Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy Flashcards

Question

What type of reactions do conjugated compounds typically undergo?

Answer 1

Conjugated compounds undergo a variety of reactions, many of which involve intermediates that **retain** some of the **resonance** **stabilization** of the conjugated system. Common intermediates include **allylic** **systems**, particularly allylic cations and radicals. Allylic cations and radicals are stabilized by delocalization.

Answer 2

The – CH2 – CH = CH2 group is called the allyl group. Many common names use this terminology.

Answer 3

When allyl bromide is heated with a good ionizing solvent, it ionizes to the **allyl** **cation**, an allyl group with a positive charge. **More-substituted** analogues are called **allylic** **cations**. All allylic cations are **stabilized** by **resonance** with the adjacent double bond, which delocalizes the positive charge over two carbon atoms.

Answer 4

We can represent a delocalized ion such as the allyl cation either by resonance forms, as shown on the left in the following figure, or by a combined structure, as shown on the right. Although the combined structure is more concise, it is sometimes confusing because it attempts to convey all the information implied by two or more resonance forms.

Answer 5

Because of its resonance stabilization, the (**primary**) **allyl** **cation** is about **as stable** as a simple secondary carbocation, such as the isopropyl cation. Most substituted allylic cations have at least one secondary carbon atom bearing part of the positive charge. They are about as stable as simple tertiary carbocations such as the tert-butyl cation.

Answer 6

Electrophilic additions to conjugated dienes usually involve allylic cations as **intermediates**. Unlike simple carbocations, an allylic cation can react with a nucleophile at **either of its positive centers**. Let’s consider the addition of HBr to buta-1,3-diene, an electrophilic addition that produces a mixture of **two constitutional isomers**. One product, 3-bromobut-1-ene, results from Markovnikov addition across one of the double bonds. In the other product, 1-bromobut-2-ene, the double bond shifts to the C2 ¬ C3 position.

Answer 7

The first product results from electrophilic addition of HBr across a double bond. This process is called a **1,2-addition** whether or not these two carbon atoms are numbered 1 and 2 in naming the compound. In the second product, the proton and bromide ion add at the ends of the conjugated system to carbon atoms with a 1,4-relationship. Such an addition is called a **1,4-addition** whether or not these carbon atoms are numbered 1 and 4 in naming the compound.

Answer 8

The mechanism is similar to other electrophilic additions to alkenes. The proton is the electrophile, adding to the alkene to give the most stable carbocation. Protonation of buta-1,3-diene gives an allylic cation, which is stabilized by resonance delocalization of the positive charge over two carbon atoms. Bromide can attack this resonance-stabilized intermediate at either of the two carbon atoms sharing the positive charge. Attack at the secondary carbon gives 1,2-addition; attack at the primary carbon gives 1,4-addition.

Answer 9

One of the interesting peculiarities of the reaction of buta-1,3-diene with HBr is the effect of temperature on the products. If the reagents are allowed to react briefly at **- 80 °C**, the **1,2-addition product predominates**. If this reaction mixture is later allowed to warm to **40 °C**, however, or if the original reaction is carried out at 40 °C, the composition favors the **1,4-addition product**.

Answer 10

This variation in product composition reminds us that the **most stable product** is not always the **major product**. Of the two products, we expect 1-bromobut-2-ene (the 1,4-product) to be more stable, since it has the more substituted double bond. This prediction is supported by the fact that this isomer predominates when the reaction mixture is warmed to 40 °C and allowed to equilibrate. A reaction-energy diagram for the second step of this reaction (Figure 15-9) helps to show why one product is favored at low temperatures and another at higher temperatures. The allylic cation is in the center of the diagram; it can react toward the left to give the 1,2-product or toward the right to give the 1,4-product. The **initial product** depends on **where bromide attacks** the resonance-stabilized allylic cation. Bromide can attack at either of the two carbon atoms that share the positive charge. Attack at the secondary carbon gives 1,2-addition, and attack at the primary carbon gives 1,4-addition.

