Spectroscopy + structure and bonding Flashcards

Question

Answer 1

An enzyme is a molecule which they able to catalyze a chemical reaction. So what happens in terms of an enzymatic pathway is we have our substrate molecule, it enters into the active site of the enzyme. The enzyme will have an induced fit around the substrate. And this places the correct chemical groups from the amino acids within the enzyme in the right positions to catalyze the chemical reaction. So the induced fit, we'll move these groups to interact with a substrate, and then a reaction will take place. And finally, the products of that reaction will be released from the enzyme. And the enzyme will be free to carry on with this process again what we can do is use what's called an allosteric site. This is any binding site on the enzyme that is not the enzyme active site. By binding a molecule to an allosteric site, we can find the molecule changes the shape of the enzyme and prevents it from performing its function. So either by moving reactive amino acids into the wrong place, or by changing the shape of the enzyme in such a way that the substrate can no longer enter into the active site. The other steps wont fully stop the reaction, it can only slow down the reaction at best so only affect is we will make less of the product.

Answer 2

So a binding site that occurs the enzyme anywhere other than the active site. So we can have what's called an allosteric inhibitor. This is a molecule that combines to the enzyme into an Allosteric site and change the shape of that enzyme. In doing so, it will change the shape of the active site and prevent our substrate from binding into it. So we have our inhibitor, it binds to the enzyme. The enzyme has induced fit from the binding of the inhibitor, and this causes the active site to change its shape. The substrate itself can no longer enter into that active site and stop it from binding. The Allosteric inhibitors can be used as a feedback control mechanism. So if we have a binding site on this first enzyme, that the product of a reaction can enter into. We can use this as a way of switching off by enzymes when we have too much of that product present. So what we do is we have our enzyme, it's churning through substrate producing products. And one of those products will then be able to enter into an enzyme and switch it off. So when we get to a certain concentration, the inhibitor binds to the enzyme and stop that enzyme from producing more molecules until the concentration of the product has dropped. gain, when there are low concentrations of the products, the molecules unbind from the enzyme and allowed to switch back on and continue its reaction. We then repeat this over

Answer 3

molecule is able to bind into the enzyme with a greater binding strength than our substrate. What this will allow is for a molecule to enter into the enzyme and essentially block that enzyme from substrate molecules coming in. So we're trying to design a molecule that can have more binding interactions than? the substrate to temporarily switch off the enzyme. Now this is called a competitive binding. andthis is because the substrate and inhibitor are in competition with each other. If we have a high amount of substrates and a low amount of inhibitor, the inhibitor will be displaced from the enzyme and function will continue. If we have high amounts of inhibitor or a low amount of substrate, then the substrate will be displaced when the enzyme and the inhibitor will whip, So it will prevent production of A product is being made for a limited amount of time until we have enough substrate and then displace that inhibitor again. What we need to have in a reversible inhibitor is for a high concentration of drug compared to the concentration of the substrate. And this will enable the drug molecule for entrance binding site. And it needs to not react with the enzyme. So this will then block the substrate from entering the enzyme and temporarily stop the reaction taking place.

Answer 4

When we're talking about reversible inhibitors, one of the key things we need to consider is how a molecule is going to recognize the active site of the enzyme. One of the ways in which we usually start by looking at to design a new inhibitor is we start by looking at the natural substrate. This is because the natural substrate will contain chemical groups in the right place to facilitate binding. So we can then do is take the natural substrate, add extra groups to it to increase the binding strength and use that as the basis for the design of our inhibitor if we know what the enzyme looks like and we know what what functional groups are present within an active site. We can design our molecule to explore different binding groups to our substrate if those groups are available. We can also produce a molecule that mimics the product of enzymatic reaction rather than the starting substrate. What will happen if we do this is the molecule enters the binding site. The enzyme will have its induced fit and around that molecule then will be unable to complete the reaction and eject it from itself, again, reducing the rate of production, rate of turnover of the natural substrates. Reversible inhibitors. Looking at binding to the active site. Then we're looking at a competitive process between the natural substrate and the drug molecule. If we're looking at a reversible inhibitor that binds to an allosteric site. We're not talking about competitive process because then there is no competition between the substrate and the inhibitor at the allosteric site. They're not competing for the same binding sites. So reversible inhibitors, we're talking about the active site. We're talking a competitive process. If we're talking about allosteric site when this process is not competitive,

