 2º-alcohol. The first uses the single step POCl3 method, which works well in this case because SN2 substitution is retarded by steric hindrance. The second method is another example in which an intermediate sulfonate ester confers halogen-like reactivity on an alcohol. In every case the anionic leaving group is the conjugate base of a strong acid.
Carbocation Rearrangements and the E1 Mechanism
Carbocation stability is always a driving force in E1 mechanisms. It is important to evaluate the structure of all carbocation intermediates to look for the possibility of 1,2-hydride or 1,2-methyl shifts to form more stable carbocation intermediates. Carbocation rearrangements are discussed more completely in chapter 7.
Exercise
1. Draw the bond-line structure(s) for the product(s) formed and specify the mechanism (E1 or E2) for each reaction below.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.08%3A_Alkene_Synthesis_by_Dehydration_of_Alcohols.txt |
Learning Objectives
• discuss the uses and sources of alkenes including catalytic cracking
Uses of ethene and propene
Produced from ethylene (ethene)
Chemical Uses
ethanol solvent; constituent of cleaning preparations; in synthesis of esters
acetaldehyde slug killer, in the form of methaldehyde (CH3CHO)4
acetic acid manufacture of vinyl acetate polymers, ethyl acetate solvent and cellulose acetate polymers
ethylene oxide “cellosolves” (industrial solvents)
ethylene glycol anti-freeze; production of DacronOR
ethylene dichloride solvent; production of vinyl chloride
vinyl chloride manufacture of poly (vinyl chloride)—PVC
vinyl acetate manufacture of poly (vinyl acetate) used in paint emulsions, plywood adhesives and textiles
polyethylene “plastic” bags; toys; packaging
Produced from propylene (propene)
Chemical Uses
isopropyl alcohol rubbing alcohol; cosmetics; synthesis of acetone
propylene oxide manufacture of polyurethanes; polyesters
cumene industrial preparation of phenol and acetone
polypropylene molded articles (e.g., kitchenware); fibres for indoor-outdoor carpeting
Catalytic Cracking to Form Ethylene
Cracking is the name given to breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high pressures and temperatures without a catalyst, or lower temperatures and pressures in the presence of a catalyst. The source of the large hydrocarbon molecules is often the naphtha fraction or the gas oil fraction from the fractional distillation of crude oil (petroleum). These fractions are obtained from the distillation process as liquids, but are re-vaporized before cracking.
The hydrocarbons are mixed with a very fine catalyst powder. These days the catalysts are zeolites (complex aluminosilicates) - these are more efficient than the older mixtures of aluminium oxide and silicon dioxide. The whole mixture is then blown rather like a liquid through a reaction chamber at a temperature of about 500°C. Because the mixture behaves like a liquid, this is known as fluid catalytic cracking (or fluidized catalytic cracking). Although the mixture of gas and fine solid behaves as a liquid, this is nevertheless an example of heterogeneous catalysis - the catalyst is in a different phase from the reactants. The catalyst is recovered afterwards, and the cracked mixture is separated by cooling and further fractional distillation.
There isnot any single unique reaction happening in the cracker. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller hydrocarbons, some of which have carbon-carbon double bonds. One possible reaction involving the hydrocarbon C15H32 might be:
Or, showing more clearly what happens to the various atoms and bonds:
This is only one way in which this particular molecule might break up. The ethene and propene are important materials for making plastics or producing other organic chemicals. The octane is one of the molecules found in petrol (gasoline).
Ethene
Cracking is the name given to breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high pressures and temperatures without a catalyst, or lower temperatures and pressures in the presence of a catalyst. The source of the large hydrocarbon molecules is often the naphtha fraction or the gas oil fraction from the fractional distillation of crude oil (petroleum). These fractions are obtained from the distillation process as liquids, but are re-vaporized before cracking.
There is not any single unique reaction happening in the cracker. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller hydrocarbons, some of which have carbon-carbon double bonds. One possible reaction involving the hydrocarbon C15H32 might be:
Or, showing more clearly what happens to the various atoms and bonds:
This is only one way in which this particular molecule might break up. The ethene and propene are important materials for making plastics or producing other organic chemicals. 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.
8.10: Additional Exercises
Physical Properties of Alkenes
8-1 Explain why cis-2-butene is less stable than trans-2-butene.
8-2 Order the following alkenes in increasing order of stability.
8-3 Why do more substituted alkenes experience more stability compared to less substituted alkenes?
8-4 Identify whether the following alkenes are mono-, di-, tri-, or tetra-substituted.
8-5 Place the following alkenes in order of increasing boiling points.
Elements of Unsaturation and the Orbital Description of Alkenes
8-6 Using the following equation, calculate the degrees of unsaturation for the following compounds.
8-7 How many of each type of bonds (sigma/pi) make up a double bond?
Alkene Synthesis by Elimination of Alkyl Halides
8-8 Identify the major product(s) of the following reactions. Include stereochemistry.
8-9 Identify the major products of the following reactions.
8-10 Give the product of the following reaction.
8-11 What is the IUPAC name of the product formed by the reaction in the previous problem (8-10)?
Alkene Synthesis by Dehydration of Alcohols
8-12 Identify the product of the following reaction.
8-13 Draw the mechanism for the reaction in previous problem (8-12).
8-14 Draw the intermediate compounds for the following reaction.
8-15 Identify the product of the following reaction.
8.11: Solutions to Additional Exercises
Physical Properties of Alkenes
8-1 When in the cis configuration, the methyl groups experience steric strain as they are in close proximity to each other. They avoid steric interactions when in the trans configuration as they are able to stay as far apart as possible.
8-2
8-3 Alkyl groups are able to stabilize their neighboring carbon atoms by donating electron density, which allows for the delocalization of electron density and an increase in stability.
8-4
a) Disubstituted
b) Trisubstituted
c) Monosubstituted
d) Tetrasubstituted
8-5
Elements of Unsaturation and the Orbital Description of Alkenes
8-6
a) 4
b) 1
c) 2
d) 2
e) 5
f) 6
8-7 One sigma and one pi bond together make a double bond.
Alkene Synthesis by Elimination of Alkyl Halides
8-8
8-9
8-10
8-11 (1E)-1-chlorobut-1-ene
8-12
8-13
8-14
8-15 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.09%3A_Uses_and_Sources_of_Alkenes.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• draw the general Electrophilic Addition Reaction (EAR) mechanism for an alkene - refer to section 9.1
• predict the products/specify the reagents for EAR of hydrohalic acids (HX) with symmetrical alkenes - refer to section 9.2
• predict the products/specify the reagents for EAR of hydrohalic acids (HX) with asymmetrical alkenes using Markovnikov's Rule for Regioselectivity - refer to section 9.3
• apply the principles of regioselectivity and stereoselectivity to the addition reactions of alkenes - refer to sections 9.3 - 9.14
• predict the products, specify the reagents, and discern most efficient reaction for hydration of alkenes (acid catalyzed hydration; or oxymercuration/demercuration; or hydroboration/oxidation) - refer to sections 9.4, 9.5, and 9.6 respectively
• discern the stereochemical differences between the EAR of chiral and achiral alkenes - refer to sections 9.7 and 9.8
• predict the products/specify the reagents for halogenation and hydrohalogenation of alkenes - refer to sections 9.9 and 9.10 respectively
• recognize organic oxidation and reduction reactions - refer to sections 9.11 and 9.12
• predict the products/specify the reagents for hydrogenation (reduction) of alkenes - refer to section 9.11
• predict the products/specify the reagents for epoxidation of alkenes - refer to section 9.12
• predict the products/specify the reagents for dihydroxylation of alkenes - refer to sections 9.13 and 9.14
• predict the products/specify the reagents for oxidative cleavage of alkenes - refer to section 9.15
• predict the products of carbene additions to alkenes - refer to section 9.16
• predict the polymer/specify the monomer for radical, chain -growth polymers of alkenes - refer to section 9.17
• discuss an example biological addition reactions - refer to section 9.18
• 9.1: Electrophilic Addition Reactions (EARs)
Electrophilic addition reactions can occur in compounds containing pi bonds like the alkenes. Depending on the structure of the alkene and the specific reagents, the reactions can be regioselective and/or stereoselective.
• 9.2: Addition of Hydrogen Halides to Symmetrical Alkenes
The regioselective reaction of the carbon-carbon double bond in alkenes with hydrohalogens (HX) is a controlled by carbocation stability. Consequently, the symmetry of the alkene must be considered for this mechanistic pathway.
• 9.3: Alkene Asymmetry and Markovnikov's Rule
The regioselectivity of electrophilic addition reactions is determined by carbocation stability and is summarized by Markovnikov's Rule.
• 9.4: Hydration- Acid Catalyzed Addition of Water
Electrophilic hydration is the act of adding electrophilic hydrogen from a non-nucleophilic strong acid (a reusable catalyst, examples of which include sulfuric and phosphoric acid) and applying appropriate temperatures to break the alkene's double bond. After a carbocation is formed, water bonds with the carbocation to form a 1º, 2º, or 3º alcohol on the alkane.
• 9.5: Hydration- Oxymercuration-Demercuration
Oxymercuration is a stereospecific, regioselective electrophilic addition reaction because there are no carbocation rearrangements due to stabilization of the reactive intermediate. The Markovnikov products are reliably synthesized by this pathway.
• 9.6: Hydration - Hydroboration-Oxidation
Hydroboration-oxidation of alkenes has been a very valuable laboratory method for the stereoselectivity and regioselectivity of alkenes that are the non-Markovnikov products for alkene hydration.
• 9.7: Stereochemistry of Reactions - Hydration of Achiral Alkenes
For achiral alkenes, the symmetrical trigonal planar geometry of the carbocation leads to equivalent synthesis of both R and S products giving a racemic (50/50).
• 9.8: Stereochemistry of Reactions - Hydration of Chiral Alkenes
Chiral alkenes form electrophilic addition products in non 50:50 ratios due to the differences in steric effects between the enantiomeric starting materials.
• 9.9: Addition of Halogens
Halogens can act as electrophiles due to polarizability of their covalent bond and react with the pi bond of alkenes. This electrophilic addition mechanism is stereospecific. The orientation of the electrophile during a stereospecific electrophilic addition reaction will determine the stereochemistry of the product(s).
• 9.10: Formation of Halohydrins
When the halogenation reaction of alkenes is performed in a nucleophilic solvent like water or alcohol, then the solvent becomes the nucleophile to give halohydrin or haloalkoxy products.
• 9.11: Reduction of Alkenes - Catalytic Hydrogenation
Catalytic hydrogenation of a carbon-carbon double bond is a reduction reaction. The alkene orientation required for interaction with the surface of the catalyst means that t his reaction is stereospecific.
• 9.12: Oxidation of Alkenes - Epoxidation
Oxidation of alkenes is introduced and the epoxidation of alkenes is discussed.
• 9.13: Dihydroxylation of Alkenes
Alkenes can react to produce glycols (two adjacent hydroxyl groups) through either an anti- or syn- addition mechanism that is stereospecific.
• 9.14: Opening of Epoxides - Acidic versus Basic Conditions
Ring-opening reactions can proceed by either SN2 or SN1 mechanisms, depending on the nature of the epoxide and on the reaction conditions.
• 9.15: Oxidative Cleavage of Alkenes
Oxidative cleavage of alkenes can occur by several different pathways. The most common reagents and pathways are discussed in this section.
• 9.16: Addition of Carbenes to Alkenes - Cyclopropane Synthesis
A carbene such as methlyene will react with an alkene breaking the pi bond to form a cyclopropane. Since methylene is highly reactive, it is prepared in situ immediately preceding the addition of the alkene.
• 9.17: Radical Chain-Growth Polymerization
All the monomers from which addition polymers are made are alkenes or functionally substituted alkenes. The free radical mechanism for chain growth polymers is explained.
• 9.18: Biological Additions of Radicals to Alkenes
Some electrophilic addition reactions that take place in nature and the role of enzymes in such processes are introduced.
• 9.19: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
• 9.20: Solutions to Additional Exercises
This section has the solutions to the additional exercises from the previous section.
09: Reactions of Alkenes
Learning Objective
• draw the general Electrophilic Addition Reaction (EAR) mechanism for an alkene
Introduction
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 above. 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 sigma 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. 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. Because the pi bond electrons lie exposed above and below the rest of the molecule, they are relatively open to reaction with other compounds.
Electrophilic Addition Reactions (EARs)
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 alkene molecule. To continue with our example, ethene undergoes 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. The pi bond is electron rich and takes the role of the nucleophile seeking out an electrophile with its full or partial positive charge.
Using a general molecule X-Y as the electrophile reacting with ethene, the atoms 'X' and 'Y' are added to the carbon chain across the vinylic carbons as shown below.
Understanding the Electrophilic Addition Mechanism
The mechanism for the reaction between ethene and a molecule X-Y begins by recognizing the electrophilic nature of X-Y. Trends in relative electronegativity help identify bonds to create partial positives within polar compounds. As shown below, 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. This polarity means that the X atom carries a slight positive charge (partial positive).
The slightly positive X atom is an electrophile and attracts the exposed pi electrons in the ethene. Now imagine what happens as they approach each other. The electrons in the half of the pi bond nearest the XY are attracted to the partial positive charge as shown below.
The two electrons in the pi bond move closer towards the X until a covalent bond is made. Simultaneously, the electrons in the X-Y bond are pushed entirely onto the Y to form the anion Y- ion and a carbocation as shown below.
EAR Mechanism
The movements of the electrons for the EAR mechanism are shown with curved arrows.
Regiochemistry and Regioselective Reactions
Regiochemistry is the orientation of the electrophile relative to the pi bond of the alkene. Electrophilic addition reactions of alkenes can be regioselective depending on the symmetry and structure of the alkene. The details will be discussed in a later sections of this chapter.
Stereoselective Reactions
Different stereoisomeric reactants produce different stereoisomeric products. Electrophilic addition reactions of alkenes can be stereoselective depending on the symmetry and structure of the alkene. The details will be discussed in a later sections of this chapter.
Exercise \(1\)
1. In the next section, we will apply the Electrophilic Addition mechanism to actual compounds. Draw the complete mechanism when hydrogen bromide gas is bubbled through a solution of cyclopentene.
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.01%3A_Electrophilic_Addition_Reactions_%28EARs%29.txt |
Learning Objective
• predict the products/specify the reagents for EAR of hydrohalic acids (HX) with symmetrical alkenes
This section looks at the reaction of symmetrical alkenes (like ethene, but-2-ene or cyclohexene) with hydrogen halides such as hydrogen chloride and hydrogen bromide. Since these alkenes have identical groups attached to each end of the carbon-carbon double bond, regioselectivity does not apply.
Addition to symmetrical alkenes
All alkenes undergo addition reactions with the hydrogen halides. A hydrogen atom joins to one of the carbon atoms originally in the double bond, and a halogen atom to the other.
For example, with ethene and hydrogen chloride, you get chloroethane:
With but-2-ene you get 2-chlorobutane:
What happens if you add the hydrogen to the carbon atom at the right-hand end of the double bond, and the chlorine to the left-hand end? You would still have the same product. The chlorine would be on a carbon atom next to the end of the chain - you would simply have drawn the molecule flipped over in space. That would be different of the alkene was unsymmetrical - that's why we have to look at them separately.
Mechanism
The addition of hydrogen halides is one of the easiest electrophilic addition reactions because it uses the simplest electrophile: the proton. Hydrogen halides provide both a electrophile (proton) and a nucleophile (halide). First, the electrophile will attack the double bond and take up a set of π electrons, attaching it to the molecule (1). This is basically the reverse of the last step in the E1 reaction (deprotonation step). The resulting molecule will have a single carbon- carbon bond with a positive charge on one of them (carbocation). The next step is when the nucleophile (halide) bonds to the carbocation, producing a new molecule with both the original hydrogen and halide attached to the organic reactant (2). The second step will only occur if a good nucleophile is used.
Mechanism of Electrophilic Addition of Hydrogen Chloride to But-2-ene
All of the halides (HBr, HCl, HI, HF) can participate in this reaction and add on in the same manner. Although different halides do have different rates of reaction, due to the H-X bond getting weaker as X gets larger (poor overlap of orbitals)s.
Reaction rates
Variation of rates when you change the halogen
Reaction rates increase in the order HF < HCl < HBr < HI. Hydrogen fluoride reacts much more slowly than the other three, and is normally ignored in talking about these reactions.
When the hydrogen halides react with alkenes, the hydrogen-halogen bond has to be broken. The bond strength falls as you go from HF to HI, and the hydrogen-fluorine bond is particularly strong. Because it is difficult to break the bond between the hydrogen and the fluorine, the addition of HF is bound to be slow.
Variation of rates when you change the alkene
This applies to unsymmetrical alkenes as well as to symmetrical ones. For simplicity the examples given below are all symmetrical ones- but they don't have to be.
Reaction rates increase as the alkene gets more complicated - in the sense of the number of alkyl groups (such as methyl groups) attached to the carbon atoms at either end of the double bond. For example:
There are two ways of looking at the reasons for this - both of which need you to know about the mechanism for the reactions.
Alkenes react because the electrons in the pi bond attract things with any degree of positive charge. Anything which increases the electron density around the double bond will help this.
Alkyl groups have a tendency to "push" electrons away from themselves towards the double bond. The more alkyl groups you have, the more negative the area around the double bonds becomes.
The more negatively charged that region becomes, the more it will attract molecules like hydrogen chloride.
The more important reason, though, lies in the stability of the intermediate ion formed during the reaction. The three examples given above produce these carbocations (carbonium ions) at the half-way stage of the reaction:
The stability of the intermediate ions governs the activation energy for the reaction. As you go towards the more complicated alkenes, the activation energy for the reaction falls. That means that the reactions become faster.
Exercise
1. Draw the bond-line structures for the products of the following reactions.
Answer
1.
9.03: Alkene Asymmetry and Markovnikov's Rule
Learning Objective
• predict the products/specify the reagents for EAR of hydrohalic acids (HX) with asymmetrical alkenes using Markovnikov's Rule for Regioselectivity
• apply the principles of regioselectivity and stereoselectivity to the addition reactions of alkenes
Addition to unsymmetrical alkenes
In terms of reaction conditions and the factors affecting the rates of the reaction, there is no difference whatsoever between these alkenes and the symmetrical ones described above. The problem comes with the orientation of the addition of the hydrogen and the halogen across the double bond.
Markovnikov's Rule
If HCl adds to an unsymmetrical alkene like propene, there are two possible ways it could add. However, in practice, there is only one major product according to Markovnikov's Rule.
Markovnikov's Rule: When HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already.
Applying Markovnikov's Rule to the reaction above, the hydrogen bonds with the CH2 group, because the CH2 group has more hydrogens than the CH group. Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count.
Regioselectivity - a closer look
If more than one reaction could occur between a set of reactants under the same conditions giving products that are constitutional isomers and if one product forms in greater amounts than the others, the overall reaction is said to be regioselective.
Say three reactions could occur between the hypothetical reactants A and B under the same conditions giving the constitutionally isomeric products C, D, and E.
There are two possibilities:
1. The three products form in equal amounts, i.e., of the total product 33% is C, another 33% D, the remaining 33% E. (These percentages are called relative yields of the products.)
If this is what is observed, the overall reaction between A and B is not regioselective.
2. One product forms in greater amounts than the others. Say, for example, the relative yields of C, D, and E are 25%, 50%, and 25%, respectively.
If this is what is observed, the overall reaction between A and B is regioselective.
eg:
Experimentally, 2 is the major product; 3 is the minor product. Thus, the overall reaction between 1 and HBr is regioselective toward 2.
If more than one reaction could occur between a set of reactants under the same conditions giving products that are constitutional isomers and if only one product is observed, the overall reaction is said to be 100% regioselective or regiospecific.
eg:
The only observed product is 5. (Relative yields of 5 and 6 are 100% and 0%, respectively.) Thus the overall reaction between 4 and HBr is regiospecific toward 5.
Regiospecificity is merely the limiting case of regioselectivity. All regiospecific reactions are regioselective, but not all regioselective reactions are regiospecific.
Exercises
1. Predict the product(s) for the following reactions:
2. In each case, suggest an alkene that would give the product shown.
Answers
1.
2. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.02%3A_Addition_of_Hydrogen_Halides_to_Symmetrical_Alkenes.txt |
Learning Objective
• apply the principles of regioselectivity and stereoselectivity to the addition reactions of alkenes
• predict the products, specify the reagents, and discern most efficient reaction for hydration of alkenes (acid catalyzed hydration; or oxymercuration/demercuration; or hydroboration/oxidation)
What Is Electrophilic Hydration?
Electrophilic hydration is the addition of hydrogen and a hydroxyl group across the two carbons of a double bond. Electrophilic hydration is the reverse of dehydration of alcohols and so begins the circular nature of organic chemistry. Alcohols can be dehydrated to form alkenes and alkenes can undergo electrophilic addition reactions to form alcohols. Electrophilic hydrogen is essentially a proton: a hydrogen atom stripped of its electrons. Electrophilic hydrogen is commonly used to help break double bonds or restore catalysts.
Electrophilic hydration of alkenes has practical applications in making alcohols for fuels and reagents for other reactions. The basic reaction under certain temperatures (given below) is the following:
In later sections, we will learn that mercury (II) sulfate and borane are also electrophiles that can react with alkenes to form hydration products. Each reaction pathway has its own regio- and stereochemical considerations. In the example below, we see that the same alkene produces different hydration products depending on the hydration pathway.
Mechanism for Acid-catalyzed Hydration of Alkenes
Temperatures for Types of Alcohol Synthesis
Heat is used to catalyze electrophilic hydration; because the reaction is in equilibrium with the dehydration of an alcohol, which requires higher temperatures to form an alkene, lower temperatures are required to form an alcohol. The exact temperatures used are highly variable and depend on the product being formed.
• Primary Alcohol: Less than 170ºC
• Secondary Alcohol: Less than 100ºC
• Tertiary Alcohol: Less than 25ºC
But...Why Does Electrophilic Hydration Work?
• An alkene placed in an aqueous non-nucleophilic strong acid immediately "reaches out" with its double bond and attacks one of the acid's hydrogen atoms (meanwhile, the bond between oxygen and hydrogen performs heterolytic cleavage toward the oxygen—in other words, both electrons from the oxygen/hydrogen single bond move onto the oxygen atom).
• A carbocation is formed on the original alkene (now alkane) in the more-substituted position, where the oxygen end of water attacks with its 4 non-bonded valence electrons (oxygen has 6 total valence electrons because it is found in Group 6 on the periodic table and the second row down: two electrons in a 2s-orbital and four in 2p-orbitals. Oxygen donates one valence electron to each bond it forms, leaving four 4 non-bonded valence electrons).
• After the blue oxygen atom forms its third bond with the more-substituted carbon, it develops a positive charge (3 bonds and 2 valence electrons give the blue oxygen atom a formal charge of +1).
• The bond between the green hydrogen and the blue oxygen undergoes heterolytic cleavage, and both the electrons from the bond move onto the blue oxygen. The now negatively-charged strong acid picks up the green electrophilic hydrogen.
• Now that the reaction is complete, the non-nucleophilic strong acid is regenerated as a catalyst and an alcohol forms on the most substituted carbon of the current alkane. At lower temperatures, more alcohol product can be formed.
What is Regiochemistry and How Does It Apply?
Regiochemistry deals with where the substituent bonds on the product. Zaitsev's and Markovnikov's rules address regiochemistry, but Zaitsev's rule applies when synthesizing an alkene while Markovnikov's rule describes where the substituent bonds onto the product. In the case of electrophilic hydration, Markovnikov's rule is the only rule that directly applies. See the following for an in-depth explanation of regiochemistry Markovnikov explanation: Radical Additions--Anti-Markovnikov Product Formation
In the mechanism for a 3º alcohol shown above, the red H is added to the least-substituted carbon connected to the nucleophilic double bonds (it has less carbons attached to it). This means that the carbocation forms on the 3º carbon, causing it to be highly stabilized by hyperconjugationelectrons in nearby sigma (single) bonds help fill the empty p-orbital of the carbocation, which lessens the positive charge. More substitution on a carbon means more sigma bonds are available to "help out" (by using overlap) with the positive charge, which creates greater carbocation stability. In other words, carbocations form on the most substituted carbon connected to the double bond. Carbocations are also stabilized by resonance, but resonance is not a large factor in this case because any carbon-carbon double bonds are used to initiate the reaction, and other double bonded molecules can cause a completely different reaction. If the carbocation does originally form on the less substituted part of the alkene, carbocation rearrangements occur to form more substituted products.
Carbocation Rearrangements - a review
• Hydride shifts: a hydrogen atom bonded to a carbon atom next to the carbocation leaves that carbon to bond with the carbocation (after the hydrogen has taken both electrons from the single bond, it is known as a hydride). This changes the once neighboring carbon to a carbocation, and the former carbocation becomes a neighboring carbon atom.
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 shifts: if no hydrogen atoms are available for a hydride shift, an entire methyl group performs the same shift.
The nucleophile attacks the positive charge formed on the most substituted carbon connected to the double bond, because the nucleophile is seeking that positive charge. In the mechanism for a 3º alcohol shown above, water is the nucleophile. When the green H is removed from the water molecule, the alcohol attached to the most substituted carbon. Hence, electrophilic hydration follows Markovnikov's rule.
Stereochemistry of Acid-catalyzed Hydration
Stereochemistry deals with how the substituent bonds on the product directionally. Dashes and wedges denote stereochemistry by showing whether the molecule or atom is going into or out of the plane of the board. Whenever the bond is a simple single straight line, the molecule that is bonded is equally likely to be found going into the plane of the board as it is out of the plane of the board. This indicates that the product is a racemic mixture.
There is no stereochemical control in acid-catalyzed hydration reactions. The carbocation intermediate has the trigonal planar geometry of sp2 hybridization which allows the subsequent reaction with water from either orientation.
Electrophilic hydration adopts a stereochemistry wherein the substituent is equally likely to bond pointing into the plane of the board as it is pointing out of the plane of the board. The 3º alcohol product could look like either of the following products:
Note: Whenever a straight line is used along with dashes and wedges on the same molecule, it could be denoting that the straight line bond is in the same plane as the board. Practice with a molecular model kit and attempting the practice problems at the end can help eliminate any ambiguity.
Is this a Reversible Synthesis?
Electrophilic hydration is reversible because an alkene in water is in equilibrium with the alcohol product. To sway the equilibrium one way or another, the temperature or the concentration of the non-nucleophilic strong acid can be changed. For example:
• Less sulfuric or phosphoric acid and an excess of water help synthesize more alcohol product.
• Lower temperatures help synthesize more alcohol product.
Is There a Better Way to Add Water to Synthesize an Alcohol From an Alkene?
A more efficient pathway does exist: see Oxymercuration - Demercuration: A Special Electrophilic Addition. Oxymercuration does not allow for rearrangements, but it does require the use of mercury, which is highly toxic. Detractions for using electrophilic hydration to make alcohols include:
• Allowing for carbocation rearrangements
• Poor yields due to the reactants and products being in equilibrium
• Allowing for product mixtures (such as an (R)-enantiomer and an (S)-enantiomer)
• Using sulfuric or phosphoric acid
Exercise
1. Draw the bond-line structure for the major product of each reaction.
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.04%3A_Hydration-_Acid_Catalyzed_Addition_of_Water.txt |
Learning Objective
• apply the principles of regioselectivity and stereoselectivity to the addition reactions of alkenes
• predict the products, specify the reagents, and discern most efficient reaction for hydration of alkenes (acid catalyzed hydration; or oxymercuration/demercuration; or hydroboration/oxidation)
Introduction
Acid-catalyzed hydration of alkenes is limited by carbocation stability. Carbocation rearrangement can occur to form a more stable ion as shown in the example below.
Alkene hydration using the oxymercuration-demercuration reaction pathway reliably produces the Markovnikov product without carbocation rearrangment as shown in the example below.
Oxymercuration-Demercuration is a two step pathway used to produce alcohols.
Oxymercuration-Demercuration Mechanism
This mechanism is similar to the previous electrophilic addition reactions. The major difference is that a mercurium ion bridge stabilizes the carbocation intermediate so that it cannot rearrange. Metals are electropositive. Mercury carries a partial positive charge in the acetate complex and is the electrophile. During the first step of this mechanism, the pi electrons form a bond to mercury while the lone pair on the mercury simultaneously bonds to the other vinyl carbon creating a mercurium ion bridge. The mercurium ion forms in conjunction with the loss of an acetate ion. The mercurium ion stabilizes the carbocation so that it does not rearrange. In the second step of this mechanism, a water molecule reacts with the most substituted carbon to open the mercurium ion bridge. The third step of this mechanism is a proton transfer to a solvent water molecule to neutralize the addition product. The fourth step of the reaction pathway is the reduction of the organomercury intermediate with sodium borohydride under basic conditions. The mechanism of the fourth step is beyond the scope of first year organic chemistry.
Notice that overall, the oxymercuration - demercuration mechanism follows Markovnikov's Regioselectivity with the OH group attached to the most substituted carbon and the H attached to the least substituted carbon. The reaction is useful, because strong acids are not required and carbocation rearrangements are avoided because no discreet carbocation intermediate forms.
Exercise
1. Show how to prepare 3-methyl2-pentanol from 3-methyl-1-pentene.
Note: Questions 2-5 have not shown the water present in the sulfuric acid solution and have indicated a second neutralization step. Some authors simply write H+/H2O as a single step.
2. Draw the bond-line structure for the product.
3. Draw the bond-line structure for the product. How does the cyclopropane group affect the reaction?
4. Draw the bond-line structure for the product. (Hint: What is different about this problem?)
5. Draw the bond-line structure for the product(s). Indicate any shifts as well as the major product:
6. In each case, predict the product(s) of these reactants of oxymercuration.
7. Propose the alkene that was the reactant for each of these products of oxymercuration.
Answer
1.
2. This reaction is electrophilic hydration.
3. The answer is additional side products, but the major product formed is still the same (the product shown). Depending on the temperatures used, the cyclopropane may open up into a straight chain, which makes it unlikely that the major product will form (after the reaction, it is unlikely that the 3º carbon will remain as such).
4. A hydride shift actually occurs from the top of the 1-methylcyclopentane to where the carbocation had formed.
5. In the first picture shown below, an alkyl shift occurs but a hydride shift (which occurs faster) is possible. Why doesn't a hydride shift occur? The answer is because the alkyl shift leads to a more stable product. There is a noticeable amount of side product that forms where the two methyl groups are, but the major product shown below is still the most significant due to the hyperconjugation that occurs by being in between the two cyclohexanes.
6.
7. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.05%3A_Hydration-_Oxymercuration-Demercuration.txt |
Learning Objectives
• apply the principles of regioselectivity and stereoselectivity to the addition reactions of alkenes
• predict the products, specify the reagents, and discern most efficient reaction for hydration of alkenes (acid catalyzed hydration; or oxymercuration/demercuration; or hydroboration/oxidation)
Hydroboration-Oxidation is a two step pathway used to produce alcohols. The reaction proceeds in an anti-Markovnikov manner, where the hydrogen (from $\ce{BH3}$ or $\ce{BHR2}$) attaches to the more substituted carbon and the boron attaches to the least substituted carbon in the alkene double bond. Furthermore, the borane acts as the electrophile by accepting two electrons in its empty p orbital from an alkene that is electron rich. This process allows boron to have an electron octet. A very interesting characteristic of this process is that it does not require any activation by a catalyst. The hydroboration mechanism has the elements of both hydrogenation and electrophilic addition and it is a stereospecific (syn addition), meaning that the hydroboration takes place on the same face of the double bond, this leads cis stereochemistry.
The Borane Complex
It is very important to understand the structure and properties of the borane molecule. Borane exists naturally as a very toxic gas and it exists as dimer of the general formula $\ce{B2H6}$ (diborane).
$\ce{BH3→B2H6} \nonumber$
Since diborane dimer ignites spontaneously in air, it commercially distributed in ether or tetrahydrofuran (THF) solutions. In these solutions, the borane can exist as a Lewis acid-base complex which allows boron to have an octet of electrons.
The Mechanism
Step #1: Hydroboration of the alkene
The addition of the borane to the alkene is initiated and proceeds as a concerted reaction because bond breaking and bond formation occur at the same time. The vacant 2p orbital of the boron takes the role of electrophile and accepts the pi electrons from the nucleophilic alkene. The boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon. Both, the boron and the hydrogen add simultaneously on the same face of the double bond (syn addition). With a concerted mechanism, there is no carbocation formation.
Transition state
* Note that a carbocation is not formed. Therefore, no rearrangement takes place.
It is important to note that reaction continues two more times until all three hydrogens on the borane have reacted with alkenes to create the trialkylborane intermediate R3B.
Step #2: Oxidation of the Trialkylborane by Hydrogen Peroxide
The hydrogen peroxide ($\ce{HOOH}$) is the nucleophile in this reaction because it is the electron donor to the newly formed trialkylborane that resulted from the previous hydroboration.
$\ce{HOOH + OH^{-} -> HOO^{-} + HOH}$
In this second part of the mechanism, a rearrangement of an R group with its pair of bonding electrons to an adjacent oxygen results in the loss of a hydroxide ion.
Two more of these reactions with hydroperoxide will occur in order give a trialkylborate
In the final step of the oxidation process, the trialkylborate reacts with aqueous $\ce{NaOH}$ to give the alcohol and sodium borate (\ce{Na3BO3}\).
$\ce{(RO3)B + 3NaOH -> 3OH + Na3BO3}$
If you need additional visuals to aid you in understanding the mechanism, click on the outside links provided at the end of this section.
Stereochemistry of the Hydroboration Step
The hydroboration reaction is among the few simple addition reactions that proceed cleanly in a syn fashion. As noted above, this is a single-step reaction. Since the bonding of the double bond carbons to boron and hydrogen is concerted, it follows that the geometry of this addition must be syn. Furthermore, rearrangements are unlikely inasmuch as a discrete carbocation intermediate is never formed. These features are illustrated for the hydroboration of α-pinene.
Since the hydroboration procedure is most commonly used to hydrate alkenes in an anti-Markovnikov fashion, we also need to know the stereoselectivity of the second oxidation reaction, which substitutes a hydroxyl group for the boron atom. Independent study has shown this reaction takes place with retention of configuration so the overall addition of water is also syn.
The hydroboration of α-pinene also provides a nice example of steric hindrance control in a chemical reaction. In the less complex alkenes used in earlier examples the plane of the double bond was often a plane of symmetry, and addition reagents could approach with equal ease from either side. In this case, one of the methyl groups bonded to C-6 (colored blue in the equation) covers one face of the double bond, blocking any approach from that side. All reagents that add to this double bond must therefore approach from the side opposite this methyl.
Exercises
1. Draw the bond-line structure of the product(s) for these following reactions?
a)
b)
c)
2. Draw the structural formulas for the alcohols that result from hydroboration-oxidation of the alkenes shown.
a)
b) (E)-3-methyl-2-pentene
3. Write out the reagents or products (A–D) shown in the following reaction schemes.
Answer
1.
a)
b)
c)
2.
a)
b)
3. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.06%3A_Hydration_-_Hydroboration-Oxidation.txt |
Learning Objective
• discern the stereochemical differences between the EAR of chiral and achiral alkenes
Stereochemistry and the Subtle Details
Organic reactions in the laboratory or in living systems can produce chiral centres. Consider reaction of 1-pentene with water (acid catalyzed). Markovnikov regiochemistry occurs and the OH adds to the second carbon. However, both R and S products occur giving a racemic (50/50) mixture of 2‑butanol. How does this occur? The proton addition to 1‑pentene results in a planar carbocation intermediate. A molecule of water is then equally likely to react from the top or the bottom of this cation to produce either (S)‑2‑pentanol or (R)‑2‑pentanol, respectively, as shown in the mechanism below.
9.08: Stereochemistry of Reactions - Hydration of Chiral Alkenes
Learning Objective
• discern the stereochemical differences between the EAR of chiral and achiral alkenes
Stereochemistry - the Subtle Details
In the previous section, the addition of water to the achiral alkene produced a racemic mixture of two enantiomeric alcohols. They are produced in equal amounts so the mixture is optically inactive. What would occur if we carried out a similar reaction on a chiral alkene? Consider (S)-3-methyl-1-pentene reacting with water (acid catalyzed). Proton addition produces a carbocation intermediate that is chiral (* denotes stereogenic centre). That intermediate does not have a plane of symmetry and therefore attack by water is not equal from the top and bottom. This ultimately produces R and S products in a non 50:50 ratio as shown in the mechanism below.
Exercise
1. Predict the products of the following reaction showing stereochemistry.
Answer
1. The products are diastereomers of one another.
9.09: Addition of Halogens
Learning Objective
• predict the products/specify the reagents for halogenation of alkenes
Introduction
As the halogen molecule, for example Br2, approaches the double bond of an alkene, electrons in the double bond repel electrons in the bromine molecule causing polarization of the halogen bond. This interaction induces a dipole moment in the halogen molecule bond allowing one of the halogens to gain a partial positive charge and take the role of electrophile. The nucleophilic pi electrons form a bond to the electrophilic halogen while the halogen molecular bond heterolytically breaks to release bromide as a leaving group. The halogen addition is not regioselective but stereoselective. Stereochemistry of this addition is analogous to the oxymercuration mechanism. In this reaction, a bromonium (halogenium) ion froms as the intermediate. The bromonium ion formation stabilizes the positive charge and prevents carbocation rearrangement. In the second step, the bromide released from the first step takes the role of thenucleophile and reacts with the cyclic bromonium ion with back side orientation. Therefore, the stereochemistry of the product is a vicinial dihalides through anti-addition.
$\ce{R_2C=CR_2 + X_2 \rightarrow R_2CX-CR_2X} \label{8.2.1}$
Halogens that are commonly used in this type of the reaction are: Br2 and Cl2. In thermodynamical terms I2 is too slow for this reaction because of the size of its atom, and F2 is too vigorous and explosive. Solvents that are used for this type of electrophilic halogenation are inert (e.g., CCl4) can be used in this reaction.
Because halogen with negative charge can attack any carbon from the opposite side of the cycle it creates a mixture of steric products.Optically inactive starting material produce optically inactive achiral products (meso) or a racemic mixture.
Electrophilic addition mechanism consists of two steps.
Before constructing the mechanism let us summarize conditions for this reaction. We will use Br2 in our example for halogenation of ethylene. Halogens can act as electrophiles due to polarizability of their covalent bond. Addition of halogens is stereospecific and produces vicinial dihalides with anti-addition. Cis starting materials will give a mixture of enantiomers and trans starting materials produce a meso compound.
Nucleophile pi electrons of alkene double bond
Electrophile halogen (Cl2 or Br2)
Regiochemistry none
Stereochemistry anti-additon
Step 1: The addition the Br-Br bond polarizes, heterolytic cleavage occurs and Br with the positive charge forms a intermediate cycle with the double bond.
Step 2: The bromide anion reacts with either carbon of the bridged bromonium ion from the back side of the ring. The ring opens up and the two halogens are in the anti-position relative to each other.
Exercise
1. What is the mechanism of adding Cl2 to the cyclohexene?
2. A reaction of Br2 molecule in an inert solvent with alkene follows?
a) syn addition
b) anti addition
c) Morkovnikov rule
3.
4.
5. Predict the product of the product of 1,2-dimethylcyclopentene reacting with Br2 with proper stereochemistry.
6. Predict the products for 1,2-dimethylcyclpentene reacting with HCl, give the proper stereochemistry. What is the relationship between the two products?
Answer
1.
2. b
3.enantiomer
4.
5.
6.
These compounds are enantiomers. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.07%3A_Stereochemistry_of_Reactions_-_Hydration_of_Achiral_Alkenes.txt |
Learning Objective
• predict the products/specify the reagents for hydrohalogenation of alkenes
Introduction
The proton is not the only electrophilic species that initiates addition reactions to the double bond of alkenes. Lewis acids like the halogens, boron hydrides and certain transition metal ions are able to accept the alkene pi-electrons. The resulting positively charged intermediates attract nucleophiles to give addition products.The electrophilic character of the halogens is well known. Fluorine adds uncontrollably with alkenes, and the addition of iodine is unfavorable, so these are not useful preparative methods. Chlorine (Cl2) and bromine (Br2) react selectively with the double bond of alkenes, so we will focus on these reactions.
