Propose structures for compounds that fit the following...

Propose structures for compounds that fit the following descriptions: (a) $\mathrm{C}_{10} \mathrm{H}_{14}$ $1 \mathrm{H} \mathrm{NMR}: 7.18 \delta(4 \mathrm{H},$ broad singlet $) ; 2.70 \delta(4 \mathrm{H},$ quartet, $J=7 \mathrm{~Hz})$ $1.20 \delta(6 \mathrm{H},$ triplet $, J=7 \mathrm{~Hz})$ $\mathrm{IR}: 745 \mathrm{~cm}^{-1}$ (b) $\mathrm{C}_{10} \mathrm{H}_{14}$ ${ }^{1} \mathrm{H} \mathrm{NMR}: 7.0 \delta(4 \mathrm{H},$ broad singlet $) ; 2.85 \delta(1 \mathrm{H},$ septet, $J=8 \mathrm{~Hz})$ $2.28 \delta(3 \mathrm{H},$ singlet $) ; 1.20 \delta(6 \mathrm{H},$ doublet, $J=8 \mathrm{~Hz})$ $\mathrm{IR}: 825 \mathrm{~cm}^{-1}$

Propose structures for compounds that fit the following descriptions:
(a) $\mathrm{C}_{10} \mathrm{H}_{14}$
$1 \mathrm{H} \mathrm{NMR}: 7.18 \delta(4 \mathrm{H},$ broad singlet $) ; 2.70 \delta(4 \mathrm{H},$ quartet, $J=7 \mathrm{~Hz})$
$1.20 \delta(6 \mathrm{H},$ triplet $, J=7 \mathrm{~Hz})$
$\mathrm{IR}: 745 \mathrm{~cm}^{-1}$
(b) $\mathrm{C}_{10} \mathrm{H}_{14}$
${ }^{1} \mathrm{H} \mathrm{NMR}: 7.0 \delta(4 \mathrm{H},$ broad singlet $) ; 2.85 \delta(1 \mathrm{H},$ septet, $J=8 \mathrm{~Hz})$
$2.28 \delta(3 \mathrm{H},$ singlet $) ; 1.20 \delta(6 \mathrm{H},$ doublet, $J=8 \mathrm{~Hz})$
$\mathrm{IR}: 825 \mathrm{~cm}^{-1}$

Transcript

00:01 Solving problem 41 of chapter 15.

00:05 Let's start with molecule a, having formula c10 h14.

00:13 So from the proton and mr, the first data we have is that at 7 .18, we have 7 .18.

00:23 We have 4 hydrogen corresponding to a broad singlet.

00:29 So around this chemical shift, we have the aromatic protons, so we have a benzene ring for sure.

00:40 And it's telling us that correspond to only four hydrogens.

00:44 So it means the molecule is substituting in two different positions.

00:49 And more likely in paraposition, because we only have one signal, so not two different chemical shift.

00:57 And so the protons are equivalent.

01:01 And also these two groups are more likely again in the paraposition.

01:08 And probably they are also symmetric because, again, we only have one chemical shift.

01:14 So the second data we have is that we have a 2 .70 for hydrogen quartet.

01:22 So 2 .70 for hydrogen that appear as a quartet.

01:29 So this means that it should be, if it's symmetric, so it should be, for example, ch2 and ch2.

01:49 So four hydrogens, and since our quartet, they are probably connected to a ch3.

01:55 But let's see if the third data we have, we confirm that.

01:59 So we have another signal 2 .20, 6 hydrogen triplet.

02:04 2 .20 6 hydrogen triplet.

02:12 So from here, since we have 6 hydrogen and we are saying that the molecule is substituted into different places and we expect to be symmetrical, so the hydrogen on this side and hydrogen on this side, it should be exactly the same.

02:29 So we can expect, like in this case, two signals from two molecular fragments that are the same.

02:40 So since hydrogen, it means we have a ch3 and ch3, and csr triplet, they should be connected to a ch2.

02:50 That match perfectly what we expect from here.

02:54 So probably we have a ch2, ch3, and so probably we have a ch2, ch3, and also on this side we have a ch2, ch3.

03:07 So in this case, let's double check.

03:11 We have four aromatic protons, one, two, three, and four that appear as only one chemical shift, a 7 .18.

03:19 So this is fine.

03:21 And then we have two alkaliic chains that are substituted in paraposition, and they are exactly the same.

03:31 So we should have a chemical shift for this plus these 2 ch2 and another chemical shift for the two methods.

03:42 So these should give are two protons and should appear as a quartet because it's always proton plus 1.

03:51 So this is 3 plus 1 form.

03:54 So this should split as a quartet.

03:57 Same for these and so are two protons plus 2 protons 4.

04:01 Protons that appears as a quartet and we have it here...

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