Concept: Practice 1: IR Structure Determination7m
Hey guys let's take a look at the following practice question so here it says based on IR data given determine the structure of the unknown, unknown compound A has a molecular formula of C4H11N it shows a peak at 2900 centimeters inverse and peaks in the fingerprint region, now when it comes to a question like this we refer to it as a structural determination question where we have to basically figure out the structure of the compound, now anytime we do a structural determination question we must always figure out what are my degrees of unsaturation or what we call the index of hydrogen deficiency so again we're going to say first we find out the IHD which is called the index of hydrogen deficiency or your degrees of unsaturation, this will help us determine if this compound has a double bond, a triple bottom or a ring because when it comes to these structures we're going to say double bond, triple bond and then ring when it comes to their IHD or degrees of unsaturation we're going to say double bond counts as 1, a triple bond counts as 2, A ring counts as 1, now all professors have their different formula that they used to figure IHD, I have my own different formula as well to figure out the IHD for a compound we're going to use the formula (2C+N-H)+2)/2 so remember everyone uses a different type of IHD formula I'm no different so use the formula that you're most accustomed to using, here we're going to say what do these letters represent? We're going to say C represents carbon and represents nitrogen, Right? C with C N with N, now H can be more than one thing, H can stand for hydrogen and also halogen because they both begin with H so we're going to use this formula first to figure out what are my degrees of unsaturation or my index of hydrogen deficiency so here we're going to say how many carbons do we have? We have 4x2=8, we have 1 nitrogen so that's +1 and we have 11 hydrogens so that's -11, +2 divided by 2 so we're going to say what? We're going to say 8+1=9 but -11 gives me this as a -2, -2+2=0, so what is this saying? this is saying that this unknown compound has no double bonds, no triple bonds, no ring so it is basically straight chains of carbons, now what we need to realize here is IHD is zero, we have a nitrogen involved which most likely means itÕs an amine, remember we have different types of amines if the nitrogen is connected to one carbon it's primary if the nitrogen is connected to two carbons it's secondary if it's connected to three carbons we're going to say it is tertiary and its most likely going to be one of these three as our answer we just have to figure out which one it is, now remember these two first two they both have NH bonds and remember NH bonds have their own signal, we're going to say that the signals are usually between 3300 and 3500 centimeters inverse and remember this one will give us a double peak because there's two Hs on the nitrogen and this one will give us a single peak because there's only one H on the nitrogen, now I never said anything about having a signal between 3300 and 3500 centimeters inverse which means that this cannot be a primary or a secondary amine which means it has to be tertiary so what is the signal really telling me? This signal is because we have SP3 carbon hydrogen bonds so the signal really was useless doesn't tell me anything and here if you're in the fingerprint region then you're definitely useless because remember in the fingerprint regions too many signals overlap so we kind of avoid that region altogether especially when it comes to structural determination, so we know that it is a tertiary amine so we're going to have the nitrogen in the middle it's going to be connected to 3 carbons, so here's 1 carbon that's a methyl here's another there's a methyl here's another carbon that's a methyl but remember there's 4 carbon is involved so make one of them an ethyl so this would be our structure of our compound, our tertiary amine and if we wanted to name this structure we'd say the longest carbon chain forms the end of the name so we call this...I'll take myself out of the image we'd say the end of the name is ethane because it's two carbons but we change the E to amine so it's Ethanamine, the smaller alkyl groups become substituent, N substituent so they become N, N Di-methyl Ethanamine and why are we saying N-N? Because they're both connected to nitrogen so that's our structure and that's the name of our structure. As you can see structural determination question, highly detailed a lot goes into figuring out what it is but you must always begin by first figuring out your index of hydrogen deficiency or your degrees of unsaturation, same exact thing remember use the formula that you're most accustomed to using to figure out what the number is to figure out if it has a double bond, triple bond or a ring if it gives you an answer of 0 that means it's all single bonded carbons none of those other three things.