The addition of chlorine and bromine to alkenes, as shown below, produces vicinal dihalo-compounds. In this reaction, we can assume that the solvent was something that is not nucleophilic, such as tetrahydrofuran (THF).
R2C=CR2 + X2 ——> R2CX-CR2X
If this same reaction is performed in a nucleophilic solvent like water or an alcohol, then the solvent becomes the nucleophile in the second step and reacts with the bromonium (or chloronium) ion to form a halohydrin as shown below.
R2C=CR2 + X2 + H2O ——> R2COH-CR2X
There are also other halogen-containing reagents that add to double bonds, such as hypohalous acids, HOX, and sulfenyl chlorides, RSCl. These reagents are unsymmetrical, so their addition to unsymmetrical double bonds may in principle take place in two ways. In practice, these addition reactions are regioselective, with one of the two possible constitutionally isomeric products being favored. The electrophilic moiety in both of these reagents is the halogen.
(CH3)2C=CH2 + HOBr ——> (CH3)2COH-CH2Br
(CH3)2C=CH2 + C6H5SCl ——> (CH3)2CCl-CH2SC6H5
Mechanisms explain the Regioselectivity
X2/H2O or X2/ROH: The regioselectivity of halohydrin formation from an alkene reaction with a halogen in a nucleophilic solvent is analogous to the oxymercuration-demercuration pathway. The halogen molecule takes the role of electrophile accepting nucleophilic pi electrons from the alkene while simultaneously forming a bond with the other vinyl carbon to create a bromonium (or chloroium) ion. The bromonium (or chloronium) ion formation stabilizes the positive charge and prevents carbocation rearrangement. The solvent takes the role of the nucleophile because it is present is a much greater percentage than the leaving group and reacts with the most substituted carbon of the cyclic bromonium (or chloronium) ion to create regiochemistry. The stereochemistry of this reaction is anti-addition because the solvent approaches the bromonium ion with back side orientation to produce the addition product. However, since the interaction of the halogen with the alkene can occur from above or below, there is no stereochemical control in this reaction and a mixture of enantiomers will be produced when applicable. The final step of this mechanism is a proton transfer to a solvent water molecule to neutralize the addition product.
HOX or RSCl: The regioselectivity of the hypohalous acids and sulfenyl chloride reactions may be explained by the same mechanism we used to rationalize the Markovnikov rule. Bonding of an electrophilic species to the double bond of an alkene forms preferentially to produce the more stable (more highly substituted) carbocation. This intermediate should then combine rapidly with a nucleophilic species to produce the addition product.
To apply this mechanism we need to determine the electrophilic moiety in each of the reagents. By using electronegativity differences we can dissect common addition reagents into electrophilic and nucleophilic moieties, as shown on the right. In the case of hypochlorous and hypobromous acids (HOX), these weak Brønsted acids (pKa's ca. 8) do not react as proton donors; and since oxygen is more electronegative than chlorine or bromine, the electrophile will be a halide cation. The nucleophilic species that bonds to the intermediate carbocation is then hydroxide ion, or more likely water (the usual solvent for these reagents), and the products are called halohydrins. Sulfenyl chlorides add in the opposite manner because the electrophile is a sulfur cation, RS(+), whereas the nucleophilic moiety is chloride anion (chlorine is more electronegative than sulfur).
Below are some examples illustrating the addition of various electrophilic halogen reagents to alkene groups. Notice the specific regiochemistry of the products, as explained above.
Exercise
1. Predict the product of the following reaction:
2. When butene is treated with NBS in the presence of water, the product shows that the bromine is on the least substituted carbon, is this Markovnikov or anti-Markovnikov?
Answer
1.
2. Since the bromine is the first addition to the alkene, this addition would be an anti-Markovnikov addition. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.10%3A_Formation_of_Halohydrins.txt |
Learning Objective
• recognize organic oxidation and reduction reactions
• predict the products/specify the reagents for hydrogenation (reduction) of alkenes
Introduction
Addition of hydrogen to a carbon-carbon double bond to form an alkane is a reduction reaction that is also called catalytic hydrogenation. Hydrogenation of a double bond is a thermodynamically favorable reaction because it forms a more stable (lower energy) product. In other words, the energy of the product is lower than the energy of the reactant; thus it is exothermic (heat is released). The heat released is called the heat of hydrogenation, which is an indicator of a molecule’s stability. Regioselectivity is not an issue because the same group (a hydrogen atom) is bonded to each of the vinyl carbons. The simplest source of two hydrogen atoms is molecular hydrogen (H2), but mixing alkenes with hydrogen does not result in any discernible reaction. Although the overall hydrogenation reaction is exothermic, a high activation energy prevents it from taking place under normal conditions. This restriction may be circumvented by the use of a catalyst, as shown in the reactions below.
The O-chem View of Oxidation and Reduction
For inorganic chemistry, the flow of electrons is easily counted with the change in oxidation numbers of the metals and non-metals. The expressions "LEO says GER" for "Loss of Electrons is Oxidation and Gain of Electrons is Reduction" or "OIL RIG" for "Oxidation Is Loss and Reduction Is Gain" can be useful guides to recognizing oxidation and reduction reactions for inorganic chemistry. However for organic chemistry, most of the reactants and products are neutral so the electron flow is more difficult to track. For organic compounds, oxidation and reduction reactions can be recognized at least three different ways.
1) Oxidation is an increase in the number of carbon to oxygen bonds or a decrease in the number of carbon to hydrogen bonds.
2) Reduction is the opposite of oxidation so it is a decrease in the number of carbon to oxygen bonds or an increase in the number of carbon to hydrogen bonds.
3) For reactions that do not involve a change in the bonding of carbon with oxygen and hydrogen, then we need to look at the differences in electronegativity. The shared electrons are assigned to the more electronegative element to determine the oxidation numbers.
The Catalyst
The reaction between hydrogen (H2) gas and an alkene (a carbon-carbon double bond) requires an active metal catalyst. A catalyst increases the reaction rate by lowering the activation energy of the reaction. Although the catalyst is not consumed in the reaction, it is required to accelerate the reaction sufficiently to be observed in a reasonable amount of time. Catalysts commonly used in alkene hydrogenation are: platinum, palladium, and nickel. The metal catalyst acts as a surface on which the reaction takes place. This increases the rate by putting the reactants in close proximity to each other, facilitating interactions between them. With this catalyst present, the sigma bond of H2 breaks, and the two hydrogen atoms instead bind to the metal (see #2 in the figure below). The $\pi$ bond of the alkene weakens as it also interacts with the metal as shown in step #3 of the diagram below.
Since both the reactants are bound to the metal catalyst, the hydrogen atoms can easily add, one at a time, to the previously double-bonded carbons as shown in steps #4 and #5 above. The position of both of the reactants bound to the catalyst makes it so the hydrogen atoms are only exposed to one side of the alkene. This explains why the hydrogen atoms add to same side of the molecule, called syn-addition.
Alkene Stability and Catalytic Hydrogenation
As shown in the reaction energy diagram below, the hydrogenation of alkenes is exothermic, and heat is released corresponding to the ΔE (colored green).
This heat of reaction can be used to evaluate the thermodynamic stability of alkenes having different numbers of alkyl substituents on the double bond. For example, the following table lists the heats of hydrogenation for three C5H10 alkenes which give the same alkane product (2-methylbutane). Since a large heat of reaction indicates a high energy reactant, these heats are inversely proportional to the stabilities of the alkene isomers. To a rough approximation, we see that each alkyl substituent on a double bond stabilizes this functional group by a bit more than 1 kcal/mole.
Alkene Isomer (CH3)2CHCH=CH2
3-methyl-1-butene
CH2=C(CH3)CH2CH3
2-methyl-1-butene
(CH3)2C=CHCH3
2-methyl-2-butene
Heat of Reaction
( ΔHº )
–30.3 kcal/mole –28.5 kcal/mole –26.9 kcal/mole
Stereochemistry of Catalytic Hydrogenation
From the mechanism shown below, we expect the addition of hydrogen to occur with syn-stereoselectivity since both reactants approach the same side of the catalyst's surface.
For example, 1,2-dimethylcyclopentene is reduced to 1,2-dimethylcyclopentane during catalytic hydrogenation.
Exercises
1. Use the catalytic hydrogenation of ethene with platinum oxide to answer the following questions.
1. 0.500 mol of ethene reacts with _______ mol of hydrogen.
2. Ethene is being _______; while _______ is being oxidized.
3. The oxidation number of carbon in ethene is _______; in ethane it is _______.
2. When 1.000 g of a certain triglyceride (fat) is treated with hydrogen gas in the presence of Adams’ catalyst, it is found that the volume of hydrogen gas consumed at 99.8 kPa and 25.0°C is 162 mL. A separate experiment indicates that the molar mass of the fat is 914 g mol−1. How many carbon-carbon double bonds does the compound contain?
3. Bromobutene reacts with hydrogen gas in the presence of a platinum catalyst. What is the name of the product?
4. Cyclohexene reacts with hydrogen gas in the presence of a palladium catalyst. What is the name of the product?
5. What is the stereochemistry of an alkene hydrogenation reaction?
6. When looking at their heats of hydrogenation, is the cis or the trans isomer generally more stable?
7. 2-chloro-4-ethyl-3methylcyclohexene reacts with hydrogen gas in the presence of a platinum catalyst. What is the name of the product?
8. Predict the products if the following alkenes were reacted with catalytic hydrogen.
Answer
1. a. 0.500 mole of hydrogen gas
b. Ethene is being reduced; while hydrogen is being oxidized.
c. The oxidation number of carbon in ethene is −2; in ethane it is −3.
2. Amount of hydrogen consumed $\begin{array}{l}=n\text{\hspace{0.17em}}\text{mol}\ \text{=}\frac{PV}{RT}\ =\frac{99.8\text{\hspace{0.17em}}\text{kPa}×0.162\text{\hspace{0.17em}}\text{L}}{8.31\text{\hspace{0.17em}}\text{kPa}\cdot {\text{mol}}^{-1}\cdot {\text{K}}^{-1}×298\text{\hspace{0.17em}}\text{K}}\ =6.53×{10}^{-3}\text{\hspace{0.17em}}\text{mol}\text{\hspace{0.17em}}{\text{H}}_{2}\end{array}$
Amount of fat used $\begin{array}{l}=\frac{\left(1.000\text{\hspace{0.17em}}\text{g}\right)×\left(1\text{\hspace{0.17em}}\text{mol}\right)}{\left(914\text{\hspace{0.17em}}\text{g}\right)}\ =1.09×{10}^{-3}\text{\hspace{0.17em}}\text{mol}\text{\hspace{0.17em}}\text{fat}\end{array}$
Ratio of moles of hydrogen consumed to moles of fat $\begin{array}{l}=6.53×{10}^{-3}:1.09×{10}^{-3}\ =6:1\end{array}$
Thus, the fat contains six carbon-carbon double bonds per molecule.
3. Bromobutane
4. Cyclohexane
5. Syn-addition
6. Trans
7. 2-chloro-4-ethyl-3methylcyclohexane
8.
9. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.11%3A_Reduction_of_Alkenes_-_Catalytic_Hydrogenation.txt |
Learning Objective
• recognize organic oxidation and reduction reactions
• predict the products/specify the reagents for epoxidation of alkenes
Oxidation - a closer look
There are a variety of oxidative reagents that can react with alkenes. These reagents oxidize the alkene to different degrees and have different synthetic applications. It can be helpful to describe the relative oxidative strength of the reagents. Some reagents are so strong that the carbon chain will be cleaved at the alkene. This reactivity can also be a useful distinction. Before we explore the specific details of these different reaction pathways, let's look at the overall patterns of functional group reactivity.
The are four levels of oxidation for alkenes. The gentlest and least oxidative is epoxide (oxacyclopropane) formation in which the vinyl carbons share a single oxygen atom as a three membered ring. Moderate oxidation will convert the alkene into a vicinal diol in which each vinyl carbon is bonded to an independent oxygen atom. The stronger oxidative reactions cleave the carbon chain at the alkene. While the overall chemical process is an oxidation reaction, the work-up (second step) of the reaction can be performed under reductive or gentle conditions or a strong, oxidative cleavage reaction can occur with the strongest reagents. These four reaction pathways are summarized below.
Epoxide (Oxacyclopropane) Synthesis by Peroxycarboxylic Acid
Oxacyclopropane rings, also called epoxide rings, are useful reagents that may be opened by further reaction to form anti vicinal diols. One way to synthesize oxacyclopropane rings is through the reaction of an alkene with a peroxycarboxylic acid, such as MCPBA (m-chloroperoxybenzoic acid). Oxacyclopropane synthesis by peroxycarboxylic acid requires an alkene and a peroxycarboxylic acid as well as an appropriate solvent. The peroxycarboxylic acid has the unique property of having an electropositive oxygen atom on the COOH group. The reaction is initiated by the electrophilic oxygen atom reacting with the nucleophilic carbon-carbon double bond. The mechanism involves a concerted reaction with a four-part, circular transition state. The result is that the originally electropositive oxygen atom ends up in the oxacyclopropane ring and the COOH group becomes COH.
Mechanism
The mechanism is a concerted reaction between the alkene and peroxyacid. As seen with other concerted reactions, it is stereospecific: a cis-alkene will produce a cis-epoxide and a trans alkene will produce a trans-epoxide.
Peroxycarboxylic acids are generally unstable. An exception is MCPBA, shown in the mechanism above. Often abbreviated MCPBA, it is a stable crystalline solid. Consequently, MCPBA is popular for laboratory use. However, MCPBA can be explosive under some conditions. Peroxycarboxylic acids are sometimes replaced in industrial applications by monoperphthalic acid, or the monoperoxyphthalate ion bound to magnesium, which gives magnesium monoperoxyphthalate (MMPP). In either case, a nonaqueous solvent such as chloroform, ether, acetone, or dioxane is used. This is because in an aqueous medium with any acid or base catalyst present, the epoxide ring is hydrolyzed to form a vicinal diol, a molecule with two OH groups on neighboring carbons. (For more explanation of how this reaction leads to vicinal diols, see below.) However, in a nonaqueous solvent, the hydrolysis is prevented and the epoxide ring can be isolated as the product. Reaction yields from this reaction are usually about 75%. The reaction rate is affected by the nature of the alkene, with more nucleophilic double bonds resulting in faster reactions.
Example \(1\)
Since the transfer of oxygen is to the same side of the double bond, the resulting oxacyclopropane ring will have the same stereochemistry as the starting alkene. A good way to think of this is that the alkene is rotated so that some constituents are coming forward and some are behind. Then, the oxygen is inserted on top. (See the product of the above reaction.) One way the epoxide ring can be opened is by an acid catalyzed oxidation-hydrolysis. Oxidation-hydrolysis gives a vicinal diol, a molecule with OH groups on neighboring carbons. For this reaction, the dihydroxylation is anti since, due to steric hindrance, the ring is attacked from the side opposite the existing oxygen atom. Thus, if the starting alkene is trans, the resulting vicinal diol will have one S and one R stereocenter. But, if the starting alkene is cis, the resulting vicinal diol will have a racemic mixture of S, S and R, R enantiomers.
Exercise \(1\)
1. Predict the product of the reaction of cis-2-hexene with MCPBA (meta-chloroperoxybenzoic acid)
a) in acetone solvent.
b) in an aqueous medium with acid or base catalyst present.
2. Predict the product of the reaction of trans-2-pentene with magnesium monoperoxyphthalate (MMPP) in a chloroform solvent.
3. Predict the product of the reaction of trans-3-hexene with MCPBA in ether solvent.
4. Predict the reaction of propene with MCPBA.
a) in acetone solvent
b) after aqueous work-up.
5. Predict the reaction of cis-2-butene in chloroform solvent.
Answer
1. a) Cis-2-methyl-3-propyloxacyclopropane
b) Racemic (2R,3R)-2,3-hexanediol and (2S,3S)-2,3-hexanediol
2. Trans-3-ethyl-2-methyloxacyclopropane.
3. Trans-3,4-diethyloxacyclopropane.
4. a) 1-ethyl-oxacyclopropane
b) Racemic (2S)-1,2-propandiol and (2R)-1,2-propanediol
5. Cis-2,3-dimethyloxacyclopropane | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.12%3A_Oxidation_of_Alkenes_-_Epoxidation.txt |
Learning Objective
• predict the products/specify the reagents for dihydroxylation of alkenes
Dihydroxylation of alkenes
Alkenes can be dihydroxylated by two different stereochemical pathways: anti-dihydroxylation or syn-dihydroxylation. The opening of epoxides follows the anti-dihydroxylation mechanism, while potassium permanganate or osmium tetroxide produce the syn-dihydroxylated products. The osmium tertroxide reaction can also take place by a two-step process: 1) OsO4 in pyridine followed by 2) H2S or NaHSO3. It is important to note that different professors will emphasize different reagent systems to accomplish the same chemical reaction. In these situations, it can be helpful to recognize the role of each reagent to discern patterns.
Anti Dihydroxylation
Epoxides may be cleaved by aqueous acid to give glycols that are often diastereomeric with those prepared by the syn-hydroxylation reaction described above. Proton transfer from the acid catalyst generates the conjugate acid of the epoxide, which is attacked by nucleophiles such as water in the same way that the cyclic bromonium ion described above undergoes reaction. The result is anti-hydroxylation of the double bond, in contrast to the syn-stereoselectivity of the earlier method. In the following equation this procedure is illustrated for a cis-disubstituted epoxide, which, of course, could be prepared from the corresponding cis-alkene. This hydration of an epoxide does not change the oxidation state of any atoms or groups. The mechanism for the ring opening of epoxides depends on the reaction conditions and is discussed in more detail in the next section of this chapter.
Syn Dihydroxylation
Osmium tetroxide oxidizes alkenes to give glycols through syn addition. A glycol, also known as a vicinal diol, is a compound with two -OH groups on adjacent carbons.
Dihydroxylated products (glycols) are obtained by reaction with aqueous potassium permanganate (pH > 8) or osmium tetroxide in pyridine solution. Both reactions appear to proceed by the same mechanism (shown below); the metallocyclic intermediate may be isolated in the osmium reaction. In basic solution the purple permanganate anion is reduced to the green manganate ion, providing a nice color test for the double bond functional group. From the mechanism shown here we would expect syn-stereoselectivity in the bonding to oxygen, and regioselectivity is not an issue.
When viewed in context with the previously discussed addition reactions, the hydroxylation reaction might seem implausible. Permanganate and osmium tetroxide have similar configurations, in which the metal atom occupies the center of a tetrahedral grouping of negatively charged oxygen atoms. How, then, would such a species interact with the nucleophilic pi-electrons of a double bond? A possible explanation is that an empty d-orbital of the electrophilic metal atom extends well beyond the surrounding oxygen atoms and initiates electron transfer from the double bond to the metal, in much the same fashion noted above for platinum. Back-bonding of the nucleophilic oxygens to the antibonding π*-orbital completes this interaction. The result is formation of a metallocyclic intermediate, as shown above.
The reaction with $OsO_4$ is a concerted process that has a cyclic intermediate and no rearrangements. Vicinal syn dihydroxylation complements the epoxide-hydrolysis sequence which constitutes an anti dihydroxylation of an alkene. When an alkene reacts with osmium tetroxide, stereocenters can form in the glycol product. Cis alkenes give meso products and trans alkenes give racemic mixtures.
$OsO_4$ is formed slowly when osmium powder reacts with gasoues $O_2$ at ambient temperature. Reaction of bulk solid requires heating to 400 °C:
$Os_{(s)} + 2O_{2\;(g)} \rightarrow OS_4$
Since Osmium tetroxide is expensive and highly toxic, the reaction with alkenes has been modified. Catalytic amounts of OsO4 and stoichiometric amounts of an oxidizing agent such as hydrogen peroxide are now used to eliminate some hazards. Also, an older reagent that was used instead of OsO4 was potassium permanganate, $KMnO_4$. Although syn diols will result from the reaction of KMnO4 and an alkene, potassium permanganate is less useful since it gives poor yields of the product because of overoxidation.
Chemical Highlight
Antitumor drugs have been formed by using dihydroxylation. This method has been applied to the enantioselective synthesis of ovalicin, which is a class of fungal-derived products called antiangiogenesis agents. These antitumor products can cut off the blood supply to solid tumors. A derivative of ovalicin, TNP-470, is chemically stable, nontoxic, and noninflammatory. TNP-470 has been used in research to determine its effectiveness in treating cancer of the breast, brain, cervix, liver, and prostate.
Exercise
1. Give the major product.
2. What is the product in the dihydroxylation of (Z)-3-hexene?
3. What is the product in the dihydroxylation of (E)-3-hexene?
4. Draw the intermediate of this reaction.
5. Fill in the missing reactants, reagents, and product.
Answer
1. A syn-1,2-ethanediol is formed. There is no stereocenter in this particular reaction. The OH groups are on the same side.
2. Meso-3,4-hexanediol is formed. There are 2 stereocenters in this reaction.
3. A racemic mixture of 3,4-hexanediol is formed. There are 2 stereocenters in both products.
4. A cyclic osmic ester is formed.
5. The Diels-Alder cycloaddition reaction is needed in the first box to form the cyclohexene. The second box needs a reagent to reduce the intermediate cyclic ester (not shown). The third box has the product: 1,2-cyclohexanediol.
• Shivam Nand | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.13%3A_Dihydroxylation_of_Alkenes.txt |
Learning Objective
• predict the products/specify the reagents for dihydroxylation of alkenes
Epoxide ring-opening reactions - SN1 vs. SN2, regioselectivity, and stereoselectivity
The nonenzymatic ring-opening reactions of epoxides provides an oppourtunity to review the nucelophilic substitution mechansims. Ring-opening reactions can proceed by either SN2 or SN1 mechanisms, depending on the nature of the epoxide and on the reaction conditions. If the epoxide is asymmetric, the structure of the product will vary according to which mechanism dominates. When an asymmetric epoxide undergoes solvolysis in basic methanol, ring-opening occurs by an SN2 mechanism, and the less substituted carbon reacts with the nucleophile under steric considerations and produces product B in the example below.
Conversely, when solvolysis occurs in acidic methanol, the reaction occurs by a mechanism with substantial SN1 character, and the more substituted carbon reacts with the nucleophile under electrostatic considerations and produces product A in the example below.
These are both good examples of regioselective reactions. In a regioselective reaction, two (or more) different constitutional isomers are possible as products, but one is formed preferentially (or sometimes exclusively).
Let us examine the basic, SN2 case first. The leaving group is an alkoxide anion, because there is no acid available to protonate the oxygen prior to ring opening. An alkoxide is a poor leaving group, and thus the ring is unlikely to open without a 'push' from the nucleophile.
The nucleophile itself is potent: a deprotonated, negatively charged methoxide ion. When a nucleophilic substitution reaction involves a poor leaving group and a powerful nucleophile, it is very likely to proceed by an SN2 mechanism.
What about the electrophile? There are two electrophilic carbons in the epoxide, but the best target for the nucleophile in an SN2 reaction is the carbon that is least hindered. This accounts for the observed regiochemical outcome. Like in other SN2 reactions, nucleophilic reactions take place with backside orientation relative to the leaving group, resulting in inversion at the electrophilic carbon.
Probably the best way to depict the acid-catalyzed epoxide ring-opening reaction is as a hybrid, or cross, between an SN2 and SN1 mechanism. First, the oxygen is protonated, creating a good leaving group (step 1 below) . Electrostatic considerations have greater importance with a protonated intermediate. As the carbon-oxygen bond begins to break (step 2), positive charge builds on the more substituted carbon with greater carbocation stability.
Unlike in an SN1 reaction, the nucleophile reacts with the electrophilic carbon (step 3) before a complete carbocation intermediate has a chance to form.
Reaction takes place preferentially from the backside (like in an SN2 reaction) because the carbon-oxygen bond is still to some degree in place, and the oxygen blocks reaction from the front side. Notice, however, the regiochemical outcome is different from the base-catalyzed reaction. In the acid-catalyzed process, the nucleophile reacts with the more substituted carbon because it is this carbon that holds a greater degree of positive charge and electrostatics (carbocation stability) take a dominant role in determining the mechanism.
Example \(1\)
Predict the major product(s) of the ring opening reaction that occurs when the epoxide shown below is treated with:
1. ethanol and a small amount of sodium hydroxide
2. ethanol and a small amount of sulfuric acid
Hint: be sure to consider both regiochemistry and stereochemistry!
Solution
Addition of HX
Epoxides can also be opened by other anhydrous acids (HX) to form a trans halohydrin. When both the epoxide carbons are either primary or secondary the halogen anion will attack the less substituted carbon and an SN2 like reaction. However, if one of the epoxide carbons is tertiary, the halogen anion will primarily attack the tertialy cabon in a SN1 like reaction.
Example \(1\)
Exercise
1. Given the following, predict the product assuming only the epoxide is affected. (Remember stereochemistry)
2. Predict the product of the following, similar to above but a different nucleophile is used and not in acidic conditions. (Remember stereochemistry)
3. Epoxides are often very useful reagents to use in synthesis when the desired product is a single stereoisomer. If the following alkene were reacted with an oxyacid to form an epoxide, would the result be a enantiomerically pure? If not, what would it be?
Answer
1.
Note that the stereochemistry has been inverted
2,
3.
First, look at the symmetry of the alkene. There is a mirror plane, shown here.
Then, think about the mechanism of epoxidation with an oxyacid, take for example mCPBA. The mechanism is concerted, so the original cis stereochemistry is not changed. This leads to "two" epoxides.
However, these two mirror images are actually identical due to the mirror plane of the cis geometry. It is a meso compound, so the final result is a single stereoisomer, but not a single enantiomer. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.14%3A_Opening_of_Epoxides_-_Acidic_versus_Basic_Conditions.txt |
Learning Objective
• predict the products/specify the reagents for oxidative cleavage of alkenes
Overview
Oxidative cleavage can occur by several different reaction pathways. The cleavage can be strong or gentle depending on the reaction conditions and/or the work-up of the initial reaction product. Both alkenes and alkynes can undergo cleavage reactions. This section will focus on alkenes.
Gentle cleavage of alkenes occurs by two primary reaction pathways: ozolysis with a reductive work-up or syn-dihydroxylation followed by oxidation with perioidc acid. Gentle cleavage will leave terminal carbons partially oxidized to aldehydes. Strong cleavage of alkenes will fully oxidize terminal carbons to carboxylic acids. Internal carbons become ketones by either reaction pathway.
Why so many reactions?
At first glance, it may seem silly to have more than more reaction pathway for the same functional group conversion. As this course proceeds, it will become necessary to target reactions to a single functional group of an organic compound with multiple functional groups. A particular reaction pathway can be advantageous when the reactivity of the entire molecule is considered, not just a single functional group.
Gentle Cleavage: Syn-Dihydroxylation followed by Periodic Acid
Alkenes can also be gently cleavage in a two-step reaction sequence in which the alkene first undergoes syn-dihydroxylation using cold, slightly basic KMnO4 or OsO4/H2O2 followed by oxidation with periodic acid (HIO4). Both reaction sequences are shown below using 1-methylcyclohexene as an example.
Gentle Cleavage: Ozonolysis with a Reduction Work-up
Ozonolysis is a method of oxidatively cleaving alkenes or alkynes using ozone (\(O_3\)), a reactive allotrope of oxygen. The process allows for carbon-carbon double or triple bonds to be replaced by double bonds with oxygen. This reaction is often used to identify the structure of unknown alkenes by breaking them down into smaller, more easily identifiable pieces. Ozonolysis also occurs naturally and would break down repeated units used in rubber and other polymers. On an industrial scale, azelaic acid and pelargonic acids are produced from ozonolysis.
Ozonolysis Reaction Mechanism
The gaseous ozone is first passed through the desired alkene solution in either methanol or dichloromethane. The first intermediate product is an ozonide molecule which is then further reduced to carbonyl products. This results in the breaking of the carbon-carbon double bond and is replaced by a carbon-oxygen double bond instead.
Step 1:
The first step in the mechanism of ozonolysis is the initial electrophilic addition of ozone to the carbon-carbon double bond to form the molozonide intermediate. Due to low stablility of molozonide, it continues reacting and breaks apart to form a carbonyl and a carbonyl oxide molecule.
Step 2:
The electrons of the carbonyl and the carbonyl oxide form the stable ozonide intermediate which can then undergo an oxidative or reductive work-up to form the products of interest. A reductive workup converts the ozonide molecule into the desired carbonyl products with aldehyde on terminal carbons. An oxidative workup converts the ozonide molecule into the desired carbonyl products with carboxylic acids on terminal carbons. The two reaction workup conditions are summarized below.
Exercises
Answers
Strong Cleavage from Strong Oxidative Reactions
When the reaction conditions for potassium permanganate are warm, then the oxidation reaction is stronger and cleavage occurs at the alkene with any terminal carbons fully oxidizing to carboxylic acids. When the ozonolysis reaction is followed by an oxidative work-up, then any terminal carbons will oxidize fully to the carboxylic acid as shown in the example below. The example for 1-methylcycohexene is shown below.
Exercise
5. What would you expect the products to be from the reaction of cis-2-pentene with m-chloro-peroxybenzoic acid? Show the stereochemistry of the final product.
6. Give a reaction scheme with starting alkenes and required reagents to produce the following compounds.
Answer
5.
6.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.15%3A_Oxidative_Cleavage_of_Alkenes.txt |
Learning Objective
• predict the products of carbene additions to alkenes
Introduction
Carbenes were once only thought of as short lived intermediates. The reactions of this section only deal with these short lived carbenes which are mostly prepared in situ, in conjunction with the main reaction. However, there do exist so called persistent carbenes. These persistent carbenes are stabilized by a variety of methods often including aromatic rings or transition metals. In general a carbene is neutral and has 6 valence electrons, 2 of which are non bonding. These electrons can either occupy the same sp2 hybridized orbital to form a singlet carbene (with paired electrons), or two different sp2 orbitals to from a triplet carbene (with unpaired electrons). The chemistry of triplet and singlet carbenes is quite different but can be oversimplified to the statement: singlet carbenes usually retain stereochemistry while triplet carbenes do not. The carbenes discussed in this section are singlet and thus retain stereochemistry.
The reactivity of a singlet carbene is concerted and similar to that of electrophilic or nucleophilic addition (although, triplet carbenes react like biradicals, explaining why sterochemistry is not retained). The highly reactive nature of carbenes leads to very fast reactions in which the rate determining step is generally carbene formation.
Preparation of methylene
The preparation of methylene starts with the yellow gas diazomethane, CH2N2. Diazomethane can be exposed to light, heat or copper to facilitate the loss of nitrogen gas and the formation of the simplest carbene methylene. The process is driven by the formation of the nitrogen gas which is a very stable molecule.
Carbene reaction with alkenes
A carbene such as methlyene will react with an alkene which will break the double bond and result with a cyclopropane. The reaction will usually leave stereochemistry of the double bond unchanged. As stated before, carbenes are generally formed along with the main reaction; hence the starting material is diazomethane not methylene.
In the above case cis-2-butene is converted to cis-1,2-dimethylcyclopropane. Likewise, below the trans configuration is maintained.
Additional Types of Carbenes and Carbenoids
In addition to the general carbene with formula R2C there exist a number of other compounds that behave in much the same way as carbenes in the synthesis of cyclopropane. Halogenated carbenes are formed from halomethanes. An example is dicholorcarbene, Cl2C. These halogenated carbenes will form cyclopropanes in the same manner as methylene but with the interesting presence of two halogen atoms in place of the hydrogen atoms.
Carbenoids are substances that form cyclopropanes like carbenes but are not technically carbenes. One common example is the stereospecific Simmon-Smith reaction which utilizes the carbenoid ICH2ZnI. The carbenoid is formed in situ via the mixing of a Zn-Cu couple with CH2I2.Since this reacts thesame as a carbene, the same methods can be applied to determine the product. An example of this is given as problem 5.
Exercise \(1\)
1. Knowing that cycloalkenes react much the same as regular alkenes what would be the expected structure of the product of cyclohexene and diazomethane facilitated by copper metal?
2. What would be the result of a Simmons-Smith reaction that used trans-3-pentene as a reagent?
3. What starting material could be used to form cis-1,2-diethylcyclopropane?
4. What would the following reaction yield?
5. Draw the product of this reaction. What type of reaction is this?
6. Predict the following products. Will they be the same product?
Answer
1. The product will be a bicyclic ring, Bicyclo[4.1.0]heptane.
2. The stereochemistry will be retained making a cyclopropane with trans methyl and ethyl groups. Trans-1-ethyl-2-methylcyclopropane
3. The cis configuration will be maintained from reagent to product so we would want to start with cis-3-hexene. A Simmons Smith reagent, or methylene could be used as the carbene or carbenoid.
4. The halogenated carbene will react the same as methylene yielding, cis-1,1-dichloro-2,3dimethylcyclopropane.
5. This is a Simmons-Smith reaction which uses the carbenoid formed by the CH2I2 and Zu-Cu. The reaction results in the same product as if methylene was used and retains stereospecificity. Iodine metal and the Zn-Cu are not part of the product. The product is trans-1,2-ethyl-methylcyclopropane.
6. No they will not be the same product, they will be isomers of each other. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.16%3A_Addition_of_Carbenes_to_Alkenes_-_Cyclopropane_Synthesis.txt |
Learning Objective
• predict the polymer/specify the monomer for radical, chain -growth polymers of alkenes
Introduction
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.
Exercise
1. Propose the monomer units in the following polymers:
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.17%3A__Radical_Chain-Growth_Polymerization.txt |
Learning Objective
• discuss an example biological addition reactions
Radical mechanisms for flavin-dependent reactions
Flavin coenzymes, like their nicotinamide adenine dinucleotide counterparts, can act as hydride acceptors and donors. In these redox reactions, two electrons are transferred together in the form of a hydride ion. Flavin, however, is also capable of mediating chemical steps in which a single unpaired electron is transferred - in other words, radical chemistry. This is due to the ability of the flavin system to form a stabilized radical intermediate called a semiquinone, formed when FADH2 (or FMNH2) donates a single electron, or when FAD (or FMN) accepts a single electron.
This single-electron transfer capability of flavins is critical to their metabolic role as the entry point of electrons into the electron transport phase of respiration. Electrons 'harvested' from the oxidation of fuel molecules are channeled, one by one, by FMNH2 into the electron transport chain, where they eventually reduce molecular oxygen. NADH is incapable of single electron transfer - all it can do is transfer two electrons, in the form of a hydride, to FMN; the regenerated FMNH2 is then able to continue sending single electrons into the transport chain.
You will learn more details about this process in a biochemistry class.
Because flavins are capable of single-electron as well as two-electron chemistry, the relevant mechanisms of flavoenzyme-catalyzed reactions are often more difficult to determine. Recall the dehydrogenation reaction catalyzed by acyl-CoA dehydrogenase (section 16.5C) - it involves the transfer of two electrons and two protons (ie. a hydrogen molecule) to FAD. Both electrons could be transferred together, with the FAD coenzyme simply acting as a hydride acceptor (this is the mechanism we considered previously). However, because the oxidizing coenzyme being used is FAD rather than NAD+, it is also possible that the reaction could proceed by a single-electron, radical intermediate process. In the alternate radical mechanism proposed below, for example, the enolate intermediate first donates a single electron to FAD, forming a radical semiquinone intermediate (step 2). The second electron is transferred when the semiquinone intermediate abstracts a hydrogen from Cb in a homolytic fashion (step 3).
Scientists are still not sure which mechanism - the hydride transfer mechanism that we saw in section 16.5B or the single electron transfer detailed above - more accurately depicts what is going on in this reaction.
The conjugated elimination catalyzed by chorismate synthase (section 14.3B) is another example of a reaction where the participation of flavin throws doubt on the question of what is the relevant mechanism. This could simply be a conjugated E1' reaction, with formation of an allylic carbocation intermediate. The question plaguing researchers studying this enzyme, however, is why FADH2 is required. This is not a redox reaction, and correspondingly, FADH2 is not used up in the course of the transformation - it just needs to be bound in the active site in order for the reaction to proceed. Given that flavins generally participate in single-electron chemistry, this is an indication that radical intermediates may be involved. Recently an alternative mechanism, involving a flavin semiquinone intermediate, has been proposed (J. Biol. Chem 2004, 279, 9451). Notice that a single electron is transferred from substrate to coenzyme in step 2, then transferred back in step 4. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.18%3A_Biological_Additions_of_Radicals_to__Alkenes.txt |
Addition of Hydrogen Halides to Alkenes
9-1 Give the IUPAC name for the product of the following reaction.
9-2 Draw the reaction mechanism of the previous problem (9-1).
9-3 Identify the product of the following reactions.
9-4 Identify the products of the following reactions.
Addition of Water: Hydration of Alkenes
9-5 Identify the product of the following reactions.
9-6 Identify the product of the following reaction.
9-7 Propose a plausible route of synthesis for the following product starting with 1-butanol.
9-8 Which of the following alkenes can be used to obtain 3,4-dimethylpentan-2-ol through a hydration reaction using dilute acid?
a) (2Z)-3,4-dimethylpent-2-ene
b) 3,4-dimethylpent-1-ene
c) 2,3,4-trimethylpent-2-ene
d) 2,4-dimethyl-3-methylidenepentane
Hydration by Oxymercuration-Demercuration
9-9 Explain why hydration of an alkene by Oxymercuration-Demercuration gives the Markovnikov product.
9-10 Identify the products of the following reactions.
9-11 Identify the product of the following reaction.
9-12 Propose a possible route of synthesis for the following ether starting with 2-ethylbutan-1-ol.
9-13 Give the IUPAC name of the product of the following reaction.
a) 2-methylpropan-2-ol
b) 2,2-dimethylbutane
c) 2-methoxy-2-methylpropane
d) 1,1-dimethylcyclopropane
Hydroboration of Alkenes
9-14 Identify the products of the following reactions.
9-15 Identify the products of the following reactions.
9-16 Give the IUPAC name for the product of the following reaction.
Addition of Halogens to Alkenes
9-17 Identify the products of the following reactions.
9-18 Identify the product of the following reaction.
9-19 What is the product of the following reaction?
a) (2Z)-2-bromo-3-methylpent-2-ene
b) 2-bromo-3-methylpentane
c) 2,3-dibromo-3-methylpentane
d) (2E)-4-bromo-3-methylpent-2-ene
9-20 What reagents can be used in each step to obtain the following products?
9-21 Explain why you do not obtain a mixture of cis- and trans-brominated products when you react Br2/CCl4 with cyclopentene.
Formation of Halohydrins
9-22 Identify the product of the following reaction, making sure to include stereochemistry.
9-23 Draw the mechanism of the reaction in the previous problem (9-22).
9-24 Give the IUPAC name for the product of the following reaction.
9-25 Identify the product of the following reaction.
Catalytic Hydrogenation of Alkenes
9-26 Identify the products of the following reactions.
9-27 Suggest a possible route of synthesis, that includes a catalytic hydrogenation step, to obtain the following product.
9-28 Identify the product of the following reaction.
9-29 Identify the product of the following reaction, making sure to include stereochemistry.
Addition of Carbenes to Alkenes
9-30 Identify the product of the following reaction.
9-31 Identify the product(s) of the following reaction.
9-32 Identify the product of the reaction when (2E)-3-methylhex-2-ene reacts with the carbene product from the previous problem (9-31), then reacts with Br2 and hv.
9-33 Propose a possible route of synthesis for the following reaction.
9-34 Identify the product of the following reaction.
Epoxidation of Alkenes and Acid-Catalyzed Opening of Epoxides
9-35 Identify the product of the following reaction.
9-36 Identify the product of the following reactions, specifying stereochemistry where appropriate.
9-37 Identify the products of the following reaction, including stereochemistry.
9-38 What is the product of the following reaction?
a) benzoic acid
b) 1-cyclohexylethan-1-ol
c) 1-cyclohexylethan-1-one
d) cyclohexanol
9-39 What is the product of the following reaction?
a) 4-chlorobutane-1,3-diol
b) 3-chlorobutan-1-ol
c) 2-chlorobutane-1,4-diol
d) 2,4-dichlorobutan-1-ol
9-40 Draw the arrows for the following epoxidation reaction to show the movement of electrons.