Answer 11

The transition state for 1,2-addition has a **lower energy** than transition state for 1,4-addition. This is not surprising, because 1,2-addition results from bromide attack at the a **more substituted secondary** carbon, which bears more of the positive charge because it is better stabilized than the primary carbon. Because the 1,2-addition has a lower activation energy than the 1,4-addition, the 1,2-addition takes place **faster (at all temperatures).** Attack by bromide on the allylic cation is a **strongly exothermic** process, so the **reverse reaction** has a **large activation energy**. At -80 °C, few collisions take place with this much energy, and the rate of the reverse reaction is **practically zero**. Under these conditions, the product that is formed faster predominates. Because the kinetics of the reaction determine the results, this situation is called **kinetic control** of the reaction. The 1,2-product, favored under these conditions, is called the **kinetic product**.

Answer 12

At 40 °C, a **significant fraction** of molecular collisions have enough energy for **reverse reactions** to occur. Notice that the **activation energy** for the reverse of the 1,2-addition is **less than** that for the reverse of the 1,4-addition. Although the 1,2-product is **still formed faster**, it also reverts to the allylic cation **faster than** the **1,4-product** does. At 40 °C, an equilibrium is set up, and the relative energy of each species determines its concentration. The 1,4-product is the **most stable** species, and **it predominates**. Since thermodynamics determine the results, this situation is called **thermodynamic control** (or **equilibrium control**) of the reaction. The 1,4-product, favored under these conditions, is called the **thermodynamic product**.

Answer 13

In general, reactions that **do not reverse** easily are **kinetically controlled** because no equilibrium is established. In kinetically controlled reactions, the product with the **lowest-energy transition state** predominates. Reactions that are **easily reversible** are **thermodynamically controlled** unless something happens to prevent equilibrium from being attained. In thermodynamically controlled reactions, the **lowest-energy product** predominates.

Answer 14

Abstraction of allylic hydrogens is **preferred** because the allylic free radical is **resonance-stabilized**. The bond-dissociation enthalpies required to generate several free radicals are compared below. Notice that the allyl radical (a primary free radical) is actually 13 kJ/mol (3 kcal/mol) **more stable** than the tertiary butyl radical.

Answer 15

The allylic cyclohex-2-enyl radical has its unpaired electron delocalized over **two secondary carbon atoms**, so it is even more stable than the unsubstituted allyl radical. The second propagation step may occur at **either** of the radical carbons, but in this symmetrical case, either position gives 3-bromocyclohexene as the product. Less symmetrical compounds often **give mixtures** of products resulting from an **allylic shift**: In the product, the **double bond** can appear at **either** of the positions it occupies in the resonance forms of the allylic radical. An allylic shift in a radical reaction is similar to the 1,4-addition of an electrophilic reagent such as HBr to a diene.

Answer 16

In the allylic bromination just shown, bromine substitutes for a hydrogen atom. The key to getting substitution is to have a **low concentration** of bromine, together with **light or free radicals** to initiate the reaction. Free radicals are highly reactive, and even a small concentration of radicals can produce a fast chain reaction. **Simply adding** bromine might **raise** the concentration too high, resulting in ionic addition of bromine across the double bond. A convenient bromine source for allylic bromination is **N-bromosuccinimide (NBS)**, a brominated derivative of succinimide. Succinimide is a cyclic imide (diamide) of the four-carbon diacid succinic acid. NBS provides a **fairly constant, low concentration** of Br₂ because it reacts with HBr liberated in the substitution, **converting it** back into Br₂. This reaction also removes the HBr by-product, preventing it from adding across the double bond by its own free-radical chain reaction.

Answer 17

The NBS reaction is carried out in a clever way. The allylic compound is **dissolved** in **carbon tetrachloride**, and **one equivalent** of NBS is added. **NBS is denser** than CCl₄ and not very soluble in it, so **it sinks** to the bottom of the CCl₄ solution. The reaction is initiated using a sunlamp for illumination or a radical initiator such as a peroxide. The NBS **gradually appears to rise** to the top of the CCl₄ layer. It is actually converted to succinimide, which is less dense than CCl₄. Once all the solid succinimide has risen to the top, the sunlamp is turned off, the solution is filtered to remove the succinimide, and the CCl₄ is evaporated to recover the product.

Answer 18

Let’s take a closer look at the electronic structure of allylic systems, using the allyl radical as our example. One resonance form shows the radical electron on C1, with a pi bond between C2 and C3. The other shows the radical electron on C3 and a pi bond between C1 and C2. These two resonance forms imply that there is half a pi bond between C1 and C2 and half a pi bond between C2 and C3, with the radical electron half on C1 and half on C3.