Answer 5

Irreversible inhibitors are molecules which can bind to the enzyme permanently. They do this by forming a permanent covalent bond between the drug molecule and a group or groups within the enzyme itself. So often this involves a bond being formed between the drug molecule, between serine or cysteine amino acids within the active site of the enzyme. So what will happen is a drug molecule will enter into the enzyme active site and we'll get a reaction take place. So e.g. in this instance, the OH group here is acting as a nucleophilic group. And when it does, it forms a permanent covalent bond with the drug molecule itself. Now, if this is in the active site, then this will permanently block the active site of this enzyme. But the downside of this, that means that we don't have the option of increasing the concentration of substrate to get rid of this molecule. This molecule is here basically forever. So that particular enzyme molecule is permanently switched off. So it's not a competative process in this instance because there is no way for the substrate to displace the drug molecule from the active site once it has permanently bound.

Answer 6

reversible inhibition is preferred over irreversible inhibition. There are several reasons for this. The first is the inhibitors often contained very reactive functional groups. So these functional groups will not just react with their intended target. They will also react with the functional groups of other protein molecules they happen to encounter whilst they're traveling to their target site. This can increase the severity of side effects or give additional side effects over irreversible inhibitors. Generally speaking. Irreversible inhibitors are not competitive because they permanently bind to the active site, which means thatwe cannot remove. So if this process goes wrong, then this could lead to the buildup of multiple substrates within an enzymatic pathway, e.g. which could then also need to separate toxic side effects. So not only do we have the side effects from the compound itself, but also side-effects from other pathways that havebeen inhibited. Therefore, we've got buildup or other substrates within those pathways. Generally reversible inhibitors are preferred because we can restore normal function by increasing the levels of the naturally occurring substrate. Irreversible inhibitors. Only way for us to return to normal function is towait for more of the enzyme to be produced.

Answer 7

So allosteric binding sites give us a second target for drug action. So we can target the active site or we can target an allosteric sites. Now, allosteric site to allow us to control the enzyme process by making the enzyme changes shape to prevent a substrate molecule from entering into it. So if we bind to a allosteric site one enzyme, we typically get a conformational change, and this normally stops the substrate from entering into the active site. Or it can change the positions of functional groups within the active site, stopping a reaction from taking place. Now, competitive inhibition could only take place at the active site of an enzyme because you have competition process between the substrate and the inhibitor. But binding allosteric sites is inherently not competitive. You're not going to have multiple molecules that are competing for the same binding site here. That as one example, which is this compound here, is able to bind to a binding site in the first enzyme is involved in purine biosynthesis, synthesis of guanine and adenine in DNA and RNA. Now this compound combined to allosteric site and block production of purines and then consequently block production of DNA because it means the base building block that built DNA are not available.

Answer 8

Noncompetitive inhibitors are a bit different. These bind to an allosteric sites on enzymes, but they don't stop the substrate from binding. So the substrate is still able to bind with an inhibitor in the allosteric site. Non-competitive inhibitor will do is it will cause the enzyme to have an induced fit. That changes the position of key reactive amino acids within the active site. They wont move the amino acids to allow binding, but they will move the amino acids that facilitate the reaction. So they move the reactive amino acids, So what theywill do is they will bind to an allosteric site They don't stop the substrate from binding, but they do stop the reaction from taking place because they'vemoved the key functional groups that are responsible for catalyzing thereaction so that they are in the wrong positions to react with the substrate.