Syn Dihydroxylation of Alkenes
9-41 Identify the product of the following reaction, including stereochemistry.
9-42 Give the IUPAC name for the product(s) of the following reaction. Include stereochemistry.
9-43 Identify the product of the following reaction.
9-44 Suggest a possible route of synthesis for the following compound starting with cyclopentanol.
Oxidative Cleavage of Alkenes
9-45 Identify the products of the following reactions.
9-46 Identify the products of the following reactions.
9-47 Identify the product of the following reaction.
Polymerization of Alkenes
9-48 Identify the alkene monomer that composes the following polymer.
9-49 Draw the mechanism for the acid catalyzed formation of the polymer in the previous problem (9-48).
9-50 Draw the resulting polymer of the following reaction. Draw the chain four monomers in length.
9.20: Solutions to Additional Exercises
9-1
9-2
9-3
9-4
9-5
9-6
9-7
9-8 B.
Hydration by Oxymercuration-Demercuration
9-9 When an alkene is going through oxymercuration, it proceeds through a three-membered mercurinium ion intermediate. This does not allow for rearrangement as no carbocation is formed. In order to open the intermediate ring, the water molecule will attack the most substituted carbon, thus giving the Markovnikov regioselectivity in the final product.
9-10
9-11
9-12
9-13 C.
Hydroboration of Alkenes
9-14
9-15
9-16 2-chloro-3-methylbutane
Addition of Halogens to Alkenes
9-17
9-18
9-19 B.
9-20
A: HBr, ROOR
B: NaNH2
9-21 You do not obtain a mixture of cis- and trans-brominated products from this reaction (only trans products) due to the intermediate the reaction goes through. No carbocation is formed, which would allow the Br- to attack from two possible sides of the carbocation. Instead, a bromonium ion is formed and in order to add the second Br, it needs to attack one side of the bromonium ion intermediate, causing the product to always have a trans configuration.
Formation of Halohydrins
9-22
9-23
9-24 Buta-1,3-diene
9-25
9-26
9-27
9-28
9-29
9-30
9-31
9-32
9-33
9-34
9-35
9-36
9-37
9-38 C.
9-39 A.
9-40
9-41
9-42
9-43
9-44
9-45
9-46
9-47
9-48
9-49
9-50 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/09%3A_Reactions_of_Alkenes/9.19%3A__Additional_Exercises.txt |
learning objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• apply bonding theories to the structure of alkynes and distinguish between internal and terminal triple bonds - refer to section 10.1
• predict relative physical properties of alkynes, such as relative boiling points and solubilities - refer to section 10.1
• predict the products and specify the reagents for the synthesis of alkynes from the double elimination of dihaloalkanes refer to section 10.2
• predict the products and specify the reagents for the Electrophilic Addition Reactions (EARs) of alkynes with HX and X2 - refer to section 10.3
• predict the products and specify the reagents for the Markovnikov-products of alkyne hydration - refer to section 10.4
• predict the products and specify the reagents for the anti-Markovnikov-products of alkyne hydration - refer to section 10.5
• predict the products and specify the reagents for the full or partial reduction of alkynes - refer to section 10.6
• predict the products and specify the reagents for the oxidation of alkynes - refer to section 10.7
• explain why alkynes are more acidic than alkanes and alkenes - refer to section 10.8
• predict the products and specify the reagents to generate nucleophilic acetylide ions and heavy metal acetylides - refer to section 10.8
• predict the products and specify the reagents to synthesize larger alkynes with acetylide ions - refer to section 10.9
• use retrosynthetic analysis to design a multi-step synthesis with correct regiochemistry and stereochemistry using the reactions studied to date - refer to section 10.10
Please note: IUPAC nomenclature and important common names of alkynes were explained in Chapter 3.
10: Alkynes
Learning Objective
• apply bonding theories to the structure of alkynes and distinguish between internal and terminal triple bonds
Alkynes: Terminal vs Internal
Alkynes are organic molecules with carbon-carbon triple bonds. They are unsaturated hydrocarbons with the empirical formula of CnH2n-2. The simplest alkyne is ethyne which has the common name acetylene. Acetylene is a common name to memorize.
It is important to distinguish between terminal and internal alkynes because they can undergo different patterns of reactivity.
Electronic Structure
The sp hybridization of the carbon-carbon triple bond results in the perpendicular orientation of the sigma bond and two pi bonds. The close proximity of the electrons in this geometry orientation creates molecules with less stability. The structure of the carbon-carbon triple bond strongly influences the chemical reactivity of alkynes and the acidity of terminal alkynes. Because of its linear configuration (the bond angle of a sp-hybridized carbon is 180º), a ten-membered carbon ring is the smallest that can accommodate this function without excessive strain.
Physical Properties
Alkynes are nonpolar, unsaturated hydrocarbons with physical properties similar to alkanes and alkenes. Alkynes dissolve in organic solvents, have slight solubility in polar solvents, and are insoluble in water. Compared to alkanes and alkenes, alkynes have slightly higher boiling points. For example, ethane has a boiling point of -88.6 C, while ethene is -103.7 C and ethyne has a higher boiling point of -84.0 ?C.
Exercise
1. Arrange ethane, ethene, and acetylene in order of decreasing carbon-carbon length.
2. How many pi bonds and sigma bonds are involved in the structure of ethyne?
3. What contribute to the weakness of the pi bonds in an alkyne?
4. Arrange the following hydrocarbons in order of decreasing boiling point: 1-heptyne, 1-hexyne, 2-methyl-1-hexyne.
5. Predict the solvent with greater 2-butyne solubility. a) water or 1-octanol? b) water or acetone? c) ethanol or hexane?
Answer
1. relative carbon-carbon bond length: ethane < ethene < acetylene
2. There are three sigma bonds and two pi bonds.
3. The sigma bond and two pi bonds are all perpendicular to each other in the triple bond creating electron repulsion between the three pairs of bonding electrons in the triple bond.
4. 1-heptyne (99.7C) > 2-methyl-1-hexyne (91C) > 1-hexyne (71C)
5. a) 1-octanol b) acetone c) hexane
• www.ucc.ie/academic/chem/dolc...t/alkynes.html
• www.cliffsnotes.com/WileyCDA/...eId-22631.html
Contributors and Attributions
• Bao Kha Nguyen, Garrett M. Chin | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/10%3A_Alkynes/10.01%3A_Structure_and_Physical_Properties.txt |
Introduction
To synthesize alkynes from dihaloalkanes we use dehydrohalogenation. The majority of these reactions take place using alkoxide bases (other strong bases can also be used) with high temperatures. This combination results in the majority of the product being from the E2 mechanism. Recall that the E2 mechanism is a concerted reaction (occurs in 1 step). However, in this 1 step there are 3 different changes in the molecule. This is the reaction between 2-bromo-2-methylpropane and sodium hydroxide.
Now, if we apply this concept using 2 halides on vicinal or geminal carbons, the E2 reaction will take place twice resulting in the formation of 2 pi bonds and an alkyne as shown in the examples below where the strong base is symbolized B-.
Double E2 of a Vicinal Dihalide Double E2 of a Geminal Dihalilde
It is important to note that the reaction of terminal haloalkanes requires 3 equivalents of base instead of 2 because of the relative acidity of alkynes that is discussed in a later section of this chapter.
The mechanism of a reaction between 2,3-dibromopentane with sodium amide in liquid ammonia is shown below where liquid ammonia is not part of the reaction, but is used as a solvent.
Notice the intermediate of the alkyne synthesis. It is stereospecifically in its anti form. Because the second proton and halogen are pulled off the molecule this is unimportant to the synthesis of alkynes.
Preparation of Alkynes from Alkenes
Lastly, we will briefly look at how to prepare alkynes from alkenes. This is a simple process using first halogenation of the alkene bond to form the dihaloalkane, and next, using the double elimination process form the alkyne.
This first process is gone over in much greater detail in the page on halogenation of an alkene. In general, chlorine or bromine is used with an inert halogenated solvent like chloromethane to create a vicinal dihalide from an alkene. The vicinal dihalide formed is then reacted with a strong base and heated to produce an alkyne. The two-step reaction pathway is shown below.
In The Lab
Due to the strong base and high temperatures needed for this reaction to take place, the triple bond may change positions. An example of this is when reactants that should form a terminal alkyne, form a 2-alkyne instead. The use of NaNH2 in liquid NH3 is used in order to prevent this from happening due to its lower reacting temperature. Even so, most chemists will prefer to use nucleophilic substitution instead of elimination when trying to form a terminal alkyne.
Exercise \(1\)
Question 1: Why would we need 3 bases for every terminal dihaloalkane instead of 2 in order to form an alkyne?
Question 2: What are the major products of the following reactions:
a.) 1,2-Dibromopentane with sodium amide in liquid ammonia
b.) 1-Pentene first with Br2 and chloromethane, followed by sodium ethoxide (Na+ -O-CH2CH3)
Question 3: What would be good starting molecules for the synthesis of the following molecules:
Question 4: Use a 6 carbon diene to synthesize a 6 carbon molecule with 2 terminal alkynes.
Answer
Answer 1: Remember that hydrogen atoms on terminal alkynes make the alkyne acidic. One of the base molecules will pull off the terminal hydrogen instead of one of the halides like we want.
Answer 2:
a.) 1-Pentyne
b.) 1-Pentyne
Answer 3:
Answer 4: Bromine or chlorine can be used with different inert solvents for the halogenation. This can be done using many different bases. Liquid ammonia is used as a solvent and needs to be followed by an aqueous work-up. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/10%3A_Alkynes/10.02%3A_10.2_Synthesis_of_Alkynes_-_Elimination_Reactions_of_Dihalides.txt |
Learning Objective
• predict the products and specify the reagents for the Electrophilic Addition Reactions (EARs) of alkynes with HX and X2
/*<![CDATA[*/ MathJax.Hub.Config({ extensions: ['TeX/AMSmath.js', 'TeX/AMSsymbols.js', 'TeX/color.js', 'TeX/noUndefined.js', 'TeX/cancel.js', 'TeX/mhchem.js', 'TeX/noErrors.js'], tex2jax: { inlineMath: [ ['$','$'], ["\$","\$"] ], displayMath: [ ['$','$'], ["\$","\$"] ]}, processUpdateDelay: 0, processUpdateTime: 0, showProcessingMessages: false, jax: ['input/TeX', 'output/HTML-CSS'], displayAlign: 'left', displayIndent: '20px', showMathMenu: true, }); /*]]>*/Addition by Electrophilic Reagents
Since the most common chemical transformation of a carbon-carbon double bond is an addition reaction, we might expect the same to be true for carbon-carbon triple bonds. Indeed, most of the alkene addition reactions also take place with alkynes with similar regio- and stereoselectivity.
When the addition reactions of electrophilic reagents, such as strong Brønsted acids and halogens, to alkynes are studied we find a curious paradox. The reactions are even more exothermic than the additions to alkenes, and yet the rate of addition to alkynes is slower by a factor of 100 to 1000 than addition to equivalently substituted alkenes. The reaction of one equivalent of bromine with 1-penten-4-yne, for example, gave 4,5-dibromo-1-pentyne as the chief product.
HC≡C-CH2-CH=CH2 + Br2 → HC≡C-CH2-CHBrCH2Br
Although these electrophilic additions to alkynes are sluggish, they do take place and generally display Markovnikov Rule regioselectivity and anti-stereoselectivity. One problem, of course, is that the products of these additions are themselves substituted alkenes and can therefore undergo further addition. Because of their high electronegativity, halogen substituents on a double bond act to reduce its nucleophilicity, and thereby decrease the rate of electrophilic addition reactions. Consequently, there is a delicate balance as to whether the product of an initial addition to an alkyne will suffer further addition to a saturated product. Although the initial alkene products can often be isolated and identified, they are commonly present in mixtures of products and may not be obtained in high yield. The following reactions illustrate many of these features. In the last example, 1,2-diodoethene does not suffer further addition inasmuch as vicinal-diiodoalkanes are relatively unstable.
As a rule, electrophilic addition reactions to alkenes and alkynes proceed by initial formation of a pi-complex, in which the electrophile accepts electrons from and becomes weakly bonded to the multiple bond. Such complexes are formed reversibly and may then reorganize to a reactive intermediate in a slower, rate-determining step. Reactions with alkynes are more sensitive to solvent changes and catalytic influences than are equivalent alkenes.
Why are the reactions of alkynes with electrophilic reagents more sluggish than the corresponding reactions of alkenes? After all, addition reactions to alkynes are generally more exothermic than additions to alkenes, and there would seem to be a higher π-electron density about the triple bond ( two π-bonds versus one ). Two factors are significant in explaining this apparent paradox. First, although there are more π-electrons associated with the triple bond, the sp-hybridized carbons exert a strong attraction for these π-electrons, which are consequently bound more tightly to the functional group than are the π-electrons of a double bond. This is seen in the ionization potentials of ethylene and acetylene.
Acetylene HC≡CH + Energy → [HC≡CH •(+) + e(–) ΔH = +264 kcal/mole
Ethylene H2C=CH2 + Energy → [H2C=CH2] •(+) + e(–) ΔH = +244 kcal/mole
Ethane H3C–CH3 + Energy → [H3C–CH3] •(+) + e(–) ΔH = +296 kcal/mole
As defined by the preceding equations, an ionization potential is the minimum energy required to remove an electron from a molecule of a compound. Since pi-electrons are less tightly held than sigma-electrons, we expect the ionization potentials of ethylene and acetylene to be lower than that of ethane, as is the case. Gas-phase proton affinities show the same order, with ethylene being more basic than acetylene, and ethane being less basic than either. Since the initial interaction between an electrophile and an alkene or alkyne is the formation of a pi-complex, in which the electrophile accepts electrons from and becomes weakly bonded to the multiple bond, the relatively slower reactions of alkynes becomes understandable.
A second factor is presumed to be the stability of the carbocation intermediate generated by sigma-bonding of a proton or other electrophile to one of the triple bond carbon atoms. This intermediate has its positive charge localized on an unsaturated carbon, and such vinyl cations are less stable than their saturated analogs. Indeed, we can modify our earlier ordering of carbocation stability to include these vinyl cations in the manner shown below. It is possible that vinyl cations stabilized by conjugation with an aryl substituent are intermediates in HX addition to alkynes of the type Ar-C≡C-R, but such intermediates are not formed in all alkyne addition reactions.
Carbocation
Stability
CH3(+) RCH=CH(+) < RCH2(+) RCH=CR(+) < R2CH(+) CH2=CH-CH2(+) < C6H5CH2(+) R3C(+)
Methyl 1°-Vinyl 2°-Vinyl 1°-Allyl 1°-Benzyl
Application of the Hammond postulate indicates that the activation energy for the generation of a vinyl cation intermediate would be higher than that for a lower energy intermediate. This is illustrated for alkenes versus alkynes by the following energy diagrams.
Despite these differences, electrophilic additions to alkynes have emerged as exceptionally useful synthetic transforms.
Addition of Hydrogen Halide to an Alkyne
Summary: Reactivity order of hydrogen halides: HI > HB r> HCl > HF.
Follows Markovnikov’s rule:
• Hydrogen adds to the carbon with the greatest number of hydrogens, the halogen adds to the carbon with fewest hydrogens.
• Protination occurs on the more stable carbocation. With the addition of HX, haloalkenes form.
• With the addition of excess HX, you get anti addition forming a geminal dihaloalkane.
Addition of a HX to an Internal Alkyne
As shown in Figure 2 below, the $\pi$ electrons react with the hydrogen of the HBr and because the alkyne carbons are equivalent it does not matter which carbon adds the hydrogen. Once the hydrogen is covalently bonded to one of the carbons, the bromide will react with the carbocation intermediate to form a vinyl halide as shown in the example of forming 2-bromobutene from 2-butyne reacting with HBr. The reaction below assumes a 1:1 mole ratio of the alkyne and HBr.
Now, what happens if there is excess HBr?
Addition due to excess HX yields a geminal dihaloalkane
Here, the electrophilic addition proceeds with the same steps used to achieve the product in Addition of a HX to an Internal Alkyne. The $\pi$ electrons react with the hydrogen (shown in blue) adding it to the carbon on the left because the lone pair electrons of the bromine can help stabilize the carbocation intermediate that reacts with the bromide ions to form a geminal dihalide.
Addition of HX to Terminal Alkyne
For terminal alkynes, the carbon atoms sharing the triple bond are not equivalent. The addition of HX to terminal alkynes occurs in a Markovnikov-manner in which the halide attaches to the most substituted carbon. The pi electrons react with the hydrogen and it bonds to the terminal carbon. The bromide reacts with the resulting carbocation intermediate to form the vinyl halide. The overall reaction and mechanism are shown below.
Addition due to excess HBr present
Similar to the addition of excess HBr to internal alkynes, both halides will add to the same carbon to form a geminal dihalide.
HBr Addition With Radical Mechanism
Most hydrogen halide reactions with terminal alkynes occur in a Markovnikov-manner in which the halide attaches to the most substituted carbon since it is the most positively polarized. However, there are two specific reactions among alkynes where anti-Markovnikov reactions take place: the radical addition of HBr and Hydroboration Oxidation reactions. For alkynes, an anti-Markovnikov addition takes place for terminal alkynes.
The Br of the Hydrogen Bromide (H-Br) attaches to the less substituted 1-carbon of the terminal alkyne shown below in an anti-Markovnikov manner while the Hydrogen proton attaches to the second carbon. As mentioned above, the first carbon is the less substituted carbon since it has fewer bonds attached to carbons and other substituents. The H-Br reagent must also be reacted with heat or some other radicial initiator such as a peroxide in order for this reaction to proceed in this manner. This presence of the radical or heat leads to the anti-Markovnikov addition since it produces the most stable reaction.
The product of a terminal alkyne that is reacted with a peroxide (or light) and H-Br is a 1-bromoalkene.
Regioselectivity: The Bromine can attach in a syn or anti manner which means the resulting alkene can be both cis and trans. Syn addition is when both Hydrogens attach to the same face or side of the double bond (i.e. cis) while the anti addition is when they attach on opposite sides of the bond (trans).
Halogenation of Alkynes
The additon of X2 to alkynes is analogous to the addition of X2 to alkenes. The halogen molecule becomes polarized by the approach of the nucleophilic alkyne. The pi electrons of the alkyne react with the bromine to form a carbon-bromine bond and cyclic halonium ion with halide as the leaving group. The formation of the cyclic halonium ion requires anti-addition of the nucleophilic halide to produce a vicinal dihalide alkene as shown in the reaction below.
Exercise
1. Draw the structure, and give the IUPAC name, of the product formed in each of the reactions listed below.
1. $\ce{\sf{CH3-C#C-CH3->[\displaystyle{\textrm{1 equiv}}][\displaystyle{\textrm{HCl}}]}}$
2. $\ce{\sf{CH3-C#C-CH3->[\displaystyle{\textrm{excess}}][\displaystyle{\textrm{HCl}}]}}$
3. $\ce{\sf{CH3-C#C-CH3->[\displaystyle{\textrm{1 equiv}}][\displaystyle{\textrm{Br}_2}]}}$
4. $\ce{\sf{CH3-C#C-CH3->[\displaystyle{\textrm{excess}}][\displaystyle{\textrm{Br}_2}]}}$
5. $\ce{\sf{CH3CH2-C#C-H->[\displaystyle{\textrm{1 equiv}}][\displaystyle{\textrm{HCl}}]}}$
6. $\ce{\sf{CH3CH2-C#C-H->[\displaystyle{\textrm{excess}}][\displaystyle{\textrm{HCl}}]}}$
7. $\ce{\sf{CH3CH2CH2-C#C-H->[\displaystyle{\textrm{1 equiv}}][\displaystyle{\textrm{Br}_2}]}}$
8. $\ce{\sf{CH3CH2CH2-C#C-H->[\displaystyle{\textrm{excess}}][\displaystyle{\textrm{Br}_2}]}}$
Answer
1. (Z)-2-chloro-2-butene
2. 2,2-dichlorobutane
3. (E)-2,3-dibromo-2-butene
4. 2,2,3,3-tetrabromobutane
5. 2-chloro-1-butene
6. 2,2-dichlorobutane
7. (E)-1,2-dibromo-1-pentene
8. 1,1,2,2-tetrabromopentane | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/10%3A_Alkynes/10.03%3A_Reactions_of_Alkynes_-_Addition_of_HX_and_X.txt |
Learning Objective
• predict the products and specify the reagents for the Markovnikov-products of alkyne hydration
Reaction: Hydration of Alkynes (Markovnikov's Rule)
Hydration of alkynes begins similar to the hydration of alkenes through the addition of the first water molecule. However, this first hydration reaction forms an enol, an alcohol bonded to a vinyl carbon. Enols immediately undergo a special type of isomerization reaction called tautomerization to form carbonyl groups - aldehydes or ketones. To keep things simple, this reaction is called "enol-keto" tautomerization with the understanding that aldehydes form on terminal alkyne carbons. As with alkenes, hydration (addition of water) to alkynes requires a strong acid, usually sulfuric acid with a mercuric sulfate catalyst as shown below.
Enol-Keto Tautomers
Tautomers are defined as rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom (colored red here) and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers ( acetone, for example, is 99.999% keto tautomer ). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples. The three examples shown below illustrate these reactions for different substitutions of the triple-bond. The tautomerization step is indicated by a red arrow. For terminal alkynes the addition of water follows the Markovnikov rule, as in the second example below, and the final product ia a methyl ketone ( except for acetylene, shown in the first example ). For internal alkynes ( the triple-bond is within a longer chain ) the addition of water is not regioselective. If the triple-bond is not symmetrically located ( i.e. if R & R' in the third equation are not the same ) two isomeric ketones will be formed.
With the addition of water, alkynes can be hydrated to form enols that spontaneously tautomerize to ketones. The reaction is catalyzed by mercury ions and follows Markovnikov’s Rule A useful functional group conversion for multiple -step syntheses is to hydrate terminal alkynes to produce methyl ketones.
Hydration of Alkyne Mechanism
The first step is an acid/base reaction where the π electrons of the triple bond acts as a Lewis base and reacts with the proton therefore protonating the carbon with the most hydrogen substituents as expected by Markovnikov's Rule. In the second step, the nucleophilic water molecule reacts with the electrophilic carbocation to produce an oxonium ion. The oxonium ion is deprotonated by a base to produce an enol which immediately tautomerizes into a ketone. The hydration reaction for propyne is shown below with its mechanism to illustrate the electron flow of the mechanism.
Exercise
1. Draw the structure of the product formed when each of the substances below is treated with H2O/H2SO4 in the presence of HgSO4.
1. CH3CHCCH
2. Draw the structure of the keto form of the compound shown below. Which form would you expect to be the most stable?
3. What alkyne would you start with to gain the following products, in an oxidation reaction? Keep in mind resonance.
4. Propose a reaction scheme for the following compound starting from the alkyne and showing required reagents and intermediates.
Answer
1.
2. ; The keto form should be the most stable.
3.
4.
10.05: Hydration of Alkynes for Anti-Markovnikov Products
Learning Objective
• predict the products and specify the reagents for the anti-Markovnikov-products of alkyne hydration
Introduction
The hydroboration-oxidation of alkynes is similar to the reaction with alkenes. However, there is one important difference. The alkyne has two pi bonds and both are capable of reacting with borane (BH3). To limit the reactivity to only one of the pi bonds within the alkyne, a dialkyl borane reagent (R2BH) is used. Replacing two of the hydrogens on the borane with alkyl groups also creates steric hindrance so that the hydroboration reaction produces the regioselective,
anti-Markovnikov product. Disiamylborane (Sia2BH) and 9-borabicyclo[3.3.1]nonane (9-BBN) are two common reagents for the hydroboration step. Their structures are shown below.
The oxidation reagents ( a basic hydrogen peroxide solution) are the same for both alkenes and alkynes, The overall reaction is shown below.
Mechanism
The hydroboration reaction of alkynes has the same stereo- and regiochemistry as the alkene reaction. The primary difference is the steric hindrance of the two isoamyl groups of the dialkyl borane creates anti-Markovnikov regeioselectivity. The hydrogen and boron bond with the same orientation to the alkyne carbon with syn-addition stereochemistry to form the enol. The enol immediately tautomerizes to the keto form which is an aldehyde for terminal alkynes. The hydration of 1-propyne is shown below along with the reaction mechanism.
Exercise
1. Draw the bond-line structure(s) for the product(s) of each reaction.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/10%3A_Alkynes/10.04%3A_Hydration_of_Alkynes_for_Markovnikov_Products.txt |
Learning Objective
• predict the products and specify the reagents for the full or partial reduction of alkynes
Introduction and Overview
Alkynes can undergo reduction reactions similar to alkenes. These reactions are also called hydrogenation reactions. With the presence of two pi bonds within the carbon-carbon triple bonds, the reduction reactions can be partial or complete depending on the reagents. Since partial reduction of an alkyne produces an alkene, the stereochemistry of the addition mechanism determines whether the cis- or trans- alkene is formed. The three most significant alkyne reduction reactions are summarized below.
Hydrogenation and the Relative Stability of Hydrocarbons
Like alkenes, alkynes readily undergo catalytic hydrogenation partially to cis- or trans- alkenes or fully to alkanes depending on the reaction employed.
The catalytic addition of hydrogen to 2-butyne provides heat of reaction data that reflect the relative thermodynamic stabilities of these hydrocarbons, as shown above. From the heats of hydrogenation, shown in blue in units of kcal/mole, it would appear that alkynes are thermodynamically less stable than alkenes to a greater degree than alkenes are less stable than alkanes. The standard bond energies for carbon-carbon bonds confirm this conclusion. Thus, a double bond is stronger than a single bond, but not twice as strong. The difference ( 63 kcal/mole ) may be regarded as the strength of the π-bond component. Similarly, a triple bond is stronger than a double bond, but not 50% stronger. Here the difference ( 54 kcal/mole ) may be taken as the strength of the second π-bond. The 9 kcal/mole weakening of this second π-bond is reflected in the heat of hydrogenation numbers ( 36.7 - 28.3 = 8.4 ).
Catalytic Hydrogenation of an Alkyne
Alkynes can be fully hydrogenated into alkanes with the help of a platinum, paladium, or nickel catalyst. Because the reaction is catalyzed on the surface of the metal, it is common for these catalysts to dispersed on carbon (Pd/C) or finely dispersed as nickel (Raney-Ni). The full reduction of 2-butyne is shown below as an example.
Hydrogenation of an Alkyne to a Cis-Alkene
Since alkynes are thermodynamically less stable than alkenes, we expect addition reactions of alkynes to be more exothermic and relatively faster than equivalent reactions of alkenes. For catalytic hydrogenation, the Pt, Pd, or Ni catalysts are so effective in promoting addition of hydrogen to both double and triple carbon-carbon bonds that the alkene intermediate formed by hydrogen addition to an alkyne cannot be isolated. A less efficient catalyst, Lindlar's catalyst permits alkynes to be converted to alkenes without further reduction to an alkane. Lindlar’s Catalyst transforms an alkyne to a cis-alkene because the hydrogenation reaction is occurring on the surface of the metal. Both hydrogen atoms are added to the same side of the alkyne as shown in the syn-addition mechanism for hydrogenation of alkenes in the previous chapter.
Lindlar's catalyst is prepared by deactivating (or poisoning) a conventional palladium catalyst. Lindlar’s catalyst has three components: palladium-calcium carbonate, lead acetate and quinoline. The quinoline serves to prevent complete hydrogenation of the alkyne to an alkane. This approach is similar to the one used for hydration of alkynes using a dialkyl borane for hydroboration. A strong reagent is modified into a less reactive form.
Hydrogenation of an Alkyne to a Trans-Alkene
Alkynes can be reduced to trans-alkenes with the use of sodium dissolved in an ammonia solvent. A sodium radical donates an electron to one of the p-orbitals in the carbon-carbon triple bond. This reaction forms an anion that can be protonated by a hydrogen atom in the ammonia solvent which prompts another sodium radical to donate an electron to the second p-orbital. The resulting anion is also protonated by a hydrogen from the ammonia solvent to produce a trans-alkene according to the mechanism shown below.
Mechanism for Hydrogenation of Alkynes to trans-Alkenes
Exercise
1. Using any alkyne how would you prepare the following compounds: pentane, trans-4-methyl-2-pentene, cis-4-methyl-2-pentene.
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/10%3A_Alkynes/10.06%3A_10.6_Reduction_of_Alkynes.txt |
Learning Objective
• predict the products and specify the reagents for the oxidation of alkynes
Alkynes, similar to alkenes, can be oxidized gently or strongly depending on the reaction environment. Since alkynes are less stable than alkenens, the reactions conditions can be gentler. For examples, alkynes form vicinal dicarbonyls in neutral permanganate solution. For the alkene reaction to vicinal dialcohols, the permanganate reaction requires a lightly basic environment for the reaction to occur. During strong oxidation with ozone or basic potassium permanganate, the alkyne is cleaved into two products. Because at least one of the reaction products is a carboxylic acid, it is important to consider the acid-base chemistry of the product in the reaction solution. Carboxylic acids are deprotonated in basic solutions to carboxylates. A second reaction step is required to protonate the carboxylate to the neutral form of the carboxylic acid. The generic reactions are summarized below for the different oxidative conditions - gentle or strong.
Gentle Alkyne Oxidation
Strong Alkyne Oxidation - Oxidative Cleavage
Exercise
1. Draw the bond-line structures for the product(s) of the following reactions.
Answer
1.
10.08: Acidity of Terminal Alkynes and Acetylide Ions
Learning Objectives
• explain why alkynes are more acidic than alkanes and alkenes
• predict the products and specify the reagents to generate nucleophilic acetylide ions and heavy metal acetylides
Acidity of Terminal Alkynes and Acetylilde Ion Formation
Terminal alkynes are much more acidic than most other hydrocarbons. Removal of the proton leads to the formation of an acetylide anion, RC=C:-. The origin of the enhanced acidity can be attributed to the stability of the acetylide anion, which has the unpaired electrons in an sp hybridized orbital. The stability results from occupying an orbital with a high degree of s-orbital character. There is a strong correlation between s-character in the orbital containing the non-bonding electrons in the anion and the acidity of hydrocarbons. The enhanced acidity with greater s-character occurs despite the fact that the homolytic C-H BDE is larger.
Table 9.7.1: Akynes
Compound Conjugate Base Hybridization "s Character" pKa C-H BDE (kJ/mol)
CH3CH3 CH3CH2- sp3 25% 50 410
CH2CH2 CH2CH- sp2 33% 44 473
HCCH HCC- sp 50% 25 523
Consequently, acetylide anions can be readily formed by deprotonation using a sufficiently strong base. Amide anion (NH2-), in the form of NaNH2 is commonly used for the formation of acetylide anions.
Exercise
1. Given that the pKa of water is 14.00, would you expect hydroxide ion to be capable of removing a proton from each of the substances listed below? Justify your answers, briefly.
1. ethanol (pKa = 16)
2. acetic acid (pKa = 4.72)
3. acetylene (pKa = 25)
Answer
Answers:
1.
1. No, The pKa of ethanol is similar to the pKa of water so proton exchange is comparable for both protonation and deprotonation between alcohols and water. Alcohols can be considered "alkylated water" and share many similarities in both physical properties and chemical reactivity.
2. Yes, very well. There is a difference of 11 pKa units between the pKa of water and the pKa of acetic acid. The equilibrium lies well to the right with acetate as the predominate form of the original acetic acid.
3. No, hardly at all. The hydroxide ion is too weak a base to remove a proton from acetylene. The equilibrium lies so far to the left that it is considered a "No Reaction".
10.09: Synthesis of Larger Alkynes from Acetylides
Learning Objective
• predict the products and specify the reagents to synthesize larger alkynes with acetylide ions
Nucleophilic Substitution Reactions of Acetylides
Acetylide anions are strong bases and strong nucleophiles. Therefore, they are able to displace halides and other leaving groups in substitution reactions. The product is a substituted alkyne with a longer, continuous carbon chain.
Because the ion is a very strong base, the substitution reaction follows the SN2 mechanism and is most efficient with methyl or primary halides without substitution near the reaction center.
Secondary, tertiary or even bulky primary substrates will give elimination by the E2 mechanism.
Nucleophilic Addition of Acetylides to Carbonyls
Acetylide anions will add to aldehydes and ketones to form alkoxides that are subsequently protonated to form propargyl alcohols.
With aldehydes and non-symmetric ketones, in the absence of chiral catalyst, the product will be a racemic mixture of the two enantiomers.
Exercise
1. The pKa of ammonia is 35. Estimate the equilibrium constant for the deprotonation of pent-1-yne by amide, as shown above.
2. Give the possible reactants for the following formations:
3. Propose a synthetic route to produce 2-pentene from propyne and an alkyl halide.
Answer
1. Assuming the pKa of pent-1-yne is about 25, then the difference in pKas is 10. Since pentyne is more acidic, the formation of the acetylide will be favored at equilibrium, so the equilibrium constant for the reaction is about 1010.
2.
3. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/10%3A_Alkynes/10.07%3A_Oxidation_of_Alkynes.txt |
Learning Objective
• use retrosynthetic analysis to design a multi-step synthesis with correct regiochemistry and stereochemistry using the reactions studied to date
Introduction
The study of organic chemistry introduces students to a wide range of interrelated reactions. Alkenes, for example, may be converted to structurally similar alkanes, alcohols, alkyl halides, epoxides, glycols and boranes; cleaved to smaller aldehydes, ketones and carboxylic acids; and enlarged by carbocation and radical additions as well as cycloadditions. Most of these reactions are shown in the Alkene Reaction Map below. All of these products may be subsequently transformed into a host of new compounds incorporating a wide variety of functional groups. Consequently, the logical conception of a multi-step synthesis for the construction of a designated compound from a specified starting material becomes one of the most challenging problems that may be posed. Functional group reaction maps like the one below for alkenes can be helpful in designing multi-step syntheses. It can be helpful to build and design your own reaction maps for each functional group studied.
Alkene Reaction Map
Please note: The reagents for each chemical transformation have been intentionally omitted so that this map can be used as a study tool. The answers are provided at the end of this section as part of the exercises.
Simple Multi-Step Syntheses
A one or two step sequence of simple reactions is not that difficult to deduce. For example, the synthesis of meso-3,4-hexanediol from 3-hexyne can occur by more than one multi-step pathway.
One approach would be to reduce the alkyne to cis or trans-3-hexene before undertaking glycol formation. Permanaganate or osmium tetroxide hydroxylation of cis-3-hexene would form the desired meso isomer.
From trans-3-hexene, it would be necessary to first epoxidize the alkene with a peracid followed by ring opening with acidic or basic hydrolysis.
Longer multi-step syntheses require careful analysis and thought, since many options need to be considered. Like an expert chess player evaluating the long range pros and cons of potential moves, the chemist must appraise the potential success of various possible reaction paths, focusing on the scope and limitations constraining each of the individual reactions being employed. The skill is acquired by practice, experience, and often trial and error.
Thinking it Through with 3 Examples
The following three examples illustrate strategies for developing multi-step syntheses from the reactions studied in the first ten chapters of this text. It is helpful to systematically look for structural changes beginning with the carbon chain and brainstorm relevant functional group conversion reactions. Retro-synthesis is the approach of working backwards from the product to the starting material.
In the first example, we are asked to synthesize 1-butanol from acetylene.
The carbon chain doubles in size indicating an acetylide SN2 reaction with an alkyl halide. Primary alcohol formation from an anti-Markovnikov alkene hydration reaction (hydroboration-oxidation) is more likely than a substitution reaction. Applying retro-synthesis, we work backwards from the alcohol to the alkene to the alkyne from an acetylide reaction that initially builds the carbon chain.
Retro-Synthesis
Working forwards, we specify the reagents needed for each transformation identified from the retro-synthesis. The ethylbromide must also be derived from acetylene so multiple reaction pathways are combined as shown below.
In the second example, we are asked to synthesize 1,2-dibromobutane from acetylene.
Once again there is an increase in the carbon chain length indicating an acetylide SN2 reaction with an alkyl halide similar to the first example. The hydrohalogenation can be subtle to discern because the hydrogen atoms are not shown in bond-line structures. Comparing the chemical formulas of 1-butyne with 1,2-dibromobutane, there is a difference of two H atoms and two Br atoms indicating hydrohalogenation and not halogenation. The addition of both bromine atoms to the same carbon atom also supports the idea that hydrohalogenation occurs on an alkyne and not an alkene. The formation of the geminal dihalide also indicates hydrohalogenation instead of halogenatioin because halogenation produces vicinal dihalides. With this insight, the retro-synthesis indicates the following series of chemical transformations.
Retro-Synthesis
Working forwards, we specify the reagents needed for each transformation.
In the third example, we are asked to produce 6-oxoheptanal from methylcyclohexane.
Counting the carbons, the starting material and product both contain seven carbon atoms and there is a cleavage reaction of an alkene under reductive conditions. One important missing aspect of this reaction is a good leaving group (LG). Alkanes are chemically quite boring. We can burn them as fuel or perform free-radical halogenation to create alkyl halides with excellent leaving groups. With these observations, the following retro-synthesis is reasonable.
Retro-Synthesis
Working forwards, we specify the reagents needed for each reaction. For the initial free-radical halogenation of the alkane, we have the option of chlorine (Cl2) or bromine (Br2). Because methylcyclohexane has several different classifications of carbons, the selectivity of Br2 is more important than the faster reactivity of Cl2. An strong base with heat can be used for the second step to follow an E2 mechanism and form 1-methylcyclohexene. The aldhyde group on the final product indicates gentle oxidative cleavage by any of several reaction pathways. These reactions can be combined in to the following multi-step synthesis.
Reaction Maps to Build Functional Group Conversion Mastery
After working through the examples above, we can see how important it is to memorize all of the functional group reactions studied in the first ten chapters. We can apply the knowledge of these reactions to the wisdom of multi-step syntheses.
Please note: The reagents for each chemical transformation have been intentionally omitted so that these maps can be used as a study tools. The answers are provided at the end of this section as part of the exercises.
Alkyne Reaction Map
Exercise
1. Starting at 3-hexyne predict synthetic routes to achieve:
a) trans-3-hexene
b) 3,4-dibromohexane
c) 3-hexanol.
2. Starting with acetylene and any alkyl halides propose a synthesis to make
a) pentanal
b) hexane.
Answer
1.
2.
10.11: Additional Exercises
Alkyne Reactions
10-1 Predict the product of these following reactions:
a)
b)
c)
d)
10-2 Using acetylene as the starting material, show how you would synthesize the following compounds
a)
b) but-2-yne
c)
d)
10-3 Identify the reagents needed to turn hex-1-yne into the following compounds
a) hexane
b) oct-3-yne
c) cis-hept-2-ene
d) trans-hept-2-ene
e) 2,2-dibromohexane
f) 1-bromohexene
10-4 Show how you would accomplish the following synthetic transformations.
a)
b)
c)
d)
e)
f)
10-5 Deduce the structure of each unknown from the information given.
a) Upon catalytic hydrogenation, unknown A yields pentane. Ozonolysis of A yields butanoic acid, HOOC(CH2)2CH3 and CO2. Draw the structure of compound A
b) Upon catalytic hydrogenation, unknown B yields pentane. Ozonolysis of B yields acetaldehyde, CH3CHO, and propionaldehyde, CH3CH2CHO.
10-6 Use compound A from the previous problem (10-5) and any additional reagents you may need to synthesize the following compound.
10.12: Solutions to Additional Exercises
10-1
a)
b)
c)
d)
10-2
a)
b)
c)
d)
10-3
a)
b)
c)
d)
e)
f)
10-4
a)
b)
c)
d)
e)
f)
10-5
a)
b)
10-6 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/10%3A_Alkynes/10.10%3A_An_Introduction_to_Multiple_Step_Synthesis.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• correlate regions of the electromagnetic spectrum to spectroscopic techniques - refer to section 11.1
• explain how an IR spectrometer works and the IR region interacts with organic compounds - refer to section 11.2
• explain the role of asymmetry in IR absorption - refer to section 11.3
• interpret IR spectra - refer to section 11.4, 11.5, and 11.6
• expalin how a mass spectrometer works - refer to section 11.7
• explain the source of the base peak and molecular ion in a mass spectrum - refer to section 11.7
• correlate bond strength to fragmentation patterns - refer to section 11.8
• use fragmentation patterns to elucidate structural features of organic compounds - refer to section 11.9
• explain how high-resolution mass can be used to determine chemical formulas - refer to section 11.10
• 11.1: The Electromagnetic Spectrum and Spectroscopy
Spectroscopy is an experimental method used by chemists to elucidate structural information. The interaction between a compound or sample and a selected region of the electromagnetic spectrum can be measured both qualitatively and quantitatively.