Answer 19

Remember that **no resonance** form has an **independent existence**: A compound has characteristics of all its resonance forms at the same time, but it **does not “resonate”** among them. The **p orbitals** of **all three carbon** atoms must be **parallel** to have simultaneous pi bonding overlap between C1 and C2 and between C2 and C3. The geometric structure of the allyl system is shown in Figure 15-10. The allyl cation, the allyl radical, and the allyl anion all have this same geometric structure, differing only in the number of pi electrons.

Answer 20

Just as the four p orbitals of buta-1,3-diene overlap to form four molecular orbitals, the **three atomic p orbitals** of the allyl system overlap to form **three molecular orbitals**, shown in Figure 15-11. These three MOs share several important features with the MOs of the butadiene system. The first MO is **entirely bonding**, the second has **one node**, and the third has **two nodes** and (because it is the highest-energy MO) is entirely **antibonding**. As with butadiene, we expect that half of the MOs will be bonding, and half antibonding; but with an odd number of MOs, they **cannot be symmetrically divided**. One of the MOs must appear at the middle of the energy levels, neither **bonding nor antibonding**: It is a **nonbonding molecular orbital**. Electrons in a nonbonding orbital have the same energy as in an isolated p orbital. The structure of the nonbonding orbital (π₂) may seem strange because there is zero electron density on the center p-orbital (C2). This is the case because π₂ must have one node, and the only symmetrical position for one node is in the center of the molecule, crossing C2. We can tell from its structure that π₂ must be nonbonding, because C1 and C3 both have zero overlap with C2. The total is zero bonding, implying a nonbonding orbital.

Answer 21

Allylic halides and tosylates show enhanced reactivity toward nucleophilic displacement reactions by the SN2 mechanism. For example, allyl bromide reacts with nucleophiles by the SN2 mechanism about 40 times faster than n-propyl bromide. Figure 15-13 shows how this rate enhancement can be explained by allylic delocalization of electrons in the transition state. The transition state for the SN2 reaction looks like a trigonal carbon atom with a p orbital perpendicular to the three substituents. The electrons of the attacking nucleophile are forming a bond using one lobe of the p orbital while the leaving group’s electrons are leaving from the other lobe. When the substrate is allylic, the transition state receives resonance stabilization through conjugation with the p orbitals of the pi bond. This stabilization lowers the energy of the transition state, resulting in a lower activation energy and an enhanced rate.

Answer 22

Allylic halides are so reactive that they couple with Grignard and organolithium reagents, a reaction that does not work well with unactivated halides.

Answer 23

In 1928, German chemists Otto Diels and Kurt Alder discovered that alkenes and alkynes with electron-withdrawing groups add to conjugated dienes to form six-membered rings. The Diels–Alder reaction has proven to be a useful synthetic tool, providing one of the best ways to make six-membered rings with diverse functionality and controlled stereochemistry. Diels and Alder were awarded the Nobel Prize in 1950 for their work.

Answer 24

The Diels–Alder reaction is called a **[4 + 2] cycloaddition** because a ring is formed by the interaction of four pi electrons in the diene with two pi electrons of the alkene or alkyne. Since the electron-poor alkene or alkyne is prone to react with a diene, it is called a **dienophile** (“lover of dienes”). In effect, the Diels–Alder reaction converts two pi bonds into two sigma bonds. We can symbolize the Diels–Alder reaction by using three arrows to show the movement of three pairs of electrons. This electron movement is **concerted**, with three pairs of electrons moving simultaneously. The Diels–Alder reaction is like a nucleophile–electrophile reaction. The diene is electron-rich, and the dienophile is electron-poor. Simple dienes such as buta-1,3-diene are sufficiently electron-rich to be effective dienes for the Diels–Alder reaction. The presence of electron-donating ( – D) groups, such as alkyl groups or alkoxy ( – OR) groups, may further enhance the reactivity of the diene.

Answer 25

The mechanism of the Diels–Alder reaction is a concerted cyclic movement of six electrons: four in the diene and two in the dienophile. For the three pairs of electrons to move simultaneously, the transition state must have a geometry that allows overlap of the two end p orbitals of the diene with those of the dienophile. Figure 15-15 shows the required geometry of the transition state. The geometry of the Diels–Alder transition state explains why some isomers react differently from others, and it enables us to predict the stereochemistry of the products.

