How would the ¹H NMR spectra for the four compounds...

Question

How would the ¹H NMR spectra for the four compounds with molecular formula C₃H₆Br₂ differ?

Accepted Answer

Hello everyone. Let's work through this problem together. It says differentiate two isomers a primary alcohol and an ether with the molecular formula C three H 80 using proton N M R. And we have four answer choices that differentiate between the number of signals each isomer can produce along with the multiplicity of those signals for each isomer. And answer choice D says that the two isomers cannot be differentiated using proton N M R. So since we are given the molecular formula first, we can calculate the index of hydrogen deficiency because we want to, we need to draw the isomers right in order to determine the number of signals and the multiplicity of those signals. But finding the index of hydrogen deficiency can help us to determine the structure of the form of the compound with the when given the molecular formula. So the formula for index of hydrogen deficiency is two multiplied by the number of carbons plus two plus the number of nitrogens minus the number of hydrogens minus the number of halogens all divided by two plugging in the numbers given to us, we have two multiplied by three carbons plus two. There are no nitrogen or halogen atoms. So we subtract eight, our number of hydrogens all divided by two. And in the numerator, we get zero. So our index of hydrogen deficiency is zero. So that means we have no characteristics of uns saturation. So no double bonds, triple bonds or rings. So this should be pretty simple to determine the structure, especially since the problem tells us that we have an a primary alcohol and an ether. So let's draw the primary alcohol. So we know since it's primary, that oxygen is only going to be attached to one carbon in or rather the carbon that has the alcohol group is only attached to one other carbon. So we know it has to be a terminal alcohol group. So let's draw our three carbon chain with our terminal alcohol group. And let's give the carbons, their octet with hydrogen atoms. So three hydrogen atoms, two hydrogen atoms, two hydrogen atoms. So that adds up to eight hydrogens. So this satisfies our molecular formula and this primary alcohol is propan one all OK, let's determine what our ether looks like. So we know in an ether, we have the oxygen atom and then on either side, we have carbon chains, right hydro hydrocarbon groups. So our compound is not symmetrical. So let's just add one carbon to the left side and the other two carbons to the right side and add the hydrogen atoms to satisfy octet. So the carbon on the left gets three hydrogens to form a methyl group. And the carbon chain on the right side, we use the rest of our hydrogens and that is an ethyl group. So this ether is an ethyl methyl either. OK. So we have determined the structures of our isomers. Now, we want to determine, based on the answer choices given to us, we want to determine the number of signals and the multiplicity of those signals that would be found in the proton and M R of these structures. So first, how do we determine the number of signals? We determine the number of unique non equivalent proton groups? OK. So starting with the alcohol on carbon three, that carbon has three hydrogens and they are all equivalent to one another because they are on the same carbon. Then moving to carbon two, we have two methylene protons for signal B and moving to carbon one, we have two more methylene protons, carbon C or proton C and the alcohol group has a hydrogen. So that will be proton D. So for the primary alcohol, we have four signals. OK. Going to our ether doing the same thing for the methyl group, we have those three protons that are chemically equivalent. So that will be signal A. Then on the other side of the ether, we have two methylene protons for signal B and three methyl protons for signal C, right. And those methyl protons are not equivalent to the other methyl protons because of that other carbon in between the oxygen and that methyl group, right. So that makes the methyl groups not chemically equivalent. So we see that this compound has three signals. So let's go back to our answer choices and eliminate some answer choices. So we can differentiate these two isomers because they have different numbers of signals. So we can eliminate answer choice D which says we cannot differentiate between them. And we can eliminate answer choice A which tells us that one of them produces two signals and one produces three signals. And answer choice is B and C say three and four signals. So those are both correct so far. So let's now look at multiplicity and see if we can find our correct answer. So remember, multiplicity is the splitting of a signal due to neighboring protons. And if all of those neighboring protons are equivalent to one another, we can use the function N plus one to determine the number of peaks that, that splitt signal will have. If those neighboring protons are not all equival equivalent to one another, then we can use the expression N plus one multiplied by M plus one to determine the number of peaks. OK. And we will see examples of both the D, both of these in this problem. So starting with our primary alcohol for signal A, we have two neighboring protons, those methylene protons. So two plus one equals three. So signal A will be a triplet oops. OK. Here's that example of the non equivalent neighbors. So signal B, there are three protons on the left, those methyl protons and two methylene protons on the right. So those methyl protons are not equivalent to the methylene protons, right? And we can see that because we labeled them to have different signals, right. If they were all equivalent, they would all be labeled either proton A or proton C. So we take three plus one for those methyl protons multiplied by two plus one for the methylene protons. So that is four multiplied by three, which equals 12 peaks. So we can call this signal a multiple, all right. Signal C has the two protons on the left from signal B two plus one equals three. So that is another triplet and we do not count the neighboring alcohol proton. OK. And lastly, we have that alcohol proton, which is going to be a single it, all right. So we have a triplet, multiple triplet and singlet. Let's look at our ether. So since we have that oxygen atom in the center, there are no neighboring protons for signal A, so zero plus one equals one, we will have a singlet. And for proton B, signal B, we have the three methyl protons on the other side in the ethyl group. So three plus one is four, that signal will be a quartet. And for signal C, we have those two neighboring protons from signal B two plus one equals three. So that will be a triplet. So we have a singlet quartet and triplet and then a triplet multiple triplet singlet. So looking at our answer choices, I see that answer choice C is incorrect because it mentions doublets. And we did not have any doublets in our N M R spectra. So that makes B the correct answer for this problem. And it reads using proton N M R. One of the isomers produces three signals, quartet singlet triplet and the other isomer produces four signals, singlet, triplet multiple and triplet. That is it for this problem? I hope this video was helpful and thanks for watching.

How would the ¹H NMR spectra for the four compounds with molecular formula C₃H₆Br₂ differ?

Key Concepts

Nuclear Magnetic Resonance (NMR) Spectroscopy

Chemical Shifts

Multiplicity and Coupling

Watch next

How would the 1H NMR spectra for the four compounds with molecular formula C3H6Br2 differ?

Key Concepts

Nuclear Magnetic Resonance (NMR) Spectroscopy

Chemical Shifts

Multiplicity and Coupling

Watch next

How would the ¹H NMR spectra for the four compounds with molecular formula C₃H₆Br₂ differ?