• 11.2: Infrared (IR) Spectroscopy
The infrared region of the electromagnetic spectrum causes asymmetric bonds to stretch, bend, and/or vibrate. This interaction can be measured to help elucidate chemical structures.
• 11.3: IR-Active and IR-Inactive Vibrations
Asymmetry and polarity increase the strength of IR absorption (infrared active). Symmetrical carbon-carbon double and triple bonds will not absorb IR light and are called "infrared inactive".
• 11.4: Interpretting IR Spectra
The analysis and interpretation of the IR spectra for several compounds are explained.
• 11.5: Infrared Spectra of Some Common Functional Groups
One of the most common applications of infrared spectroscopy is the identification of organic compounds. The IR spectra for the major classes of organic molecules are shown and discussed.
• 11.6: Summary and Tips to Distinguish between Carbonyl Functional Groups
This summary includes the minimum information that needs to be memorized to interpret IR spectra in first year organic chemistry along with some tips on how to distinguish between the different functional groups that all contain at least one carbonyl structural feature.
• 11.7: Mass Spectrometry - an introduction
Mass spectrometry is an analytic method that employs ionization and mass analysis of compounds in order to determine the mass, formula and structure of the compound being analyzed. A mass analyzer is the component of the mass spectrometer that takes ionized masses and separates them based on charge to mass ratios and outputs them to the detector where they are detected and later converted to a digital output.
• 11.8: Fragmentation Patterns in Mass Spectrometry
When interpreting fragmentation patterns, as you might expect, the weakest carbon-carbon bonds are the ones most likely to break.
• 11.9: Useful Patterns for Structure Elucidation
Pattern recognition is a chemistry student's best friend, especially when analyzing and interpreting mass spectra.
• 11.10: Determination of the Molecular Formula by High Resolution Mass Spectrometry
High resolution mass spectrometry can determine molecular formulas by distinguishing between the masses of compounds based on the isotopic distribution of each element in the compound.
11: Infrared Spectroscopy and Mass Spectrometry
Objectives
After completing this section, you should be able to
1. write a brief paragraph discussing the nature of electromagnetic radiation.
2. write the equations that relate energy to frequency, frequency to wavelength and energy to wavelength, and perform calculations using these relationships.
3. describe, in general terms, how absorption spectra are obtained.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• electromagnetic radiation
• electromagnetic spectrum
• hertz (Hz)
• infrared spectroscopy
• photon
• quantum
Study Notes
From your studies in general chemistry or physics, you should be familiar with the idea that electromagnetic radiation is a form of energy that possesses wave character and travels through space at a speed of 3.00 × 108m · s−1. However, such radiation also displays some of the properties of particles, and on occasion it is more convenient to think of electromagnetic radiation as consisting of a stream of particles called photons.
In spectroscopy, the frequency of the electromagnetic radiation being used is usually expressed in hertz (Hz), that is, cycles per second. Note that 1 Hz = s−1.
A quantum is a small, definite quantity of electromagnetic radiation whose energy is directly proportional to its frequency. (The plural is “quanta.”) If you wish, you can read about the properties of electromagnetic radiation and the relationships among wavelength, frequency and energy, or refer to your general chemistry textbook if you still have it.
Note also that in SI units, Planck’s constant is 6.626 × 10−34J · s.
The electromagnetic spectrum
Electromagnetic radiation, as you may recall from a previous chemistry or physics class, is composed of electrical and magnetic waves which oscillate on perpendicular planes. Visible light is electromagnetic radiation. So are the gamma rays that are emitted by spent nuclear fuel, the x-rays that a doctor uses to visualize your bones, the ultraviolet light that causes a painful sunburn when you forget to apply sun block, the infrared light that the army uses in night-vision goggles, the microwaves that you use to heat up your frozen burritos, and the radio-frequency waves that bring music to anybody who is old-fashioned enough to still listen to FM or AM radio.
Just like ocean waves, electromagnetic waves travel in a defined direction. While the speed of ocean waves can vary, however, the speed of electromagnetic waves – commonly referred to as the speed of light – is essentially a constant, approximately 300 million meters per second. This is true whether we are talking about gamma radiation or visible light. Obviously, there is a big difference between these two types of waves – we are surrounded by the latter for more than half of our time on earth, whereas we hopefully never become exposed to the former to any significant degree. The different properties of the various types of electromagnetic radiation are due to differences in their wavelengths, and the corresponding differences in their energies: shorter wavelengths correspond to higher energy.
High-energy radiation (such as gamma- and x-rays) is composed of very short waves – as short as 10-16 meter from crest to crest. Longer waves are far less energetic, and thus are less dangerous to living things. Visible light waves are in the range of 400 – 700 nm (nanometers, or 10-9 m), while radio waves can be several hundred meters in length.
The notion that electromagnetic radiation contains a quantifiable amount of energy can perhaps be better understood if we talk about light as a stream of particles, called photons, rather than as a wave. (Recall the concept known as ‘wave-particle duality’: at the quantum level, wave behavior and particle behavior become indistinguishable, and very small particles have an observable ‘wavelength’). If we describe light as a stream of photons, the energy of a particular wavelength can be expressed as:
$E = \dfrac{hc}{\lambda} \tag{12.5.1}$
where E is energy in kcal/mol, λ (the Greek letter lambda) is wavelength in meters, c is 3.00 x 108 m/s (the speed of light), and h is 9.537 x 10-14 kcal•s•mol-1, a number known as Planck’s constant.
Because electromagnetic radiation travels at a constant speed, each wavelength corresponds to a given frequency, which is the number of times per second that a crest passes a given point. Longer waves have lower frequencies, and shorter waves have higher frequencies. Frequency is commonly reported in hertz (Hz), meaning ‘cycles per second’, or ‘waves per second’. The standard unit for frequency is s-1.
When talking about electromagnetic waves, we can refer either to wavelength or to frequency - the two values are interconverted using the simple expression:
$\lamabda \nu = c \tag{12.5.2}$
where ν (the Greek letter ‘nu’) is frequency in s-1. Visible red light with a wavelength of 700 nm, for example, has a frequency of 4.29 x 1014 Hz, and an energy of 40.9 kcal per mole of photons. The full range of electromagnetic radiation wavelengths is referred to as the electromagnetic spectrum.
Notice in the figure above that visible light takes up just a narrow band of the full spectrum. White light from the sun or a light bulb is a mixture of all of the visible wavelengths. You see the visible region of the electromagnetic spectrum divided into its different wavelengths every time you see a rainbow: violet light has the shortest wavelength, and red light has the longest.
Example
Visible light has a wavelength range of about 400-700 nm. What is the corresponding frequency range? What is the corresponding energy range, in kcal/mol of photons?
Solution
Molecular spectroscopy – the basic idea
In a spectroscopy experiment, electromagnetic radiation of a specified range of wavelengths is allowed to pass through a sample containing a compound of interest. The sample molecules absorb energy from some of the wavelengths, and as a result jump from a low energy ‘ground state’ to some higher energy ‘excited state’. Other wavelengths are not absorbed by the sample molecule, so they pass on through. A detector on the other side of the sample records which wavelengths were absorbed, and to what extent they were absorbed.
Here is the key to molecular spectroscopy: a given molecule will specifically absorb only those wavelengths which have energies that correspond to the energy difference of the transition that is occurring. Thus, if the transition involves the molecule jumping from ground state A to excited state B, with an energy difference of ΔE, the molecule will specifically absorb radiation with wavelength that corresponds to ΔE, while allowing other wavelengths to pass through unabsorbed.
By observing which wavelengths a molecule absorbs, and to what extent it absorbs them, we can gain information about the nature of the energetic transitions that a molecule is able to undergo, and thus information about its structure. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/11%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.01%3A_The_Electromagnetic_Spectrum_and_Spectroscopy.txt |
Introduction
Photon energies associated with the infrared (from 1 to 15 kcal/mole) are not large enough to excite electrons, but may induce vibrational excitation of covalently bonded atoms and groups.
The covalent bonds in molecules are not rigid sticks or rods, such as found in molecular model kits, but are more like stiff springs that can be stretched and bent. The mobile nature of organic molecules was noted in the chapter concerning conformational isomers. We must now recognize that, in addition to the facile rotation of groups about single bonds, molecules experience a wide variety of vibrational motions, characteristic of their component atoms. Consequently, virtually all organic compounds will absorb infrared radiation that corresponds in energy to these vibrations. Infrared spectrometers, similar in principle to the UV-Visible spectrometer described elsewhere, permit chemists to obtain absorption spectra of compounds that are a unique reflection of their molecular structure.
Vibrational Spectroscopy
A molecule composed of n-atoms has 3n degrees of freedom, six of which are translations and rotations of the molecule itself. This leaves 3n-6 degrees of vibrational freedom (3n-5 if the molecule is linear). Vibrational modes are often given descriptive names, such as stretching, bending, scissoring, rocking and twisting. The four-atom molecule of formaldehyde, the gas phase spectrum of which is shown below, provides an example of these terms. If a ball & stick model of formaldehyde is not displayed to the right of the spectrum, press the view ball&stick model button on the right. We expect six fundamental vibrations (12 minus 6), and these have been assigned to the spectrum absorptions. To see the formaldehyde molecule display a vibration, click one of the buttons under the spectrum, or click on one of the absorption peaks in the spectrum.
Covalent bonds in organic molecules are not rigid sticks – rather, they behave more like springs. At room temperature, organic molecules are always in motion, as their bonds stretch, bend, and twist. These complex vibrations can be broken down mathematically into individual vibrational modes, a few of which are illustrated below.
The energy of molecular vibration is quantized rather than continuous, meaning that a molecule can only stretch and bend at certain 'allowed' frequencies. If a molecule is exposed to electromagnetic radiation that matches the frequency of one of its vibrational modes, it will in most cases absorb energy from the radiation and jump to a higher vibrational energy state - what this means is that the amplitude of the vibration will increase, but the vibrational frequency will remain the same. The difference in energy between the two vibrational states is equal to the energy associated with the wavelength of radiation that was absorbed. It turns out that it is the infrared region of the electromagnetic spectrum which contains frequencies corresponding to the vibrational frequencies of organic bonds.
An IR Spectrum
We will use a ketone sample to illustrate this process. The sample is irradiated with infrared light and the carbonyl bond will specifically absorb light with this same frequency, which by equations 4.1 and 4.2 corresponds to a wavelength of 5.83 x 10-6 m and an energy of 4.91 kcal/mol. When the carbonyl bond absorbs this energy, it jumps up to an excited vibrational state.
The value of ΔE - the energy difference between the low energy (ground) and high energy (excited) vibrational states - is equal to 4.91 kcal/mol, the same as the energy associated with the absorbed light frequency. The molecule does not remain in its excited vibrational state for very long, but quickly releases energy to the surrounding environment in form of heat, and returns to the ground state.
With an instrument called an infrared spectrophotometer, we can 'see' this vibrational transition. In the spectrophotometer, infrared light with frequencies ranging from about 1013 to 1014 Hz is passed though our sample of cyclohexane. Most frequencies pass right through the sample and are recorded by a detector on the other side.
Our 5.15 x 1013 Hz carbonyl stretching frequency, however, is absorbed by the 2-hexanone sample, and so the detector records that the intensity of this frequency, after having passed through the sample, is something less than 100% of its initial intensity.
The vibrations of a 2-hexanone molecule are not, of course, limited to the simple stretching of the carbonyl bond. The various carbon-carbon bonds also stretch and bend, as do the carbon-hydrogen bonds, and all of these vibrational modes also absorb different frequencies of infrared light.
The power of infrared spectroscopy arises from the observation that different functional groups have different characteristic absorption frequencies. The carbonyl bond in a ketone, as we saw with our 2-hexanone example, typically absorbs in the range of 5.11 - 5.18 x 1013 Hz, depending on the molecule. The carbon-carbon triple bond of an alkyne, on the other hand, absorbs in the range 6.30 - 6.80 x 1013 Hz. The technique is therefore very useful as a means of identifying which functional groups are present in a molecule of interest. If we pass infrared light through an unknown sample and find that it absorbs in the carbonyl frequency range but not in the alkyne range, we can infer that the molecule contains a carbonyl group but not an alkyne.
Now, let's look at some actual output from IR spectroscopy experiments. Below is the IR spectrum for 2-hexanone.
There are a number of things that need to be explained in order for you to understand what it is that we are looking at. On the horizontal axis we see IR wavelengths expressed in terms of a unit called wavenumber (cm-1), which tells us how many waves fit into one centimeter. On the vertical axis we see ‘% transmittance’, which tells us how strongly light was absorbed at each frequency (100% transmittance means no absorption occurred at that frequency). The solid line traces the values of % transmittance for every wavelength – the ‘peaks’ (which are actually pointing down) show regions of strong absorption. For some reason, it is typical in IR spectroscopy to report wavenumber values rather than wavelength (in meters) or frequency (in Hz). The ‘upside down’ vertical axis, with absorbance peaks pointing down rather than up, is also a curious convention in IR spectroscopy. We wouldn’t want to make things too easy for you!
A calculator for interconverting these frequency and wavelength values is provided on the right. Simply enter the value to be converted in the appropriate box, press "Calculate" and the equivalent number will appear in the empty box.
Infrared spectra may be obtained from samples in all phases (liquid, solid and gaseous). Liquids are usually examined as a thin film sandwiched between two polished salt plates (note that glass absorbs infrared radiation, whereas NaCl is transparent). If solvents are used to dissolve solids, care must be taken to avoid obscuring important spectral regions by solvent absorption. Perchlorinated solvents such as carbon tetrachloride, chloroform and tetrachloroethene are commonly used. Alternatively, solids may either be incorporated in a thin KBr disk, prepared under high pressure, or mixed with a little non-volatile liquid and ground to a paste (or mull) that is smeared between salt plates.
Frequency - Wavelength Converter
Frequency in cm-1
Wavelength in μ
Gas Phase Infrared Spectrum of Formaldehyde, H2C=O
1. View CH2 Asymmetric Stretch
View CH2 Symmetric Stretch
View C=O Stretch
View CH2 Scissoring
View CH2 Rocking
View CH2 Wagging
Ball&Stick Model
Spacefill Model
Stick Model
Motion Off
The exact frequency at which a given vibration occurs is determined by the strengths of the bonds involved and the mass of the component atoms. For a more detailed discussion of these factors Click Here. In practice, infrared spectra do not normally display separate absorption signals for each of the 3n-6 fundamental vibrational modes of a molecule. The number of observed absorptions may be increased by additive and subtractive interactions leading to combination tones and overtones of the fundamental vibrations, in much the same way that sound vibrations from a musical instrument interact. Furthermore, the number of observed absorptions may be decreased by molecular symmetry, spectrometer limitations, and spectroscopic selection rules. One selection rule that influences the intensity of infrared absorptions, is that a change in dipole moment should occur for a vibration to absorb infrared energy. Absorption bands associated with C=O bond stretching are usually very strong because a large change in the dipole takes place in that mode.
Some General Trends:
1. Stretching frequencies are higher than corresponding bending frequencies. (It is easier to bend a bond than to stretch or compress it.)
2. Bonds to hydrogen have higher stretching frequencies than those to heavier atoms.
3. Triple bonds have higher stretching frequencies than corresponding double bonds, which in turn have higher frequencies than single bonds.(Except for bonds to hydrogen).
The general regions of the infrared spectrum in which various kinds of vibrational bands are observed are outlined in the following chart. Note that the blue colored sections above the dashed line refer to stretching vibrations, and the green colored band below the line encompasses bending vibrations. The complexity of infrared spectra in the 1450 to 600 cm-1 region makes it difficult to assign all the absorption bands, and because of the unique patterns found there, it is often called the fingerprint region. Absorption bands in the 4000 to 1450 cm-1 region are usually due to stretching vibrations of diatomic units, and this is sometimes called the group frequency region.
To illustrate the usefulness of infrared absorption spectra, examples for five C4H8O isomers are presented below their corresponding structural formulas. Try to associate each spectrum (A - E) with one of the isomers in the row above it.
Answers | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/11%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.02%3A_Infrared_%28IR%29_Spectroscopy.txt |
Some bonds absorb infrared light more strongly than others, and some bonds do not absorb at all. In order for a vibrational mode to absorb infrared light, it must result in a periodic change in the dipole moment of the molecule. Such vibrations are said to be infrared active. In general, the greater the polarity of the bond, the stronger its IR absorption. The carbonyl bond is very polar, and absorbs very strongly. The carbon-carbon triple bond in most alkynes, in contrast, is much less polar, and thus a stretching vibration does not result in a large change in the overall dipole moment of the molecule. Alkyne groups absorb rather weakly compared to carbonyls.
Some kinds of vibrations are infrared inactive. The stretching vibrations of completely symmetrical double and triple bonds, for example, do not result in a change in dipole moment, and therefore do not result in any absorption of light (but other bonds and vibrational modes in these molecules do absorb IR light).
11.04: Interpretting IR Spectra
Guided IR Spectrum Interpretation
Now, let’s take a look at the IR spectrum for 1-hexanol. There is a very broad ‘mountain’ centered at about 3400 cm-1. This signal is characteristic of the O-H stretching mode of alcohols, and is a dead giveaway for the presence of an alcohol group. The breadth of this signal is a consequence of hydrogen bonding between molecules.
In the spectrum of octanoic acid we see, as expected, the characteristic carbonyl peak, this time at 1709 cm-1. We also see a low, broad absorbance band that looks like an alcohol, except that it is displaced slightly to the right (long-wavelength) side of the spectrum, causing it to overlap to some degree with the C-H region. This is the characteristic carboxylic acid O-H single bond stretching absorbance.
The spectrum for 1-octene shows two peaks that are characteristic of alkenes: the one at 1642 cm-1 is due to stretching of the carbon-carbon double bond, and the one at 3079 cm-1 is due to stretching of the s bond between the alkene carbons and their attached hydrogens.
Alkynes have characteristic IR absorbance peaks in the range of 2100-2250 cm-1 due to stretching of the carbon-carbon triple bond, and terminal alkenes can be identified by their absorbance at about 3300 cm-1, due to stretching of the bond between the sp-hybridized carbon and the terminal hydrogen.
It is possible to identify other functional groups such as amines and ethers, but the characteristic peaks for these groups are considerably more subtle and/or variable, and often are overlapped with peaks from the fingerprint region. For this reason, we will limit our discussion here to the most easily recognized functional groups, which are summarized in this table.
As you can imagine, obtaining an IR spectrum for a compound will not allow us to figure out the complete structure of even a simple molecule, unless we happen to have a reference spectrum for comparison. In conjunction with other analytical methods, however, IR spectroscopy can prove to be a very valuable tool, given the information it provides about the presence or absence of key functional groups. IR can also be a quick and convenient way for a chemist to check to see if a reaction has proceeded as planned. If we were to run a reaction in which we wished to convert cyclohexanone to cyclohexanol, for example, a quick comparison of the IR spectra of starting compound and product would tell us if we had successfully converted the ketone group to an alcohol.
More examples of IR spectra
To illustrate the usefulness of infrared absorption spectra, examples for five C4H8O isomers are presented below their corresponding structural formulas. Try to associate each spectrum with one of the isomers in the row above it. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/11%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.03%3A_IR-Active_and_IR-Inactive_Vibrations.txt |
Common Group Frequencies Summary
When analyzing an IR spectrum, it is helpful to overlay the diagram below onto the spectrum with our mind to help recognize functional groups.
The region of the infrared spectrum from 1200 to 700 cm-1 is called the fingerprint region. This region is notable for the large number of infrared bands that are found there. Many different vibrations, including C-O, C-C and C-N single bond stretches, C-H bending vibrations, and some bands due to benzene rings are found in this region. The fingerprint region is often the most complex and confusing region to interpret, and is usually the last section of a spectrum to be interpreted. However, the utility of the fingerprint region is that the many bands there provide a fingerprint for a molecule.
Group Frequencies - a closer look
Detailed information about the infrared absorptions observed for various bonded atoms and groups is usually presented in tabular form. The following table provides a collection of such data for the most common functional groups. Following the color scheme of the chart, stretching absorptions are listed in the blue-shaded section and bending absorptions in the green shaded part. More detailed descriptions for certain groups (e.g. alkenes, arenes, alcohols, amines & carbonyl compounds) may be viewed by clicking on the functional class name. Since most organic compounds have C-H bonds, a useful rule is that absorption in the 2850 to 3000 cm-1 is due to sp3 C-H stretching; whereas, absorption above 3000 cm-1 is from sp2 C-H stretching or sp C-H stretching if it is near 3300 cm-1.
Stretching Vibrations
Bending Vibrations
Functional Class
Range (cm-1)
Intensity
Assignment
Range (cm-1)
Intensity
Assignment
Alkanes
2850-3000 str CH3, CH2 & CH
2 or 3 bands
1350-1470
1370-1390
720-725
med
med
wk
CH2 & CH3 deformation
CH3 deformation
CH2 rocking
Alkenes
3020-3100
1630-1680
1900-2000
med
var
str
=C-H & =CH2 (usually sharp)
C=C (symmetry reduces intensity)
C=C asymmetric stretch
880-995
780-850
675-730
str
med
med
=C-H & =CH2
(out-of-plane bending)
cis-RCH=CHR
Alkynes
3300
2100-2250
str
var
C-H (usually sharp)
C≡C (symmetry reduces intensity)
600-700 str C-H deformation
Arenes
3030
1600 & 1500
var
med-wk
C-H (may be several bands)
C=C (in ring) (2 bands)
(3 if conjugated)
690-900 str-med C-H bending &
ring puckering
Alcohols & Phenols
3580-3650
3200-3550
970-1250
var
str
str
O-H (free), usually sharp
O-H (H-bonded), usually broad
C-O
1330-1430
650-770
med
var-wk
O-H bending (in-plane)
O-H bend (out-of-plane)
Amines
3400-3500 (dil. soln.)
3300-3400 (dil. soln.)
1000-1250
wk
wk
med
N-H (1°-amines), 2 bands
N-H (2°-amines)
C-N
1550-1650
660-900
med-str
var
NH2 scissoring (1°-amines)
NH2 & N-H wagging
(shifts on H-bonding)
Aldehydes & Ketones
2690-2840(2 bands)
1720-1740
1710-1720
med
str
str
str
str
str
str
C-H (aldehyde C-H)
C=O (saturated aldehyde)
C=O (saturated ketone)
aryl ketone
α, β-unsaturation
cyclopentanone
cyclobutanone
1350-1360
1400-1450
1100
str
str
med
α-CH3 bending
α-CH2 bending
C-C-C bending
Carboxylic Acids & Derivatives
2500-3300 (acids) overlap C-H
1705-1720 (acids)
1210-1320 (acids)
str
str
med-str
str
str
str
str
str
str
O-H (very broad)
C=O (H-bonded)
O-C (sometimes 2-peaks)
C=O
C=O (2-bands)
O-C
C=O
O-C (2-bands)
C=O (amide I band)
1395-1440
1590-1650
1500-1560
med
med
med
C-O-H bending
N-H (1°-amide) II band
N-H (2°-amide) II band
Nitriles
Isocyanates,Isothiocyanates,
Diimides, Azides & Ketenes
2240-2260
2100-2270
med
med
C≡N (sharp)
-N=C=O, -N=C=S
-N=C=N-, -N3, C=C=O
Hydrocarbons
Hydrocarbons compounds contain only C-H and C-C bonds, but there is plenty of information to be obtained from the infrared spectra arising from C-H stretching and C-H bending.
In alkanes, which have very few bands, each band in the spectrum can be assigned:
• C–H stretch from 3000–2850 cm-1
• C–H bend or scissoring from 1470-1450 cm-1
• C–H rock, methyl from 1370-1350 cm-1
• C–H rock, methyl, seen only in long chain alkanes, from 725-720 cm-1
Figure 3. shows the IR spectrum of octane. Since most organic compounds have these features, these C-H vibrations are usually not noted when interpreting a routine IR spectrum. Note that the change in dipole moment with respect to distance for the C-H stretching is greater than that for others shown, which is why the C-H stretch band is the more intense.
In alkenes compounds, each band in the spectrum can be assigned:
• C=C stretch from 1680-1640 cm-1
• =C–H stretch from 3100-3000 cm-1
• =C–H bend from 1000-650 cm-1
Figure 4. shows the IR spectrum of 1-octene. As alkanes compounds, these bands are not specific and are generally not noted because they are present in almost all organic molecules.
In alkynes, each band in the spectrum can be assigned:
• –C?C– stretch from 2260-2100 cm-1
• –C?C–H: C–H stretch from 3330-3270 cm-1
• –C?C–H: C–H bend from 700-610 cm-1
The spectrum of 1-hexyne, a terminal alkyne, is shown below.
In aromatic compounds, each band in the spectrum can be assigned:
• C–H stretch from 3100-3000 cm-1
• overtones, weak, from 2000-1665 cm-1
• C–C stretch (in-ring) from 1600-1585 cm-1
• C–C stretch (in-ring) from 1500-1400 cm-1
• C–H "oop" from 900-675 cm-1
Note that this is at slightly higher frequency than is the –C–H stretch in alkanes. This is a very useful tool for interpreting IR spectra. Only alkenes and aromatics show a C–H stretch slightly higher than 3000 cm-1.
Figure 6. shows the spectrum of toluene.
Functional Groups Containing the C-O Bond
Alcohols have IR absorptions associated with both the O-H and the C-O stretching vibrations.
• O–H stretch, hydrogen bonded 3500-3200 cm-1
• C–O stretch 1260-1050 cm-1 (s)
Figure 7. shows the spectrum of ethanol. Note the very broad, strong band of the O–H stretch.
The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears:
• C=O stretch - aliphatic ketones 1715 cm-1
- ?, ?-unsaturated ketones 1685-1666 cm-1
Figure 8. shows the spectrum of 2-butanone. This is a saturated ketone, and the C=O band appears at 1715.
If a compound is suspected to be an aldehyde, a peak always appears around 2720 cm-1 which often appears as a shoulder-type peak just to the right of the alkyl C–H stretches.
• H–C=O stretch 2830-2695 cm-1
• C=O stretch:
• aliphatic aldehydes 1740-1720 cm-1
• alpha, beta-unsaturated aldehydes 1710-1685 cm-1
Figure 9. shows the spectrum of butyraldehyde.
The carbonyl stretch C=O of esters appears:
• C=O stretch
• aliphatic from 1750-1735 cm-1
• ?, ?-unsaturated from 1730-1715 cm-1
• C–O stretch from 1300-1000 cm-1
Figure 10. shows the spectrum of ethyl benzoate.
The carbonyl stretch C=O of a carboxylic acid appears as an intense band from 1760-1690 cm-1. The exact position of this broad band depends on whether the carboxylic acid is saturated or unsaturated, dimerized, or has internal hydrogen bonding.
• O–H stretch from 3300-2500 cm-1
• C=O stretch from 1760-1690 cm-1
• C–O stretch from 1320-1210 cm-1
• O–H bend from 1440-1395 and 950-910 cm-1
Figure 11. shows the spectrum of hexanoic acid.
Organic Nitrogen Compounds
• N–O asymmetric stretch from 1550-1475 cm-1
• N–O symmetric stretch from 1360-1290 cm-1
Organic Compounds Containing Halogens
Alkyl halides are compounds that have a C–X bond, where X is a halogen: bromine, chlorine, fluorene, or iodine.
• C–H wag (-CH2X) from 1300-1150 cm-1
• C–X stretches (general) from 850-515 cm-1
• C–Cl stretch 850-550 cm-1
• C–Br stretch 690-515 cm-1
The spectrum of 1-chloro-2-methylpropane are shown below.
For more Infrared spectra Spectral database of organic molecules is introduced to use free database. Also, the infrared spectroscopy correlation table is linked on bottom of page to find other assigned IR peaks.
Exercise
1. What functional groups give the following signals in an IR spectrum?
A) 1700 cm-1
B) 1550 cm-1
C) 1700 cm-1 and 2510-3000 cm-1
2. How can you distinguish the following pairs of compounds through IR analysis?
A) CH3OH (Methanol) and CH3CH2OCH2CH3 (Diethylether)
B) Cyclopentane and 1-pentene.
C)
3. The following spectra is for the accompanying compound. What are the peaks that you can I identify in the spectrum?
Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 2 December 2016)
4. What absorptions would the following compounds have in an IR spectra?
Answer
1.
2.
A) A OH peak will be present around 3300 cm-1 for methanol and will be absent in the ether.
B) 1-pentene will have a alkene peak around 1650 cm-1 for the C=C and there will be another peak around 3100 cm-1 for the sp2 C-H group on the alkene
C) Cannot distinguish these two isomers. They both have the same functional groups and therefore would have the same peaks on an IR spectra.
3.
Frequency (cm-1) Functional Group
3200 C≡C-H
2900-3000 C-C-H, C=C-H
2100 C≡C
1610 C=C
(There is also an aromatic undertone region between 2000-1600 which describes the substitution on the phenyl ring.)
4.
A)
Frequency (cm-1) Functional Group
2900-3000 C-C-H, C=C-H
1710 C=O
1610 C=C
1100 C-O
B)
Frequency (cm-1) Functional Group
3200 C≡C-H
2900-3000 C-C-H, C=C-H
2100 C≡C
1710 C=O
C)
Frequency (cm-1) Functional Group
3300 (broad) O-H
2900-3000 C-C-H, C=C-H
2000-1800 Aromatic Overtones
1710 C=O
1610 C=C | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/11%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.05%3A_Infrared_Spectra_of_Some_Common_Functional_Grou.txt |
Learning Objectives
After completing this section, you should be able to
1. sketch a simple diagram to show the essential features of a mass spectrometer.
2. identify peaks in a simple mass spectrum, and explain how they arise.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• parent peak (molecular ion peak)
• relative abundance
• mass spectroscopy
• molecular ion (M+·)
• mass-to-charge ratio (m/z)
Study Notes
You may remember from general first-year chemistry how mass spectroscopy has been used to establish the atomic mass and abundance of isotopes.
Mass spectrometers are large and expensive, and usually operated only by fully trained personnel, so you will not have the opportunity to use such an instrument as part of this course. Research chemists often rely quite heavily on mass spectra to assist them in the identification of compounds, and you will be required to interpret simple mass spectra both in assignments and on examinations. Note that in most attempts to identify an unknown compound, chemists do not rely exclusively on the results obtained from a single spectroscopic technique. A combination of chemical and physical properties and spectral evidence is usually employed.
The Mass Spectrometer
In order to measure the characteristics of individual molecules, a mass spectrometer converts them to ions so that they can be moved about and manipulated by external electric and magnetic fields. The three essential functions of a mass spectrometer, and the associated components, are:
1. The ions are sorted and separated according to their mass and charge. The Mass Analyzer
2. The separated ions are then measured, and the results displayed on a chart. The Detector
Because ions are very reactive and short-lived, their formation and manipulation must be conducted in a vacuum. Atmospheric pressure is around 760 torr (mm of mercury). The pressure under which ions may be handled is roughly 10-5 to 10-8 torr (less than a billionth of an atmosphere). Each of the three tasks listed above may be accomplished in different ways. In one common procedure, ionization is effected by a high energy beam of electrons, and ion separation is achieved by accelerating and focusing the ions in a beam, which is then bent by an external magnetic field. The ions are then detected electronically and the resulting information is stored and analyzed in a computer. A mass spectrometer operating in this fashion is outlined in the following diagram. The heart of the spectrometer is the ion source. Here molecules of the sample (black dots) are bombarded by electrons (light blue lines) issuing from a heated filament. This is called an EI (electron-impact) source. Gases and volatile liquid samples are allowed to leak into the ion source from a reservoir (as shown). Non-volatile solids and liquids may be introduced directly. Cations formed by the electron bombardment (red dots) are pushed away by a charged repellor plate (anions are attracted to it), and accelerated toward other electrodes, having slits through which the ions pass as a beam. Some of these ions fragment into smaller cations and neutral fragments. A perpendicular magnetic field deflects the ion beam in an arc whose radius is inversely proportional to the mass of each ion. Lighter ions are deflected more than heavier ions. By varying the strength of the magnetic field, ions of different mass can be focused progressively on a detector fixed at the end of a curved tube (also under a high vacuum).
When a high energy electron collides with a molecule it often ionizes it by knocking away one of the molecular electrons (either bonding or non-bonding). This leaves behind a molecular ion (colored red in the following diagram). Residual energy from the collision may cause the molecular ion to fragment into neutral pieces (colored green) and smaller fragment ions (colored pink and orange). The molecular ion is a radical cation, but the fragment ions may either be radical cations (pink) or carbocations (orange), depending on the nature of the neutral fragment. An animated display of this ionization process will appear if you click on the ion source of the mass spectrometer diagram.
Below is typical output for an electron-ionization MS experiment (MS data below is derived from the Spectral Database for Organic Compounds, a free, web-based service provided by AIST in Japan.
The sample is acetone. On the horizontal axis is the value for m/z (as we stated above, the charge z is almost always +1, so in practice this is the same as mass). On the vertical axis is the relative abundance of each ion detected. On this scale, the most abundant ion, called the base peak, is set to 100%, and all other peaks are recorded relative to this value. For acetone, the base peak corresponds to a fragment with m/z = 43 - . The molecular weight of acetone is 58, so we can identify the peak at m/z = 58 as that corresponding to the molecular ion peak, or parent peak. Notice that there is a small peak at m/z = 59: this is referred to as the M+1 peak. How can there be an ion that has a greater mass than the molecular ion? Simple: a small fraction - about 1.1% - of all carbon atoms in nature are actually the 13C rather than the 12C isotope. The 13C isotope is, of course, heavier than 12C by 1 mass unit. In addition, about 0.015% of all hydrogen atoms are actually deuterium, the 2H isotope. So the M+1 peak represents those few acetone molecules in the sample which contained either a 13C or 2H. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/11%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.07%3A_Mass_Spectrometry_-_an_introduction.txt |
Objectives
After completing this section, you should be able to
1. suggest possible molecular formulas for a compound, given the m/z value for the molecular ion, or a mass spectrum from which this value can be obtained.
2. predict the relative heights of the M+·, (M + 1)+·, etc., peaks in the mass spectrum of a compound, given the natural abundance of the isotopes of carbon and the other elements present in the compound.
3. interpret the fragmentation pattern of the mass spectrum of a relatively simple, known compound (e.g., hexane).
4. use the fragmentation pattern in a given mass spectrum to assist in the identification of a relatively simple, unknown compound (e.g., an unknown alkane).
Study Notes
When interpreting fragmentation patterns, you may find it helpful to know that, as you might expect, the weakest carbon-carbon bonds are the ones most likely to break. You might wish to refer to the table of bond dissociation energies when attempting problems involving the interpretation of mass spectra.
This page looks at how fragmentation patterns are formed when organic molecules are fed into a mass spectrometer, and how you can get information from the mass spectrum.
The Origin of Fragmentation Patterns
When the vaporized organic sample passes into the ionization chamber of a mass spectrometer, it is bombarded by a stream of electrons. These electrons have a high enough energy to knock an electron off an organic molecule to form a positive ion. This ion is called the molecular ion - or sometimes the parent ion and is often given the symbol M+ or . The dot in this second version represents the fact that somewhere in the ion there will be a single unpaired electron. That's one half of what was originally a pair of electrons - the other half is the electron which was removed in the ionization process.
The molecular ions are energetically unstable, and some of them will break up into smaller pieces. The simplest case is that a molecular ion breaks into two parts - one of which is another positive ion, and the other is an uncharged free radical.
The uncharged free radical will not produce a line on the mass spectrum. Only charged particles will be accelerated, deflected and detected by the mass spectrometer. These uncharged particles will simply get lost in the machine - eventually, they get removed by the vacuum pump.
The ion, X+, will travel through the mass spectrometer just like any other positive ion - and will produce a line on the stick diagram. All sorts of fragmentations of the original molecular ion are possible - and that means that you will get a whole host of lines in the mass spectrum. For example, the mass spectrum of pentane looks like this:
Note
The pattern of lines in the mass spectrum of an organic compound tells you something quite different from the pattern of lines in the mass spectrum of an element. With an element, each line represents a different isotope of that element. With a compound, each line represents a different fragment produced when the molecular ion breaks up.
In the stick diagram showing the mass spectrum of pentane, the line produced by the heaviest ion passing through the machine (at m/z = 72) is due to the molecular ion. The tallest line in the stick diagram (in this case at m/z = 43) is called the base peak. This is usually given an arbitrary height of 100, and the height of everything else is measured relative to this. The base peak is the tallest peak because it represents the commonest fragment ion to be formed - either because there are several ways in which it could be produced during fragmentation of the parent ion, or because it is a particularly stable ion.
Using Fragmentation Patterns
This section will ignore the information you can get from the molecular ion (or ions). That is covered in three other pages which you can get at via the mass spectrometry menu. You will find a link at the bottom of the page.
Example: Pentane
Let's have another look at the mass spectrum for pentane:
What causes the line at m/z = 57?
How many carbon atoms are there in this ion? There cannot be 5 because 5 x 12 = 60. What about 4? 4 x 12 = 48. That leaves 9 to make up a total of 57. How about C4H9+ then?
C4H9+ would be [CH3CH2CH2CH2]+, and this would be produced by the following fragmentation:
The methyl radical produced will simply get lost in the machine.
The line at m/z = 43 can be worked out similarly. If you play around with the numbers, you will find that this corresponds to a break producing a 3-carbon ion:
The line at m/z = 29 is typical of an ethyl ion, [CH3CH2]+:
The other lines in the mass spectrum are more difficult to explain. For example, lines with m/z values 1 or 2 less than one of the easy lines are often due to loss of one or more hydrogen atoms during the fragmentation process.
Example: Pentan-3-one
This time the base peak (the tallest peak - and so the commonest fragment ion) is at m/z = 57. But this is not produced by the same ion as the same m/z value peak in pentane.
If you remember, the m/z = 57 peak in pentane was produced by [CH3CH2CH2CH2]+. If you look at the structure of pentan-3-one, it's impossible to get that particular fragment from it.
Work along the molecule mentally chopping bits off until you come up with something that adds up to 57. With a small amount of patience, you'll eventually find [CH3CH2CO]+ - which is produced by this fragmentation:
You would get exactly the same products whichever side of the CO group you split the molecular ion. The m/z = 29 peak is produced by the ethyl ion - which once again could be formed by splitting the molecular ion either side of the CO group.
Peak Heights and Stability
The more stable an ion is, the more likely it is to form. The more of a particular sort of ion that's formed, the higher its peak height will be. We'll look at two common examples of this.
Carbocations (carbonium ions)
Summarizing the most important conclusion from the page on carbocations:
Order of stability of carbocations
primary < secondary < tertiary
Applying the logic of this to fragmentation patterns, it means that a split which produces a secondary carbocation is going to be more successful than one producing a primary one. A split producing a tertiary carbocation will be more successful still. Let's look at the mass spectrum of 2-methylbutane. 2-methylbutane is an isomer of pentane - isomers are molecules with the same molecular formula, but a different spatial arrangement of the atoms.
Look first at the very strong peak at m/z = 43. This is caused by a different ion than the corresponding peak in the pentane mass spectrum. This peak in 2-methylbutane is caused by:
The ion formed is a secondary carbocation - it has two alkyl groups attached to the carbon with the positive charge. As such, it is relatively stable. The peak at m/z = 57 is much taller than the corresponding line in pentane. Again a secondary carbocation is formed - this time, by:
You would get the same ion, of course, if the left-hand CH3 group broke off instead of the bottom one as we've drawn it. In these two spectra, this is probably the most dramatic example of the extra stability of a secondary carbocation.
Acylium ions, [RCO]+
Ions with the positive charge on the carbon of a carbonyl group, C=O, are also relatively stable. This is fairly clearly seen in the mass spectra of ketones like pentan-3-one.
