An aromatic hydrocarbon with a molecular formula of C₁₃

Question

An aromatic hydrocarbon with a molecular formula of C₁₃H₂₀ has an ¹H NMR spectrum with a signal at ~7 ppm that integrates to 5H. It also has two singlets; one of the singlets has 1.5 times the area of the second. What is the structure of the aromatic hydrocarbon?

Accepted Answer

Welcome back everyone. To another video propose the structure of an aromatic hydrocarbon with a molecular formula of C 16 H 26. The proton NMR spectrum of the aromatic compound has three signals 7.1 parts per million for H singlet, 2.4 parts per million for H singlet and 0.9 parts per million 18 H singlet. What we always have to do as our first step is essentially calculate the hydrogen deficiency index. In this case, we only have carbon atoms and hydrogen atoms. So we can say that we are simply going to calculate one half multiplied by two multiplied by the number of carbon atoms plus two minus the number of hydrogen atoms. And in this case, the number of carbon atoms would be 16. So we would calculate one half multiplied by two, multiplied by 16 plus two minus 26 which corresponds to the number of hydrogen atoms. Now we have 34 minus 26 that's eight divided by two, which gives us four because the problem states that we have an aromatic hydrocarbon, the hydrogen deficiency index of four essentially represents the benzene ring where basically one ring three double bonds. That means there is no additional un saturation in our structure and we are simply ready to draw benzene. Now, we only have three signals for so many carbon atoms and hydrogen atoms. In particular, because we're considering the H and Mr spectrum, meaning there must be a lot of symmetry starting with 7.1 P PM, which is the signal, which is the most downfield, we can essentially state that this signal will, will represent the protons bonded to our rank. And because we have 4h S, it means that all of the protons are actually equivalent, right based on that symmetry. And it means that our rank must be paradise substituted. Now, how do we know that? Well, basically if our ring has four hydrogen atoms, because we have four, a singlet, we will definitely have two subscriptions. But due to that singlet, we can say that all of our protons are equivalent to each other. So there will be a horizontal line of symmetry with respect to which our protons will be equivalent to each other as well as the vertical line of symmetry. So that the protons on each side of the vertical line are also equivalent. And if we combine the two symmetries together, that implies that each proton is exactly same relative to its P PM value. So this tells us that we have our para that substituted benzene. Now, if we consider our next shift, we have 2.4 P PM for H slat. And then we also have 0.9 p pm, 18 H slat. So we can start with the 0.9 p pm shift because we have a shift at around one P pm. This is a very typical of a methyl group. And essentially because it says we have 18 Hagens total. Let's remember that a third beetle group would be a common group for nine H slat. The reason why is simply because if we think about t beal group, we have three metal groups bonded to a questionary carbon so that there are no neighbors and we get a singlet, right, singlet means that there are no neighboring hydrogens. And if we count all of the hydrogen atoms from each metal group, we get nine. And based on that symmetry, we are expecting to have two of them. So now, essentially what we're going to do is take our structure. And of course, we want to make sure that we include our 4h slut 2.4 P PM. That means those hydrogen atoms are bonded to the carbon atom, which is directly bonded to the benzene ring. So if we consider only one chain for now, we want to make sure that we have two hygen due to the symmetry of the ring, right, we have four total. So one chain will have two hydrogen atoms. This could be a siege two group, right? Because we will have carbon with two hydrogen atoms. And then we can essentially use our third beetle group. So essentially our carbon with three metal groups. And we can check if this structure is valid. Well, essentially assuming that this chain is correct, we will definitely have nine H syle for our T beetle group. And now from the ch two group, the two hydrogen atoms will also correspond to a syle because there are no neighboring hydrogens. So that is sensible. And now what we're going to do, we're just going to draw another symmetrical chain. So our ch two group, as well as our third beetle group, we can modify our bond angle slightly to make sure that everything is consistent. Now, let's draw that tuttle group and now we can also check if our molecular formula is correct. So now we have a total of six carbon atoms in the ring. Then we have 789, 1011, 1213, 1415, 16. So that's indeed C 16. Now, if we consider our hygiene within the ring, that would be four, we have four in the ring, then we have two more for the siege two group and nine more from the metal groups. So we can say if we take nine plus two, that's 11 multiplied by two, that's 22 plus four, that's 26. So we have a total of 26 hygen assigned, meaning it corresponds to our structure. And now looking at the structure, we have two equivalent turtle groups due to the lines of symmetry. So we will have a total of 18 hygen which will correspond to a Syle because there are no neighbors, then we will have four hygen syle for the two equivalents, two groups. And finally, for a single corresponds to the four hygen atoms that are equivalent and directly bonded to the benzene ring. So we are going to label our structure as our final answer. Thank you for watching.

An aromatic hydrocarbon with a molecular formula of C₁₃H₂₀ has an ¹H NMR spectrum with a signal at ~7 ppm that integrates to 5H. It also has two singlets; one of the singlets has 1.5 times the area of the second. What is the structure of the aromatic hydrocarbon?

Key Concepts

Aromatic Hydrocarbons

Nuclear Magnetic Resonance (NMR) Spectroscopy

Integration and Signal Patterns in NMR

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An aromatic hydrocarbon with a molecular formula of C13H20 has an 1H NMR spectrum with a signal at ~7 ppm that integrates to 5H. It also has two singlets; one of the singlets has 1.5 times the area of the second. What is the structure of the aromatic hydrocarbon?

Key Concepts

Aromatic Hydrocarbons

Nuclear Magnetic Resonance (NMR) Spectroscopy

Integration and Signal Patterns in NMR

Watch next

An aromatic hydrocarbon with a molecular formula of C₁₃H₂₀ has an ¹H NMR spectrum with a signal at ~7 ppm that integrates to 5H. It also has two singlets; one of the singlets has 1.5 times the area of the second. What is the structure of the aromatic hydrocarbon?