Answer 26

* *s*-cis Conformation of the Diene * syn Stereochemistry * The Endo Rule

Answer 27

The diene must be in the *s*-cis conformation to react. When the diene is in the *s*-tran conformation, the end *p* orbitals are too far apart to overlap with the p orbitals of the dienophile. The *s*-trans conformation usually has a lower energy than the *s*-cis, but this energy difference is not enough to prevent most dienes from undergoing Diels-Alder reactions. For example, the *s*-trans conformation of butadiene is only 9.6 kJ/mol (2.3 kcal/mol) lower in energy than the *s*-cis conformation.

Answer 28

Structural features that aid or hinder the diene in achieving the s-cis conformation affect its ability to participate in Diels–Alder reactions. Figure 15-16 shows that dienes with functional groups that hinder the s-cis conformation react more slowly than butadiene. Dienes with functional groups that hinder the s-trans conformation react faster than butadiene.

Answer 29

Because cyclopentadiene is fixed in the s-cis conformation, it is highly reactive in the Diels–Alder reaction. It is so reactive, in fact, that at room temperature, cyclopentadiene slowly reacts with itself to form dicyclopentadiene. Cyclopentadiene is regenerated by heating the dimer above 200 °C. At this temperature, the Diels–Alder reaction reverses, and the more volatile cyclopentadiene monomer distills over into a cold flask. The monomer can be stored indefinitely at dry-ice temperatures.

Answer 30

The Diels–Alder reaction is a syn addition with respect to both the diene and the dienophile. The dienophile adds to one face of the diene, and the diene adds to one face of the dienophile. As you can see from the transition state in Figure 15-15, there is no opportunity for any of the substituents to change their stereochemical positions during the course of the reaction. Substituents that are on the same side of the diene or dienophile will be cis on the newly formed ring. The following examples show the results of this syn addition.

Answer 31

When the dienophile has a pi bond in its electron-withdrawing group (as in a carbonyl group or a cyano group), the p orbitals in that electron-withdrawing group approach one of the central carbon atoms (C2 or C3) of the diene. This proximity results in **secondary** **overlap**: an overlap of the p orbitals of the electron-withdrawing group with the p orbitals of C2 and C3 of the diene (Figure 15-17). Secondary overlap helps to stabilize the transition state. The influence of secondary overlap was first observed in reactions using cyclopentadiene to form bicyclic ring systems. In the bicyclic product (called norbornene), the electron-withdrawing substituent occupies the stereochemical position closest to the central atoms of the diene. This position is called the **endo** **position** because the substituent seems to be inside the pocket formed by the six-membered ring of norbornene. This stereochemical preference for the electron-withdrawing substituent to appear in the endo position is called the **endo** **rule**.

Answer 32

Even when the diene and dienophile are both unsymmetrically substituted, the Diels–Alder reaction usually gives a single product (or a major product) rather than a random mixture. We can usually predict the major product by considering how the substituents polarize the diene and the dienophile in their charge-separated resonance forms. If we then arrange the reactants to connect the most negatively charged carbon in the (electron-rich) diene with the most positively charged carbon in the (electron-poor) dienophile, we can usually predict the correct orientation. The following examples show that an electron-donating substituent (D) on the diene and an electron-withdrawing substituent (W) on the dienophile usually show either a 1,2- or 1,4-relationship in the product. In most cases, we don’t even need to draw the charge-separated resonance forms to determine which orientation of the reactants is preferred. We can predict the major products of unsymmetrical Diels–Alder reactions simply by remembering that the electron-donating groups of the diene and the electron-withdrawing groups of the dienophile usually bear either a 1,2-relationship or a 1,4-relationship in the products, but not a 1,3-relationship.

Answer 33

The Diels–Alder reaction is a **cycloaddition**: Two molecules combine in a one-step, **concerted** **reaction** to form a new ring. Cycloadditions such as the Diels–Alder are one class of **pericyclic** **reactions**, which involve the concerted forming and breaking of bonds within a closed ring of interacting orbitals. Figure 15-15 (page 686) shows the closed loop of interacting orbitals in the Diels–Alder transition state. Each carbon atom of the new ring has one orbital involved in this closed loop.