Answer 9

*Enzymes and substrates have some degree of flexibility in their interactions. Enzymes can work on a range of different substrates *Drugs capable of forming covalent bonds with enzyme likely able to act on different targets that have same functional groups *This can lead to a range of side effects *The whole point of a suicide molecules is that it is totally inert and harmless in the body UNTIL it binds to the target enzyme – rendering it inactive. Delivery of a drug which will only have an effect once at the correct enzyme site. This prevents unnecessary binding at wrong targets *Suicide substrates are harmless molecules until converted within the enzyme active site to a highly reactive species – this means it will prevent unnecessary binding to targets where it is not needed

Answer 10

Receptors are protein molecules embedded within the cell membrane, with part of the structure on the outside of the cell. They carry messages from outside of the cell to the inside of the cell *Receptors are crucial for communication pathways within the body. *Neurons carry message from the central nervous system,but stop just beside the target cell. Then neurotransmitters are released which bind to the target cell receptors and cause some sort of message to be transported. Therefore this process relies on a chemical messenger (neurotransmitter)

Answer 11

*Receptor surface is a complicated shape which contains selective binding sites for chemical messengers. *Similar in concept to enzyme active sites but the binding substrate does not undergo a chemical reaction. *So what does happen? *Binding of messenger to receptor results in an induced fit. Change in the shape of the binding site leads to change in shape of protein, leading to transmission of the message

Answer 12

*3 main types of membrane-bound receptors: *Ion channel receptors *G-protein-coupled receptors Kinase-linked receptors

Answer 13

How does a lock gate work: -Outside of receptor – we have a substrate binding site -When there is nothing attached to this site, the lock gate is closed -Prevents movement through the tunnel -When a messenger molecule binds, it opens the lock gate Transport can then happen through the cell membrane

Answer 14

A G coupled binding receptor has a binding site for the chemical messenger on the outside of the receptor -On the inside it has a binding site for a G protein -This G protein contains multiple protein sub units -When a messenger binds to a receptor, the induced fit actually opens the binding site for the G protein inside of the cell -G protein binds to the inside -It is then processed to release multiple G protein subunits -These multiple G protein subunits then go on and bind to membrane bound enzymes elsewhere in the cellular membrane This opens up the active sites of the enzyme, allowing it to process substrates to form a chemical reaction

Answer 15

Messenger molecule binds to the receptor -The receptor has its induced fit -This opens up the active site on the inside of the cell The receptor is then able to catalyse reactions on the inside of the cell directly – doesn’t require a separate enzyme to catalyse reaction

Answer 16

Agonists mimic the natural substrate and need to bind to the receptor in such a way to activate it. *Need to consider 3 main points: *Drug must have correct binding groups as the original substrate *Binding groups must be in the correct position as the original substrate *Drug must be correct size for binding site *Binding Groups *Van der Waals interaction *Hydrogen bonding *Electrostatic interaction - We can choose the binding sites where we want the substrate and the receptor to bind – we can make use of van der waals, hydrogen bonding and electrostatic interaction

Answer 17

-We should consider if all of the binding groups are required or not -For example, we can see if we can change the 6 membered aromatic ring into one that is not aromatic -We can also get rid of the hydrogen bonding group aswell -If we remove all of these binding interactions, we will not have an induced fit anymore and we wont have our strong binding interactions -The substate may enter the active site but it wont interact with the receptor anymore -If we get rid of only some of the binding sites, then we may only have partial success – may only partially activate the receptor basically, when we remove the binding sites of the receptor, we are going to have less success of activating that receptor

Answer 18

*The presence of all required binding groups orientated in the correct position is critical for agonist action. *If the agonist does not fit into the binding site then activation of the receptor will not occur. *Molecules must be of the correct size to fit the space available. Molecules must be correct shape to allow binding groups to align with binding regions within the target.

