4.3: Espectrometría ultravioleta y espectrometría visible

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<div id="section_5"> Mientras que la interacción con la luz infrarroja provoca que las moléculas experimentan transiciones vibracionales, la longitud de onda más corta, radiación de energía más alta en el UV (200-400 nm) y el rango visible (400-700 nm) del espectro electromagnético hace que muchas moléculas orgánicas se sometan a transiciones electrónicas. Lo que esto significa es que cuando la energía de la luz visible o UV es absorbida por una molécula, uno de sus electrones salta de una energía más baja a un orbital molecular de energía superior. <h3>4.3A: Transiciones electrónicas</h3> Tomemos como nuestro primer ejemplo el caso sencillo de hidrógeno molecular, H2. Como recordarás de la sección 2.1A, la imagen orbital molecular para la molécula de hidrógeno consiste en una unión MO σ, y una energía mayor antienlazante σ* MO. Cuando la molécula está en el estado fundamental, ambos electrones están emparejados en el enlace orbital de la más baja energía -  este es el Orbital Molecular Ocupado Más Alto (HOMO). El orbital antienlazante σ*, a su vez, es el Orbital Molecular Desocupado Más Bajo (LUMO). <img alt="image024.png" class="internal default" height="212" src="/@api/deki/files/33825/=image023.png" style="" width="535" /> Si la molécula se expone a luz de una longitud de onda con una energía igual a ΔE, la brecha de energía HOMO-LUMO, esta longitud de onda será absorbida y la energía utilizada para topar uno de los electrones desde el HOMO a LUMO - en otras palabras, del orbital σ a el σ*. Esto se conoce como una transición σ - σ*. ΔE para esta transición electrónica es 258 kcal / mol, correspondiente a la luz con una longitud de onda de 111 nm. Cuando una molécula con doble enlace tal como eteno (su nombre común es etileno) absorbe la luz, se somete a una transición  π - π*. Debido a que los intervalos de energía π- π* son más estrechos que las brechas σ - σ*, eteno absorbe luz a 165 nm - una longitud de onda más larga que el hidrógeno molecular.  <img alt="image026.png" class="internal default" height="162" src="/@api/deki/files/33823/=image025.png" width="597" /> Las transiciones electrónicas The electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded by standard UV spectrophotometers, which generally have a range of 220 – 700 nm.  Where UV-vis spectroscopy becomes useful to most organic and biological chemists is in the study of molecules with conjugated pi systems.  In these groups, the energy gap for π -π* transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer.  Molecules or parts of molecules that absorb light strongly in the UV-vis region are called chromophores. Let’s revisit the MO picture for 1,3-butadiene, the simplest conjugated system (see <a href="Core/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_02:_Introduction_to_organic_structure_and_bonding_II/Section_2.2:_Molecular_orbital_theory:_conjugation_and_aromaticity" title="Organic Chemistry/Organic Chemistry With a Biological Emphasis/Chapter 2: Introduction to organic structure and bonding II/Section 2.1: Molecular orbital theory: conjugation and aromaticity">section 2.1B</a>).  Recall that we can draw a diagram showing the four pi MO’s that result from combining the four 2pz atomic orbitals. The lower two orbitals are bonding, while the upper two are antibonding. <img alt="image028.png" class="internal default" height="183" src="/@api/deki/files/33821/=image027.png" width="553" /> Comparing this MO picture to that of ethene, our isolated pi-bond example, we see that the HOMO-LUMO energy gap is indeed smaller for the conjugated system. 1,3-butadiene absorbs UV light with a wavelength of 217 nm. As conjugated pi systems become larger, the  energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer.   The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a ΔE of 111 kcal/mol. <img alt="image030.png" class="internal default" height="221" src="/@api/deki/files/33819/=image029.png" width="605" /> In molecules with extended pi systems, the HOMO-LUMO energy gap becomes so small that absorption occurs in the visible rather then the UV region of the electromagnetic spectrum.  Beta-carotene, with its system of 11 conjugated double bonds,  absorbs light with wavelengths in the blue region of the visible spectrum while allowing other visible wavelengths – mainly those in the red-yellow region - to be transmitted. This is why carrots are orange. <img alt="image032.png" class="internal default" height="136" src="/@api/deki/files/33817/=image031.png" width="572" /> The conjugated pi system in 4-methyl-3-penten-2-one gives rise to a strong UV absorbance at 236 nm due to a π - π* transition.  However, this molecule also absorbs at 314 nm.  This second absorbance is due to the transition of a non-bonding (lone pair) electron on the oxygen up to a π* antibonding MO: <img alt="image034.png" class="internal default" height="203" src="/@api/deki/files/33815/=image033.png" width="654" /> This is referred to as an n - π* transition.  The nonbonding (n) MO’s are higher in energy than the highest bonding p orbitals, so the energy gap for an n - π* transition is smaller that that of a π - π* transition – and thus the n - π* peak is at a longer wavelength.  