When compound A (C₅H₁₂O) is treated with...

Question

When compound A (C₅H₁₂O) is treated with HBr, it forms compound B (C₅H₁₁Br). The ¹H NMR spectrum of compound A has a 1H singlet, a 3H doublet, a 6H doublet, and two 1H multiplets. The ¹H NMR spectrum of compound B has a 6H singlet, a 3H triplet, and a 2H quartet. Identify compounds A and B.

Accepted Answer

Welcome back everyone to another video compound B with a molecular formula of C six H 13 BR is produced when compound A with a molecular formula of C six H 14 0 is combined with hydroponic acid. We're given their signals in H and Mr spectrum and we are going to analyze those signals. What are the structures of the compounds? And we are given for answer choices. ABC and D compound A would be an alcohol and compound B would be an alkyl bromide. We have a variation of our answer choices. Now, what we're going to do is just analyze all of the peaks that we have. So starting with A, we can say that according to the context, we have one H slat and in parentheses, we are going to state the number of neighboring hydrogen. If it is a singlet, according to the M plus one rule, we have to recall that N would be equal to zero. So that would be zero neighbors, then we have to a multi plat meaning there are multiple neighbors, right? The peak is complicated due to the fact that there might be hygiene atoms that are neighboring in different environments. Then we have one, H multiplied another one multiplied three H double blood, which essentially means that there is only one neighboring hydrogen based on the M plus one rule. And six H double blood, again, we have one neighbor. So if we count our signals, we have a total of six, meaning we can just consider our answer choices and label non equivalent hydrogens starting with a compound. A will have four equivalent metal groups due to the vertical line of symmetry. Then we will have pro dance B which are again, symmetrical protein C in the middle and protein D bonded to the oxygen atom. So we would have four signals instead of six. Meaning we can exclude option A option B, we have two equivalent metal groups bonded to the same carbon atom ch hydrogen, another ch hydrogen. Then we have our metal group, then we have our siege two and then the hydrogen atom bonnet to the oxygen atom ABC DEF that gives us six. So B is a possibility for us. Now, for C, there are two equivalent metal groups ch then we have once again two equivalent metal groups and hygiene of the alcohol that gives us a total of four signals. So we cannot have that because we need sex. And for D compound A is exactly the same compound as for choice A. So we have four equivalent metal groups. Then we have our ch, another ch and our alcohol So that gives us a total of four signals, meaning the correct answer should be b but we just want to verify it. So one and glad essentially corresponds to our alcohol group because that hygen has no neighboring hydrogens. So you can say that would be our one slat, then we have two H multi plat and basically a two H multi corresponds to e because those two Hagens are actually diao topic. So they split each other's signals. We noticed that there is only one neighboring hydrogen. But because those hydrogens at position er diao topic, the signal becomes slightly complicated and we get a multi plat, then we have one, each multiply two times. So CNB and that's because if we consider each of them, we notice that they are neighboring hydrogens in different environments. For example, for B, we have two metal groups and then the hydrogen C, right? So they're in different environments. On the other hand, if we consider C, there are three different environments de and B, so we get multiplex. And now considering we age the blood, that would be the because we have our metal group and one neighboring hygen and six of blood, we have two metal groups with one neighboring hygen bee. So that would be correct. We verified that compound A will be our primary alcohol. And now let's consider our alkyl bromide. Now considering the alky bromide, we have one H sub dot Sat essentially means that there should be six neighbors, right? According to the M plus one rule. So let's identify that peak, one of that would be proton bonded to carbon A because we have one hydrogen and there are two methyl groups with six hydrogen atoms total. Now our next speak is six H and glad that means there are zero neighbors and we can identify this very easily. There are two metal groups which are equivalent and these six hydrogens have zero neighbors because there's bromine and another carbon bonded well done. And now we also have six, the black. That means we have one neighbor. And essentially, that would correspond to our other two metal groups because there are six hygen total. And there is a one neighboring hygen from a well done. So we have verified that the correct answer should be b we can also check the molecular formulas because C six H 14 0 A saturated alcohol. Since we have twice as many hydrogens and additional two hygen are basically two M plus two hygen relative to the N number of carbons, we can verify that our alcohol is saturated. And for the alkyl bromide, we have C six H 13 BR we once again can replace our bromine with hydrogen. And we will notice that our alky bromite must be saturated. Because if we replace bromine with hydrogen, we will get two M plus two hydrogens relative to the N number of carbons. And of course, when we think about these structures. We can say that our final answer would be option B, compound A is a primary alcohol and compound B is our alkyl hide at the tertiary position. Thank you for watching.

Key Concepts

NMR Spectroscopy

Substitution Reactions

Molecular Formula and Structure

When compound A (C5H12O) is treated with HBr, it forms compound B (C5H11Br). The 1H NMR spectrum of compound A has a 1H singlet, a 3H doublet, a 6H doublet, and two 1H multiplets. The 1H NMR spectrum of compound B has a 6H singlet, a 3H triplet, and a 2H quartet. Identify compounds A and B.

Key Concepts

NMR Spectroscopy

Substitution Reactions

Molecular Formula and Structure