Answer 19

*Antagonists bind to the receptor in such a way that does not activate it. *Antagonist binding will inhibit binding of the natural messenger *Using a molecule that perfectly fits the active site is one approach. *Use natural binding regions to mimic binding of natural substrate. *Must have enhanced binding over natural substrate – must want to bind more than natural substrate otherwise the substrate will just come and bind instead Perfect fit to active site results in minimal induced fit and no biological activity *Another approach is to utilise alternative binding regions within the active site. This can inhibit binding of the natural substrate and result in an alternative induced fit, deactivating the receptor Natural Substrate – Uses specific binding regions resulting in induced fit and activation of receptor Antagonist – Uses natural plus extra binding regions causing induced fit which does not activate the receptor

Answer 20

. In enzymes – allosteric inhibitors bind in places which aren’t the active site The same is true of receptors but we use different terminology Umbrella effect – doesn’t change the active site but does block it

Answer 21

Partial agonists only partially activate the receptor -This is because when they bind, they bind in a way that is not ideal for the receptor -So, we may get a partially complete induced fit rather than a full induced fit -This results in a decrease in the effects of the receptor -An example of this is the opening of ion channels -If we take an ion channel receptor with a receptor binding site and introduce a partial agonist to it, that partial agonist may only open the receptor (and therefore ion channel) partially This may still allow the movement of ions into the cell but this will happen at a much reduced rate – it will take a lot longer for ions to go in and out

Answer 22

A – natural equilibrium, some activation but mainly deactivated B – agonist pushes equilibrium mainly to the activated form C – antagonist, pushes equilibrium mainly to the unactivated form but there is still some small activation present D – inverse agonist – turns off all activity associated with equilibrium E – partial agonist, gives activity higher than inverse agonist but less than full agonist

Answer 23

.*Enzymes *Receptors *Other *Transport proteins *Structural proteins Protein-protein interactions

Answer 24

-Transport proteins are molecules that are able to carry polar molecules across the cellular membrane via active transport -What happens is that we have our polar molecule and this can enter into thetransport protein and the transport protein can actively move and that molecule can be brought into the cell we can develop molecules that will bind to that transport protein instead and stop the movement of ions

Answer 25

Structural proteins – proteins which make up cellular or biological structures One example is the viral capsid (also called protein coat) Protein coat opens and releases the genetic material into the host cell Pocket factor helps to stabilise the pocket, this is displaced when the cell binds

Answer 26

.*Several important biological processes involve the interaction between two proteins. *Targeting these protein-protein interactions could inhibit the related biological process. *The regions of interaction in proteins can be relatively large, containing 40-50 amino acid residues. *To target these regions would require large drug molecules which would have difficulty in crossing the cell membrane. *Research has showed that often a few key amino acids are important for a large percentage of the binding interaction. Drugs could be designed to target these key amino acids.

Answer 27

Qualitative analytical chemistry “What components are present in your sample?” *NMR, Mass spectrometry, GC-MS, Microanalysis, IR-Spectrophotometry, Atomic Spectroscopy *Quantitative analytical chemistry “How much of each component is in your sample?” *Titrations, Gravimetric analysis, electroanalytical methods, optical (spectroscopic) methods Qualitative What compounds are present within your sample. What we have but not how much of it. Quantitative - How much of a molecule we have but can't tell us the identity. Bare minimum of 2 analytic techniques are required normally for credibility

Answer 28

*Accuracy is defined as closeness to a standard or true value. *Precision is defined as closeness of a series of measurements. *The manufacture of medicines requires precision to ensure each tablet contains the same quantity of active ingredient and accuracy to ensure the correct quantity. S1 Acc and precise S2 Accurate ( not precise as large gaps between repeated measurements) S3 Precise S4 – nor accurate or precise – Measurements have large spread and away from the true value We need accuracy to ensure to make sure the amount in each tablet is correct and close to acceptable and true value.