In general, n - π* transitions are weaker (less light absorbed) than those due to π - π* transitions. <div> </div> <div> <div> <div class="exercise"> Exercise 4.3 How large is the π - π* transition in 4-methyl-3-penten-2-one? <a href="Core/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Solution_Manual/Chapter_04_Solutions" title="Organic Chemistry/Organic Chemistry With a Biological Emphasis/Solution Manual/Chapter 4 Solutions">Solution</a> </div> </div> </div> <div> <div class="exercise"> Exercise 4.4 Which of the following molecules would you expect absorb at a longer wavelength in the UV region of the electromagnetic spectrum? Explain your answer. <img alt="image036.png" class="internal default" height="142" src="/@api/deki/files/33813/=image035.png" width="280" /> <div> </div> <div> </div> <div><a href="Core/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Solution_Manual/Chapter_04_Solutions" title="Organic Chemistry/Organic Chemistry With a Biological Emphasis/Solution Manual/Chapter 4 Solutions">Solution</a></div> </div> </div> </div> <div id="section_6"> <h3>4.3B: Looking at UV-vis spectra</h3> We have been talking in general terms about how molecules absorb UV and visible light – now let's look at some actual examples of data from a UV-vis absorbance spectrophotometer. The basic setup is the same as for IR spectroscopy: radiation with a range of wavelengths is directed through a sample of interest, and a detector records which wavelengths were absorbed and to what extent the absorption occurred.  Below is the absorbance spectrum of an important biological molecule called nicotinamide adenine dinucleotide, abbreviated NAD+ (we'll learn what it does in <a href="Core/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_16:_Oxidation_and_reduction_reactions/Section_16.04:_Hydrogenation////dehydrogenation_reactions_of_carbonyls,_imines,_and_alcohols" title="Organic Chemistry/Organic Chemistry With a Biological Emphasis/Chapter 16: Oxidation and reduction reactions/Section 16.4: Hydrogenation//dehydrogenation reactions of carbonyls, imines, and alcohols">section 16.4</a>)  This compound absorbs light in the UV range due to the presence of conjugated pi-bonding systems. <img alt="image038.png" class="internal default" height="405" src="/@api/deki/files/33811/=image037.png" width="700" /> You’ll notice that this UV spectrum is much simpler than the IR spectra we saw earlier: this one has only one peak, although many molecules have more than one.  Notice also that the convention in  UV-vis spectroscopy is to show the  baseline at the bottom of the graph with the peaks pointing up.  Wavelength values on the x-axis are generally measured in nanometers (nm) rather than in cm-1 as is the convention in IR spectroscopy.  Peaks in UV spectra tend to be quite broad, often spanning well over 20 nm at half-maximal height.  Typically, there are two things that we look for and record from a UV-Vis spectrum..  The first is λmax, which is the wavelength at maximal light absorbance.  As you can see, NAD+ has λmax, = 260 nm.  We also want to record how much light is absorbed at λmax.  Here we use a unitless number called absorbance, abbreviated 'A'.  This contains the same information as the 'percent transmittance' number used in IR spectroscopy, just expressed in slightly different terms.  To calculate absorbance at a given wavelength, the computer in the spectrophotometer simply takes the intensity of light at that wavelength before it passes through the sample (I0), divides this value by the intensity of the same wavelength after it passes through the sample (I), then takes the log10 of that number: \[A = log \dfrac{I_0}{I}\] You can see that the absorbance value at 260 nm (A260) is about 1.0 in this spectrum.  <div> <div class="exercise"> Exercise 4.5 Express A = 1.0 in terms of percent transmittance (%T, the unit usually used in IR spectroscopy (and sometimes in UV-vis as well). <a href="Core/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Solution_Manual/Chapter_04_Solutions" title="Organic Chemistry/Organic Chemistry With a Biological Emphasis/Solution Manual/Chapter 4 Solutions">Solution</a> </div> </div> Here is the absorbance spectrum of the common food coloring Red #3:   <img alt="image040.png" class="internal default" height="426" src="/@api/deki/files/33809/=image039.png" width="682" /> Here, we see that the extended system of conjugated pi bonds causes the molecule to absorb light in the visible range.  Because the  λmax of 524 nm falls within the green region of the spectrum, the compound appears red to our eyes. Now, take a look at the spectrum of another food coloring, Blue #1:   <img alt="image042.png" class="internal default" height="437" src="/@api/deki/files/33807/=image041.png" width="680" /> Here, maximum absorbance is at 630 nm, in the orange range of the visible spectrum, and the compound appears blue.  </div> <div id="section_7"> <h3>4.3C: Applications of UV spectroscopy in organic and biological chemistry</h3> UV-vis spectroscopy has many different applications in organic and biological chemistry.  One of the most basic of these applications is the use of the Beer - Lambert Law to determine the concentration of a chromophore.  You most likely have performed a Beer – Lambert experiment in a previous chemistry lab.  The law is simply an application of the observation that, within certain ranges, the absorbance of a chromophore at a given wavelength varies in a linear fashion with its concentration: the higher the concentration of the molecule, the greater its absorbance. If we divide the observed value of A at λmax by the concentration of the sample (c, in mol/L), we obtain the molar absorptivity, or extinction coefficient (ε), which is a characteristic value for a given compound.  \[ ε = \dfrac{A}{c}\] The absorbance will also depend, of course, on the path length - in other words, the distance that the beam of light travels though the sample.  In most cases, sample holders are designed so that the path length is equal to 1 cm, so the units for molar absorptivity are  mol * L-1cm-1.  If we look up the value of e for our compound at λmax, and we measure absorbance at this wavelength, we can easily calculate the concentration of our sample.   As an example, for NAD+ the literature value of ε at 260 nm is 18,000 mol * L-1cm-1.   In our NAD+ spectrum we observed A260 = 1.0, so using equation 4.4 and solving for concentration we find that our sample is 5.6 x 10-5 M.   <div> <div class="exercise"> Exercise 4.6 The literature value of ε for 1,3-pentadiene in hexane is 26,000 mol * L-1cm-1 at its maximum absorbance at 224 nm.  You prepare a sample and take a UV spectrum, finding that A224 = 0.850.   What is the concentration of your sample? <a href="Core/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Solution_Manual/Chapter_04_Solutions" title="Organic Chemistry/Organic Chemistry With a Biological Emphasis/Solution Manual/Chapter 4 Solutions">Solution</a> </div> </div> The bases of DNA and RNA are good chromophores: <img alt="image044.png" class="internal default" height="178" src="/@api/deki/files/33805/=image043.png" width="617" /> Biochemists and molecular biologists often determine the concentration of a DNA sample by assuming an average value of  ε = 0.020 ng-1×mL for double-stranded DNA at its λmax of 260 nm (notice that concentration in this application is expressed in mass/volume rather than molarity:  ng/mL is often a convenient unit for DNA concentration when doing molecular biology). <div> <div class="exercise"> Exercise 4.7 50 mL of an aqueous sample of double stranded DNA is dissolved in 950 mL of water.  This diluted solution has a maximal absorbance of 0.326 at 260 nm.  What is the concentration of the original (more concentrated) DNA sample, expressed in mg/mL? <a href="Core/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Solution_Manual/Chapter_04_Solutions" title="Organic Chemistry/Organic Chemistry With a Biological Emphasis/Solution Manual/Chapter 4 Solutions">Solution</a> </div> </div> Because the extinction coefficient of double stranded DNA is slightly lower than that of single stranded DNA, we can use UV spectroscopy to monitor a process known as DNA melting.   If  a short stretch of double stranded DNA is gradually heated up, it will begin to ‘melt’, or break apart, as the temperature increases (recall that two strands of DNA are held together by a specific pattern of hydrogen bonds formed by ‘base-pairing’).   <img alt="image046.png" class="internal default" height="357" src="/@api/deki/files/33803/=image045.png" width="557" /> As melting proceeds, the absorbance value for the sample increases, eventually reaching a high plateau as all of the double-stranded DNA breaks apart, or ‘melts’.  The mid-point of this process, called the ‘melting temperature’, provides a good indication of how tightly the two strands of DNA are able to bind to each other.  In <a href="Core/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_16:_Oxidation_and_reduction_reactions/Section_16.08:_Observing_the_progress_of_hydrogenation_and_dehydrogenation_reactions_by_UV_spectroscopy" title="Organic Chemistry/Organic Chemistry With a Biological Emphasis/Chapter 16: Oxidation and reduction reactions/Section 16.8: Observing the progress of hydrogenation and dehydrogenation reactions by UV spectroscopy">section 16.8</a> we will see how the Beer - Lambert Law and UV spectroscopy provides us with a convenient way to follow the progress of many different enzymatic redox (oxidation-reduction) reactions.  In biochemistry, oxidation of an organic molecule often occurs concurrently with reduction of nicotinamide adenine dinucleotide (NAD+, the compound whose spectrum we saw earlier in this section) to NADH:   <img alt="image048.png" class="internal default" height="190" src="/@api/deki/files/33801/=image047.png" width="703" /> Both NAD+ and NADH absorb at 260 nm.  However NADH, unlike NAD+, has a second absorbance band with λmax = 340 nm and ε = 6290 mol L-1cm-1.  The figure below shows the spectra of both compounds superimposed, with the NADH spectrum offset slightly on the y-axis: <img alt="image050.png" class="internal default" height="397" src="/@api/deki/files/33799/=image049.png" width="610" /> By monitoring the absorbance of a reaction mixture at 340 nm, we can 'watch' NADH being formed as the reaction proceeds, and calculate the rate of the reaction. UV spectroscopy is also very useful in the study of proteins.  Proteins absorb light in the UV range due to the presence of the aromatic amino acids tryptophan, phenylalanine, and tyrosine, all of which are chromophores.  <img alt="image052.png" class="internal default" height="212" src="/@api/deki/files/33797/=image051.png" width="700" /> Biochemists frequently use UV spectroscopy to study conformational changes in proteins - how they change shape in response to different conditions. When a protein undergoes a conformational shift (partial unfolding, for example), the resulting change in the environment around an aromatic amino acid chromophore can cause its UV spectrum to be altered. </div> <div> <h2 style="text-align: justify;">Contributors</h2> template.SpanishSoderberg()template.TransVazquez()template.HideTOC() </div>

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