Answer 34

A concerted pericyclic reaction has a single transition state, whose activation energy may be supplied by heat (thermal induction) or by ultraviolet light (photochemical induction). Some pericyclic reactions proceed only under thermal induction, and others proceed only under photochemical induction. Some pericyclic reactions take place under both thermal and photochemical conditions, but the two sets of conditions give different products.

Answer 35

For many years, pericyclic reactions were poorly understood and unpredictable. Around 1965, Robert B. Woodward and Roald Hoffmann developed a theory for predicting the results of pericyclic reactions by considering the symmetry of the molecular orbitals of the reactants and products. Their theory, called conservation of orbital symmetry, says that the molecular orbitals of the reactants must flow smoothly into the MOs of the products without any drastic changes in symmetry. In that case, there will be bonding interactions to help stabilize the transition state. Without these bonding interactions in the transition state, the activation energy is much higher, and the concerted cyclic reaction cannot occur. Conservation of symmetry has been used to develop "rules" to predict which pericyclic reactions are feasible and what products will result. These rules are often called **Woodward-Hoffmann rules**.

Answer 36

Butadiene, with four atomic p orbitals, has four molecular orbitals: two bonding MOs (filled) and two antibonding MOs (vacant). Ethylene, with two atomic p orbitals, has two MOs: a bonding MO (filled) and an antibonding MO (vacant).

Answer 37

In the Diels–Alder reaction, the diene acts as the electron-rich nucleophile, and the dienophile acts as the electron-poor electrophile. If we imagine the diene contributing a pair of electrons to the dienophile, the highest-energy electrons of the diene require the least activation energy for such a donation. The electrons in the highest-energy occupied orbital, called the **Highest Occupied Molecular Orbital (HOMO)**, are the important ones because they are the most weakly held. The HOMO of butadiene is π₂, and its symmetry determines the course of the reaction.

Answer 38

The orbital in ethylene that receives these electrons is the lowest-energy orbital available, the Lowest Unoccupied Molecular Orbital (LUMO). In ethylene, the LUMO is the π\* antibonding orbital. If the electrons in the HOMO of butadiene can flow smoothly into the LUMO of ethylene, a concerted reaction can take place.

Answer 39

Figure 15-19 shows that the HOMO of butadiene has the correct symmetry to overlap in phase with the LUMO of ethylene. Having the correct symmetry means the orbitals that form the new bonds can overlap constructively: plus with plus and minus with minus. These bonding interactions stabilize the transition state and promote the concerted reaction. This favorable result predicts that the reaction is **symmetry-allowed**. The Diels–Alder reaction is common, and this theory correctly predicts a favorable transition state.

Answer 40

If a cycloaddition produces an overlap of positive-phase orbitals with negative-phase orbitals (destructive overlap), antibonding interactions are generated. Antibonding interactions raise the activation energy, so the reaction is classified as symmetry-forbidden. The thermal [2 + 2] cycloaddition of two ethylenes to give cyclobutane is a symmetry-forbidden reaction. This [2 + 2] cycloaddition requires the HOMO of one of the ethylenes to overlap with the LUMO of the other. Figure 15-20 shows that an antibonding interaction results from this overlap, raising the activation energy. For a cyclobutane molecule to result, one of the MOs would have to change its symmetry. Orbital symmetry would not be conserved, so the reaction is symmetry-forbidden. Such a symmetry-forbidden reaction can occasionally be made to occur, but it cannot occur in the concerted pericyclic manner shown in the figure.

Answer 41

When ultraviolet light rather than heat is used to induce pericyclic reactions, our predictions generally must be reversed. For example, the [2 + 2] cycloaddition of two ethylenes is photochemically “allowed.” When a photon with the correct energy strikes ethylene, one of the pi electrons is excited to the next higher molecular orbital (Figure 15-21). This higher orbital, formerly the LUMO, is now occupied: It is the new HOMO\*, the HOMO of the excited molecule. The HOMO\* of the excited ethylene molecule has the same symmetry as the LUMO of a ground-state ethylene. An excited molecule can react with a ground-state molecule to give cyclobutane (Figure 15-22). The [2 + 2] cycloaddition is therefore photochemically allowed but thermally forbidden. In most cases, *photochemically allowed* reactions are *thermally forbidden*, and thermally allowed reactions are photochemically forbidden.