Answer 29

*Level of detail in monographs is required to ensure repeatability of measurement *Same person *Same instrument *Results in increased precision *Analytical method is useless without reproducibility *Different person *Different instrument *Same result Monographs enable to conduct the same experiment to try and replicate the same result. Standard procedures ensure the experiment is carried out the same way every single time. Monographs are easily repeatable in a different lab. We need to ensure medicines are made the same way as its on a global level. To ensure all patients are safe . If something is reproducible it can be repeated by a different person with slightly different equipment but still get the same result

Answer 30

After the drug has come to market… *During manufacture – Quality control: *Is it of the same purity *Are the impurities the same *Is the solid state structure the same (or is a polymorph produced) After production – Quality control: Does the compound degrade over time Does the solid state structure of the drug change over time

Answer 31

Spectroscopy describe the interaction between matter Right to left means energy of the rays is increased. The more left we go, the worse it is for human health. Spectroscopy is generally safe for humans to interaction with.

Answer 32

Our light passes through the monochromator, we choose a wavelength, the light goes through the sample and enters it. The light passes through the sample, and a detector reads how intense that light is, we can detect the intensity of light at each wavelength and how much light has been absorbed at each wavelength. Incident light – Light going in to the sample before being absorbed Transmitted light – light coming out the other side of the sample – Transmitted light Reference spectrum = reference sample – Subtracted by proper sample as they are mixed.

Answer 33

*Irradiation of organic compounds with UV/vis light can lead to promotion of an electron from either bonding molecular orbital or lone pair to another orbital (excited state). This causes the incident photons at that specific frequency to be absorbed. *A functional group that absorbs UV/vis light is called a ‘chromophore’ Definition: A chromophore is the part of a molecule which absorbs UV or visible light When light interacts with matter, it can lead to the matter absorbing that light, Causing an electron in that matter to gain energy. When it gains energy that light is absorbed. This means the electron is promoted to an excited state. A functional group which can absorb UV or visible light is called a chromophore. The absorption is normally specific to the energy coming in, we have to have functional groups where the energy change of that electron corresponds to the same energy provided by the wave length /wavelight we are using. Majority of molecule that have single soluble alternating bonds can absorb light. Unsaturated molecules are for example the molecules listed above. Metal ions are also very good at absorbing UV or infrared light If there is a double bond, strong chance it will absorb UV light Different chromophores absorb at different wavelengths (energies), providing information about the molecule. The most intense absorption is called lmax. Increasing conjugation lowers the HOMO-LUMO band gap shifting lmax to higher wavelengths (lower energy) NB:Chromophores absorb at different wavelengths because of the Gap in energy between the highest occupied molecular orbital and the lowest unoccupied molecular orbital is different. The more alternating double and single bonds we have, the more absorbance is shifted towards a longer wavelength. Longer wavelength means lower energy needed to absorb, so the less energy it takes to absorb, less energy required to promote an electron from the highest occupied molecular orbital to the lowest unoccupied molecular orbital. More conjugation = more double bonds and single bonds = Absorption shifted to the right of the spectrum More we go to the right, the more we go to the visible region of the spectrum, this is because it takes less energy to promote an electron up into excited state. The most intense absorption in the spectrum is called the landomax. So for example the blue lines we take the highest listed.

Answer 34

We can use this formula to determine how much compound we have in a solution. Beer-lambert law. A – Abosrbance – Number we would read of a spectrometer A(1% 1cm) (specific absorbance) = The absorbance that would be given by a 1 % W% solution so 1G in 100ml solution. C – concentration L – amount of distance it takes for the light to pass through the sample. Unless we are told otherwise this value is always equal to 1.