Concept: Practice 2: IR Structure Determination5m
Hey guys, let's take a look at the following practice question. So, here it says, based on ir data given determine the structure of the unknown, unknown compound B has molecular formula C4H11N, it shows a single peak at approximately 3,400 centimeters inverse as well as the peak at 2,900 centimeters inverse and in the fingerprint region, compound B also possesses a branched alkyl group. So, remember, when it comes to structural determination what we always need to do first is to figure out the ihd, the index of hydrogen deficiency, or what's called degrees of unsaturation. Now remember, this formula, there's different formulas that all professors use, so this is my version 2C plus N minus H, in brackets, plus 2 divided by 2. Remember, C represents carbons, N represents nitrogen, H represents hydrogen's and halogens. So, here we have 4 carbon. So, that's 2 times 4, which is 8 plus 1 nitrogen minus 11 hydrogens. Remember here, that this is going to be basically 9 minus 11, which gives us negative 2 inside the brackets, negative 2 plus 2 is 0. So, this compound has no double bonds, no triple bonds, no rings. So, it's just a single bonded group of carbons. Now, here I tell us I have a single peak at approximately 3,400 centimeters inverse. Remember, for a secondary amine, we say nitrogen is connected to two carbons and it has one H connected to it because there's one H connected to that nitrogen that's going to give us a single peak, if we had a primary amine, nitrogen will be connected to only one carbon and because it has two H's on it it would have a double peak. So, if you're, we're dealing with a secondary amine. So, we can draw the
N in the middle, its secondary so we know it's connected to two carbons and its it's a 1H, here I say that there's a branch group on this compound, which would mean this. So, this would represent our branched alkyl group. So, here we'd say the name of this structure, if we wanted to name it, we say, this is 1,2,3, that's the larger alkyl chain, this is the smaller alkyl chain so the whole thing becomes a substituent.
So, here we're going to say that the name of this structure is going to be what> it's going to be, took myself out of the image, it's going to be propane chain. Remember, it's an amine. So, you change the e from a propane to amine. So, it's propanamine, we're going to say the smaller alkyl group is called an N substituent so it'd be N methyl and the nitrogen is on carbon number 2. So, it's actually going to be N methyl 2 propanamine. So again, it's a single piece of the secondary amine, we figured out the ihd was 0. So, there's no double bond, no triple bond, no ring, branched alkyl changes means here that we have a branch group, which is really this, coming off of that chain, which linked in to change the three carbons. So, this would be our structure, again structural determination when we move over to NMR is difficult, even IR structural determination is difficult but you have to remember the steps to take, always find ihd first, since we're dealing at IR here remember, when they give me a signal which functional group are they alluding to, knowing that is the key to figure out what the compound looks like.
Concept: Practice 3: IR Structure Determination4m
Hey guys, let's take a look at the following question. So, here it says, based on ir data given determine the structure of the unknown, unknown compound c has molecular formula C6H10O3, it shows Peaks at 2900, 1850, 1740 centimeters inverse and in the fingerprint region. So, remember anytime we're given an structural determination question we must always first begin by figuring out the ihd also called your degrees of unsaturation. Remember, all of us use different formulas, my particular formula is 2c plus N minus H, in brackets, plus 2 divided by 2, remember, C represents carbon, N represents a nitrogen, H represents hydrogen's and halogens. So, we have six carbons here. So, at 6 times 2, which gives me 12, there;s no nitrogen so we can ignore the N minus 10 hydrogen's, in brackets, plus 2 divided by 2, oxygen is not part of my formula therefore I ignore it as well. So, here we're going to say 12 minus 10 gives me 2 plus 2 divided by 2, that's 4 divided by 2, which gives me an ihd of 2. Now, here we say we have 1740, which means that we have a carbonyl but more specifically it means we have an ester because an ester falls within this region because an ester is 1730 to 1750 centimeters inverse, next, we're going to say this 2900 really isn't telling me much because this represents sp3 carbon-hydrogen bonds. So, here we're going to say that signal there is kind of useless. Now, here we have an 1850 signal, which also really doesn't tell me much at all. So, what we're going to have to do is try to think of the possible shapes that this compound could take. Now, when we get to NMR, we'll be more specific in terms of what this structure could be but because I only give you IR information this opens up the possibility of having multiple answers for this particular question.
So, ihd of 2, we know that one of the ihd comes from the fact that we have a carbonyl of the ester, we could say here, we have 3 oxygens so that carbonyl has both oxygens and remember to be an ester those oxygens on the N have to be connected to carbons. Now, I can't do straight chains for them because here it wouldn't give me an ihd of 2. So, that's second ihd that I'm getting, I can assume maybe it's coming from a ring, this K takes care of for my carbons but how many carbons do we have? we have 6 carbons. So, I need to draw two more. So, we could say, potentially maybe we have two methyl groups on one of these carbons, this molecule has a lot of symmetry. So, this could be a possible answer. Remember, I'm only giving us IR so it opens up the possibility to multiple answers. So, maybe both those methyl groups are now on the same carbon, maybe one is over here and one is over here, molecule still have symmetry but it's different, when we get to NMR will be given more information, more signals, which will narrow, narrow it down to one particular answer, for a question like this or basically in a sense guessing which one is the best structure based on the information that I've provided you because I make it so vague more than one possible answer exists.