Answer 42

We now study ultraviolet (UV) spectroscopy, which detects the electronic transitions of conjugated systems and provides information about the length and structure of the conjugated part of a molecule. UV spectroscopy gives more specialized information than does IR or NMR, and it is less commonly used than the other techniques.

Answer 43

Ultraviolet frequencies correspond to shorter wavelengths and much higher energies than infrared. The UV region is a range of frequencies just beyond the visble: ultra, meaning beyond, and violet, the highest-frequency visible light. Wavelengths of the UV region are given in units of nanometers (nm; 10^-9).

Answer 44

Common UV spectrometers operate in the range of 200 to 400 nm (2 x 10^-5 to 4 x 10^-5 cm), corresponding to photon energies of about 300 to 600 kJ/mol (70 to 140 kcal/mol). These spectrometers often extend into the visible region (longer wavelength, lower energy) and are called UV-visible spectrometers. UV-visible energies correspond to electronic trainsitions: the energy needed to excite an electron from one molecular orbital to another.

Answer 45

*The wavelengths of UV light absorbed by a molecule are determined by the electronic energy differences between orbitals in the molecule*. Sigma bonds are very stable, and the electrons in sigma bonds are usually unaffected by UV wavelengths of light above 200 nm. Pi bonds have electrons that are more easily excited into higher energy orbitals. Conjugated systems are particularly likely to have low-lying vacant orbitals, and electronic transitions into these orbitals produce characteristic ultraviolet absorptions.

Answer 46

Ethylene, for example, has two pi orbitals: the bonding orbital (π, the HOMO) and the antibonding orbital (π\*, the LUMO). The ground state has two electrons in the bonding orbital and none in the antibonding orbital. A photon with the right amount of energy can excite an electron from the bonding orbital (π) to the antibonding orbital (π\*). This transition from a p bonding orbital to a π\* antibonding orbital is called π --\> π\* trainsition (Figure 15-23).

Answer 47

The π --\> π\* transition of ethylene requires absorption of light at 171 nm (686 kJ/mol). Most UV spectrometers cannot detect this absorption because it is obscured by the absorption caused by oxygen in the air. In conjugated systems, however, there are electronic trainsitions with lower energies that correspond to wavelengths longer than 200 nm. Figure 15-24 compares the MO energies of ethylene with those of butadiene to show that the HOMO and LUMO of butadiene are closer in energy than those of ethylene. The HOMO of butadiene is higher in energy than the HOMO of ethylene, and the LUMO of butadiene is lower in energy than the LUMO of ethylene. Both differences reduce the relative energy of the π₂ --\> π₃^* transition. The resulting absorption is at 217 nm (540 kJ/mol). which can be measured using standard UV spectrometer.

Answer 48

Just as conjugated dienes absorb at longer wavelengths than simple alkenes, conjugated trienes absorb at even longer wavelengths. In general, the energy difference between HOMO and LUMO decreases as the length of conjugation increases. In hexa-1,3,5-triene, for example (Figure 15-25), the HOMO is π3 and the LUMO is π4\* . The HOMO in hexa-1,3,5-triene is slightly higher in energy than for buta-1,3-diene, and the LUMO is slightly lower in energy. Once again, the narrowing of the energy between the HOMO and the LUMO gives a lower-energy, longer-wavelength absorption. The principal π --\> π\* transition in hexa-1,3,5-triene occurs at 258 nm (452 kJ/mol).

Answer 49

We can summarize the effects of conjugation on the wavelength of UV absorption by stating a general rule: *A compound that contains a longer chain of conjugated double bonds absorbs light at a longer wavelength*. The trend of longer wavelengths for longer conjugated chains continues, and at seven conjugated double bonds, the absorption surpasses 400 nm and enters in the visible portion of the spectrum.

Answer 50

Because they have no interaction with each other, isolated double bonds do not contribute to shifting the UV absorption to longer wavelengths. Both their reactions and their UV absorptions are like those of simple alkenes. For example, penta-1,4-diene absorbs at 178 nm, a value that is typical of simple alkenes rather than conjugated dienes.

Answer 51

To measure the ultraviolet (or UV–visible) spectrum of a compound, the sample is dissolved in a solvent (often ethanol) that does not absorb above 200 nm. The sample solution is placed in a quartz cell, and some of the solvent is placed in a **reference cell**. An ultraviolet spectrometer operates by comparing the amount of light transmitted through the sample (the **sample beam**) with the amount of light in the **reference beam**. The reference beam passes through the reference cell to compensate for any absorption of light by the cell and the solvent.