Answer 35

Auxochromes are chemical groups that are bound to the chromophore, so there are replicated to the group that can absorb uv or visible light, they can change how that group absorbs light, usually by making it making it absorb more light in certain circumstances. But they also don’t absorb light themselves. If we take an hexane group and attach hetero atoms to it we can change the way it absorbs light. Auxochrome doesn’t absorb itself but can change the electronic configuration of a chromophore . Hetero atoms usually lone pairs of electrons, the lone pairs of electron are able to enter bonds between hetero atoms and the carbon and temporarily convert its electro configurations. This means it can absorb more light or light in a different way,

Answer 36

Changing absorbance of a molecule with pH can enable UV spectroscopy to be used to ascertain the pKa values of the ionisable group that is responsible for the spectroscopic shift If we have a molecule that has the ability to pick up or lose protons and ions, we can use the PK to calculate the value of that group. Example = Phenol – Phenol absorbs 270 NM in uv range – specific absorbtion is 172 – If we take phenol and put it in alkaline conditioner, we can remove the proton from oxygen and convert to something called a Phenolate. This is a charged specie, which means the oxygen had a negative charge – This therefor changes how the benzine ring interacts or aborbs UV light. Above diagram therefore shows Phenolate is able to absorb more UV light then in the phenol state. To convert back we just need to add acid. Which will convert it back to an uncharged state. What can PKA of a drug determine for example where in the gastro tract the drug can be dissolved. A drug that needs to be dissolved in the stomach can have a different PKA to lets say a drug that needs to be dissolved in the intenstines. PKA is the point where 50 percent of the drug is ionized and 50 percent of the drug is unionized. Depending on how the drug is ioninsed, whether we adjust the PH above or below its value can effect the ionization status of the drug. Being able to calculate pka of the drug and under what PH the drub is soluble is very important in pharmacy.. By using UV spectroscopy and another equation we can determine the PKA of the functional group. *Wavelength selected where there is the greatest difference between ionised and un-ionised forms *A rough estimate of the pKa for the drug (± 1) is needed to ensure meaningful selection of the pH for the solution used to obtain ‘A’ *Note: In the rare cases that the ionisation process (i.e. increase in pH) leads to a decrease in absorbance or a shift to lower wavelengths (a hypsochromic shift) then the log term in the above equation is subtracted from the pH value: (equation in SC) We need to have a rough idea of what the PH of that functional group is, so we can pick a buffer with the right PH match. Not closely but the correct ballpark for its pka. Because we want the molecule to be partially ionized and partially unionized in that buffer. So we don’t want it to be the same prediction ( a strong acid or a strong base).

Answer 37

Non acidic – ionized Acidic – unionized AI value – 1.224 – ionized form AU value – 0.02 – absorbance of unionized form PH 8.5 Absorption values (buffer) 0.349 Final answer is 8.924

Answer 38

Spectroscopy concerns the interaction of electromagnetic radiation with atoms and molecules to generate structural information. Spectroscopy is the study of matter using light. The study how matter can interact with electromagnetic radiation So infrared spectroscopy is very good at telling us what functional groups we have in a molecule. with infrared spectroscopy we are trying to excite molecules causing them to vibrate, we do this by hitting them with infrared light that has the correct energy to induce this vibration. Molecules can vibrate in different ways, and this is determined by the correct wavelength. We require exactly the right level of energy to excite the molecules. Compared to UV the peaks are ballpark and hard to determine the exact wavelength. If we don’t have the correct energy, groups will not absorb, and the molecules will not vibrate.

Answer 39

Example – Spectrum for hexane. We can already identify it’s a different graph to what a UV spectroscopy graph looks like. Peaks are shown upside down ( historic reasons). Here we are looking at transmittance and absorbance. 100 percent transmittance means a high level of light has passed trough the sample and therefore not been absorbed. I Lower the peak comes the more intense absorption. Peaks are important to help us identify functional group X axis is also flipped, instead of starting from low numbers upwards, we start from high numbers on the left and low numbers on the right. Right – long wavelength. left - short wavelength On the left we refer to this as high frequency or fast vibration. As we move to the left this is where functional groups like alcohol absorb. As we move to the right, molecules tend to have roughly similar masses. We can therefore most likely see carbon- carbon bonds absorb because there is almost no difference in the masses of the atoms. As we move to the left we have shorter wavelength which means higher. wave number. Right is higher wavelength and shorter wave number.