The base peak, at m/z=57, is due to the [CH3CH2CO]+ ion. We've already discussed the fragmentation that produces this.
Note
The more stable an ion is, the more likely it is to form. The more of a particular ion that is formed, the higher will be its peak height.
Using mass spectra to distinguish between compounds
Suppose you had to suggest a way of distinguishing between pentan-2-one and pentan-3-one using their mass spectra.
pentan-2-one CH3COCH2CH2CH3
pentan-3-one CH3CH2COCH2CH3
Each of these is likely to split to produce ions with a positive charge on the CO group. In the pentan-2-one case, there are two different ions like this:
• [CH3CO]+
• [COCH2CH2CH3]+
That would give you strong lines at m/z = 43 and 71. With pentan-3-one, you would only get one ion of this kind:
• [CH3CH2CO]+
In that case, you would get a strong line at 57. You don't need to worry about the other lines in the spectra - the 43, 57 and 71 lines give you plenty of difference between the two. The 43 and 71 lines are missing from the pentan-3-one spectrum, and the 57 line is missing from the pentan-2-one one.
The two mass spectra look like this:
As you've seen, the mass spectrum of even very similar organic compounds will be quite different because of the different fragmentation patterns that can occur. Provided you have a computer data base of mass spectra, any unknown spectrum can be computer analyzed and simply matched against the data base.
Exercise
5. Caffeine has a mass of 194.19 amu, determined by mass spectrometry, and contains C, N, H, O. What is a molecular formula for this molecule?
6. The following are the spectra for 2-methyl-2-hexene and 2-heptene, which spectra belongs to the correct molecule. Explain.
A:
B:
Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 2 December 2016)
Answer
5. C8H10N4O2
C = 12 × 8 = 96
N = 14 × 4 = 56
H = 1 × 10 = 10
O = 2 × 16 = 32
96+56+10+32 = 194 g/mol
6. The (A) spectrum is 2-methyl-2-hexene and the (B) spectrum is 2-heptene. Looking at (A) the peak at 68 m/z is the fractioned molecule with just the tri-substituted alkene present. While (B) has a strong peak around the 56 m/z, which in this case is the di-substituted alkene left behind from the linear heptene. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/11%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.08%3A_Fragmentation_Patterns_in_Mass_Spectrometry.txt |
The Nature of Mass Spectra
A mass spectrum will usually be presented as a vertical bar graph, in which each bar represents an ion having a specific mass-to-charge ratio (m/z) and the length of the bar indicates the relative abundance of the ion. The most intense ion is assigned an abundance of 100, and it is referred to as the base peak. Most of the ions formed in a mass spectrometer have a single charge, so the m/z value is equivalent to mass itself. Modern mass spectrometers easily distinguish (resolve) ions differing by only a single atomic mass unit, and thus provide completely accurate values for the molecular mass of a compound. The highest-mass ion in a spectrum is normally considered to be the molecular ion, and lower-mass ions are fragments from the molecular ion, assuming the sample is a single pure compound.
Atomic mass is given in terms of the unified atomic mass unit (symbol: μ) or dalton (symbol: Da). In recent years there has been a gradual change towards using the dalton in preference to the unified atomic mass unit. The dalton is classified as a "non-SI unit whose values in SI units must be obtained experimentally". It is defined as one twelfth of the rest mass of an unbound atom of carbon-12 in its nuclear and electronic ground state, and has a value of 1.660538782(83)x10-27 kg.
The following diagram displays the mass spectra of three simple gaseous compounds, carbon dioxide, propane and cyclopropane. The molecules of these compounds are similar in size, CO2 and C3H8 both have a nominal mass of 44 Da, and C3H6 has a mass of 42 Da. The molecular ion is the strongest ion in the spectra of CO2 and C3H6, and it is moderately strong in propane. The unit mass resolution is readily apparent in these spectra (note the separation of ions having m/z=39, 40, 41 and 42 in the cyclopropane spectrum). Even though these compounds are very similar in size, it is a simple matter to identify them from their individual mass spectra. By clicking on each spectrum in turn, a partial fragmentation analysis and peak assignment will be displayed. Even with simple compounds like these, it should be noted that it is rarely possible to explain the origin of all the fragment ions in a spectrum. Also, the structure of most fragment ions is seldom known with certainty.
Since a molecule of carbon dioxide is composed of only three atoms, its mass spectrum is very simple. The molecular ion is also the base peak, and the only fragment ions are CO (m/z=28) and O (m/z=16). The molecular ion of propane also has m/z=44, but it is not the most abundant ion in the spectrum. Cleavage of a carbon-carbon bond gives methyl and ethyl fragments, one of which is a carbocation and the other a radical. Both distributions are observed, but the larger ethyl cation (m/z=29) is the most abundant, possibly because its size affords greater charge dispersal. A similar bond cleavage in cyclopropane does not give two fragments, so the molecular ion is stronger than in propane, and is in fact responsible for the the base peak. Loss of a hydrogen atom, either before or after ring opening, produces the stable allyl cation (m/z=41). The third strongest ion in the spectrum has m/z=39 (C3H3). Its structure is uncertain, but two possibilities are shown in the diagram. The small m/z=39 ion in propane and the absence of a m/z=29 ion in cyclopropane are particularly significant in distinguishing these hydrocarbons.
Most stable organic compounds have an even number of total electrons, reflecting the fact that electrons occupy atomic and molecular orbitals in pairs. When a single electron is removed from a molecule to give an ion, the total electron count becomes an odd number, and we refer to such ions as radical cations. The molecular ion in a mass spectrum is always a radical cation, but the fragment ions may either be even-electron cations or odd-electron radical cations, depending on the neutral fragment lost. The simplest and most common fragmentations are bond cleavages producing a neutral radical (odd number of electrons) and a cation having an even number of electrons. A less common fragmentation, in which an even-electron neutral fragment is lost, produces an odd-electron radical cation fragment ion. Fragment ions themselves may fragment further. As a rule, odd-electron ions may fragment either to odd or even-electron ions, but even-electron ions fragment only to other even-electron ions. The masses of molecular and fragment ions also reflect the electron count, depending on the number of nitrogen atoms in the species.
Ions with no nitrogen
or an even # N atoms
odd-electron ions
even-number mass
even-electron ions
odd-number mass
Ions having an
odd # N atoms
odd-electron ions
odd-number mass
even-electron ions
even-number mass
This distinction is illustrated nicely by the follwing two examples. The unsaturated ketone, 4-methyl-3-pentene-2-one, on the left has no nitrogen so the mass of the molecular ion (m/z = 98) is an even number. Most of the fragment ions have odd-numbered masses, and therefore are even-electron cations. Diethylmethylamine, on the other hand, has one nitrogen and its molecular mass (m/z = 87) is an odd number. A majority of the fragment ions have even-numbered masses (ions at m/z = 30, 42, 56 & 58 are not labeled), and are even-electron nitrogen cations. The weak even -electron ions at m/z=15 and 29 are due to methyl and ethyl cations (no nitrogen atoms). The fragmentations leading to the chief fragment ions will be displayed by clicking on the appropriate spectrum. Repeated clicks will cycle the display.
4-methyl-3-pentene-2-one
N,N-diethylmethylamine
When non-bonded electron pairs are present in a molecule (e.g. on N or O), fragmentation pathways may sometimes be explained by assuming the missing electron is partially localized on that atom. A few such mechanisms are shown above. Bond cleavage generates a radical and a cation, and both fragments often share these roles, albeit unequally.
Isotopes
Since a mass spectrometer separates and detects ions of slightly different masses, it easily distinguishes different isotopes of a given element. This is manifested most dramatically for compounds containing bromine and chlorine, as illustrated by the following examples. Since molecules of bromine have only two atoms, the spectrum on the left will come as a surprise if a single atomic mass of 80 Da is assumed for Br. The five peaks in this spectrum demonstrate clearly that natural bromine consists of a nearly 50:50 mixture of isotopes having atomic masses of 79 and 81 Da respectively. Thus, the bromine molecule may be composed of two 79Br atoms (mass 158 Da), two 81Br atoms (mass 162 Da) or the more probable combination of 79Br-81Br (mass 160 Da). Fragmentation of Br2 to a bromine cation then gives rise to equal sized ion peaks at 79 and 81 Da.
bromine
vinyl chloride
methylene chloride
The center and right hand spectra show that chlorine is also composed of two isotopes, the more abundant having a mass of 35 Da, and the minor isotope a mass 37 Da. The precise isotopic composition of chlorine and bromine is:
• Chlorine: 75.77% 35Cl and 24.23% 37Cl
• Bromine: 50.50% 79Br and 49.50% 81Br
The presence of chlorine or bromine in a molecule or ion is easily detected by noticing the intensity ratios of ions differing by 2 Da. In the case of methylene chloride, the molecular ion consists of three peaks at m/z=84, 86 & 88 Da, and their diminishing intensities may be calculated from the natural abundances given above. Loss of a chlorine atom gives two isotopic fragment ions at m/z=49 & 51 Da, clearly incorporating a single chlorine atom. Fluorine and iodine, by contrast, are monoisotopic, having masses of 19 Da and 127 Da respectively. It should be noted that the presence of halogen atoms in a molecule or fragment ion does not change the odd-even mass rules given above.
To make use of a calculator that predicts the isotope clusters for different combinations of chlorine, bromine and other elements Click Here. This application was developed at Colby College.
Isotopic Abundance Calculator
C H N O SSi
Molecular Ion
100%
M + 1
M + 2
Two other common elements having useful isotope signatures are carbon, 13C is 1.1% natural abundance, and sulfur, 33S and 34S are 0.76% and 4.22% natural abundance respectively. For example, the small m/z=99 Da peak in the spectrum of 4-methyl-3-pentene-2-one (above) is due to the presence of a single 13C atom in the molecular ion. Although less important in this respect, 15N and 18O also make small contributions to higher mass satellites of molecular ions incorporating these elements.
The calculator on the right may be used to calculate the isotope contributions to ion abundances 1 and 2 Da greater than the molecular ion (M). Simply enter an appropriate subscript number to the right of each symbol, leaving those elements not present blank, and press the "Calculate" button. The numbers displayed in the M+1 and M+2 boxes are relative to M being set at 100%.
Fragmentation Patterns
The fragmentation of molecular ions into an assortment of fragment ions is a mixed blessing. The nature of the fragments often provides a clue to the molecular structure, but if the molecular ion has a lifetime of less than a few microseconds it will not survive long enough to be observed. Without a molecular ion peak as a reference, the difficulty of interpreting a mass spectrum increases markedly. Fortunately, most organic compounds give mass spectra that include a molecular ion, and those that do not often respond successfully to the use of milder ionization conditions. Among simple organic compounds, the most stable molecular ions are those from aromatic rings, other conjugated pi-electron systems and cycloalkanes. Alcohols, ethers and highly branched alkanes generally show the greatest tendency toward fragmentation.
The mass spectrum of dodecane on the right illustrates the behavior of an unbranched alkane. Since there are no heteroatoms in this molecule, there are no non-bonding valence shell electrons. Consequently, the radical cation character of the molecular ion (m/z = 170) is delocalized over all the covalent bonds. Fragmentation of C-C bonds occurs because they are usually weaker than C-H bonds, and this produces a mixture of alkyl radicals and alkyl carbocations. The positive charge commonly resides on the smaller fragment, so we see a homologous series of hexyl (m/z = 85), pentyl (m/z = 71), butyl (m/z = 57), propyl (m/z = 43), ethyl (m/z = 29) and methyl (m/z = 15) cations. These are accompanied by a set of corresponding alkenyl carbocations (e.g. m/z = 55, 41 &27) formed by loss of 2 H. All of the significant fragment ions in this spectrum are even-electron ions. In most alkane spectra the propyl and butyl ions are the most abundant.
The presence of a functional group, particularly one having a heteroatom Y with non-bonding valence electrons (Y = N, O, S, X etc.), can dramatically alter the fragmentation pattern of a compound. This influence is thought to occur because of a "localization" of the radical cation component of the molecular ion on the heteroatom. After all, it is easier to remove (ionize) a non-bonding electron than one that is part of a covalent bond. By localizing the reactive moiety, certain fragmentation processes will be favored. These are summarized in the following diagram, where the green shaded box at the top displays examples of such "localized" molecular ions. The first two fragmentation paths lead to even-electron ions, and the elimination (path #3) gives an odd-electron ion. Note the use of different curved arrows to show single electron shifts compared with electron pair shifts.
The charge distributions shown above are common, but for each cleavage process the charge may sometimes be carried by the other (neutral) species, and both fragment ions are observed. Of the three cleavage reactions described here, the alpha-cleavage is generally favored for nitrogen, oxygen and sulfur compounds. Indeed, in the previously displayed spectra of 4-methyl-3-pentene-2-one and N,N-diethylmethylamine the major fragment ions come from alpha-cleavages. Further examples of functional group influence on fragmentation are provided by a selection of compounds that may be examined by clicking the left button below. Useful tables of common fragment ions and neutral species may be viewed by clicking the right button.
Assorted Mass Spectra
View Fragment Tables
The complexity of fragmentation patterns has led to mass spectra being used as "fingerprints" for identifying compounds. Environmental pollutants, pesticide residues on food, and controlled substance identification are but a few examples of this application. Extremely small samples of an unknown substance (a microgram or less) are sufficient for such analysis. The following mass spectrum of cocaine demonstrates how a forensic laboratory might determine the nature of an unknown street drug. Even though extensive fragmentation has occurred, many of the more abundant ions (identified by magenta numbers) can be rationalized by the three mechanisms shown above. Plausible assignments may be seen by clicking on the spectrum, and it should be noted that all are even-electron ions. The m/z = 42 ion might be any or all of the following: C3H6, C2H2O or C2H4N. A precise assignment could be made from a high-resolution m/z value (next section).
Odd-electron fragment ions are often formed by characteristic rearrangements in which stable neutral fragments are lost. Mechanisms for some of these rearrangements have been identified by following the course of isotopically labeled molecular ions. A few examples of these rearrangement mechanisms may be seen by clicking the following button.
Assorted Rearrangement Fragmentations
Exercise
7. What are the masses of all the components in the following fragmentations?
Answer
7.
11.10: Determination of the Molecular Formula by High
High Resolution Mass Spectrometry
In assigning mass values to atoms and molecules, we have assumed integral values for isotopic masses. However, accurate measurements show that this is not strictly true. Because the strong nuclear forces that bind the components of an atomic nucleus together vary, the actual mass of a given isotope deviates from its nominal integer by a small but characteristic amount (remember E = mc2). Thus, relative to 12C at 12.0000, the isotopic mass of 16O is 15.9949 Da (not 16) and 14N is 14.0031 Da (not 14).
Formula
C6H12 C5H8O C4H8N2
Mass
84.0939 84.0575 84.0688
By designing mass spectrometers that can determine m/z values accurately to four decimal places, it is possible to distinguish different formulas having the same nominal mass. The table on the right illustrates this important feature, and a double-focusing high-resolution mass spectrometer easily distinguishes ions having these compositions. Mass spectrometry therefore not only provides a specific molecular mass value, but it may also establish the molecular formula of an unknown compound.
Tables of precise mass values for any molecule or ion are available in libraries; however, the mass calculator provided below serves the same purpose. Since a given nominal mass may correspond to several molecular formulas, lists of such possibilities are especially useful when evaluating the spectrum of an unknown compound. Composition tables are available for this purpose, and a particularly useful program for calculating all possible combinations of H, C, N & O that give a specific nominal mass has been written by Jef Rozenski. To use this calculator Click Here.
University of ArizonaIowa State University University of Leeds | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/11%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.09%3A__Useful_Patterns_for_Structure_Elucidation.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• explain how 1H NMR spectrometers work - refer to section 12.1
• interpret chemical shifts of 1H NMR spectra as they relate to shielding and deshielding - refer to section 12.2 and 12.14
• explain the delta scale of 1H NMR spectra - refer to section 12.3
• recognize equivalent protons within an organic compound - refer to section 12.4
• correlate functional group structural features with chemical shifts - refer to section 12.5
• determine the proton ratio from 1H NMR spectra peak integration data - refer to section 12.6
• explain and interpret spin-spin splitting patterns in 1H NMR spectra - refer to section 12.7
• explain and interpret spin-spin splitting patterns in 1H NMR spectra - refer to section 12.8
• describes examples of some uses of 1H NMR spectroscopy - refer to section 12.9
• explain how 13C NMR spectrometers work - refer to section 12.10
• interpret the chemical shifts of 13C NMR spectra to determine the structural features of organic compounds - refer to section 12.11 and 12.14
• explain how DEPT (distortionless enhancement by polarization transfer) is used to determine the number of hydrogens bonded to each carbon - refer to section 12.12
• describes some uses of 13C NMR spectroscopy - refer to section 12.13
12: Nuclear Magnetic Resonance Spectroscopy
Nuclear precession, spin states, and the resonance condition
Some types of atomic nuclei act as though they spin on their axis similar to the Earth. Since they are positively charged they generate an electromagnetic field just as the Earth does. So, in effect, they will act as tiny bar magnetics. Not all nuclei act this way, but fortunately both 1H and 13C do have nuclear spins and will respond to this technique.
NMR Spectrometer
In the absence of an external magnetic field the direction of the spin of the nuclei will be randomly oriented (see figure below left). However, when a sample of these nuclei is place in an external magnetic field, the nuclear spins will adopt specific orientations much as a compass needle responses to the Earth’s magnetic field and aligns with it. Two possible orientations are possible, with the external field (i.e. parallel to and in the same direction as the external field) or against the field (i.e. antiparallel to the external field). See figure below right.
When the same sample is placed within the field of a very strong magnet in an NMR instrument (this field is referred to by NMR spectroscopists as the applied field, abbreviated B0 ) each hydrogen will assume one of two possible spin states. In what is referred to as the +½ spin state, the hydrogen's magnetic moment is aligned with the direction of B0, while in the -½ spin state it is aligned opposed to the direction of B0.
Because the +½ spin state is slightly lower in energy, in a large population of organic molecules slightly more than half of the hydrogen atoms will occupy this state, while slightly less than half will occupy the –½ state. The difference in energy between the two spin states increases with increasing strength of B0.This last statement is in italics because it is one of the key ideas in NMR spectroscopy, as we shall soon see.
At this point, we need to look a little more closely at how a proton spins in an applied magnetic field. You may recall playing with spinning tops as a child. When a top slows down a little and the spin axis is no longer completely vertical, it begins to exhibit precessional motion, as the spin axis rotates slowly around the vertical. In the same way, hydrogen atoms spinning in an applied magnetic field also exhibit precessional motion about a vertical axis. It is this axis (which is either parallel or antiparallel to B0) that defines the proton’s magnetic moment. In the figure below, the proton is in the +1/2 spin state.
The frequency of precession (also called the Larmour frequency, abbreviated ωL) is simply the number of times per second that the proton precesses in a complete circle. A proton`s precessional frequency increases with the strength of B0.
If a proton that is precessing in an applied magnetic field is exposed to electromagnetic radiation of a frequency ν that matches its precessional frequency ωL, we have a condition called resonance. In the resonance condition, a proton in the lower-energy +½ spin state (aligned with B0) will transition (flip) to the higher energy –½ spin state (opposed to B0). In doing so, it will absorb radiation at this resonance frequency ν = ωL. This frequency, as you might have already guessed, corresponds to the energy difference between the proton’s two spin states. With the strong magnetic fields generated by the superconducting magnets used in modern NMR instruments, the resonance frequency for protons falls within the radio-wave range, anywhere from 100 MHz to 800 MHz depending on the strength of the magnet.
If the ordered nuclei are now subjected to EM radiation of the proper frequency the nuclei aligned with the field will absorb energy and "spin-flip" to align themselves against the field, a higher energy state. When this spin-flip occurs the nuclei are said to be in "resonance" with the field, hence the name for the technique, Nuclear Magentic Resonance or NMR.
The amount of energy, and hence the exact frequency of EM radiation required for resonance to occur is dependent on both the strength of the magnetic field applied and the type of the nuclei being studied. As the strength of the magnetic field increases the energy difference between the two spin states increases and a higher frequency (more energy) EM radiation needs to be applied to achieve a spin-flip (see image below).
Superconducting magnets can be used to produce very strong magnetic field, on the order of 21 tesla (T). Lower field strengths can also be used, in the range of 4 - 7 T. At these levels the energy required to bring the nuclei into resonance is in the MHz range and corresponds to radio wavelength energies, i.e. at a field strength of 4.7 T 200 MHz bring 1H nuclei into resonance and 50 MHz bring 13C into resonance. This is considerably less energy then is required for IR spectroscopy, ~10-4 kJ/mol versus ~5 - ~50 kJ/mol.
1H and 13C are not unique in their ability to undergo NMR. All nuclei with an odd number of protons (1H, 2H, 14N, 19F, 31P ...) or nuclei with an odd number of neutrons (i.e. 13C) show the magnetic properties required for NMR. Only nuclei with even number of both protons and neutrons (12C and 16O) do not have the required magnetic properties.
The basic arrangement of an NMR spectrometer is displayed below. A sample (in a small glass tube) is placed between the poles of a strong magnetic. A radio frequency generator pulses the sample and excites the nuclei causing a spin-flip. The spin flip is detected by the detector and the signal sent to a computer where it is processed.
Exercise
1. If in a field strength of 4.7 T, H1 requires 200 MHz of energy to maintain resonance. If atom X requires 150 MHz, calculate the amount of energy required to spin flip atom X’s nucleus. Is this amount greater than the energy required for hydrogen?
2. Calculate the energy required to spin flip at 400 MHz. Does changing the frequency to 500 MHz decrease or increase the energy required? What about 300 MHz.
Answer
1.
E = hυ
E = (6.62 × 10−34)(150 MHz)
E = 9.93 × 10−26 J
The energy is equal to 9.93x10-26 J. This value is smaller than the energy required for hydrogen (1.324 × 10−25 J).
2.
E = hυ
E = (6.62 × 10−34)(400 MHz)
E = 2.648 × 10−25 J
The energy would increase if the frequency would increase to 500 MHz, and decrease if the frequency would decrease to 300 MHz. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.01%3A_Theory_of_Nuclear_Magnetic_Resonance_%28NMR%29.txt |
Objectives
After completing this section, you should be able to
1. explain, in general terms, the origin of shielding effects in NMR spectroscopy.
2. explain the number of peaks occurring in the 1H or 13C NMR spectrum of a simple compound, such as methyl acetate.
3. describe, and sketch a diagram of, a simple NMR spectrometer.
4. explain the difference in time scales of NMR and infrared spectroscopy.
5. predict the number of peaks expected in the 1H or 13C NMR spectrum of a given compound.
Study Notes
Before you go on, make sure that you understand that each signal in the 1H NMR spectrum shown for methyl acetate is due to a different proton environment. The three protons on the same methyl group are equivalent and appear in the spectrum as one signal. However, the two methyl groups are in two different environments (one is more deshielded) and so we see two signals in the whole spectrum (aside from the TMS reference peak).
Methyl acetate has a very simple 1H NMR spectrum, because there is no proton-proton coupling, and therefore no splitting of the signals. In later sections, we discuss splitting patterns in 1H NMR spectra and how they help a chemist determine the structure of organic compounds.
The basics of an NMR experiment
Given that chemically nonequivalent protons have different resonance frequencies in the same applied magnetic field, we can see how NMR spectroscopy can provide us with useful information about the structure of an organic molecule. A full explanation of how a modern NMR instrument functions is beyond the scope of this text, but in very simple terms, here is what happens. First, a sample compound (we'll use methyl acetate) is placed inside a very strong applied magnetic field (B0).
All of the protons begin to precess: the Ha protons at precessional frequency ωa, the Hb protons at ωb.At first, the magnetic moments of (slightly more than) half of the protons are aligned with B0, and half are aligned against B0. Then, the sample is hit with electromagnetic radiation in the radio frequency range. The two specific frequencies which match ωaandωb(i.e. the resonance frequencies) cause those Ha and Hb protons which are aligned with B0 to 'flip' so that they are now aligned against B0. In doing so, the protons absorb radiation at the two resonance frequencies. The NMR instrument records which frequencies were absorbed, as well as the intensity of each absorbance.
In most cases, a sample being analyzed by NMR is in solution. If we use a common laboratory solvent (diethyl ether, acetone, dichloromethane, ethanol, water, etc.) to dissolve our NMR sample, however, we run into a problem – there many more solvent protons in solution than there are sample protons, so the signals from the sample protons will be overwhelmed. To get around this problem, we use special NMR solvents in which all protons have been replaced by deuterium. Recall that deuterium is NMR-active, but its resonance frequency is very different from that of protons, and thus it is `invisible` in 1H-NMR. Some common NMR solvents are shown below.
The Chemical Shift
Let's look at an actual 1H-NMR plot for methyl acetate. Just as in IR and UV-vis spectroscopy, the vertical axis corresponds to intensity of absorbance, the horizontal axis to frequency (typically the vertical axis is not shown in an NMR spectrum).
We see three absorbance signals: two of these correspond to Ha and Hb, while the peak at the far right of the spectrum corresponds to the 12 chemically equivalent protons in tetramethylsilane (TMS), a standard reference compound that was added to our sample.
You may be wondering about a few things at this point - why is TMS necessary, and what is the meaning of the `ppm (δ)` label on the horizontal axis? Shouldn't the frequency units be in Hz? Keep in mind that NMR instruments of many different applied field strengths are used in organic chemistry laboratories, and that the proton's resonance frequency range depends on the strength of the applied field. The spectrum above was generated on an instrument with an applied field of approximately 7.1 Tesla, at which strength protons resonate in the neighborhood of 300 million Hz (chemists refer to this as a 300 MHz instrument). If our colleague in another lab takes the NMR spectrum of the same molecule using an instrument with a 2.4 Tesla magnet, the protons will resonate at around 100 million Hz (so we’d call this a 100 MHz instrument). It would be inconvenient and confusing to always have to convert NMR data according to the field strength of the instrument used. Therefore, chemists report resonance frequencies not as absolute values in Hz, but rather as values relative to a common standard, generally the signal generated by the protons in TMS. This is where the ppm – parts per million – term comes in. Regardless of the magnetic field strength of the instrument being used, the resonance frequency of the 12 equivalent protons in TMS is defined as a zero point. The resonance frequencies of protons in the sample molecule are then reported in terms of how much higher they are, in ppm, relative to the TMS signal (almost all protons in organic molecules have a higher resonance frequency than those in TMS, for reasons we shall explore quite soon).
The two proton groups in our methyl acetate sample are recorded as resonating at frequencies 2.05 and 3.67 ppm higher than TMS. One-millionth (1.0 ppm) of 300 MHz is 300 Hz. Thus 2.05 ppm, on this instrument, corresponds to 615 Hz, and 3.67 ppm corresponds to 1101 Hz. If the TMS protons observed by our 7.1 Tesla instrument resonate at exactly 300,000,000 Hz, this means that the protons in our ethyl acetate samples are resonating at 300,000,615 and 300,001,101 Hz, respectively. Likewise, if the TMS protons in our colleague's 2.4 Tesla instrument resonate at exactly 100 MHz, the methyl acetate protons in her sample resonate at 100,000,205 and 100,000,367 Hz (on the 100 MHz instrument, 1.0 ppm corresponds to 100 Hz). The absolute frequency values in each case are not very useful – they will vary according to the instrument used – but the difference in resonance frequency from the TMS standard, expressed in parts per million, should be the same regardless of the instrument.
Expressed this way, the resonance frequency for a given proton in a molecule is called its chemical shift. A frequently used symbolic designation for chemical shift in ppm is the lower-case Greek letter delta (δ). Most protons in organic compounds have chemical shift values between 0 and 12 ppm from TMS, although values below zero and above 12 are occasionally observed. By convention, the left-hand side of an NMR spectrum (higher chemical shift) is called downfield, and the right-hand direction is called upfield.
In our methyl acetate example we included for illustrative purposes a small amount of TMS standard directly in the sample, as was the common procedure for determining the zero point with older NMR instruments．That practice is generally no longer necessary, as modern NMR instruments are designed to use the deuterium signal from the solvent as a standard reference point, then to extrapolate the 0 ppm baseline that corresponds to the TMS proton signal (in an applied field of 7.1 Tesla, the deuterium atom in CDCl3 resonates at 32 MHz, compared to 300 MHz for the protons in TMS). In the remaining NMR spectra that we will see in this text we will not see an actual TMS signal, but we can always assume that the 0 ppm point corresponds to where the TMS protons would resonate if they were present.
Example
A proton has a chemical shift (relative to TMS) of 4.56 ppm.
1. a) What is its chemical shift, expressed in Hz, in a 300 MHz instrument? On a 200 MHz instrument?
2. b) What is its resonance frequency, expressed in Hz, in a 300 MHz instrument? On a 200 MHz instrument?
(Assume that in these instruments, the TMS protons resonate at exactly 300 or 200 MHz, respectively)
Solution
Diamagnetic shielding and deshielding
We come now to the question of why nonequivalent protons have different chemical shifts. The chemical shift of a given proton is determined primarily by its immediate electronic environment. Consider the methane molecule (CH4), in which the protons have a chemical shift of 0.23 ppm. The valence electrons around the methyl carbon, when subjected to B0, are induced to circulate and thus generate their own very small magnetic field that opposes B0. This induced field, to a small but significant degree, shields the nearby protons from experiencing the full force of B0, an effect known as local diamagnetic shielding. The methane protons therefore do not experience the full force of B0 - what they experience is called Beff, or the effective field, which is slightly weaker than B0.
Therefore, their resonance frequency is slightly lower than what it would be if they did not have electrons nearby to shield them.
Now consider methyl fluoride, CH3F, in which the protons have a chemical shift of 4.26 ppm, significantly higher than that of methane. This is caused by something called the deshielding effect. Because fluorine is more electronegative than carbon, it pulls valence electrons away from the carbon, effectively decreasing the electron density around each of the protons. For the protons, lower electron density means less diamagnetic shielding, which in turn means a greater overall exposure to B0, a stronger Beff, and a higher resonance frequency. Put another way, the fluorine, by pulling electron density away from the protons, is deshielding them, leaving them more exposed to B0. As the electronegativity of the substituent increases, so does the extent of deshielding, and so does the chemical shift. This is evident when we look at the chemical shifts of methane and three halomethane compounds (remember that electronegativity increases as we move up a column in the periodic table).
To a large extent, then, we can predict trends in chemical shift by considering how much deshielding is taking place near a proton. The chemical shift of trichloromethane is, as expected, higher than that of dichloromethane, which is in turn higher than that of chloromethane.
The deshielding effect of an electronegative substituent diminishes sharply with increasing distance:
The presence of an electronegative oxygen, nitrogen, sulfur, or sp2-hybridized carbon also tends to shift the NMR signals of nearby protons slightly downfield:
Table 2 lists typical chemical shift values for protons in different chemical environments.
Armed with this information, we can finally assign the two peaks in the the 1H-NMR spectrum of methyl acetate that we saw a few pages back. The signal at 3.65 ppm corresponds to the methyl ester protons (Hb), which are deshielded by the adjacent oxygen atom. The upfield signal at 2.05 ppm corresponds to the acetate protons (Ha), which is deshielded - but to a lesser extent - by the adjacent carbonyl group.
Finally, a note on the use of TMS as a standard in NMR spectroscopy: one of the main reasons why the TMS proton signal was chosen as a zero-point is that the TMS protons are highly shielded: silicon is slightly less electronegative than carbon, and therefore donates some additional shielding electron density. Very few organic molecules contain protons with chemical shifts that are negative relative to TMS.
Exercise
3. 2-cholorobutene shows 4 different hydrogen signals. Explain why this is.
Answer
3. The same colors represent the same signal. 4 different colors for 4 different signals. The hydrogen on the alkene would give two different signals. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.02%3A_NMR_Spectra_-_an_introduction_and_overview.txt |
Objectives
After completing this section, you should be able to
1. describe the delta scale used in NMR spectroscopy.
2. perform calculations based on the relationship between the delta value (in ppm), the observed chemical shift (in Hz), and the operating frequency of an NMR spectrometer (in Hz).
Key Terms
Make certain that you can define, and use in context, the key terms below.
• chemical shift
• delta scale
• upfield/downfield
Study Notes
Although the calculations described in this section will help you understand the principles of NMR, it is the actual delta values, not the calculations, which are of greatest importance to the beginning organic chemist. Thus, we shall try to focus on the interpretation of NMR spectra, not the mathematical aspects of the technique.
In Section 13.9 we discuss 1H NMR chemical shifts in more detail. Although you will eventually be expected to associate the approximate region of a 1H NMR spectrum with a particular type of proton, you are expected to use a general table of 1H NMR chemical shifts such as the one shown in Section 13.9.
Chemical Shifts
The NMR spectra is displayed as a plot of the applied radio frequency versus the absorption. The applied frequency increases from left to right, thus the left side of the plot is the low field, downfield or deshielded side and the right side of the plot is the high field, upfield or shielded side (see the figure below). The concept of shielding will be explained shortly.
The position on the plot at which the nuclei absorbs is called the chemical shift. Since this has an arbitrary value a standard reference point must be used. The two most common standards are TMS (tetramethylsilane, (Si(CH3)4) which has been assigned a chemical shift of zero, and CDCl3 (deuterochloroform) which has a chemical shift of 7.26 for 1H NMR and 77 for 13C NMR.
The scale is commonly expressed as parts per million (ppm) which is independent of the spectrometer frequency. The scale is the delta (δ) scale.
The range at which most NMR absorptions occur is quite narrow. Almost all 1H absorptions occur downfield within 10 ppm of TMS. For 13C NMR almost all absorptions occurs within 220 ppm downfield of the C atom in TMS.
Shielding in NMR
Structural features of the molecule will have an effect on the exact magnitude of the magnetic field experienced by a particular nucleus. This means that H atoms which have different chemical environments will have different chemical shifts. This is what makes NMR so useful for structure determination in organic chemistry. There are three main features that will affect the shielding of the nucleus, electronegativity, magnetic anisotropy of π systems and hydrogen bonding.
Electronegativity
The electrons that surround the nucleus are in motion so they created their own electromagnetic field. This field opposes the the applied magnetic field and so reduces the field experienced by the nucleus. Thus the electrons are said to shield the nucleus. Since the magnetic field experienced at the nucleus defines the energy difference between spin states it also defines what the chemical shift will be for that nucleus. Electron with-drawing groups can decrease the electron density at the nucleus, deshielding the nucleus and result in a larger chemical shift. Compare the data in the table below.
Compound, CH3X CH3F CH3OH CH3Cl CH3Br CH3I CH4 (CH3)4Si
Electronegativity of X 4.0 3.5 3.1 2.8 2.5 2.1 1.8
Chemical shift δ (ppm) 4.26 3.4 3.05 2.68 2.16 0.23 0
As can be seen from the data, as the electronegativity of X increases the chemical shift, δ increases. This is an effect of the halide atom pulling the electron density away from the methyl group. This exposes the nuclei of both the C and H atoms, "deshielding" the nuclei and shifting the peak downfield.
The effects are cumulative so the presence of more electron withdrawing groups will produce a greater deshielding and therefore a larger chemical shift, i.e.
Compound CH4 CH3Cl CH2Cl2 CHCl3
δ (ppm) 0.23 3.05 5.30 7.27
These inductive effects are not only felt by the immediately adjacent atoms, but the deshielding can occur further down the chain, i.e.
NMR signal -CH2-CH2-CH2Br
δ (ppm) 1.25 1.69 3.30
Magnetic Anisotropy: Pi Electron Effects
The π electrons in a compound, when placed in a magnetic field, will move and generate their own magnetic field. The new magnetic field will have an effect on the shielding of atoms within the field. The best example of this is benzene (see the figure below).
This effect is common for any atoms near a π bond, i.e.
Proton Type Effect Chemical shift (ppm)
C6H5-H highly deshielded 6.5 - 8
C=C-H deshielded 4.5 - 6
C≡C-H shielded* ~2.5
O=C-H very highly deshielded 9 - 10
* the acetylene H is shielded due to its location relative to the π system
Hydrogen Bonding
Protons that are involved in hydrogen bonding (i.e.-OH or -NH) are usually observed over a wide range of chemical shifts. This is due to the deshielding that occurs in the hydrogen bond. Since hydrogen bonds are dynamic, constantly forming, breaking and forming again, there will be a wide range of hydrogen bonds strengths and consequently a wide range of deshielding. This as well as solvation effects, acidity, concentration and temperature make it very difficult to predict the chemical shifts for these atoms.
Experimentally -OH and -NH can be identified by carrying out a simple D2O exchange experiment since these protons are exchangeable.
• run the normal H-NMR experiment on your sample
• add a few drops of D2O
• re-run the H-NMR experiment
• compare the two spectra and look for peaks that have "disappeared"
Exercise
4. The following peaks were from a H1 NMR spectra from a 400 MHz spectrometer. Convert to δ units
A. CHCl3 1451 Hz
B. CH3Cl 610 Hz
C. CH3OH 693 Hz
D. CH2Cl2 1060 Hz
5. Butan-2-one shows a chemical shift around 2.1 on a 300 MHz spectrometer in the H1 NMR spectrum.
A. How far downfield is this peak from TMS in Hz?
B. If the spectrum was done with a 400 MHz instrument, would a different chemical shift be seen?
C. On this new 400 MHz spectrum, what would be the difference in Hz from the chemical shift and TMS?
Answer
4.
A. 3.627 ppm
B. 1.525 ppm
C. 1.732 ppm
D. 2.65 ppm
5.
A. Since TMS is at 0 δ = 0 Hz for reference, the difference between the two would be 630 Hz
B. No not a different chemical shift, but a different frequency would be seen, 840 Hz
C. 840 Hz | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.03%3A_Chemical_Shifts_and_Shielding.txt |
Objectives
After completing this section, you should be able to
1. identify those protons which are equivalent in a given chemical structure.
2. use the 1H NMR spectrum of a simple organic compound to determine the number of equivalent sets of protons present.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• diastereotopic
• enantiotopic
• homotopic
Study Notes
It is important at this stage to be able to identify equivalent protons in any organic compound given the structure of that compound. Once you know the number of different groups of equivalent protons in a compound, you can predict the number (before coupling) and relative strength of signals. Look at the following examples and make sure you understand how the number and intensity ratio of signals are derived from the structure shown.
Structure Number of Signals Ratio of Signals
CH3OCH2CH2Br 3 A : B : C 3 : 2 : 2
1
3 A : B : C 2 : 2 : 6 (or 1 : 1 : 3)
3 A : B : C 2 : 4 : 2 (or 1 : 2 : 1)
4 A : B : C : D 3 : 2 : 2 : 3
5 A : B : C : D : E 3 : 1 : 1 : 1 : 1
If all protons in all organic molecules had the same resonance frequency in an external magnetic field of a given strength, the information in the previous paragraph would be interesting from a theoretical standpoint, but would not be terribly useful to organic chemists. Fortunately for us, however, resonance frequencies are not uniform for all protons in a molecule. In an external magnetic field of a given strength, protons in different locations in a molecule have different resonance frequencies, because they are in non-identical electronic environments. In methyl acetate, for example, there are two ‘sets’ of protons. The three protons labeled Ha have a different - and easily distinguishable – resonance frequency than the three Hb protons, because the two sets of protons are in non-identical environments: they are, in other words, chemically nonequivalent.
On the other hand, the three Ha protons are all in the same electronic environment, and are chemically equivalent to one another. They have identical resonance frequencies. The same can be said for the three Hb protons.
The ability to recognize chemical equivalancy and nonequivalency among atoms in a molecule will be central to understanding NMR. In each of the molecules below, all protons are chemically equivalent, and therefore will have the same resonance frequency in an NMR experiment.
You might expect that the equitorial and axial hydrogens in cyclohexane would be non-equivalent, and would have different resonance frequencies. In fact, an axial hydrogen is in a different electronic environment than an equitorial hydrogen. Remember, though, that the molecule rotates rapidly between its two chair conformations, meaning that any given hydrogen is rapidly moving back and forth between equitorial and axial positions. It turns out that, except at extremely low temperatures, this rotational motion occurs on a time scale that is much faster than the time scale of an NMR experiment.
In this sense, NMR is like a camera that takes photographs of a rapidly moving object with a slow shutter speed - the result is a blurred image. In NMR terms, this means that all 12 protons in cyclohexane are equivalent.