Answer 52

The spectrometer (Figure 15-26) has a source that emits all frequencies of UV light (above 200 nm). This light passes through a monochromator, which uses a diffraction grating or a prism to spread the light into a spectrum and select one wavelength. This single wavelength of light is split into two beams, with one beam passing through the sample cell and the other passing through the reference (solvent) cell. The detector continously measures the intensity ratio of the reference beam (I_r) compared with sample beam (I_s). As the sepctrometer scans the wavelengths in the UV region, a printer draws a graph (called the spectrum) of the absorbance of the sample as a function of the wavelength.

Answer 53

The **absorbance**, A, of the sample at a particular wavelength is governed by *Beer's* *Law*. Beer's Law: A = Log (I_r/I_s) = εcL Where: * c = sample cocentration in moles per liter * L = path length of light through the cell in centimeters * ε = the molar absorptivity (or molar extinction coefficient) of the sample.

Answer 54

Molar absoprtivity (ε) is a measure of how strongly the sample absorbs light at that wavelength.

Answer 55

If the sample absorbs light at a particular wavelength, the sample beam (I_s) is less intense than the reference beam (I_r), and the ratio I_r/I_s is greater than 1. The ratio is equal to 1 when there is no absorption. The absorbance (the logarithm of the ratio) is therefore greater than zero when the sample absorbs, and is equal to zero when it does not. A UV spectrum is a plot of A, the absorbance of the sample, as a function of the wavelength.

Answer 56

UV–visible spectra tend to show broad peaks and valleys. The spectral data that are most characteristic of a sample are as follows: 1. The wavelength(s) of maximum absorbance, called λ_max 2. The value of the molar absorptivity ε at each maximum Since UV–visible spectra are broad and lacking in detail, they are rarely printed as actual spectra. The spectral information is given as a list of the value or values of lmax together with the molar absorptivity for each value of λ_max.

Answer 57

The UV spectrum of isoprene (2-methylbuta-1,3-diene) is shown in Figure 15-27. This spectrum could be summarized as follows: λ_max = 222 nm ε = 20,000 The value of λ_max is read directly from the spectrum, but the molar absorptivity ε must be calculated from the concentration of the solution and the path length of the cell. For an isoprene concentration of 4 x 10^-5 M and a 1-cm cell, the molar absorptivity is found by rearranging Beer's Law ( A = εcL ) ε = A/cL = .8/(4 x 10^-5) = 20,000

Answer 58

Molar absorptivities in the range of 5000 to 30,000 are typical for the π --\> π\* transitions of conjugated polyene systems. Such large molar absorptivities are helpful, since spectra may be obtained with very small amounts of sample. On the other hand, samples and solvents for UV spectroscopy must be extremely pure. A minute impurity with a large molar absorptivity can easily obscure the spectrum of the desired compound.

Answer 59

The human eye can see the part of the electromagnetic spectrum with wavelengths from about 400 nm (blue-violet) to a little over 700 nm (red). Wavelengths less than 400 nm are the higher-energy ultraviolet (“beyond violet”) region, and the lower-energy wavelengths over 700 nm are the infrared (“below red”) region. White light contains light of all the wavelengths, so we see all the colors of the rainbow when we use a prism or diffraction grating to separate white light into its component wavelengths.

Answer 60

Colored objects do not emit their own light. They reflect part of the ambient light that falls on them. A white object reflects all of the light, but a colored object absorbs some wavelengths and reflects the rest of the spectrum. The color that we see is what is reflected, that is, the complete spectrum minus the portion that is absorbed. Therefore, we observe the opposite, or complement, of the light that is absorbed. For example, ß-carotene, the orange color in carrots, has 11 conjugated double bonds. It absorbs strongly at 454 nm, in the blue portion of the spectrum, reflecting the orange complement.

Answer 61

Not all color in nature arises from conjugated organic compounds. Minerals, for example, are colored by metal atoms in various oxidation states. However, virtually all of the color present in plants and animals is because of conjugated organic molecules. Figure 15-28 shows examples from some of the classes of colored compounds in living organisms. Note the extensive conjugation in each of these compounds. These brightly colored compounds have large extinction coefficients, so they require only a tiny amount to produce an intense color.