Answer 40

Region 1 - Alcohols and amines absorb between 3000 and 3600 Region 2 – triple bonds (carbon- carbon or nitrogen carbon) absorb between 2000/2400 Region 3 – Anything containing a carbon/oxygen double bond absorbs between 1550 and 1750 wavenumbers. 3 key regions most often used within the examination of drug molecules; quiet a lot of based drug compounds will have functional groups within these 3 regions. We ignore anything less then 1500, this is fingerprint region. 3 regions directly correspond to function groups that are importance in the production of medicine. We can use these regions to check if a chemical reaction has happened properly or determine whether a ketone has converted to an alcohol properly.

Answer 41

1) 2000/2400 2 Spectrum A – Alcohols absorb between 3000/3600 – matches with compound 2 3. Ketones 1550/1750 wavenumbers matches with compound 3

Answer 42

If we have a molecule within it a 2 atoms that differ in mass an electronegativity, we will have something called a dipole setup. A dipole within a molecule is where electrons within a bond ( for example a carbon/oxygen bond ) they do not lay in the center of the bond because the oxygen is highly electronegative. This means the oxygen is very good at pulling electrons towards it within a bond that shares another atom. So the oxygen carbon bond, the electrons will sit very slightly towards the oxygen and not perfectly in the middle. This gives the bond a slight negative charge towards oxygen and slight positive charge towards the carbon. So if a bond has a dipole associated with it, we can look at it with IR spectroscopy. This movement of the atoms will cause a movement of the dipole. . If we hit this particular molecule with infrared spectroscopy, this will cause the molecule to vibrate and as it vibrates the dipole will move. This cases the absorbance of the infrared radiation, so in our infrared spectrum if we have a group that contains a dipole, its able to absorb the infrared light and vibrate, and its this that makes it show up within the spectrum. It just so happens that the energy required to do this ties up with the energy provided by the infrared light, so by light with the correct wavelength being the infrared region. Different functional groups require different energy of light meaning they require different wavelengths of light. This is why different functional groups absorb in different places. We can use these differences to work out what functional groups we have and use it to monitor changes in functional groups.

Answer 43

So what happens when we are looking at these molecules is we hit it with IR light, and we get a change in dipole. The change in dipole position is reflected in slide 16 (alternate in 15/16) Only groups that have a dipole and experience a change in dipole can be seen using IR – groups that don’t have a change in dipole ( as seen with the bottom figure) do not vibrate and are called infrared inactive, which means they are not seen in the infrared spectrum. An example of this , is a carbon-carbon single bond in the middle of a long chain. If the chain is the same on both sides the dipole will not move within the molecule. So the functional groups we are interested in seeing which are common within organic chemistry ( such as alcohols ketones carbocylic acid, nitro and amine groups ) will show up within IR because they have a dipole that is able to move and vibrate. In larger molecules such as polymers, that can contain long carbon carbon chains are not typically seen in an IR.

Answer 44

Key terms – Intensity – How far the peak goes down. The further it drops the more infrared light the functional group has absorbed. The position of which the group has absorbed is the wave number it absorbs at. Eg absorbing 1710 wavenumbers corresponds to the carbon group instead of for e.g 3300 typical absorbance of an alcohol group We will look at why the intensity is different and they absorb in the groups that they do. In the example above, the molecule above we have 1 carbon oxygen different bond this absorbs at 1710 wavelength with a high intensity (50 Percent approx.) but we have 8 carbon hydrogen bond. The absorbance is really small. We will look to understand why this is. And understand where they absorb on the spectrum.