Each the molecules in the next figure contains two sets of protons, just like our previous example of methyl acetate, and again in each case the resonance frequency of the Ha protons will be different from that of the Hb protons.
Notice how the symmetry of para-xylene results in there being only two different sets of protons.
Most organic molecules have several sets of protons in different chemical environments, and each set, in theory, will have a different resonance frequency in 1H-NMR spectroscopy.
When stereochemistry is taken into account, the issue of equivalence vs nonequivalence in NMR starts to get a little more complicated. It should be fairly intuitive that hydrogens on different sides of asymmetric ring structures and double bonds are in different electronic environments, and thus are non-equivalent and have different resonance frequencies. In the alkene and cyclohexene structures below, for example, Ha is trans to the chlorine substituent, while Hb is cis to chlorine.
What is not so intuitive is that diastereotopic hydrogens (section 3.10) on chiral molecules are also non-equivalent:
However, enantiotopic and homotopic hydrogens are chemically equivalent.
Example
How many different sets of protons do the following molecules contain? (count diastereotopic protons as non-equivalent).
Solution
Exercise
6. How many non-equivalent hydrogen are in the following molecules; how many different signals will you see in a H1 NMR spectrum.
A. CH3CH2CH2Br
B. CH3OCH2C(CH3)3
C. Ethyl Benzene
D. 2-methyl-1-hexene
Answer
6. A. 3; B. 3; C. 5; D. 7 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.04%3A_H_NMR_Spectroscopy_and_Proton_Equivalence.txt |
Objectives
After completing this section, you should be able to
1. state the approximate chemical shift (δ) for the following types of protons:
1. aromatic.
2. vinylic.
3. those bonded to carbon atoms which are in turn bonded to a highly electronegative element.
4. those bonded to carbons which are next to unsaturated centres.
5. those bonded to carbons which are part of a saturated system.
2. predict the approximate chemical shifts of each of the protons in an organic compound, given its structure and a table of chemical shift correlations.
Study Notes
You should not attempt to memorize the chemical shifts listed in the table of this section, although it is probable that you will need to refer to it quite frequently throughout the remainder of this course. To fulfil Objective 1, above, you should be familiar with the information presented in the figure of chemical shift ranges for organic compounds. If you have an approximate idea of the chemical shifts of some of the most common types of protons, you will find the interpretation of 1H NMR spectra less arduous than it might otherwise be. Notice that we shall not try to understand why aromatic protons are deshielded or why alkynyl protons are not deshielded as much as vinylic protons. These phenonomena can be explained, but the focus is on the interpretation of 1H NMR spectra, not on the underlying theory.
1H NMR Chemical Shifts
Chemical shift is associated with the Larmor frequency of a nuclear spin to its chemical environment. Tetramethylsilan[TMS;(CH3)4Si] is generally used for standard to determine chemical shift of compounds: δTMS=0ppm. In other words, frequencies for chemicals are measured for a 1H or 13C nucleus of a sample from the 1H or 13C resonance of TMS. It is important to understand trend of chemical shift in terms of NMR interpretation. The proton NMR chemical shift is affect by nearness to electronegative atoms (O, N, halogen.) and unsaturated groups (C=C,C=O, aromatic). Electronegative groups move to the down field (left; increase in ppm). Unsaturated groups shift to downfield (left) when affecting nucleus is in the plane of the unsaturation, but reverse shift takes place in the regions above and below this plane. 1H chemical shift play a role in identifying many functional groups. Figure 1. indicates important example to figure out the functional groups.
Chemical shift values are in parts per million (ppm) relative to tetramethylsilane.
Hydrogen type
Chemical shift (ppm)
RCH3
0.9 - 1.0
RCH2R
1.2 - 1.7
R3CH
1.5 – 2.0
2.0 – 2.3
1.5 – 1.8
RNH2
1 - 3
ArCH3
2.2 – 2.4
2.3 – 3.0
ROCH3
3.7 – 3.9
3.7 – 3.9
ROH
1 - 5
3.7 – 6.5
5 - 9
ArH
6.0 – 8.7
9.5 – 10.0
10 - 13
Exercise
7. The following have one H1 NMR peak. In each case predict approximately where this peak would be in a spectra.
8. Identify the different equivalent protons in the following molecule and predict their expected chemical shift.
Answer
7. A. 5.20 δ; B. 1.50 δ; C. 6.40 δ; D. 1.00 δ
8. There are 6 different protons in this molecule
The shifts are (close) to the following: (a) 2 δ; (b) 6 δ; (c) 6.5 δ; (d) 7 δ; (e) 7.5 δ; (f) 7 δ | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.05%3A_Functional_Groups_and_Chemical_Shifts_in_H_NMR__Spe.txt |
Objectives
After completing this section, you should be able to
1. explain what information can be obtained from an integrated 1H NMR spectrum, and use this information in the interpretation of such a spectrum.
2. use an integrated 1H NMR spectrum to determine the ratio of the different types of protons present in an organic compound.
Study Notes
The concept of peak integration is that the area of a given peak in a 1H NMR spectrum is proportional to the number of (equivalent) protons giving rise to the peak. Thus, a peak which is caused by a single, unique proton has an area which measures one third of the area of a peak resulting from a methyl (CH3) group in the same spectrum.
In practice, we do not have to measure these areas ourselves: it is all done electronically by the spectrometer, and an integration curve is superimposed on the rest of the spectrum. The integration curve appears as a series of steps, with the height of each step being proportional to the area of the corresponding absorption peak, and consequently, to the number of protons responsible for the absorption.
As it can be difficult to decide precisely where to start and stop when measuring integrations, you should not expect your ratios to be exact whole numbers.
Signal integration
The computer in an NMR instrument can be instructed to automatically integrate the area under a signal or group of signals. This is very useful, because in 1H-NMR spectroscopy the area under a signal is proportional to the number of hydrogens to which the peak corresponds. The two signals in the methyl acetate spectrum, for example, integrate to approximately the same area, because they both correspond to a set of three equivalent protons.
Take a look next at the spectrum of para-xylene (IUPAC name 1,4-dimethylbenzene):
This molecule has two sets of protons: the six methyl (Ha) protons and the four aromatic (Hb) protons. When we instruct the instrument to integrate the areas under the two signals, we find that the area under the peak at 2.6 ppm is 1.5 times greater than the area under the peak at 7.4 ppm. This (along with the actual chemical shift values, which we'll discuss soon) tells us which set of protons corresponds to which NMR signal.
The integration function can also be used to determine the relative amounts of two or more compounds in a mixed sample. If we have a sample that is a 50:50 (mole/mole) mixture of benzene and acetone, for example, the acetone signal should integrate to the same value as the benzene sample, because both signals represent six equivalent protons. If we have a 50:50 mixture of acetone and cyclopentane, on the other hand, the ratio of the acetone peak area to the cylopentane peak area will be 3:5 (or 6:10), because the cyclopentane signal represents ten protons.
Example \(1\)
You take a 1H-NMR spectrum of a mixed sample of acetone (CH3(CO)CH3) and dichloromethane (CH2Cl2). The integral ratio of the two signals (acetone : dichloromethane) is 2.3 to 1. What is the molar ratio of the two compounds in the sample?
Example \(2\)
You take the 1H-NMR spectrum of a mixed sample of 36% para-xylene and 64% acetone in CDCl3 solvent (structures are shown earlier in this chapter). How many peaks do you expect to see? What is the expected ratio of integration values for these peaks? (set the acetone peak integration equal to 1.0)
Exercise
9. Predict how many signals the following molecule would have? Sketch the spectra and estimate the integration of the peaks.
Answer
9. There will be two peaks. Ideal general spectrum shown with integration. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.06%3A_Integration_of_H_NMR_Absorptions-_Proton_Counting.txt |
Objectives
After completing this section, you should be able to
1. explain the spin-spin splitting pattern observed in the 1H NMR spectrum of a simple organic compound, such as chloroethane or 2-bromopropane.
2. interpret the splitting pattern of a given 1H NMR spectrum.
3. determine the structure of a relatively simple organic compound, given its 1H NMR spectrum and other relevant information.
4. use coupling constants to determine which groups of protons are coupling with one another in a 1H NMR spectrum.
5. predict the splitting pattern which should be observed in the 1H NMR spectrum of a given organic compound.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• coupling constant
• multiplet
• quartet
• triplet
• doublet
Study Notes
From what we have learned about 1H NMR spectra so far, we might predict that the spectrum of 1,1,2-trichloroethane, CHCl2CH2Cl, would consist of two peaks—one, at about 2.5-4.0 δ, expected for CH2-halogen compounds and one shifted downfield because of the presence of an additional electronegative chlorine atom on the second carbon. However, when we look at the spectrum it appears to be much more complex. True, we see absorptions in the regions we predicted, but these absorptions appear as a group of two peaks (a doublet) and a group of three peaks (a triplet). This complication, which may be disturbing to a student who longs for the simple life, is in fact very useful to the organic chemist, and adds greatly to the power of NMR spectroscopy as a tool for the elucidation of chemical structures. The split peaks (multiplets) arise because the magnetic field experienced by the protons of one group is influenced by the spin arrangements of the protons in an adjacent group.
Spin-spin coupling is often one of the more challenging topics for organic chemistry students to master. Remember the n + 1 rule and the associated coupling patterns.
The source of spin-spin coupling
The 1H-NMR spectra that we have seen so far (of methyl acetate and para-xylene) are somewhat unusual in the sense that in both of these molecules, each set of protons generates a single NMR signal. In fact, the 1H-NMR spectra of most organic molecules contain proton signals that are 'split' into two or more sub-peaks. Rather than being a complication, however, this splitting behavior actually provides us with more information about our sample molecule.
Consider the spectrum for 1,1,2-trichloroethane. In this and in many spectra to follow, we show enlargements of individual signals so that the signal splitting patterns are recognizable.
The signal at 3.96 ppm, corresponding to the two Ha protons, is split into two subpeaks of equal height (and area) – this is referred to as a doublet. The Hb signal at 5.76 ppm, on the other hand, is split into three sub-peaks, with the middle peak higher than the two outside peaks - if we were to integrate each subpeak, we would see that the area under the middle peak is twice that of each of the outside peaks. This is called a triplet.
The source of signal splitting is a phenomenon called spin-spin coupling, a term that describes the magnetic interactions between neighboring, non-equivalent NMR-active nuclei. In our 1,1,2 trichloromethane example, the Ha and Hb protons are spin-coupled to each other. Here's how it works, looking first at the Ha signal: in addition to being shielded by nearby valence electrons, each of the Ha protons is also influenced by the small magnetic field generated by Hb next door (remember, each spinning proton is like a tiny magnet). The magnetic moment of Hb will be aligned with B0 in (slightly more than) half of the molecules in the sample, while in the remaining half of the molecules it will be opposed to B0. The Beff ‘felt’ by Ha is a slightly weaker if Hb is aligned against B0, or slightly stronger if Hb is aligned with B0. In other words, in half of the molecules Ha is shielded by Hb (thus the NMR signal is shifted slightly upfield) and in the other half Ha is deshielded by Hb(and the NMR signal shifted slightly downfield). What would otherwise be a single Ha peak has been split into two sub-peaks (a doublet), one upfield and one downfield of the original signal. These ideas an be illustrated by a splitting diagram, as shown below.
Now, let's think about the Hbsignal. The magnetic environment experienced by Hb is influenced by the fields of both neighboring Ha protons, which we will call Ha1 and Ha2. There are four possibilities here, each of which is equally probable. First, the magnetic fields of both Ha1 and Ha2 could be aligned with B0, which would deshield Hb, shifting its NMR signal slightly downfield. Second, both the Ha1 and Ha2 magnetic fields could be aligned opposed to B0, which would shield Hb, shifting its resonance signal slightly upfield. Third and fourth, Ha1 could be with B0 and Ha2 opposed, or Ha1opposed to B0 and Ha2 with B0. In each of the last two cases, the shielding effect of one Ha proton would cancel the deshielding effect of the other, and the chemical shift of Hb would be unchanged.
So in the end, the signal for Hb is a triplet, with the middle peak twice as large as the two outer peaks because there are two ways that Ha1 and Ha2 can cancel each other out.
Now, consider the spectrum for ethyl acetate:
We see an unsplit 'singlet' peak at 1.833 ppm that corresponds to the acetyl (Ha) hydrogens – this is similar to the signal for the acetate hydrogens in methyl acetate that we considered earlier. This signal is unsplit because there are no adjacent hydrogens on the molecule. The signal at 1.055 ppm for the Hc hydrogens is split into a triplet by the two Hb hydrogens next door. The explanation here is the same as the explanation for the triplet peak we saw previously for 1,1,2-trichloroethane.
The Hbhydrogens give rise to a quartet signal at 3.915 ppm – notice that the two middle peaks are taller then the two outside peaks. This splitting pattern results from the spin-coupling effect of the three Hc hydrogens next door, and can be explained by an analysis similar to that which we used to explain the doublet and triplet patterns.
Example
1. Explain, using left and right arrows to illustrate the possible combinations of nuclear spin states for the Hc hydrogens, why the Hb signal in ethyl acetate is split into a quartet.
2. The integration ratio of doublets is 1:1, and of triplets is 1:2:1. What is the integration ratio of the Hb quartet in ethyl acetate? (Hint – use the illustration that you drew in part a to answer this question.)
Solution
By now, you probably have recognized the pattern which is usually referred to as the n + 1 rule: if a set of hydrogens has n neighboring, non-equivalent hydrogens, it will be split into n + 1 subpeaks. Thus the two Hb hydrogens in ethyl acetate split the Hc signal into a triplet, and the three Hc hydrogens split the Hb signal into a quartet. This is very useful information if we are trying to determine the structure of an unknown molecule: if we see a triplet signal, we know that the corresponding hydrogen or set of hydrogens has two `neighbors`. When we begin to determine structures of unknown compounds using 1H-NMR spectral data, it will become more apparent how this kind of information can be used.
Three important points need to be emphasized here. First, signal splitting only occurs between non-equivalent hydrogens – in other words, Ha1 in 1,1,2-trichloroethane is not split by Ha2, and vice-versa.
Second, splitting occurs primarily between hydrogens that are separated by three bonds. This is why the Ha hydrogens in ethyl acetate form a singlet– the nearest hydrogen neighbors are five bonds away, too far for coupling to occur.
Occasionally we will see four-bond and even 5-bond splitting, but in these cases the magnetic influence of one set of hydrogens on the other set is much more subtle than what we typically see in three-bond splitting (more details about how we quantify coupling interactions is provided in section 5.5B). Finally, splitting is most noticeable with hydrogens bonded to carbon. Hydrogens that are bonded to heteroatoms (alcohol or amino hydrogens, for example) are coupled weakly - or not at all - to their neighbors. This has to do with the fact that these protons exchange rapidly with solvent or other sample molecules.
Below are a few more examples of chemical shift and splitting pattern information for some relatively simple organic molecules.
Multiplicity in Proton NMR
The number of lines in a peak is always one more (n+1) than the number of hydrogens on the neighboring carbon. This table summarizes coupling patterns that arise when protons have different numbers of neighbors.
# of lines
ratio of lines
term for peak
# of neighbors
1
-
singlet
0
2
1:1
doublet
1
3
1:2:1
triplet
2
4
1:3:3:1
quartet
3
5
1:4:6:4:1
quintet
4
6
1:5:10:10:5:1
sextet
5
7
1:6:15:20:15:6:1
septet
6
8
1:7:21:35:35:21:7:1
octet
7
9
1:8:28:56:70:56:28:8:1
nonet
8
Example
How many proton signals would you expect to see in the 1H-NMR spectrum of triclosan (a common antimicrobial agent found in detergents)? For each of the proton signals, predict the splitting pattern. Assume that you see only 3-bond coupling.
Solutions
Example
Predict the splitting pattern for the 1H-NMR signals corresponding to the protons at the locations indicated by arrows (the structure is that of the neurotransmitter serotonin).
Solutions
Coupling constants
Chemists quantify the spin-spin coupling effect using something called the coupling constant, which is abbreviated with the capital letter J. The coupling constant is simply the difference, expressed in Hz, between two adjacent sub-peaks in a split signal. For our doublet in the 1,1,2-trichloroethane spectrum, for example, the two subpeaks are separated by 6.1 Hz, and thus we write 3Ja-b = 6.1 Hz.
The superscript 3 tells us that this is a three-bond coupling interaction, and the a-b subscript tells us that we are talking about coupling between Ha and Hb. Unlike the chemical shift value, the coupling constant, expressed in Hz, is the same regardless of the applied field strength of the NMR magnet. This is because the strength of the magnetic moment of a neighboring proton, which is the source of the spin-spin coupling phenomenon, does not depend on the applied field strength.
When we look closely at the triplet signal in 1,1,2-trichloroethane, we see that the coupling constant - the `gap` between subpeaks - is 6.1 Hz, the same as for the doublet. This is an important concept! The coupling constant 3Ja-b quantifies the magnetic interaction between the Ha and Hb hydrogen sets, and this interaction is of the same magnitude in either direction. In other words, Ha influences Hb to the same extent that Hb influences Ha. When looking at more complex NMR spectra, this idea of reciprocal coupling constants can be very helpful in identifying the coupling relationships between proton sets.
Coupling constants between proton sets on neighboring sp3-hybridized carbons is typically in the region of 6-8 Hz. With protons bound to sp2-hybridized carbons, coupling constants can range from 0 Hz (no coupling at all) to 18 Hz, depending on the bonding arrangement.
For vinylic hydrogens in a trans configuration, we see coupling constants in the range of 3J = 11-18 Hz, while cis hydrogens couple in the 3J = 6-15 Hz range. The 2-bond coupling between hydrogens bound to the same alkene carbon (referred to as geminal hydrogens) is very fine, generally 5 Hz or lower. Ortho hydrogens on a benzene ring couple at 6-10 Hz, while 4-bond coupling of up to 4 Hz is sometimes seen between meta hydrogens.
Fine (2-3 Hz) coupling is often seen between an aldehyde proton and a three-bond neighbor. Table 4 lists typical constant values.
Exercise
10. Predict the splitting patterns of the following molecules:
11. Draw the following according to the criteria given.
A. C3H5O; two triplet, 1 doublet
B. C4H8O2; three singlets
C. C5H12; one singlet
12. The following spectrum is for C3H8O. Determine the structure.
A triplet; B singlet; C sextet; D triplet
Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 3 December 2016)
Answer
10.
A. H: Doublet. H: Septet
B. H: Doublet, H: Triplet
C. H: Singlet, H: Quartet, H: Triplet
11.
These are just some drawings, more may be possible.
12.
Note: Remember, chemically equivalent protons do not couple with one another to give spin-spin splitting. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.07%3A_Spin-Spin_Splitting_in_H_NMR__Spectra.txt |
Objectives
After completing this section, you should be able to
1. explain how multiple coupling can give rise to complex-looking 1H NMR spectra.
2. predict the splitting pattern expected in the 1H NMR spectrum of an organic compound in which multiple coupling is possible.
3. interpret 1H NMR spectra in which multiple coupling is evident.
Key Terms
Make certain that you can define, and use in context, the key term below.
• tree diagram
Study Notes
We saw the effects of spin-spin coupling on the appearance of a 1H NMR signal. These effects can be further complicated when that signal is coupled to several different protons. For example, BrCH2CH2CH2Cl would produce three signals. The hydrogens at C1 and C3 would each be triplets because of coupling to the two hydrogens on C2. However, the hydrogen on C2 “sees” two different sets of neighbouring hydrogens, and would therefore produce a triplet of triplets.
Another effect that can complicate a spectrum is the “closeness” of signals. If signals accidently overlap they can be difficult to identify. In the example above, we expected a triplet of triplets. However, if the coupling is identical (or almost identical) between the hydrogens on C2 and the hydrogens on both C1 and C3, one would observe a quintet in the 1H NMR spectrum. [You can try this yourself by drawing a tree diagram of a triplet of triplets assuming, first, different coupling constants, and then, identical coupling constants.] Keep this point in mind when interpreting real 1H NMR spectra.
Also, when multiplets are well separated, they form patterns. However, when multiplets approach each other in the spectrum they sometimes become distorted. Usually, the inner peaks become larger than the outer peaks. Note the following examples:
Aromatic ring protons quite commonly have overlapping signals and multiplet distortions. Sometimes you cannot distinguish between individual signals, and one or more messy multiplets often appear in the aromatic region.
It is much easier to rationalize the observed 1H NMR spectrum of a known compound than it is to determine the structure of an unknown compound from its 1H NMR spectrum. However, rationalizations can be a useful learning technique as you try to improve your proficiency in spectral interpretation. Remember that when a chemist tries to interpret the 1H NMR spectrum of an unknown compound, he or she usually has additional information available to make the task easier. For example, the chemist will almost certainly have an infrared spectrum of the compound and possibly a mass spectrum too. Details of how the compound was synthesized may be available, together with some indication of its chemical properties, its physical properties, or both.
In examinations, you will be given a range of information (IR, MS, UV data and empirical formulae) to aid you with your structural determination using 1H NMR spectroscopy. For example, you may be asked to determine the structure of C6H12O given the following spectra:
Infrared spectrum: 3000 cm−1 and 1720 cm−1 absorptions are both strong
1H NMR δ (ppm) Protons Multiplicity
0.87 6 doublet
1.72 1 broad multiplet
2.00 3 singlet
2.18 2 doublet
To answer this question, you note that the infrared spectrum of C6H12O shows \$\ce{\sf{C-H}}\$ stretching (3000 cm−1) and \$\ce{\sf{C-O}}\$ stretching (1720 cm−1). Now you have to piece together the information from the 1H NMR spectrum. Notice the singlet with three protons at 2.00 ppm. This signal indicates a methyl group that is not coupled to other protons. It could possibly mean the presence of a methyl ketone functional group.
The signal at 1.72 ppm is a broad multiplet, suggesting that a carbon with a single proton is beside carbons with several different protons.
The doublet signal at 2.18 ppm implies that a \$\ce{\sf{-CH2-}}\$ group is attached to a carbon having only one proton.
The six protons showing a doublet at 0.87 ppm indicate two equivalent methyl groups attached to a carbon with one proton.
Whenever you see a signal in the 0.7-1.3 ppm range that is a multiplet of three protons (3, 6, 9) it is most likely caused by equivalent methyl groups.
Using trial and error, and with the above observations, you should come up with the correct structure.
Complex coupling
In all of the examples of spin-spin coupling that we have seen so far, the observed splitting has resulted from the coupling of one set of hydrogens to just one neighboring set of hydrogens. When a set of hydrogens is coupled to two or more sets of nonequivalent neighbors, the result is a phenomenon called complex coupling. A good illustration is provided by the 1H-NMR spectrum of methyl acrylate:
First, let's first consider the Hc signal, which is centered at 6.21 ppm. Here is a closer look:
With this enlargement, it becomes evident that the Hc signal is actually composed of four sub-peaks. Why is this? Hc is coupled to both Ha and Hb , but with two different coupling constants. Once again, a splitting diagram (or tree diagram) can help us to understand what we are seeing. Ha is trans to Hc across the double bond, and splits the Hc signal into a doublet with a coupling constant of 3Jac = 17.4 Hz. In addition, each of these Hc doublet sub-peaks is split again by Hb (geminal coupling) into two more doublets, each with a much smaller coupling constant of 2Jbc = 1.5 Hz.
The result of this `double splitting` is a pattern referred to as a doublet of doublets, abbreviated `dd`.
The signal for Ha at 5.95 ppm is also a doublet of doublets, with coupling constants 3Jac= 17.4 Hz and 3Jab = 10.5 Hz.
The signal for Hb at 5.64 ppm is split into a doublet by Ha, a cis coupling with 3Jab = 10.4 Hz. Each of the resulting sub-peaks is split again by Hc, with the same geminal coupling constant 2Jbc = 1.5 Hz that we saw previously when we looked at the Hc signal. The overall result is again a doublet of doublets, this time with the two `sub-doublets` spaced slightly closer due to the smaller coupling constant for the cis interaction. Here is a blow-up of the actual Hbsignal:
Example
Construct a splitting diagram for the Hb signal in the 1H-NMR spectrum of methyl acrylate. Show the chemical shift value for each sub-peak, expressed in Hz (assume that the resonance frequency of TMS is exactly 300 MHz).
Solution
When constructing a splitting diagram to analyze complex coupling patterns, it is usually easier to show the larger splitting first, followed by the finer splitting (although the reverse would give the same end result).
When a proton is coupled to two different neighboring proton sets with identical or very close coupling constants, the splitting pattern that emerges often appears to follow the simple `n + 1 rule` of non-complex splitting. In the spectrum of 1,1,3-trichloropropane, for example, we would expect the signal for Hb to be split into a triplet by Ha, and again into doublets by Hc, resulting in a 'triplet of doublets'.
Ha and Hc are not equivalent (their chemical shifts are different), but it turns out that 3Jab is very close to 3Jbc. If we perform a splitting diagram analysis for Hb, we see that, due to the overlap of sub-peaks, the signal appears to be a quartet, and for all intents and purposes follows the n + 1 rule.
For similar reasons, the Hc peak in the spectrum of 2-pentanone appears as a sextet, split by the five combined Hb and Hd protons. Technically, this 'sextet' could be considered to be a 'triplet of quartets' with overlapping sub-peaks.
Example
What splitting pattern would you expect for the signal coresponding to Hb in the molecule below? Assume that Jab ~ Jbc. Draw a splitting diagram for this signal, and determine the relative integration values of each subpeak.
Solution
In many cases, it is difficult to fully analyze a complex splitting pattern. In the spectrum of toluene, for example, if we consider only 3-bond coupling we would expect the signal for Hb to be a doublet, Hd a triplet, and Hc a triplet.
In practice, however, all three aromatic proton groups have very similar chemical shifts and their signals overlap substantially, making such detailed analysis difficult. In this case, we would refer to the aromatic part of the spectrum as a multiplet.
When we start trying to analyze complex splitting patterns in larger molecules, we gain an appreciation for why scientists are willing to pay large sums of money (hundreds of thousands of dollars) for higher-field NMR instruments. Quite simply, the stronger our magnet is, the more resolution we get in our spectrum. In a 100 MHz instrument (with a magnet of approximately 2.4 Tesla field strength), the 12 ppm frequency 'window' in which we can observe proton signals is 1200 Hz wide. In a 500 MHz (~12 Tesla) instrument, however, the window is 6000 Hz - five times wider. In this sense, NMR instruments are like digital cameras and HDTVs: better resolution means more information and clearer pictures (and higher price tags!)
Practice Unknowns
1. Given the information below, draw the structures of compounds A through D.
1. An unknown compound A was prepared as follows:
Mass spectrum:
base peak m/e = 39
parent peak m/e = 54
1H NMR spectrum:
δ (ppm) Relative Area Multiplicity
1.0 2 triplet
5.4 1 quintet
2. Unknown compound B has the molecular formula C7H6O2.
Infrared spectrum:
3200 cm−1 (broad) and 1747 cm−1 (strong) absorptions
1H NMR spectrum:
δ (ppm) Protons
6.9 2
7.4 2
9.8 1
10.9 1
Hint: Aromatic ring currents deshield all proton signals just outside the ring.
3. Unknown compound C shows no evidence of unsaturation and contains only carbon and hydrogen.
Mass spectrum:
parent peak m/e = 68
1H NMR spectrum:
δ (ppm) Relative Area Multiplicity
1.84 3 triplet
2.45 1 septet
Hint: Think three dimensionally!
4. Unknown compound D (C15H14O) has the following spectral properties.
Infrared spectrum:
3010 cm−1 (medium)
1715 cm−1 (strong)
1610 cm−1 (strong)
1500 cm−1 (strong)
1H NMR spectrum:
δ (ppm) Relative Area Multiplicity
3.00 2 triplet
3.07 2 triplet
7.1-7.9 10 Multiplets
Answers
Exercise
13. In the following molecule, the C2 is coupled with both the vinyl, C1, and the alkyl C3. Draw the splitting tree diagram.
Answer
13. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.08%3A_More_Complex_Spin-Spin_Splitting_Patterns.txt |
There will be cases in which you already know what the structure might be. In these cases:
• You should draw attention to pieces of data that most strongly support your expected structure. This approach will demonstrate evaluative understanding of the data; that means you can look at data and decide what parts are more crucial than others.
• You should also draw attention to negative results: that is, peaks that might be there if this spectrum matched another, possible structure, but that are in fact missing.
One of the most complicated problems to deal with is the analysis of a mixture. This situation is not uncommon when students run reactions in lab and analyse the data.
• Sometimes the spectra show a little starting material mixed in with the product.
• Sometimes solvents show up in the spectrum.
• As you might expect, the minor component usually shows up as smaller peaks in the spectrum. If there are fewer molecules present, then there are usually fewer protons to absorb in the spectrum.
• In this case, you should probably make two completely separate sets of data tables for your analysis, one for each compound, or else one for the main compound and one for impurities.
Remember that integration ratios are really only meaningful within a single compound. If your NMR sample contains some benzene (C6H6) and some acetone (CH3COCH3), and there is a peak at 7.15 that integrates to 1 proton and a peak at 2.10 ppm integrating to 6 protons, it might mean there are 6 protons in acetone and 1 in benzene, but you can tell that isn't true by looking at the structure. There must be six times as many acetone molecules as benzene molecules in the sample.
There are six protons in the benzene, and they should all show up near 7 ppm. There are six protons in acetone, and they should all show up near 2 ppm. Assuming that small integral of 1H for the benzene is really supposed to be 6H, then the large integral of 6H for the acetone must also represent six times as many hydrogens, too. It would be 36 H. There are only six hydrogens in acetone, so it must represent six times as many acetone molecules as there are benzenes.
Similarly, if you have decided that you can identify two sets of peaks in the 1H spectrum, analysing them in different tables makes it easy to keep the integration analysis completely separate too ; 1 H in one table will not be the same size integral as 1 H in the other table unless the concentrations of the two compounds in the sample are the same.
However, comparing the ratio of two integrals for two different compounds can give you the ratio of the two compounds in solution, just as we could determine the ratio of benzene to acetone in the mixture described above.
We will look at two examples of sample mixtures that could arise in lab. Results like these are pretty common events in the labIn the first example, a student tried to carry out the following reaction, a borohydride reduction of an aldehyde. The borohydride should give a hydride anion to the C=O carbon; washing with water should then supply a proton to the oxygen, giving an alcohol.
Her reaction produced the following spectrum.
(simulated data)
From this data, she produced the table below.
Notice how she calculated that ratio. She found a peak in molecule 1, the aldehyde, that she was pretty sure corresponded to the aldehydic hydrogen, the H attached to the C=O; in other words, the CH=O. She found another peak from molecule 2, the alcohol, that she was pretty sure represented the two hydrogens on the carbon attached to oxygen, the CH2-O.
The integrals for those two peaks are equal. They are both 2H in her table. However, she notes that within each molecule, the first integral really represents 1H and the second represents 2H. That means there must be twice as many of molecule 1 as there are molecule 2. That way, there would be 2 x CH=O, and its integral would be the same as the 1 x CH2-O in the other molecule.
One way to approach this kind of problem is to:
• choose one peak from each of the two compounds you want to compare.
• decide how many hydrogens each peak is supposed to represent in a molecule. Is it supposed to be a CH2, a CH, a CH3?
• divide the integral value for that peak by that number of hydrogens it is supposed to represent in a molecule.
• compare the two answers (integral A / ideal # H) vs (integral B / ideal # H).
• the ratio of those two answers is the ratio of the two molecules in the sample.
So there is twice as much aldehyde as alcohol in the mixture. In terms of these two compounds alone, she has 33% alcohol and 66% aldehyde. That's ( 1/(1+2) ) x100% for the alcohol, and ( 2/(1+2) ) x100% for the aldehyde. That calculation just represents the amount of individual component divided by the total of the components she wants to compare.
There are a number of things to take note of here.
• Her reaction really didn't work very well. She still has majority starting material, not product.
• She will get a good grade on this lab. Although the experiment didn't work well, she has good data, and she has analyzed it very clearly.
• She has separated her data table into different sections for different compounds. Sometimes that makes it easier to analyze things.
• She has noted the actual integral data (she may have measured the integral with a ruler) and also converted it into a more convenient ratio, based on the integral for a peak that she felt certain about.
• She went one step further, and indicated the internal integration ratio within each individual compound.
• She calculated the % completion of the reaction using the integral data for the reactant and product, and she made clear what part of the data she used for that calculation. A similar procedure could be done if a student were just trying to separate two components in a mixture rather than carry out a reaction.
• She also calculated the overall purity of the mixture, including a solvent impurity that she failed to remove.
• However, CHCl3 is not included in her analysis of purity. CHCl3 really isn't part of her sample; it was just present in the NMR solvent, so it doesn't represent anything in the material she ended up with at the end of lab.
Another student carried out a similar reaction, shown below. He also finished the reaction by washing with water, but because methanol is soluble in water, he had to extract his product out of the water. He chose to use dichloromethane for that purpose.
He obtained the following data.
From this data, he constructed the following table.
There are some things to learn about this table, too.
• Does the integration ratio really match the integral data? Or is this just wishful thinking?
• This table might reflect what he wants to see in the data. But what else could be in the data?
• CHCl3 is often seen in NMR spectra if CDCl3 is used for the NMR sample. It's there, at 7.2 ppm.
• "Leftover" or residual solvent is very common in real lab data. There it is, CH2Cl2 from the extraction, at 5.4 ppm.
• What about water? Sometimes people don't dry their solutions properly before evaporating the solvent. There is probably water around 1.5 to 1.6 ppm here.
This student might not get a very good grade; the sample does not even show up in the spectrum, so he lost it somewhere. But his analysis is also poor, so he will really get a terrible grade.
Example
Three students performed a synthesis of a fragrant ester, ethyl propanoate, CH3CH2CO2CH2CH3. During their reactions, they each used a different solvent. The students were able to see peaks in the NMR spectrum for ethyl propanoate, as well as peaks for chloroform (CHCl3, in the CDCl3 they used to make their NMR samples).
• See the first student's spectrum.
• See the second student's spectrum.
• See the third student's spectrum.
They were also able to determine that they had some leftover solvent in their samples by consulting a useful table of solvent impurities in NMR (which they found in Goldberg et. al., Organometallics 2010, 29, 2176-2179).
1. What is the ratio of leftover solvent to ethyl propanoate in each sample?
2. What is the percent of each sample that is leftover solvent
Exercise
14. How can H1 NMR determine products? For example, how can you tell the difference between the products of this reaction?
Answer
14. Yes, you are able to determine the difference in the spectra. For the 2-chloro compound will have multiple quartets while the 1-chloro compound will only have a quintet and a triplet for the signals in the ring. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.09%3A_Uses_of_H_NMR_Spectroscopy.txt |
The basics of 13C-NMR spectroscopy
The magnetic moment of a 13C nucleus is much weaker than that of a proton, meaning that NMR signals from 13C nuclei are inherently much weaker than proton signals. This, combined with the low natural abundance of 13C, means that it is much more difficult to observe carbon signals: more sample is required, and often the data from hundreds of scans must be averaged in order to bring the signal-to-noise ratio down to acceptable levels. Unlike 1H-NMR signals, the area under a 13C-NMR signal cannot be used to determine the number of carbons to which it corresponds. This is because the signals for some types of carbons are inherently weaker than for other types – peaks corresponding to carbonyl carbons, for example, are much smaller than those for methyl or methylene (CH2) peaks. Peak integration is generally not useful in 13C-NMR spectroscopy, except when investigating molecules that have been enriched with 13C isotope (see section 5.6B).
The resonance frequencies of 13C nuclei are lower than those of protons in the same applied field - in a 7.05 Tesla instrument, protons resonate at about 300 MHz, while carbons resonate at about 75 MHz. This is fortunate, as it allows us to look at 13C signals using a completely separate 'window' of radio frequencies. Just like in 1H-NMR, the standard used in 13C-NMR experiments to define the 0 ppm point is tetramethylsilane (TMS), although of course in 13C-NMR it is the signal from the four equivalent carbons in TMS that serves as the standard. Chemical shifts for 13C nuclei in organic molecules are spread out over a much wider range than for protons – up to 200 ppm for 13C compared to 12 ppm for protons (see Table 3 for a list of typical 13C-NMR chemical shifts). This is also fortunate, because it means that the signal from each carbon in a compound can almost always be seen as a distinct peak, without the overlapping that often plagues 1H-NMR spectra. The chemical shift of a 13C nucleus is influenced by essentially the same factors that influence a proton's chemical shift: bonds to electronegative atoms and diamagnetic anisotropy effects tend to shift signals downfield (higher resonance frequency). In addition, sp2 hybridization results in a large downfield shift. The 13C-NMR signals for carbonyl carbons are generally the furthest downfield (170-220 ppm), due to both sp2 hybridization and to the double bond to oxygen.
Example \(1\)
How many sets of non-equivalent carbons are there in each of the molecules shown in exercise 5.1?
Example \(2\)
How many sets of non-equivalent carbons are there in:
1. toluene
2. 2-pentanone
3. para-xylene
4. triclosan
Because of the low natural abundance of 13C nuclei, it is very unlikely to find two 13C atoms near each other in the same molecule, and thus we do not see spin-spin coupling between neighboring carbons in a 13C-NMR spectrum. There is, however, heteronuclear coupling between 13C carbons and the hydrogens to which they are bound. Carbon-proton coupling constants are very large, on the order of 100 – 250 Hz. For clarity, chemists generally use a technique called broadband decoupling, which essentially 'turns off' C-H coupling, resulting in a spectrum in which all carbon signals are singlets. Below is the proton-decoupled13C-NMR spectrum of ethyl acetate, showing the expected four signals, one for each of the carbons.
While broadband decoupling results in a much simpler spectrum, useful information about the presence of neighboring protons is lost. However, another modern NMR technique called DEPT (Distortionless Enhancement by Polarization Transfer) allows us to determine how many hydrogens are bound to each carbon. For example, a DEPT experiment tells us that the signal at 171 ppm in the ethyl acetate spectrum is a quaternary carbon (no hydrogens bound, in this case a carbonyl carbon), that the 61 ppm signal is from a methylene (CH2) carbon, and that the 21 ppm and 14 ppm signals are both methyl (CH3) carbons. The details of the DEPT experiment are beyond the scope of this text, but DEPT information will often be provided along with 13C spectral data in examples and problems.
12.11: Chemical Shifts and Interpreting C NMR Spectra
13C NMR Chemical Shifts
The Carbon NMR is used for determining functional groups using characteristic shift values. 13C chemical shift is affect by electronegative effect and steric effect. If an H atoms in an alkane is replace by substituent X, electronegative atoms (O, N, halogen), ?-carbon and ?-carbon shift to downfield (left; increase in ppm) while ?-carbon shifts to upfield. The steric effect is observed in acyclic and clyclic system, which leads to downshifted chemical shifts. Figure 9 shows typical 13C chemical shift regions of the major chemical class.
Spin-Spin splitting
Comparing the 1H NMR, there is a big difference thing in the 13C NMR. The 13C-13Cspin-spin splitting rarely exit between adjacent carbons because 13C is naturally lower abundant (1.1%)
• 13C-1H Spin coupling: 13C-1H Spin coupling provides useful information about the number of protons attached a carbon atom. In case of one bond coupling (1JCH), -CH, -CH2, and CH3 have respectively doublet, triplet, quartets for the 13C resonances in the spectrum. However, 13C-1H Spin coupling has an disadvantage for 13C spectrum interpretation. 13C-1H Spin coupling is hard to analyze and reveal structure due to a forest of overlapping peaks that result from 100% abundance of 1H.
• Decoupling: Decoupling is the process of removing 13C-1H coupling interaction to simplify a spectrum and identify which pair of nuclei is involved in the J coupling. The decoupling 13C spectra shows only one peak(singlet) for each unique carbon in the molecule(Fig 10.). Decoupling is performed by irradiating at the frequency of one proton with continuous low-power RF.
Fig 10. Decoupling in the 13C NMR | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.10%3A_C_NMR_Spectroscopy.txt |
Distortions Enhancement by Polarization Transfer (DEPT)
DEPT is used for distinguishing between a CH3 group, a CH2 group, and a CH group. The proton pulse is set at 45°, 90°, or 135° in the three separate experiments. The different pulses depend on the number of protons attached to a carbon atom. Fig 11. is an example about DEPT spectrum.