Answer 62

Dyes are intensely colored compounds used in fabrics, plastics, inks, and other products. Dyes were originally extracted from plants or animals and used to color cloth. For example, red carmine (page 2) was extracted from cochineal insects, and blue indigo (the dye used in blue jeans) was extracted from plant material. Both of these dyes are now synthesized in large quantities. The Romans extracted the indigo derivative Tyrian purple (imperial purple) from a sea snail and used the dye to color the robes of emperors and high-ranking senators.

Answer 63

The era of synthetic dyes is credited to Sir William Henry Perkin, who accidentally synthesized the purple dye mauveine in 1856 (at age 18) while trying to make quinine. Mauveine was an inexpensive substitute for Tyrian purple. Chemists soon developed many other synthetic dyes, and by the late 1800s the dye industry had become one of the major chemical industries in Europe. Commoners could now wear all the colors that were once reserved for royalty. The dye industry also provided the commercial funding and motivation for much of the early research in organic chemistry, especially the chemistry of compounds derived from benzene.

Answer 64

Industrial chemists have developed thousands of commercial dyes for every imaginable use, such as fabrics, hair color, inks, toys, and foods. Many dyes are toxic, so the US government has regulated their use in foods since 1906, when the Pure Food and Drug Act created a group of dyes that were approved for coloring foods. Like all unnecessary food additives, food dyes have come under intense scrutiny to determine which ones are safest, and whether any of them cause significant side effects. Only seven dyes are currently approved for unrestricted food use in the US. Two of these food dyes are Indigo Carmine and Sunset Yellow shown below.

Answer 65

Many of the recent advances in biology and medicine have resulted from applications of biochemistry and molecular biology based on knowledge of the physiology at the cellular and molecular levels. To understand cellular processes, we need to detect compounds at micromolar and lower concentrations, often in an intact cell, without vaporizing or destroying the sample. UV–visible spectroscopy is nondestructive and exceptionally sensitive, and it can measure small concentrations of highly conjugated metabolites like ATP, as well as macromolecules like proteins and nucleic acids in aqueous solutions. If we know the wavelength at maximum absorption (λ_max) and the molar absorptivity (ε) of the molecule of interest, we can use Beer’s law to calculate minuscule concentrations of these biomolecules, often in complex mixtures.

Answer 66

Proteins, DNA, ATP, and many other biomolecules contain conjugated systems of pi bonds with strong characteristic absorptions in the UV region of the spectrum. Common heteroatoms with nonbonding electrons, like oxygen and nitrogen, are frequently part of these conjugated pi systems, contributing to their unique characteristics.

Answer 67

Proteins are polymers of twenty common amino acids (Chapter 24), plus occasional rare amino acids. Four of the twenty common amino acids contain conjugated ring systems that strongly absorb UV light, making almost all proteins detectable and quantifiable by UV analysis. The four common UV-absorbing amino acids (shown below) are phenylalanine, tyrosine, histidine, and tryptophan. The standard wavelength for measuring protein absorbance is 280 nm, where most of the absorption is due to tryptophan and tyrosine, with a small contribution from phenylalanine.

Answer 68

Clinical chemists analyze blood and urine to determine the concentrations of hormones, metabolites, and other substances to diagnose illnesses rapidly and accurately. The central instrument in a modern blood analyzer is a UV–visible spectrometer. Depending on the UV absorption of the biomolecule to be measured, the spectrometer may detect the substance directly, or it may detect a specific “color-developing reagent” that changes its UV absorption when it reacts with the molecule of interest.

Answer 69

Alkaline Phosphatase removes the phosphate group from many types of biomolecules. Abnormal levels of alkaline phosphatase can indicate liver and bone disorders, among other conditions. Alkaline phosphatase gives a UV spectrum similar to other proteins, so other proteins would interfere with a direct analysis, but we can measure its distinctive ability to remove phosphate groups. We add the colorless compound indoxyl phosphate, which loses its phosphate group when it reacts with the alkaline phosphatase enzyme. The product is indoxyl, which quickly dimerizes to blue indigo. By setting up the UV–visible spectrometer to measure the amount of indigo produced in a certain time period, we can calculate the amount of alkaline phosphatase present in the blood.

Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy Flashcards

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