Answer 45

Bond strength we look at the different between single doiuble and triple bonds As we increase bond strength to stiffness we go from light to left Right is the fingerprint region

Answer 46

The intensity we see in the infrared spectrum is directly related to the amount of movement or change in dipole moment we get when the bond vibrates – so chemical groups that have a larger dipole will give us a greater intensity compared to groups which have a smaller dipole. Highly polar groups (carbon bound to an electronegative atom such as oxygen or hydrogen ) typically give the most absorbance. In the example above (Ethanol) we see the O-H bond has the biggest absorbance in the infrared spectrum. Whereas the absorbance for 5 carbon hydrogen bonds give approx. the same intensity, ( we have 5 bonds contributing to the 2 peaks within that infrared spectrum) However there are 5. his is because when we introduce an electronegative atom we make a really strong dipole in for example the O-H bond. So when it vibrates it experiences a large change in dipole moment,, which means it able to absorb infrared light and gives us much larger absorbance on the infrared spectrum. In contrast the dipole between a carbon hydrogen bond is restively very small, so we see much less of an absorbance for each hydrogen carbon bond in the infrared spectrum than we would for something that’s more polar. We can not only use the position the functional groups absorb at to try and identify what they are, but we can also look at the relative intensity's peaks, and the shapes of the peaks as well to identify a functional group. For example, Alcohol groups normally give very broad absorbance whereas groups such as carbon oxygen double bonds give very sharp absorbance,

Answer 47

Example -- These 2 molecules contain triple bonds. Terminal alkyne –this is when we have carbon carbon triple bond that is not in the middle of a molecule, it is the last carbon in the chain that has a bond with the carbon next to it. Nitrile – carbon nitrogen triple bond. They both give very different absorption reading in the infrared spectrum. Triple bonds 2000/2400 wavenumbers But we have an electronegative group ( nitrogen)(right) which experiences a much larger change in dipole when the carbon triple bond. IR is also very useful for chemical synthesis, it tells us if there has been a change within our functional group or not, also if we want to convert 1 functional group to another it can tell us if our reaction has been successful. The peak on the left is different, this peak is missing on the nitrogen. This is because terminal alkyne has a single carbon hydrogen bond, which absorbs in the for example alcohol region, the molecule on the right does not have such single bond characteristic so there is no peak detected, or it has a very small peak

Answer 48

These are a series of alkines. They all contain carbon carbon triple bonds. But the bottom one it is a terminal alkine, Top molecule – carbon-Carbon triple bond , entirely symmetric on either side. The middle molecule is a carbon carbon triple bonds i, but to both sides of this the molecule is other wise identical, what this means is when we hit this molecule with infrared light we don’t see a change in dipole moment, we have no electronegative groups on either side of the molecule, this means as this molecule vibrates the dipole stays purely in the middle of the molecule. In the infrared spectrum we don’t see an absorbance as there is no change in dipole so it cant absorb infrared light. No peak at the top molecule either as the alkyne is symmetrical, no difference in polarity and no movement in dipole as the molecule vibrates. We can only see functional groups that experience a change in dipole moments. (middle) – However we see electronegativity either side of this alkyne we can start to see a small peak in the infrared spectrum, so the triple bond was moved so its no longer symmetrical which creates a difference on either side, so when the alkyne vibrates we get a small change in dipole moments, which means we get a tiny peak in the infrared sectum. But because there's no electronegative actens in the molecule, and the alkyne group is still within molecule rather then the end as a result he peak is very small because the change in dipole moment is small. When we move alkyne to the end of the molecule, we get a large increase in the absorbance. Because one side we have CH on the other side we have CH2. So when this group vibrates, its able to vibrate more strongly which gives a larger change in dipople position, this gives us the stronger absorbance in the infrared spectrum. At the start of the infrared spectrum, we can also see a CH and no peak in the other 2. so we get 2 different peaks. If we are dealing with an alkyne or even an alkeane that is entirely symmetric, we cant see absorption in that region on the infrared spectrum that is because as the molecule vibrates it will not experience a change in dipole moment. If we move it away from the middle we see some but not enough compared to the bottom. Essentially it all comes down to how much the dipole can change its position when the molecule is hit with infrared light.

Answer 49

Calibration ensures reproducibility across different instruments in different environments.

Spectroscopy + structure and bonding Flashcards

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