Problem: Match the following functional group choices with the supplied infrared spectra data3m
Problem: Match the following functional group choices with the supplied infrared spectra data3m
Problem: Match the following functional group choices with the supplied infrared spectra data.3m
Problem: Match the following functional group choices with the supplied infrared spectra data.4m
A C-D(carbon–deuterium) bond is electronically much like a C-H bond, and it has a similar stiffness, measured by the spring constant, k. The deuterium atom has twice the mass (m) of a hydrogen atom, however.
(a) The infrared absorption frequency is approximately proportional to when one of the bonded atoms is much heavier than the other, and m is the lighter of the two atoms (H or D in this case). Use this relationship to calculate the IR absorption frequency of a typical C-D bond. Use 3000 cm-1 as a typical C-H absorption frequency.
(b) A chemist dissolves a sample in deuterochloroform (CDCl3), then decides to take the IR spectrum and simply evaporates most of the CDCl3.What functional group will appear to be present in this IR spectrum as a result of the CDCl3 impurity?
Convert the following infrared wavelengths to cm-1.
(a) 6.24 typical for an aromatic
(b) 3.38 typical for an aromatic C-H bond
(c) 5.58 typical for a ketone carbonyl
(d) 5.75 typical for an ester carbonyl
(e) 4.52 typical for a nitrile
(f) 3.03 typical for an alcohol O-H
Spectra are given for three compounds. Each compound has one or more of the following functional groups: alcohol, amine, ketone, aldehyde, and carboxylic acid. Determine the functional group(s) in each compound, and assign the major peaks above 1600 cm-1.
For each hydrocarbon spectrum, determine whether the compound is an alkane, an alkene, an alkyne, or an aromatic hydrocarbon, and assign the major peaks above (to the left of) 1600 cm-1. More than one unsaturated group may be present.
a. Which compound will have the stretching vibration for its carbonyl group at the highest frequency: acetyl chloride, methyl acetate, or acetamide?
b. Which one will have the stretching vibration for its carbonyl group at the lowest frequency?
Match the compound to the appropriate carbonyl IR absorption band:
acyl chloride 1800 cm-1
ester 1640 cm-1
amide 1730 cm-1
There are three carbon–oxygen bonds in methyl acetate.
a. What are their relative lengths?
b. What are the relative infrared (IR) stretching frequencies of these bonds?
How could IR spectroscopy be used to distinguish between the following compounds?
a. a ketone and an aldehyde d. cis-2-hexene and trans-2-hexene
b. a cyclic ketone and an open-chain ketone e. cyclohexene and cyclohexane
c. benzene and cyclohexene f. a primary amine and a tertiary amine
a. An oxygen-containing compound shows an absorption band at ~1700 cm−1 and no absorption bands at ~3300 cm−1, ~2700 cm−1, or ~1100 cm−1. What class of compound is it?
b. A nitrogen-containing compound shows no absorption band at ~3400 cm−1 and no absorption bands between 1700 cm−1 and 1600 cmWhat class of compound is it?
What changes occur in molecules after being irradiated with IR absorption?
Match each compound to the correct IR spectrum, and give a brief reason for your selection.
For the following reaction, which of the following is consistent with the IR spectrum of the product?
A) Absorption at 2250 cm –1 should disappear and absorption around 3400 cm –1 should appear.
B) Absorption at 2250 cm –1 should disappear.
C) None of these.
D) Absorption at 3200-3400 cm –1 and 1720cm –1 should appear
E) Absorption at 2250 cm –1 should disappear, new absorption at 2600-2800 cm –1 and 1720 cm –1 should appear.
Write structures for all compounds with molecular formula C 4H6O that would not be expected to exhibit infrared absorption in the 3200–3550-cm -1 and 1620–1780-cm -1 regions.
There are four amides with the formula C3H7NO. (a) Write their structures. ( b) One of these amides has a melting and a boiling point that are substantially lower than those of the other three. Which amide is this? Explain your answer. (c) Explain how these amides could be differentiated on the basis of their IR spectra.
Write structural formulas for four compounds with the formula C 3H6O and classify each according to its functional group. Predict IR absorption frequencies for the functional groups you have drawn.
Circle the molecule within each given set that most likely produces the given Infrared (IR) spectrum.
PRACTICE: State how the following pairs of molecules can be differentiated using Infrared (IR) spectroscopy.