Fig 11. DEPT spectrum of n-isobutlybutrate
While broadband decoupling results in a much simpler spectrum, useful information about the presence of neighboring protons is lost. However, another modern NMR technique called DEPT (Distortionless Enhancement by Polarization Transfer) allows us to determine how many hydrogens are bound to each carbon. For example, a DEPT experiment tells us that the signal at 171 ppm in the ethyl acetate spectrum is a quaternary carbon (no hydrogens bound, in this case a carbonyl carbon), that the 61 ppm signal is from a methylene (CH2) carbon, and that the 21 ppm and 14 ppm signals are both methyl (CH3) carbons. The details of the DEPT experiment are beyond the scope of this text, but DEPT information will often be provided along with 13C spectral data in examples and problems.
Below are two more examples of 13C NMR spectra of simple organic molecules, along with DEPT information.
Example 13.5.2
Give peak assignments for the 13C-NMR spectrum of methyl methacrylate, shown above.
Solution
One of the greatest advantages of 13C-NMR compared to 1H-NMR is the breadth of the spectrum - recall that carbons resonate from 0-220 ppm relative to the TMS standard, as opposed to only 0-12 ppm for protons. Because of this, 13C signals rarely overlap, and we can almost always distinguish separate peaks for each carbon, even in a relatively large compound containing carbons in very similar environments. In the proton spectrum of 1-heptanol, for example, only the signals for the alcohol proton (Ha) and the two protons on the adjacent carbon (Hb) are easily analyzed. The other proton signals overlap, making analysis difficult.
In the 13C spectrum of the same molecule, however, we can easily distinguish each carbon signal, and we know from this data that our sample has seven non-equivalent carbons. (Notice also that, as we would expect, the chemical shifts of the carbons get progressively smaller as they get farther away from the deshielding oxygen.)
This property of 13C-NMR makes it very helpful in the elucidation of larger, more complex structures.
Example 13.5.3
13C-NMR (and DEPT) data for some common biomolecules are shown below (data is from the Aldrich Library of 1H and 13C NMR). Match the NMR data to the correct structure, and make complete peak assignments.
• spectrum a: 168.10 ppm (C), 159.91 ppm (C), 144.05 ppm (CH), 95.79 ppm (CH)
• spectrum b: 207.85 ppm (C), 172.69 ppm (C), 29.29 ppm (CH3)
• spectrum c: 178.54 ppm (C), 53.25 ppm (CH), 18.95 ppm (CH3)
• spectrum d: 183.81 ppm (C), 182. 63 ppm (C), 73.06 ppm (CH), 45.35 ppm (CH2)
Solution
13C NMR Chemical Shifts
The Carbon NMR is used for determining functional groups using characteristic shift values. 13C chemical shift is affect by electronegative effect and steric effect. If an H atoms in an alkane is replace by substituent X, electronegative atoms (O, N, halogen), ?-carbon and ?-carbon shift to downfield (left; increase in ppm) while ?-carbon shifts to upfield. The steric effect is observed in acyclic and clyclic system, which leads to downshifted chemical shifts. Figure 9 shows typical 13C chemical shift regions of the major chemical class.
Spin-Spin splitting
Comparing the 1H NMR, there is a big difference thing in the 13C NMR. The 13C-13Cspin-spin splitting rarely exit between adjacent carbons because 13C is naturally lower abundant (1.1%)
• 13C-1H Spin coupling: 13C-1H Spin coupling provides useful information about the number of protons attached a carbon atom. In case of one bond coupling (1JCH), -CH, -CH2, and CH3 have respectively doublet, triplet, quartets for the 13C resonances in the spectrum. However, 13C-1H Spin coupling has an disadvantage for 13C spectrum interpretation. 13C-1H Spin coupling is hard to analyze and reveal structure due to a forest of overlapping peaks that result from 100% abundance of 1H.
• Decoupling: Decoupling is the process of removing 13C-1H coupling interaction to simplify a spectrum and identify which pair of nuclei is involved in the J coupling. The decoupling 13C spectra shows only one peak(singlet) for each unique carbon in the molecule(Fig 10.). Decoupling is performed by irradiating at the frequency of one proton with continuous low-power RF.
Fig 10. Decoupling in the 13C NMR | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.12%3A_C_NMR_Spectroscopy_and_DEPT.txt |
The interpretation of 13C NMR spectra does not form a part of Chemistry 350; hence, you may omit Section 13.7. Interested students may wish to read this section for enrichment purposes.
Features of a C-13 NMR spectrum
Butane shows two different peaks in the 13C NMR spectrum, below. Note that: the chemical shifts of these peaks are not very different from methane. The carbons in butane are in a similar environment to the one in methane.
• there are two distinct carbons in butane: the methyl, or CH3, carbon, and the methylene, or CH2, carbon.
• the methyl carbon absorbs slightly upfield, or at lower shift, around 10 ppm.
• the methylene carbon absorbs at slightly downfield, or at higher shift, around 20 ppm.
• other factors being equal, methylene carbons show up at slightly higher shift than methyl carbons.
In the 13C NMR spectrum of pentane (below), you can see three different peaks, even though pentane just contains methyl carbons and methylene carbons like butane. As far as the NMR spectrometer is concerned, pentane contains three different kinds of carbon, in three different environments. That result comes from symmetry.
Symmetry is an important factor in spectroscopy. Nature says:
• atoms that are symmetry-inequivalent can absorb at different shifts.
• atoms that are symmetry-equivalent must absorb at the same shift.
To learn about symmetry, take a model of pentane and do the following:
• make sure the model is twisted into the most symmetric shape possible: a nice "W".
• choose one of the methyl carbons to focus on.
• rotate the model 180 degrees so that you are looking at the same "W" but from the other side.
• note that the methyl you were focusing on has simply switched places with the other methyl group. These two carbons are symmetry-equivalent via two-fold rotation.
Animation NMR1. A three-dimensional model of pentane. Grab the model with the mouse and rotate it so that you are convinced that the second and fourth carbons are symmetry-equivalent, but the third carbon is not.
By the same process, you can see that the second and fourth carbons along the chain are also symmetry-equivalent. However, the middle carbon is not; it never switches places with the other carbons if you rotate the model. There are three different sets of inequivalent carbons; these three groups are not the same as each other according to symmetry.
Example \(1\)
Determine how many inequivalent carbons there are in each of the following compounds. How many peaks do you expect in each 13C NMR spectrum?
Practically speaking, there is only so much room in the spectrum from one end to the other. At some point, peaks can get so crowded together that you can't distinguish one from another. You might expect to see ten different peaks in eicosane, a twenty-carbon alkane chain, but when you look at the spectrum you can only see seven different peaks. That may be frustrating, because the experiment does not seem to agree with your expectation. However, you will be using a number of methods together to minimize the problem of misleading data.
The C-13 NMR spectrum for ethanol
This is a simple example of a C-13 NMR spectrum. Don't worry about the scale for now - we'll look at that in a minute.
Note
Note: The NMR spectra on this page have been produced from graphs taken from the Spectral Data Base System for Organic Compounds (SDBS) at the National Institute of Materials and Chemical Research in Japan.
There are two peaks because there are two different environments for the carbons. The carbon in the CH3 group is attached to 3 hydrogens and a carbon. The carbon in the CH2 group is attached to 2 hydrogens, a carbon and an oxygen. The two lines are in different places in the NMR spectrum because they need different external magnetic fields to bring them in to resonance at a particular radio frequency.
The C-13 NMR spectrum for a more complicated compound
This is the C-13 NMR spectrum for 1-methylethyl propanoate (also known as isopropyl propanoate or isopropyl propionate).
This time there are 5 lines in the spectrum. That means that there must be 5 different environments for the carbon atoms in the compound. Is that reasonable from the structure?
Well - if you count the carbon atoms, there are 6 of them. So why only 5 lines? In this case, two of the carbons are in exactly the same environment. They are attached to exactly the same things. Look at the two CH3 groups on the right-hand side of the molecule.
You might reasonably ask why the carbon in the CH3 on the left is not also in the same environment. Just like the ones on the right, the carbon is attached to 3 hydrogens and another carbon. But the similarity is not exact - you have to chase the similarity along the rest of the molecule as well to be sure.
The carbon in the left-hand CH3 group is attached to a carbon atom which in turn is attached to a carbon with two oxygens on it - and so on down the molecule. That's not exactly the same environment as the carbons in the right-hand CH3 groups. They are attached to a carbon which is attached to a single oxygen - and so on down the molecule. We'll look at this spectrum again in detail on the next page - and look at some more similar examples as well. This all gets easier the more examples you look at.
For now, all you need to realize is that each line in a C-13 NMR spectrum recognizes a carbon atom in one particular environment in the compound. If two (or more) carbon atoms in a compound have exactly the same environment, they will be represented by a single line.
Note
You might wonder why all this works, since only about 1% of carbon atoms are C-13. These are the only ones picked up by this form of NMR. If you had a single molecule of ethanol, then the chances are only about 1 in 50 of there being one C-13 atom in it, and only about 1 in 10,000 of both being C-13.
But you have got to remember that you will be working with a sample containing huge numbers of molecules. The instrument can pick up the magnetic effect of the C-13 nuclei in the carbon of the CH3 group and the carbon of the CH2 group even if they are in separate molecules. There's no need for them to be in the same one. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.13%3A_Uses_of_C_NMR_Spectroscopy.txt |
Additional NMR Examples
For each molecule, predict the number of signals in the 1H-NMR and the 13C-NMR spectra (do not count split peaks - eg. a quartet counts as only one signal). Assume that diastereotopic groups are non-equivalent.
P5.2: For each of the 20 common amino acids, predict the number of signals in the proton-decoupled 13C-NMR spectrum.
P5.3: Calculate the chemical shift value (expressed in Hz, to one decimal place) of each sub-peak on the 1H-NMR doublet signal below. Do this for:
a) a spectrum obtained on a 300 MHz instrument
b) a spectrum obtained on a 100 MHz instrument
P5.4: Consider a quartet signal in an 1H-NMR spectrum obtained on a 300 MHz instrument. The chemical shift is recorded as 1.7562 ppm, and the coupling constant is J = 7.6 Hz. What is the chemical shift, expressed to the nearest 0.1 Hz, of the furthest downfield sub-peak in the quartet? What is the resonance frequency (again expressed in Hz) of this sub-peak?)
P5.5: One easily recognizable splitting pattern for the aromatic proton signals from disubstituted benzene structures is a pair of doublets. Does this pattern indicate ortho, meta, or para substitution?
P5.6 :Match spectra below to their corresponding structures A-F.
Structures:
Spectrum 1
δ
splitting
integration
4.13
q
2
2.45
t
2
1.94
quintet
1
1.27
t
3
Spectrum 2
δ
splitting
integration
3.68
s
3
2.99
t
2
1.95
quintet
1
Spectrum 3
δ
splitting
integration
4.14
q
1
2.62
s
1
1.26
t
1.5
Spectrum 4
δ
splitting
integration
4.14
q
4
3.22
s
1
1.27
t
6
1.13
s
9
Spectrum 5
δ
splitting
integration
4.18
q
1
1.92
q
1
1.23
t
1.5
0.81
t
1.5
Spectrum 6
δ
splitting
integration
3.69
s
1.5
2.63
s
1
P5.7: Match spectra 7-12 below to their corresponding structures G-L .
Structures:
Spectrum 7:
δ
splitting
integration
9.96
d
1
5.88
d
1
2.17
s
3
1.98
s
3
Spectrum 8:
δ
splitting
integration
9.36
s
1
6.55
q
1
2.26
q
2
1.99
d
3
0.96
t
3
Spectrum 9:
δ
splitting
integration
9.57
s
1
6.30
s
1
6.00
s
1
1.84
s
3
Spectrum 10:
δ
splitting
integration
9.83
t
1
2.27
d
2
1.07
s
9
Spectrum 11:
δ
splitting
integration
9.75
t
1
2.30
dd
2
2.21
m
1
0.98
d
6
Spectrum 12:
δ
splitting
integration
8.08
s
1
4.13
t
2
1.70
m
2
0.96
t
3
P5.8: Match the 1H-NMR spectra 13-18 below to their corresponding structures M-R .
Structures:
Spectrum 13:
δ
splitting
integration
8.15
d
1
6.33
d
1
Spectrum 14: 1-723C (structure O)
δ
splitting
integration
6.05
s
1
2.24
s
3
Spectrum 15:
δ
splitting
integration
8.57
s (b)
1
7.89
d
1
6.30
d
1
2.28
s
3
Spectrum 16:
δ
splitting
integration
9.05
s (b)
1
8.03
s
1
6.34
s
1
5.68
s (b)
1
4.31
s
2
Spectrum 17:
δ
splitting
integration
7.76
d
1
7.57
s (b)
1
6.44
d
1
2.78
q
2
1.25
t
3
Spectrum 18:
δ
splitting
integration
4.03
s
1
2.51
t
1
2.02
t
1
P5.9: Match the 1H-NMR spectra 19-24 below to their corresponding structures S-X.
Structures:
Spectrum 19:
δ
splitting
integration
9.94
s
1
7.77
d
2
7.31
d
2
2.43
s
3
Spectrum 20:
δ
splitting
integration
10.14
s
2
8.38
s
1
8.17
d
2
7.75
t
1
Spectrum 21:
δ
splitting
integration
9.98
s
1
7.81
d
2
7.50
d
2
Spectrum 22:
δ
splitting
integration
7.15-7.29
m
2.5
2.86
t
1
2.73
t
1
2.12
s
1.5
Spectrum 23:
δ
splitting
integration
7.10
d
1
6.86
d
1
3.78
s
1.5
3.61
s
1
2.12
s
1.5
Spectrum 24:
δ
splitting
integration
7.23-7.30
m
1
3.53
s
1
P5.10: Match the 1H-NMR spectra 25-30 below to their corresponding structures AA-FF.
Structures:
Spectrum 25:
δ
splitting
integration
9.96
s
1
7.79
d
2
7.33
d
2
2.72
q
2
1.24
t
3
Spectrum 26:
δ
splitting
integration
9.73
s
1
7.71
d
2
6.68
d
2
3.06
s
6
Spectrum 27:
δ
splitting
integration
7.20-7.35
m
10
5.12
s
1
2.22
s
3
Spectrum 28:
δ
splitting
integration
8.08
s
1
7.29
d
2
6.87
d
2
5.11
s
2
3.78
s
3
Spectrum 29:
δ
splitting
integration
7.18
d
1
6.65
m
1.5
3.2
q
2
1.13
t
3
Spectrum 30:
δ
splitting
integration
8.32
s
1
4.19
t
2
2.83
t
2
2.40
s
3
P5.11: Match the 1H-NMR spectra 31-36 below to their corresponding structures GG-LL
Structures:
Spectrum 31:
δ
splitting
integration
6.98
d
1
6.64
d
1
6.54
s
1
4.95
s
1
2.23
s
3
2.17
s
3
Spectrum 32:
δ
splitting
integration
7.08
d
1
6.72
d
1
6.53
s
1
4.81
s
1
3.15
7-tet
1
2.24
s
3
1.22
d
6
Spectrum 33:
δ
splitting
integration
7.08
d
2
6.71
d
2
6.54
s
1
3.69
s
3
3.54
s
2
Spectrum 34:
δ
splitting
integration
9.63
s
1
7.45
d
2
6.77
d
2
3.95
q
2
2.05
s
3
1.33
t
3
Spectrum 35:
δ
splitting
integration
9.49
s
1
7.20
d
2
6.49
d
2
4.82
s
2
1.963
s
3
Spectrum 36:
δ
splitting
integration
9.58
s(b)
1
9.31
s
1
7.36
d
1
6.67
s
1
6.55
d
1
2.21
s
3
2.11
s
3
P5.12: Use the NMR data given to deduce structures.
a ) Molecular formula: C5H8O
1H-NMR:
δ
splitting
integration
9.56
s
1
6.25
d (J~1 Hz)
1
5.99
d (J~1 Hz)
1
2.27
q
2
1.18
t
3
13C-NMR
δ
DEPT
194.60
CH
151.77
C
132.99
CH2
20.91
CH2
11.92
CH3
b) Molecular formula: C7H14O2
1H-NMR:
δ
splitting
integration
3.85
d
2
2.32
q
2
1.93
m
1
1.14
t
3
0.94
d
6
13C-NMR
δ
DEPT
174.47
C
70.41
CH2
27.77
CH
27.64
CH2
19.09
CH3
9.21
CH3
c) Molecular formula: C5H12O
1H-NMR:
δ
splitting
integration
3.38
s
2H
2.17
s
1H
0.91
s
9H
13C-NMR
δ
DEPT
73.35
CH2
32.61
C
26.04
CH3
d) Molecular formula: C10H12O
1H-NMR:
δ
splitting
integration
7.18-7.35
m
2.5
3.66
s
1
2.44
q
1
1.01
t
1.5
13C-NMR
δ
DEPT
208.79
C
134.43
C
129.31
CH
128.61
CH
126.86
CH
49.77
CH2
35.16
CH2
7.75
CH3
P5.13:
13C-NMR data is given for the molecules shown below. Complete the peak assignment column of each NMR data table.
a)
δ
DEPT
carbon #
161.12
CH
65.54
CH2
21.98
CH2
10.31
CH3
b)
δ
DEPT
carbon #
194.72
C
149.10
C
146.33
CH
16.93
CH2
14.47
CH3
12.93
CH3
c)
δ
DEPT
carbon #
171.76
C
60.87
CH2
58.36
C
24.66
CH2
14.14
CH3
8.35
CH3
d)
δ
DEPT
carbon #
173.45
C
155.01
C
130.34
CH
125.34
C
115.56
CH
52.27
CH3
40.27
CH2
e)
δ
DEPT
carbon #
147.79
C
129.18
CH
115.36
CH
111.89
CH
44.29
CH2
12.57
CH3
P5.14: You obtain the following data for an unknown sample. Deduce its structure.
1H-NMR:
13C-NMR:
Mass Spectrometry:
P5.15:You take a 1H-NMR spectrum of a sample that comes from a bottle of 1-bromopropane. However, you suspect that the bottle might be contaminated with 2-bromopropane. The NMR spectrum shows the following peaks:
δ
splitting
integration
4.3
septet
0.0735
3.4
triplet
0.661
1.9
sextet
0.665
1.7
doublet
0.441
1.0
triplet
1.00
How badly is the bottle contaminated? Specifically, what percent of the molecules in the bottle are 2-bromopropane?
Challenge problems
C5.1: All of the 13C-NMR spectra shown in this chapter include a signal due to CDCl3, the solvent used in each case. Explain the splitting pattern for this signal.
C5.2: Researchers wanted to investigate a reaction which can be catalyzed by the enzyme alcohol dehydrogenase in yeast. They treated 4'-acylpyridine (1) with living yeast, and isolated the alcohol product(s) (some combination of 2A and 2B).
a) Will the products 2A and 2B have identical or different 1H-NMR spectra? Explain.
b) Suggest a 1H-NMR experiment that could be used to determine what percent of starting material (1) got turned into product (2A and 2B).
c) With purified 2A/2B, the researchers carried out the subsequent reaction shown below to make 3A and 3B, known as 'Mosher's esters'. Do 3A and 3B have identical or different 1H-NMR spectra? Explain.
d) Explain, very specifically, how the researchers could use 1H-NMR to determine the relative amounts of 2A and 2B formed in the reaction catalyzed by yeast enzyme.
12.15: Sample NMR Spectra
Sample 1H-NMR Spectra
List of Animated 1H-NMR Spectra
Bromoethane 1-bromopropane 2-propanol 3-bromopropene propanal
Phenol acetone propanoic acid ethyl acetate 2-propenamide
For all spectra click on a peak to highlight the protons responsible for the peak.
More spectra can be found at Animated Spectra
To see the integratals, right click on the spectra to open the menu, go to "view" and check the integrate" box. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.14%3A__More_NMR_Examples.txt |
The development of nuclear magnetic resonance spectroscopy subsequent to the initial discoveries by Purcell1 and Bloch2 in 1946 is now recognized as one of the most important events in the last fifty years for the advancement of organic chemistry. Nuclear magnetic resonance (NMR) techniques are throwing new light on many difficult organic problems. With the possible exception of gas-liquid chromatography, no new experimental method has been so rapidly accepted or proved so widely applicable. It is the purpose of this book to present the elements of NMR spectroscopy in a form suitable for practical use by organic chemists. Examples of applications will be mainly drawn from high resolution proton resonance spectroscopy, but the principles so illustrated will often be useful in dealing with other types of NMR spectroscopy.
An NMR spectrometer consists basically of a magnet, radio-frequency (rf) transmitter or oscillator, and a suitable rf detector. When a sample of a material comprised of atoms having nuclei with certain magnetic properties (to be described later) is placed in the magnet pole gap and subjected to the rf field of the oscillator, absorption of rf energy (resonance) occurs at particular combinations of the oscillator frequency and the magnetic field strength and an rf signal is picked up by the detector. Customarily, the detector output is measured at constant oscillator frequency as a function of the magnetic field strength, although there are advantages to the alternative procedure of maintaining the magnetic field constant and varying the oscillator frequency.3
Information of chemical interest arises from the fact that nuclei of atoms in different chemical environments are also generally in quite different magnetic environments and come into resonance with a fixed frequency oscillator at different values of the applied magnetic field. Figure 1-1 shows a nuclear magnetic resonance spectrogram of a typical organic molecule, N-ethylethylenimine.
Besides the obvious utility of such a spectrogram to serve as a fingerprint of the compound in question, much more information m be gleaned from the spectrum than is provided by a qualitative examination of its major features. For example, it can be stated with a high degree of certainty that an ethyl group is present and also that the methyl of the ethyl group is rotating about the bond connecting it to the methylene considerably more than 15 times per second. In addition, the spectrum shows the imine nitrogen to have a configuration such that the nitrogen atom and the three carbon atoms to which it is directly attached do not lie in a plane. Furthermore, we can say that the nitrogen atom is not undergoing configurational inversion of the kind shown in the following equation at a rate approaching or greater than 80 times per second.
Clearly, the NMR spectrum of the compound is a veritable treasure trove of useful information not easily obtainable in any other way.
We shall now consider the connection between the structure of an organic molecule and its NMR spectrum with the intention of ultimately getting at the principles which enable one to derive the kind of inferences made above. At the outset, we shall show how magnetic nuclei can absorb rf energy and produce an rf signal in the detector. Only the "crossed-coil" nuclear induction apparatus developed by Bloch, Hansen, and Packard will be considered, since it provides the basis for the commercially available high-resolution spectrometers.
1 E. M. Purcell, H. C. Torrey, and R. V. Pound, Phys. Rev., 69, 37 (1946).
2 F. Bloch, W. W. Hansen, and M. E. Packard, Phys. Rev., 69, 127 (1946).
3 B. Baker and L. W. Burd, Rev. Sci. Instr., 28, 313 (1957).
1.02: Nuclear Magnetic Resonance Spectrometers
A block diagram of an NMR spectrometer utilizing an electromagnet is shown in Fig. 1-2. For high-resolution spectra, the magnet will have pole faces up to 12 in. in diameter, a pole gap of about 1.75 in., and a field of up to 14,000 gauss. The magnet is energized by a highly stable d-c power supply. If a fixed-frequency rf oscillator is employed, one "sweeps" through the resonance by varying the total magnetic field through injection of the linearly varying output from a "sweep generator" into coils either wound around the magnet pole faces or located within the pole gap. The output of the generator is synchronized with the trace along the X axis of an oscilloscope or suitable graphic recorder.
The sample is placed within the pole gap and subjected to the rf alternating magnetic field produced by passing a high-frequency a-c current through the oscillator coil. The detector serves to pick up changes in the magnetization of the nuclei induced by the rf oscillator, and the detector signal is fed to the Y axis of the oscilIoscope or graphic recorder. A nuclear resonance spectrogram is thus a plot of detector signal against magnetic field at constant oscillator frequency.
The state of affairs in the immediate vicinity of the sample is shown in Fig. 1-3. It will be seen that the oscillator coil is oriented with its axis perpendicular to the principal magnetic field. The receiver coil is tuned to the oscillator frequency but is oriented with its axis perpendicular to both the direction of the principal magnetic field and the axis of the oscillator coil. This arrangement is used to minimize the overloading of the necessarily sensitive receiver which would result from direct coupling between the oscillator and receiver coils. Therefore, a nuclear resonance signal arises from an indirect coupling between the oscillator and receiver coils produced by the sample itself. The requirements for such coupling can be described more precisely as follows. The magnetic field of the oscillator alternates through the sample along one direction. The receiver coil responds to a magnetic field which alternates perpendicularly to the field produced by the oscillator coil. The signal results from an alternating magnetization which is induced in the sample by the oscillator field in a direction perpendicular to the axis of the oscillator coil. | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/01%3A_Introduction_-_The_Nuclear_Resonance_Phenomenon/1.01%3A_Introduction.txt |
Induction of an alternating magnetization in a substance like an organic compound, by an oscillatory magnetic field as described above, can be shown by isotopic substitution procedures to involve certain types of atomic nuclei which act like tiny magnets. In the ensuing discussion of magnetic properties of nuclei, we shall find it convenient to ascribe certain electromechanical properties to nuclei which are gross oversimplifications of the real state of affairs but are nonetheless very helpful in explaining how a nuclear resonance signal can arise.
In some ways, certain nuclei behave as though they are nonspinning spherical bodies with the nuclear charge distributed evenly over their surfaces. This type of nucleus does not have a magnetic moment because there is no circulation of the nuclear charge. We also say that the "nuclear quadrupole moment" is zero because, when a probing electrical charge approaches such a nucleus, it experiences an electrostatic field, the magnitude of which is independent of the direction of approach. These nuclei are said to have their "nuclear spin" value equal to zero and, not having a magnetic moment, they can give no nuclear resonance signal. Many nuclei of considerable importance to organic chemistry, particularly 12C and 16O, are of this type as, in fact, are all nuclei whose mass numbers A and charges Z are both even.
It is not so unfortunate as it might seem that the principal isotope of carbon can give no nuclear magnetic resonance signal, since if 12C had a sizable nuclear moment the proton NMR spectra of most organic compounds would be much more complicated than they actually are. Furthermore, 13C has a magnetic moment so that when there is vital necessity for observing a carbon resonance signal, 13C can usually be used, either at its prevailing low natural concentration or with the aid of 13C-enriched material.
A number of nuclei of particular importance to organic chemistry may be assigned nuclear spin values of 1/2. This means that they act as though they were spherical bodies possessing uniform charge distributions but spinning like tops. A spinning nucleus has circulating charge, and this generates a magnetic field so that a nuclear magnetic moment results. The spherical charge distribution ascribed to nuclei with spin of 1/2 means that a probing charge approaching them experiences the same electrostatic field regardless of the direction of approach and, therefore, as with the spherical nonspinning nuclei, the electric quadrupole moment is zero. Nuclei with a spin of 1/2 include 1H, 13C, 15N, 19F, and 31P, and, in general, such nuclei are particularly favorable for nuclear resonance experiments.
A very large number of magnetic nuclei act as though they are spinning bodies with nonspherical charge distributions and are assigned spin values of unity or larger integral multiples of 1/Z. Often such nuclei are taken to approximate ellipsoids spinning about the principal axis. A charged, elongated (prolate) ellipsoid with the charge uniformly distributed over its surface will present an anisotropic electrostatic field to an approaching unit charge so that the electrostatic work will be different in bringing up a unit charge to a given distance if the charge approaches along the spin axis or at some angle to it. By convention, the electric quadrupole moment of a nucleus ascribed the shape of a prolate ellipsoid is assigned a value greater than zero. Important examples are 2H and 14N.
Nuclei which behave like charged, flattened (oblate) ellipsoids also present an anisotropic electric field to a probing charge and by convention are assigned negative electric-quadrupole-moment values. Nuclei of this type include 17O, 33S, 35Cl, etc. In the ensuing discussion, we shall confine our attention largely to nuclei with a spin of 1/2, since, as will be seen, complications are often introduced when the electric quadrupole moment is different from zero. These complications are in themselves capable of providing useful chemical information but are not helpful to an understanding of the operation of a nuclear resonance spectrometer. | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/01%3A_Introduction_-_The_Nuclear_Resonance_Phenomenon/1.03%3A_Magnetic_Properties_of_Nuclei._Nuclear_Spin.txt |
An important property of spinning nuclei is that their magnetic moment vectors appear to have only certain specified average values in any given direction, such as along the axis of the principal magnetic field. The permitted values of the vector moment along the direction of interest can be described with the aid of a set of magnetic quantum numbers m, which are derivable from the nuclear spin I and the relation m = I, (I - 1) (I - 2), . . . , -I. Thus, if I is 1/2, the possible magnetic quantum numbers are +1/2 and -1/2, and if the magnetic moment is p, the possible values of the vector components of the moment in the direction of the principal magnetic field H will be +/mu x and -mu x, as shown below.
If I is unity, then the possible magnetic quantum numbers are +1,0, and -1, and the vector along the field direction will have possible values corresponding to the nucleus being oriented so as to have a component in the same direction as the field vector, perpendicular to the field vector, or opposite in direction to the field vector.
In the absence of a magnetic field there will be no preference for one or the other of the two possible magnetic quantum numbers for a nucleus with I equal to 1/2. In a large assemblage of such nuclei, there will be then exactly equal numbers with rn equal to +1/2 and m equal to -1/2. In a magnetic field, the nuclei will tend to assume the magnetic quantum number (+1/2) which represents alignment with the field in just the same way as compass needles tend to line up in the earth's magnetic field. Thus, in the presence of a magnetic field, m = +1/2 represents a more favorable energy state than m = -1/2 [provided the gyromagnetic ratio y (see Sec. 1-5) is positive]. However, the tendency of the nuclei to assume the magnetic quantum number +1/2 is opposed by thermal agitation. The nuclear moment, field strength, and temperature can be used to calculate the equilibrium percentages of the nuclei in each quantum state by the Boltzmann distribution law. At room temperature, even in rather high magnetic fields such as 10,000 gauss, thermal agitation is so important relative to the energy gained by alignment of the nuclei that only a very slight excess of the nuclei go into the more favorable quantum state, as shown by the following:
$N =Ap exp \frac {-\epsilon}{kT} \nonumber$ (Boltzmann equation)
$\epsilon = -\mu_H H = - \frac {\gamma h}{2 \pi} mH \nonumber$
For protons at 300°K in a field of 9,400 gauss
$\frac {N(+ \frac{1}{2})}{N(- \frac{1}{2})} = exp ( \frac {\gamma hH / 2 \pi}{kT}) = 1.0000066 \nonumber$
This situation is analogous to an assemblage of compasses on a table subjected to violent agitation. The movements of the table tend to throw the compass needles out of alignment with the earth's magnetic field, so that on the average only a very slight excess of the needles may be actually pointing north.
Nuclear magnetic resonance spectroscopy is primarily concerned with transitions of the nuclei in a magnetic field between energy levels which are expressed by the different magnetic quantum numbers. These energy changes are analogous to electronic and vibrational-rotational energy changes in other forms of spectroscopy. There is no direct magnetic interaction between the nuclei and the electrons which surround them. Thus, a problem is posed with regard to the transfer of energy from the nuclei to and from their surroundings. The energy-transfer problem may be restated in the following way. Consider an assemblage of nuclei in the absence of a magnetic field. As stated before, there will be exactly equal numbers of nuclei with the magnetic quantum numbers +1/2 and -1/2 In the presence of a magnetic field, this distribution corresponds to an infinitely high temperature because the state with the magnetic quantum number -1/2 is now energetically less favorable than the +1/2 state and only an infinitely high temperature could produce sufficient thermal agitation to keep the nuclear magnets from having some net alignment in the field direction. In order to achieve the equilibrium distribution of nuclei between the two possible spin states at a lower temperature, it is necessary that energy be lost to the surroundings by nuclear "relaxation." Relaxation is hardly expected to be a simple process, since the nuclei are not easily able to collide with one another or the surrounding electrons and convert their energy due to an external magnetic field into molecular vibrational, rotational, or translational energy. Transfer of energy back and forth among nuclei in various magnetic quantum states and their surroundings can be achieved with the aid of another property which might be ascribed to magnetic nuclei, called "nuclear precession." | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/01%3A_Introduction_-_The_Nuclear_Resonance_Phenomenon/1.04%3A_Magnetic_Quantum_Numbers.txt |
When a nucleus with a magnetic moment is placed in a magnetic field, it acts as though it were undergoing precession around the field axis at an angular velocity $\omega$o which is directly proportional to the magnetic field at the nucleus Ho. This precession is analogous to the precession of a spinning gyroscope when allowed to topple in the earth's gravitational field. The direction of precession of a gyroscope depends on the direction of its angular momentum vector, and the angular velocity depends on the magnitude of the angular momentum vector and the strength of the gravitational field to which it is subjected. For nuclei, the proportionality constant $\gamma$ between the angular velocity of precession and the field strength depends on the angular spin momentum and the magnetic moment of the nucleus. $\gamma$ is called the "gyromagnetic" ratio or, less commonly, the "magnetogyric" ratio. All nuclei of the same charge and mass number have the same gyromagnetic ratio. Thus, all protons act as though they precess at the same angular velocity when the magnetic field strength at the nucleus is the same. $\gamma$ may be either positive or negative, corresponding to different directions of precession.
1.06: Nuclear Relaxation
The property of magnetic nuclei which corresponds to precession provides a means whereby energy may be transferred back and forth between the nuclei and their surroundings. Consider a magnetic field vector arranged so as to rotate perpendicular to a magnetic field in which are immersed magnetic nuclei precessing at the angular velocity mo (Fig. 1-4). If the rotating vector has a quite different angular velocity from the precessing nuclei, the rotating field vector and the precessing nuclear magnetic vectors cannot remain in phase and there will be no effective interaction between them. On the other hand, if the rotating field vector has the same angular velocity as the precessing nuclear vectors, it will remain in phase with them and can exert a magnetic torque tending to flip over the orientation of the nuclei and thence change their magnetic quantum numbers. Of course, if the nuclear magnetic quantum numbers change, energy is transferred to or from the agency producing the rotating field vector. Thus, an assemblage of nuclear magnets immersed in a magnetic field can come to thermal equilibrium with its surroundings.
An important mechanism for relaxation of a group of nuclei at a nonequilibrium spin temperature utilizes atomic and molecular thermal motions as follows. Suppose a magnetic nucleus is surrounded by others of its type contained in atoms undergoing violent thermal motions. The thermal motions of the nuclei produce random oscillatory magnetic fields which can have frequency components with frequencies equal to the precession frequencies of the relaxing nuclei and can act as a rotating magnetic field vector so as to permit the magnetic orientation energy to be converted to thermal energy. The rate of relaxation by this mechanism depends on the temperature, the concentration of magnetic nuclei, and the viscosity of the medium. It is kinetically a first-order process and can be expressed in terms of a "relaxation time," which is the mean lifetime of the excess of nuclei in the nonrelaxed state. Thermal relaxation is often slow, on the order of seconds to weeks. A vivid example is provided by 13C of natural abundance located at the central atom of neopentane molecules. The natural abundance of 13C is SO low that such atoms will usually be connected only to nonmagnetic 12C atoms and thus are sterically shielded from other magnetic nuclei such as the methyl protons in the same or surrounding molecules. As a result, the rotating field components produced by the thermal motions of the surrounding magnetic nuclei are not very effective at aiding the relaxation of the central 13C nucleus, and the mean lifetime before relaxation is very long.
As might be expected, thermal motions of substances with unpaired electrons are particularly effective in inducing thermal relaxation, and such paramagnetic substances present as impurities may spoil high resolution spectra by making the relaxation times very short, which, as will be seen later, results in line-broadening.
In summary, transitions between states with various magnetic quantum numbers which have different energies because of an applied magnetic field may be induced by thermal motions of magnetic nuclei or paramagnetic substances or else by an external rotating magnetic field which has a frequency equal or very nearly equal to the precession frequency of the nuclei. | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/01%3A_Introduction_-_The_Nuclear_Resonance_Phenomenon/1.05%3A_Nuclear_Precession.txt |
It turns out that there are two varieties of relaxation. The first, discussed above, has to do with the establishment of thermal equilibrium between an assemblage of nuclear magnets with different quantum numbers. This is "longitudinal" relaxation, since it results in establishment of an equilibrium value of the nuclear magnetization along the magnetic field axis. Thus, an assemblage of nuclei in a very weak magnetic field, such as the earth's magnetic field, will have essentially no net magnetization of the nuclei along the field axis, since only a few more of the nuclei possess the spin quantum number +1/2 as compared with those with the value of -1/2. When this assemblage is placed in a magnetic field and relaxation takes place, there is an increase in the sample magnetization along the field axis as more of the nuclei drop into the lower energy state with magnetic quantum number of +1/2. The characteristic longitudinal relaxation time is designated as T1.
1.08: Transverse Relaxation T
The other variety of relaxation may be illustrated as follows. Consider a group of nuclei which are precessing in phase about the axis of a common magnetic field. If the nuclei were all centered on the same point, their magnetic vectors would be precessing together like a tied-up bundle of sticks. If we take the magnetic field axis to be the Z axis, the nuclei precessing in phase produce a resultant rotating magnetic vector which has a component in the XY plane. If by any process the nuclei tend to lose their phase coherence, their resultant will move toward the Z axis and the macroscopic component of magnetization in the XY plane will go to zero. This type of relaxation is commonly referred to as "transverse" relaxation, and its rate is customarily expressed in terms of the characteristic time T2. Characteristic time T2 is the time constant for the kinetically first-order decay of X, Y magnetization.
There are several factors which can contribute to transverse relaxation, and these may be classified as intrinsic in the nature of the sample or arising from the equipment used. The homogeneity of the applied magnetic field will be extremely important as an external factor. If the assemblage of nuclei under consideration is in a nonhomogeneous field, the nuclei will not have identical precession frequencies, and if they start off in phase, they will soon get out of phase because of their different precession rates. In many cases, the inhomogeneity of the applied magnetic field will be the most important factor determining T2. Nonhomogeneous magnetic fields within the sample will also decrease T2. Viscosity plays an important part here. In the liquid state, nuclei which might otherwise be expected to have the same precession frequency will not usually have instantaneously identical environments as regards nuclear magnetic dipole-dipole interaction and diamagnetic shielding involving neighboring molecules. Thus, the nucleus of one atom may have one type of molecule as a neighbor while another nucleus may have quite a different molecule as a neighbor. Such nuclei will in general be subjected to different magnetic fields and have different precession frequencies, thus permitting them to lose phase coherence. This effect will be of most importance in viscous media where the molecules move slowly with respect to one another. If the viscosity is low and the molecules tumble rapidly relative to the precession frequencies of their nuclei, the fluctuations in the local magnetic fields are effectively averaged to zero and T2 is thereby increased.
Another factor which influences T2 in solids or viscous liquids is the occurrence of what are often called "spin-spin collisions." These "collisions" occur when two identical nuclei exchange spins — one nucleus acting as a rotating field vector for the other. It can be shown by the uncertainty principle that spin-spin collisions limit the time of maintenance of phase coherence for an assemblage of identical nuclei precessing in phase and thus decrease T2. As will be seen, both T1 and T2 are vitally important in determining the character of nuclear resonance signals. | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/01%3A_Introduction_-_The_Nuclear_Resonance_Phenomenon/1.07%3A_Longitudinal_Relaxation_T.txt |
The operation of the nuclear magnetic resonance spectrometer may now be discussed in terms of the magnetic properties of nuclei outlined above. As a quick review, we note as before that the sample is subjected to (1) a large magnetic field varied by the sweep generator and directed along the Z axis and (2) an alternating magnetic field along the X axis produced by the oscillator coil (Fig. 1-5). The receiver coil is oriented so as to respond to an alternating magnetic field along the Y axis, and the oscillator must induce a component of Y magnetization in the sample if a signal is to be picked up by the receiver. At the nuclear level, the situation is as shown in Fig. 1-6. In the first place, we note that an assemblage of magnetic nuclei placed in a magnetic field undergo relaxation and reach aa equilibrium distribution in which there is a slight excess with the magnetic quantum number +1/2 This excess of nuclei with m = +1/2 combine to give a small macroscopic resultant magnetization in the Z direction. In addition, each nucleus acts as if it were precessing around the magnetic field axis with the angular velocity $\omega$o (equal to $\gamma$Ho). At the beginning of a nuclear magnetic resonance experiment, the nuclei will have no phase coherence and the excess of nuclear magnets with m = +1/2 is represented graphically by having the individual magnetic vectors distributed evenly over the surface of a cone whose axis coincides with the direction of the magnetic field. In this situation, there will be no net X, Y component of nuclear magnetization.
Consider the nuclei to be now subjected to an alternating magnetic field in the X direction produced by the rf oscillator. This field will have no net component in the Y direction, but we can consider that it is made up of two equal magnetic vectors rotating at the same velocity in opposite directions with phase relations so that they exactly cancel each other in the Y direction. One of these vectors will be rotating in the same direction as the nuclear magnets precess while the other will be rotating in the opposite direction. Of course, the field which is rotating oppositely to the direction of nuclear precession will not interact with the nuclei because it cannot stay in phase with them. However, the field which is rotating in the same direction can stay in phase with and tend to flip over the nuclear magnets, provided it has the same angular velocity (see Sec. 1-6).
In a typical nuclear magnetic resonance experiment, we change the precession frequency of the nuclei by varying the applied magnetic field and keep the oscillator frequency constant. At some value of the field, the nuclear precession frequency becomes equal to the frequency of the rotating field vector produced by the oscillator, and energy may then be transferred from the oscillator to the nuclei, causing some of them to go to the higher energy state with m = -1/2. At the same time, the rotating field vector acts to tip the vectors of the individual nuclear magnets, with which it is 90" out of phase, away from the field axis and thus causes the axis of the cone of vectors to "wobble" around the field axis at the precession frequency. This has an effect such as would result from bunching the nuclear vectors as shown in Fig. 1-7, so that the macroscopic resultant is moved away from the field axis and produces a rotating component of magnetization in the X and Y directions which, of course, precesses around the field axis with the same angular velocity as the individual nuclei. This alternating field in the Y direction induces a current in the receiver coil and thus generates an NMR signal.
As the magnetic field is increased through action of the sweep generator, the nuclei increase their precession frequencies and drop out of phase with the rotating field vector. At this point, transverse and longitudinal relaxation return the nuclear magnetization of the nuclei in the X, Y, and Z directions to the equilibrium values. As the Y magnetization decreases by transverse relaxation, the signal dies away in the receiver. | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/01%3A_Introduction_-_The_Nuclear_Resonance_Phenomenon/1.09%3A_The_Nuclear_Resonance_Signal.txt |
It should be emphasized that relaxation begins as soon as the nuclei absorb energy from the rotating field vector. Obviously, if T1 and T2 are both extremely short, the signal strength at a given field value will be very small because both longitudinal and transverse relaxation destroy the component of magnetization in the X, Y directions. On the other hand, if the relaxation times are long, other effects are noted. For example, if T1 is very long, a "saturation" effect may be noted with respect to the signal strength because the energy absorbed by the nuclei from the oscillator is not readily dissipated to the surroundings. In this situation, the nuclei reach an equilibrium distribution between their magnetic quantum states which is determined by the relaxation time T1. When T1 is short, the nuclei remain more or less at thermal equilibrium with their surroundings and the absorption of energy then depends primarily on T2.
It should be clear that if one were suddenly to turn off the oscillator in the middle of a resonance signal, the signal would not cease at once because the rate of loss of magnetization in the X, Y directions depends on both T1 and T2. This sort of effect leads to pronounced differences in the appearance of NMR signals, depending on the rate of sweep, as illustrated in Fig. 1-8. A fast sweep produces a signal peak followed by a succession of diminishing peaks, which are often called "relaxation wiggles." A slow sweep gives a more symmetrical peak with perhaps only a trace of the relaxation wiggles. The "envelope" of relaxation wiggles arises in the following way. The first peak of a plot of signal vs. magnetic field represents the point at which the precession frequency of the nuclei is equal to the oscillator frequency. As the field sweep continues from this point, the precession frequency increases and, hence, the precessing macroscopic resultant goes out of phase with the oscillator while the X, Y magnetization diminishes by longitudinal and transverse relaxation. Whenever the precessing macroscopic vector gains 360" on the rotating field vector, it comes into phase again with that vector and picks up an increment of X, Y magnetization which produces an increase in the signal strength. Repetitions of this process give a series of signal pulsations which finally cease when transverse relaxation is complete. Obviously, the slope of the decay envelope of the relaxation wiggles4 is a measure of T2. It has been shown by comparison of the decay envelopes of their relaxation wiggles that hydrogens in different chemical environments (as the methyl and phenyl groups of toluene) have quite different T2 values.
4 G. W. Nederbrogt and C. A. Reilly, J. Chem. Phys., 24, 1110 (1956); C. A. Reilly and R. L. Strombotne, J. Chem. Phys., 26, 1338 (1957).
1.11: Properties of Magnetic Nuclei of Interest to Organic Problems
Table 1-1 shows some of the magnetic properties of nuclei of particular interest to organic chemists. The comparison of relative sensitivities to detection of NMR signals for different nuclei is of particular importance. These sensitivities are computed for equal numbers of nuclei at a constant field strength, which normally is taken as high as practically possible within limitations of field homogeneity, stability, etc. The sensitivity figure given for deuterons must be corrected for their low natural abundance in order to have a measure of the relative ease of detecting deuterium NMR signals in material containing the natural proportions of the hydrogen isotopes. It is seen that protons and 19F are particularly favorable nuclei for observation. The precession frequencies of the nuclei are expressed in Table 1-1 in units of megacycles per 10,000 gauss and are seen to cover a very wide range of values. For high-resolution NMR spectroscopy, the rf oscillator must be extremely stable because oscillator stability is just as important to good resolution as is the homogeneity and stability of the magnetic field. As it is difficult to make a highly stable oscillator with a continuously variable frequency, customarily one uses separate fixed frequency units set at 3, 10, 20, 30, 40, and 60 Mc. The magnetic field is then adjusted to bring the nuclear precession frequencies to the appropriate oscillator frequency. In this way, one can cover a wide range of nuclei, keeping the magnetic field between 5,000 and 14,000 gauss. In general, the oscillator frequency is chosen which corresponds to the highest possible value of the field consistent with good homogeneity and stability because the sensitivity increases with field strength as does the separation between resonance lines corresponding to nuclei in different chemical environments.
Table 1-1. Magnetic properties of Representative Nuclei
Nucleus Z I Mc/kgauss % Natural abundance Relative sensitivity*
1H 1 1/2 42.6 99.98 1.000
2H 1 1 6.5 0.016 0.0096
13C 6 1/2 10.7 1.11 0.016
14N 7 1 3.1 99.63 0.0010
15N 7 1/2 -4.3 0.37 0.0010
17O 8 5/2 -5.8 0.04 0.029
19F 9 1/2 40.1 100 0.834
31P 15 1/2 17.2 100 0.066
*For equal numbers of nuclei at constant field.
A more mathematical treatment of nuclear resonance absorption is presented in Appendix A for the purpose of elucidating the differences between the absorption and dispersion modes and the nature of certain probe adjustments which may influence the shapes of resonance signal curves. | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/01%3A_Introduction_-_The_Nuclear_Resonance_Phenomenon/1.10%3A_Relaxation_Effects_on_NMR_Signals.txt |
A low-resolution proton NMR spectrum of ethyl alcohol at a field of 9,400 gauss and an oscillator frequency of 40 Mc is shown in Fig. 2-1. The three resonance lines correspond to protons with different precession frequencies, which come into resonance with the oscillator frequency at different values of the magnetic field. The areas under the peaks stand roughly in the ratio 1:2:3, as would be expected if each peak corresponded to the chemically different OH, CH2, and CH3 ethanol protons. These assignments have been substantiated by studies of other alcohols and substitution of deuterium for hydrogen. Thus, if the hydroxyl proton of ethyl alcohol is replaced by deuterium, the resonance peak on the left disappears. To be sure, deuterium can give a nuclear resonance signal, but reference to Table 1-1 shows that its resonance frequency would be 6.1 Mc at 9,400 gauss. Therefore, with a 40-Mc oscillator, the deuterium resonance would not be observed until the magnetic field reached 58,000 gauss, which is 106 times the total sweep shown in Fig. 2-1.
The spacing between the ethanol absorption lines is found to be directly proportional to the magnetic field, and if one operates at an oscillator frequency of 30 Mc and a field of 7,000 gauss, the resonance lines are three-fourths as far apart. Field-dependent differences between resonance line positions are called "chemical shifts" and arise because the lines of force of the applied magnetic field tend to be turned away from the nuclei by a diamagnetic shielding effect (but can also be turned in by a second-order paramagnetic effect) of the surrounding electrons. The degree of diamagnetic shielding is directly proportional to the applied field, and therefore chemical shifts are directly proportional to the magnetic field. In order that workers with NMR equipment having different oscillator frequencies and magnetic fields may have a simple basis for comparison of spectra, it is common to report resonance line positions in terms of a dimensionless parameter $\delta$, which is the proportionality constant between resonance frequency and field strength. Usually, $\delta$ is expressed relative to some standard substance for protons customarily water, benzene, or cyclohexane, each of which has only a single resonance line.
$\delta = \frac {Hsample — Hreference}{Hreference}$ x 106
2.02: Measurement of Chemical Shift
Chemical shifts can be measured in several ways. It is quite simple to use an internal standard such as is provided by a sharp resonance line from a solvent or added solute. However, a number of investigations have indicated that internal standards must be used with some care because of possible specific solvent effects and the like. In the final analysis, the best comparisons will be of chemical shifts determined as a function of concentration in a given solvent and extrapolated to infinite dilution. However, this procedure is time consuming, particularly since there is no one solvent which is useful for all types of organic molecules. The special advantage of internal standards as an aid to rapid qualitative analysis of functional groups will be discussed later.
Chemical shifts can be conveniently measured relative to an external standard by using a set of concentric tubes, the standard occupying one compartment and the substance under investigation the other. The separation between the standard and sample resonance lines can be taken directly off the recorder chart if the magnetic field sweep is highly stable and linear. Such conditions are not often met in practice and it is common to measure line separations by the audio-oscillator beat method.1 To do this, one superimposes an audio frequency on the oscillator output, so that the sample is subjected to rotating magnetic field vectors corresponding not only to the principal frequency but also to the beat frequencies which are equal to the principal frequency plus or minus integral multiples of the audio frequency. If the audio power is suitably adjusted, the spectrum of a substance with a single resonance line like benzene appears as shown in Fig. 2-2, with the superimposed audio frequency at 100 cps. The positions of the so-called "sidebands" on the principal resonance can be varied by adjustment of the audio frequency so as to coincide with other signals whose positions are to be measured. With a reasonably stable oscillator, line positions can be determined by this method to better than $\pm$1 cps. For crude measurements of line positions, it often suffices to use a substitution method in which the sweep rate is fixed, and when the desired resonance is observed, the sample is quickly removed and a standard is inserted with a resonance which falls fairly close to the one that is to be measured. | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/02%3A_The_Chemical_Shift/2.01%3A_Introduction._The_Chemical-shift_Parameter.txt |
In general, the chemical shift parameter $\delta$ will be a function of the electron density around the nucleus, since the electrons are directly involved in the diamagnetic shielding which acts to attenuate the applied magnetic field. Specific solvent and bulk diamagnetic susceptibility effects will also be important in determining $\delta$. Temperature is not usually very important unless a change in temperature causes marked changes in some type of association equilibrium. For example, the O-H resonance line of ethanol moves toward the CH2 and CH3 lines with increasing temperature probably because of changes in, the concentrations of the various hydrogen-bonded species.1 A similar effect is noted when ethanol is diluted with carbon tetrachloride.2 At low concentrations, the O-H absorption appears between that of the methylene and methyl groups. Extrapolation of the position of the O-H line as a function of concentration to infinite dilution indicates that in such circumstances the O-H line would be at higher fields than even the CH3 absorption.
In some cases, a temperature effect on chemical shifts may result from changes in amplitude of torsional vibrations with temperature. An example is 1,1,2,2-tetrafluoro-3-phenylcyclobutane, which shows substantial changes in the chemical shifts of the fluorine atoms as the temperature is increased or decreased because of changes in the amplitudes of vibration of the fluorines with respect to each other and the aromatic ring at the 3-position.3
Studies of the proton resonance absorptions of a wide variety of organic compounds have revealed that the resonance lines for similarly located hydrogens appear at comparable applied magnetic fields. This fact permits setting up a table of $\delta$ values for various types of protons as is shown in Table 2-1. The line positions for a given species of hydrogen are seen to occur over a moderate range of values as would depend on the electrical and shielding effects of substituent groups as well as bulk diamagnetic shielding influences produced by the other molecules in the sample. Recent work has shown that variations in $\delta$ because of the latter factor can be minimized by extrapolating the line positions to infinite dilution in a suitable common solvent 4a or by using internal standards.4b For best results, internal standards should be chosen to be of such nature that any bulk diamagnetic shielding effects are expected to influence the resonances of the sample and standard to the same degree. The internal standard procedure is very convenient and has been shown by Chamberlain4b to narrow the range of variations in $\delta$ values for many of the individual types of proton resonances listed in Table 2-1 to less than $\pm$0.2 X 106.
Some idea of the magnitude of solvent effects on chemical shifts is provided by the data of Table 2-2 which show the change in $\delta$ for methyl protons in a variety of compounds between pure liquid and extrapolated to infinite dilution in carbon tetrachloride.5 In general, the changes are large and irregular enough to suggest need for considerable caution in interpreting pure liquid spectra. Changes of chemical shift in some solvents appear to parallel changes in bulk diamagnetic susceptibility.
A cursory examination of Table 2-1 might lead one to believe that there is a relationship between the $\delta$ value of a proton and its acidity. This is not an unreasonable idea, since the electron density is important in determining both the diamagnetic shielding and the ease of removal of the proton by bases. Thus, a sulfonic acid proton appears at very low fields as would be expected if the electron density and diamagnetic shielding of the acidic proton were small. Carboxyl protons come at higher fields while water and alcohol protons are near the center of the table. Furthermore, the very weakly acidic amine and hydrocarbon protons come at the end of the table, corresponding to high electron densities and high diamagnetic shielding. However, closer inspection of Table 2-1 shows a number of groups, the proton $\delta$ values of which cannot be fitted into any simple acidity scale. Thus, aldehydic hydrogens appear at quite low fields, as do aromatic hydrogens. In contrast, the relatively strongly acidic acetylenic hydrogens come at rather high fields.
1 J. T. Arnold and M. E. Packard, J. Chem. Phys., 19, 1608 (1951).
2 A. D. Cohen and C. Reid, J. Chem. Phys., 25, 790 (1556).
3 Unpublished experiments by W. D. Phillips and coworkers.
4a A. A. Bothner-By and R. E. Giick, J. Clzem. Phys., 26, 1651 (1957); 4b N. F. Chamberlain, Anal. Chem., 31, 56 (1959).
5 A. L. Allred and E. G. Rochow, J. Am. Chem. Soc., 79, 5363 (1957).
2.04: Chemical Shifts for Other Nuclei
Generally speaking, the chemical shifts for most magnetic nuclei are much larger than for hydrogen. Representative $\delta$ values for 17O and 14N are given in Table 2-3; these have a greater spread than the proton $\delta$ values by a factor of about 100. The decisive factor with these nuclei is probably the greater polarizability of the electrons in the L shell as compared to those in the K shell. Chemical shifts associated with 13C have also been measured for a variety of simple organic compounds by Lauterbur and Holm,6 using 13C of natural abundance. A fairly special technique is necessary, and rather large samples are required. Nonetheless, the shifts are large, as with 14N and 17O, and many chemical applications of the 13C resonances can be foreseen.
6 P. C. Lauterbur, J. Chem. Phys., 26, 217 (1957) and Ann. N.Y. Acad. Sci., 70, 841 (1958); C. H. Holm, J. Chem. Phys., 26, 707 (1957).
2.05: Proton delta Values and Electronegativity
Chemical shifts are of considerable value in structural investigations involving nuclei in different locations. Several examples of such uses of proton chemical shifts will be described shortly, but first we shall consider possible correlations between $\delta$ and other structural parameters, since much effort has been expended in this direction. Most important are attempted correlations of $\delta$ with electronegativities and some parameters which have been shown to be useful in defining chemical reactivities. From the earlier discussions regarding the relation between chemical shift and electron density, one would expect electronegativity effects to be displayed by substituent groups, so placed as to influence the electron density around a given nucleus. Of course, the substituent or substituents can act also to change the chemical shift by effectively making a change in the solvent as well.
Dailey and Shoolery7 have studied substituent effects in ethyl derivatives by determining differences in chemical shifts between the methylene and methyl protons as a function of X for CH3CH2X. The system is a particularly desirable one because the effect of X in altering the medium would be expected to be nearly the same at the two positions. An excellent correlation was noticed between the difference in resonance line positions for CH3 and CH2 protons of ethyl halides and the Huggins electronegativity values. The relationship was nicely linear (see Fig. 2-3) although there seems to be no real a priori reason for expecting this to be the case. With the aid of the linear relationship and chemical shifts for the protons of other ethyl derivatives, apparent electronegativities were assigned to a wide variety of substituent groups. The ultimate worth of the figures so obtained is yet to be demonstrated through correlation with other properties related to electronegativities.
7 B. P. Dailey and J. N. Shoolery, J. Am. Chem. Soc., 77, 3977 (1955). Similar measurements have been made by Allred and Rochow. | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/02%3A_The_Chemical_Shift/2.03%3A_Some_Factors_Which_Influence_delta.txt |
Considerable work has been done over the last 30 years on chemical reactivities of aromatic compounds. Hammett8 has shown bow much of the resulting data for reactivities of groups connected to benzene rings carrying substituents in the meta and para positions can be correlated with the aid of parameters $\delta$ (substituent) and $\rho$ (reaction).
$\log \frac{k_1}{k_0} = \sigma \rho \nonumber$
Each substituent can usually be assigned a $\sigma$ constant the value of which depends on whether the substituent is located at a meta or para position. In turn, each reaction may be assigned a $\rho$ parameter which measures the sensitivity of its rate to changes of substituents in the meta and para positions. The reaction parameter turns out to be a function of temperature, solvent, salt concentrations, and the character of the reagents employed. Hammett's relationship holds with very considerable precision for the ionization constants of meta- and para-substituted benzoic acids and the alkaline saponification rates of the corresponding ethyl esters. With its aid, one can compute many thousands of individual rates and equilibrium constants from rather small tables of substituent and reaction parameters. Hammett's $\sigma$ constants are of particular interest to the present discussion and are taken normally to represent the difference in the logarithms of the ionization constants of the meta- or para-substituted benzoic acids and benzoic acid itself.
Numerous attempts have been made to correlate $\sigma$ with various physical properties, such as polarographic reduction potentials, infrared vibration frequencies, and wavelengths of maximum absorption in electronic spectra. Gutowsky and coworkers have published data on the chemical shifts df a large number of meta- and para-substituted fluorobenzenes which are of considerable pertinence to the correlation of $\delta$ values with chemical reactivity parameters. Figure 2-4 shows the degree of correlation of fluorine $\delta$ values with Hammett's $\sigma$ constants, and it is immediately clear that a different relationship holds from that customarily observed for chemical equilibrium and rate constants. In the first place, the points corresponding to substituents at meta positions fall along a line of substantially different slope from the best line drawn through the points for substituents located at para positions. This behavior has no exact parallel in chemical-reactivity studies.
Gutowsky interprets the data of Fig. 2-4 to mean that the resonance effects of substituent groups are more important in determining chemical shifts than are the other factors that make up the over-all electrical effect. The observed scatter of the points is expected on this basis, since the resonance contribution is already included for each substituent in the $\sigma$ constants, at least to the extent that it influences chemical reactivity. Taft10 has evaluated the resonance contributions of substituents to their $\sigma$ constants and has corroborated Gutowsky's suggestion that resonance effects are particularly important in determining the influence of substituents on fluorine $\delta$ values.
The situation for protons is not quite so simple. Dailey11 and Bothner-By4 and coworkers have measured proton chemical shifts for a variety of benzenoid compounds. Some typical data are presented in Fig. 2-5, and it is seen that there is no simple relationship between $\sigma$ and $\delta$. Furthermore, no obvious tendency exists for the meta and para points to lie along lines with different slopes as is observed with substituted fluorobenzenes.
It is possible that some of the above anomalies may be due to abnormal diamagnetic shielding effects produced by the unsaturation electrons of the benzene ring, since when an aromatic compound is placed in a magnetic field, the $\pi$ electrons circulate around the ring so as to produce a rather substantial local magnetic field directed normal to the ring which can influence the chemical shift. Waugh12 has shown with 1,4-decamethylenebenzene that the position of each methylene group with respect to the circulating current of electrons above and below the ring is quite critical in determining the chemical shift of the methylene protons. The apparently abnormal chemical shift of acetylenic protons may be due to circulation of the unsaturation electrons around the axis of the triple bond in such a way as to set up a local field which opposes the applied field along the triple bond, thus causing more shielding than is normal for an unsaturated atom.
8 L. P. Hammett, "Physical Organic Chemistry," chap. 7, McGraw-Hill Book Company, Inc., New York, 1940.
9 H. S. Gutowsky, D. W. McCall, B. R. McGarvey, and L. H. Meyer, J. Am. Chem. Soc., 74, 4809 (1952).
10 R. W. Taft, Jr., J. Am. Chem. Soc., 79, 1045 (1957).
11 P. L. Corio and B. P. Dailey, J. Am. Chern. Soc., 78, 3043 (1957).
12 J. S. Waugh and R. W. Fessenden, J. Am. Chem. Soc., 79, 846 (1957). | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/02%3A_The_Chemical_Shift/2.06%3A_delta_and_Hammett%27s_sigma_constants.txt |
For use of NMR spectra as an aid to quantitative analysis and structure determinations, it is important to consider the limitations of the relationship between the magnitude of the nuclear resonance signal and the number of nuclei involved. In the first place, for a given kind of nuclei, the area under the resonance curve is expected to be closely proportional to the number of nuclei involved in the absence of large solvent effects. Thus, one should be able to determine the amount of benzene in mixtures with cyclohexane by measuring the area under the benzene peak and employing a single calibration mixture, provided that the volume of the material in the receiver coil was the same in each case. Another way of making such an analysis would be to take the ratio between the areas under the benzene and cyclohexane peaks. With very sharp peaks, a rough analysis may also be possible on the basis of peak heights, but this is not to be regarded as a generally satisfactory procedure unless the resonance lines have exactly the same width.
In general, the width of a resonance line is usually determined by the relaxation time T2 so that the width at one-half of the full signal strength is numerically equal to 2/T2 (cf. discussion on page 97). Here, we distinguish between the intrinsic relaxation time of the sample and the relaxation time which results from use of a nonhomogeneous external magnetic field. The line width which can be ascribed to the sample itself is sometimes called the "natural line width," and for many nonviscous substances it is smaller than can be achieved even with the most homogeneous magnetic fields now available. Line widths can often be reduced by spinning the sample in the magnetic field with the aid of an air turbine or some similar device. This has the effect of averaging out the field inhomogeneities perpendicular to the direction of spinning and, as is seen in Fig. 2-6, produces a very notable sharpening of the spectrum. An apparatus which permits thermostating of a spinning sample is shown in Fig. 2-7.
Under conditions of "slow passage" through a nuclear resonance spectrum (see Appendix A), the areas under the peaks are approximately proportional to the number of nuclei and (1 + H12T1T2$\gamma$2) -1/2 where H1 is the magnitude of the rotating field vector. Consequently, when the oscillator field strength is large and the relaxation times long, so that H12T1T2$\gamma$2 >_ 1, the areas will be a function of the relaxation times and H1. This will be important not only for nuclei with different relaxation times because of different chemical locations (as for protons on the ring or side-chain of toluene), but also for those located in equivalent molecular positions and subjected to different solvent influences as might change the relaxation times.
The importance of relaxation times on resonance signal areas under slow-passage conditions can be reduced by operating at low oscillator field strengths or, better, by extrapolating the ratio of the areas at several oscillator strengths to zero. However, this does not necessarily solve the signal-area problem, since slow passage is not generally practical for the determination of high-resolution spectra of organic liquids. In analysis of multicomponent systems, appropriate corrections can be made empirically with suitable calibration graphs. A much more serious problem may arise in structure determinations when one requires ratios of numbers of nuclei in different chemical locations for a given substance. For example, t-butylbenzene as an unknown could be distinguished from tetramethylbenzene by showing that the ratio between aliphatic and aromatic hydrogens is 9:5 for one and 12:2 for the other. Such a structure proof is unambiguous, provided the differences in relaxation times between the aromatic and aliphatic hydrogens are not so great as to cause the resonance signal areas to differ greatly from the theoretical values. An example of the kind of difficulties sometimes encountered is provided by cyclobutene, which gives areas under the methylene and double-bond hydrogen resonances as high as 2.8:1 (vs. the theoretical 2: 1) even at low oscillator strengths. | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/02%3A_The_Chemical_Shift/2.07%3A_Resonance_Signal_Areas_and_Widths.txt |
a. Feist's Acid
Measurements of chemical shifts and signal areas can be extremely valuable in determinations of structures of both simple and complicated molecules. An example is Feist's acid, originally postulated by Thorpe and coworkers,13 to be a methylcyclopropenedicarboxylic acid on the basis of chemical evidence-particularly, ozonization of the diethyl ester, which gave no formaldehyde but only a substance having properties appropriate to diethyl oxaloacetoacetate. Later, Ettlinger14 showed that a more likely structure for Feist's acid was trans-3-methylenecyclopropane-1,2-dicarboxylic acid.
Feist's acid is a high-melting solid, and satisfactory high-resolution spectra are usually obtainable only with nonviscous liquids or solutions. This is because in solid or viscous materials, the intermolecular magnetic effects of the various magnetic nuclei on one another are substantial and produce local variations in the total magnetic field. These result in a spread of precession frequencies even for nuclei of the same chemical kind, and such a spread in precession frequencies causes line broadening. In a nonviscous liquid, the molecules tumble over and over at a rapid rate and the eflect of magnetic nuclei in one molecule on the precession frequencies of nuclei in another molecule is averaged to zero.
The choice of a suitable solvent for determination of the NMR spectra of Feist's acid or similar substances illustrates several points of practical importance. Ideally, the solvent would have no resonance absorption of its own or else only a single line such as might serve as a calibration point. Carbon tetrachloride and carbon disulfide are common nonhydrogenous organic solvents in which, unfortunately, most polar organic compounds have only a limited solubility. A number of substances are available with only a single proton absorption, such as chloroform, acetone, benzene, cyclohexane, dimethyl sulfoxide, dioxane, and water, which nicely cover the range between polar and nonpolar solvents. However, even with these solvents, if the sample is only partly soluble, the increase in receiver sensitivity necessary to obtain a measurable signal may well produce a troublesome, large, and broad solvent peak. This can be avoided by using suitably deuterated solvents, and many of these, such as deuterochloroform, benzene, acetic acid, acetone, and, of course, heavy water, are commercially available. In general, concentrated solutions are desirable to give a good signal-to-noise ratio and to diminish the importance of signals due to solvent impurities.
The proton resonance spectrum of Feist's acid was found to be most conveniently measured with the substance in the form of the sodium salt as obtained by dissolution in sodium deuteroxide-deuterium oxide solution.15 Use of D2O avoided a strong water peak and, in addition, gave a convenient calibration peak for the spectrum, because a twoproton water (or HDO) peak results from neutralization of the two carboxyl groups of the acid. As will be seen from Fig. 2-8, two other hydrogen resonances are observed. One of these appears in the vinyl hydrogen region as evinced by its $\delta$ value of -0.7 X 106, while the other comes at a much higher field strength ($\delta$ = +2.4 X l06). The areas of all three peaks are approximately the same, indicating that eack corresponds to two protons. The NMR spectrum of Feist's acid is decisive with respect to a choice between the Thorpe and Ettlinger structures. The Ettlinger structure is in complete agreement with the observed spectrum, while the Thorpe structure would require aliphatic methyl and alicyclic ring hydrogen resonances at high fields in the ratio of 3: 1 and no vinyl hydrogen resonance.
The study of Feist's acid gave a bonus observation in that the water peaks of the sodium deuteroxide solution (which contained some excess sodium deuteroxide) increased on standing at room temperature while the alicyclic hydrogen peaks decreased (see Fig. 2-8). This shows that the aliphatic hydrogens but not vinyl hydrogens exchanged slowly with the sodium deuteroxide-deuterium oxide solution at room temperature and, furthermore, proves that Feist's acid is not in rapid equilibrium under these conditions with molecules having the methylcyclopropenedicarboxylic acid structure because, in that case, the vinyl hydrogens would exchange as well.
Ettlinger16 has used the same technique to show that exchange of the vinyl hydrogens does occur at higher temperatures, and this is the first evidence that molecules with the Thorpe structure can be formed at all.
b. Ring-opening Reaction of 2-Chloro-3-phenylcyclobutenone
Another structure proof utilizing the same principles involves the phenylchlorobutenoic acid formed from ring opening of 2-chloro-3- phenylcyclobutenone with sodium hydroxide.17 Reasonable mechanisms leading to three isomeric acids can be written for this reaction, and a priori one might expect that formation of the conjugated isomer 2- chloro-3-phenylbutenoic acid would be favored (see Fig. 2-9). The NMR spectrum (Fig. 2-10) of the reaction product indicates otherwise. . Besides the carboxyl and phenyl hydrogen resonances which would be common to all three of the reasonably expected structures, there are two other peaks in the ratio of 1:2. The one-hydrogen resonance occurs in the vinyl hydrogen region, while the two-hydrogen resonance is at somewhat lower fields than a normal aliphatic hydrogen. We thus conclude that the product is 4-chloro-3-phenyl-3-butenoic acid formed by breaking the cyclobutenone ring between the 1- and 2-positions. The a priori expected structure is ruled out, since it would show no vinyl hydrogens and a simple three-hydrogen methyl resonance in the aliphatic hydrogen region. The line positions alone do not serve to rule out 2-chloro-3-phenyl-3-butenoic acid as a possibility, since it has both the vinyl- and aliphatic-type hydrogens, but the ratios of high-to-low field resonance lines for this compound would be reversed from what is actually observed.
c. Complex Organic Structures
The use of nuclear magnetic resonance can be profitably extended to even quite complex naturally occurring substances. An example taken from steroid chemistry is the tetracyclic ketol obtained by Johnson and coworkers18 by way of an aldol condensation and assigned the structure shown below by virtue of its conversion through the action of strong base and then by a long sequence of synthetic steps to naturally occurring steroidal hormones.
A dehydration product derived from the ltetol was found to have chemical and physical properties hard to reconcile with the given structure of the starting material. The NMR spcctra of the ketol (as the acetate) and its dehydration product are shown in Fig. 2-1 1 with most of the principal resonance lines identified as belonging to the various structural entities. There are two prominent lines in the spectrum of the ketol which are of the appropriate height and position to correspond to C-CH3 groups. However, only one such absorption is expected on the basis of the postulated structure for the ketol. It is interesting that one of the C-CH3 groups disappears on dehydration.
Reevaluation of the structure of the ketol on the basis of its NMR spectra has indicated that it and its dehydration product are best formulated as shown above.19 Synthesis of natural steroidal hormones from the ketol is possible because ring opening, recyclization, and dehydration occur under the influence of strong base.
Nuclear magnetic resonance spectra of a large number of steroidal compounds have been analyzed by Shoolery and Rogers.20 Interesting differences of about 20 cps were noted between axial and equatorial protons at the 3- and 11-positions of the steroid carbon skeleton.
A further elegant example of the use of NMR data for elucidation of complex organic structures is provided by the work of van Tamelen21 on photosantonic acid. Here, a key finding was of a vinyl-proton resonance which could not be accommodated by most of the proposed structures. The spectrum will be discussed further in Chap. 3 in connection with spin-spin splitting, since, besides its characteristic chemical shift, the vinyl proton resonance showed other features which give strong support to van Tamelen's structure for photosantonic acid.
13 F. R. GOSS, C. K. Ing~ld, and J. F. Thorpe, J. Clzem. Soc., 123, 327 (1923).
14 M. G. Ettlinger, J. Am. Chem. Soc., 74, 5805 (1952).
15 A. T. Bottini and J. D. Roberts, J. Org. Chem., 21, 1169 (1956).
16 M. G. Ettlinger and F. Kennedy, Chem. & Znd. (London), 891 (1957).
17 E. F. Silversmith, Y. Kitahara, M. C. Caserio, and J. D. Roberts, 3. Am. Chem. Soc., 80, 5840 (1958).
18 W. S. Johnson and coworkers, J. Am. Chem. Soc., 78, 6302 (1956).
19 Private communication from W. S. Johnson.
20 J. N. Shoolery and M. T. Rogers, I. Am. Chem. Soc., 80, 5121 (1958).
21 E. E. van Tamelen, S. H. Levin, G. Brenner, J. Wolinsky, and P. Aldrich, J. Am. Chem. Soc., 80, 501 (1958).
2.09: Illustrative Analysis of a Reaction Product
Nuclear magnetic resonance spectra are often useful for characterization of complex reaction mixtures. For example, in an exploratory methylation of phenyldimethylcarbinol, distillation of the reaction products gave two principal fractions with NMR spectra as shown in Fig. 2-12. These spectra can be quite well analyzed even without the aid of separate spectra of the pure component.22 The presence of resonance lines in the vinyl hydrogen region indicates the formation of some $\alpha$-methylstyrene, while the band at +100 cps shows the presence of O-CH3. The methyl absorptions of the ether are separated somewhat from those of the alcohol, and we can assign the higher of the two peaks in the first fraction to the ether, since it is roughly twice the O-CH3 resonance. Each fraction shows O-H lines which are shifted somewhat from one fraction to the other by chemical shift changes associated with solvent changes as explained earlier for ethanol. From the NMR spectra, we can conclude that the first and second fractions have approximately the compositions 53 and 23 percent methyl phenyldimethylcarbinyl ether, 35 and 77 percent phenyldimethylcarbinol, and 12 and 0 per cent $\alpha$-methylstyrene, respectively.
22 Unpublished experiments by M. C. Caserio.
2.10: Accentuation of Chemical Shifts by Paramagnetic Salts
Phillips and coworkers23 have shown how in favorable cases chemical shifts can be greatly and selectively accentuated by paramagnetic salts. For example, 1.0 M cobaltous chloride dissolved in n-propyl alcohol causes the various proton resonances to separate rather widely from one another. The O-H line remains about in its usual position, but the $\alpha$-methylene proton resonance is shifted some 200 cps to higher fields at 7,050 gauss. The shift is proportional to the applied magnetic field and, for a long-chain alcohol such as 1-hexanol, increases as one goes down the chain to higher numbered carbons until a maximum is reached at and beyond the 3-position. The differences in chemical shift for the various kinds of hydrogens depend on the paramagnetic cobaltous ion being coordinated by the alcohol molecules at the hydroxyl oxygen. It seems likely that studies of this character may give valuable information as to the conformation of organic molecules in solution.
23 W. D. Phillips, C. E. Looney, and C. K. Ikeda, I. Chem. Phys., 27, 1435 (1957). | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/02%3A_The_Chemical_Shift/2.08%3A_Chemical_Shifts_and_Organic_Structure_Determinations._General_Considerations.txt |
The spectrum of ethanol shown in Fig. 2-1 was obtained with a degree of resolution far below that routinely possible for commercial NMR spectrometers. Under high resolution, the proton spectra of ethyl derivatives show a considerably greater number of lines-the CH2 resonance being split into four principal lines and the CH3 resonance into three principal lines. Still higher resolution such as is possible with an extremely stable oscillator and a highly homogeneous magnetic field shows each of these lines to have fine structure, as shown in Fig. 3-1. We shall be concerned here only with the first-order splitting, since the explanation of the higher-order splitting is rather involved.
The mechanism by which the protons of an ethyl group produce seven major resonance lines is interesting and important to theories of structure as well as structural determinations. In the first place, one might infer that the splitting of the CH2 and CH3 lines is evidence for chemically different kinds of methylene and methyl hydrogens. However, the fine structure is not a chemical-shift phenomenon. This is proved by observing the spectrum at two different oscillator frequencies (and field strengths) whereby the principal lines of the methyl and methylene patterns move closer together or farther apart while the spacing of the principal four-three pattern of fine structure remains unaltered, shown in Fig. 3-2. Therefore, we conclude that the different lines are not due to chemically different hydrogens among the methylene and methyl groups, respectively. This conclusion is, of course, in agreement with chemical experience. It can be shown that field-independent splitting represented by the line spacing J arises from interaction between the magnetic moments of one group of hydrogens and the other. The way that this comes about will be illustrated with the aid of a simple but experimentally unrealizable example.
3.02: Spin-Spin Splitting in a Single Crystal
Consider a single crystal made up of H-D molecules, all of which are so oriented that the lines connecting the nuclei of the individual molecules make an angle $\theta$ with an applied magnetic field (Fig. 3-3). Now consider nuclear resonance absorption by deuterium nuclei located at position 1. Each deuteron precesses in a field whose magnitude is determined partly by the applied field and partly by the degree of diamagnetic shielding produced by the bonding electrons. However, this is not the whole story, since each deuterium nucleus will be connected to a proton (at position 2) which can have either of the two possible magnetic quantum numbers, +1/2 or -1/2. If the proton nuclear magnet is oriented in the direction of the magnetic field, then it will augment the field which is experienced by the adjacent deuteron and the total field at the deuteron will correspond to a higher precession frequency than if the proton had no nuclear moment. If the proton magnet is directed the other way, the field at the deuterium nucleus will be reduced from its nominal value. Therefore, the precession frequency of a given deuterium and the position of its resonance lines will depend on the magnetic quantum number of the proton to which it is bonded. In a large assemblage of H-D molecules, there will be almost precisely equal numbers of protons with the two possible spin quantum numbers, even in an intense applied field, unless the temperature is so low that thermal agitation cannot prevent an appreciable excess of the nuclei from being lined up with the applied field. Consequently, at ordinary temperatures, very nearly half of the H-D molecules will show a deuteron resonance at a lower field strength than would be the case if the proton moment were absent, while the other half of the molecules will have their deuterium resonance at a correspondingly higher field strength. The observed spectrum will then appear somewhat as in Fig. 3-4.
The proton resonance spectrum will be affected by the deuterons in an analogous way, except that, since the deuteron has I = 1 and three possible magnetic quantum numbers (+1, 0, -1 ) , the proton resonances will occur at three field strengths. Since the probability that a given deuteron will have any one of the magnetic quantum numbers is essentially one-third, the three lines will be of equal height. It can be shown theoretically that the interaction between the spins is such that the spacing of the deuteron and proton resonance lines depends on the ratios of the gyromagnetic ratios of the nuclei. A rough calculation for an H-D crystal with $\theta$ = 90° shows the predicted spacing of the deuteron resonances to be about 35 gauss, so that proton resonances would be separated by 35 X $\gamma$D/$\gamma$H or 5.3 gauss. This type of magnetic interaction between nuclei is usually called "spin-spin splitting." Since magnetic coupling between nuclei causes changes in precession frequencies depending on the magnitude of the nuclear moment but not the external field, it is clear that the magnitude of the splitting should be independent of the applied field.
Magnetic interaction among nuclei as postulated for the hypothetical crystal of H-D molecules is "direct dipole-dipole interaction" and leads to a line separation proportional to (3cos2$\theta$ -1)r-3 with $\theta$ as defined earlier and r the distance between the nuclei. In crystals, besides the intramolecular nuclear interactions, one will expect also mbstantial intermolecular dipole-dipole interactions leading to additional spKttings or broadening of the absorption lines (see earlier discussion, page 33). | textbooks/chem/Organic_Chemistry/Nuclear_Magnetic_Resonance%3A_Applications_to_Organic_Chemistry_(Roberts)/03%3A_Spin-Spin_Splitting/3.01%3A_The_High-resolution_Ethanol_Spectrum.txt |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.