How many types of protons? (^1 H NMR)

[2h2o]

Homework Statement

How many types of protons are present?

Homework Equations

compound 1:

CH₃CH₂OCH₂CH₂CH₂CH₂CH₃

and

compound 2:

CH₃CH₂CO₂CH₂CH₃

The Attempt at a Solution

Compound 1:

Arbitrarily assigning hydrogens which see a unique environment, left to right as appropriate.

CH_3(A)CH_2(B)OCH_2(C)CH_2(D)CH_2(D)CH_2(E)CH_3(A)

My solutions manual claims that there are seven types of ¹H in this molecule. I was under the impression that the high electron affinity of the oxygen will only deshield atoms ~1 sigma bond away. Is this impression false?

Compound 2:

First of all, this structural formula appears to me to describe an impossible molecule. Neither O atom has enough bonds to satisfy their "octet", yet to meet this criteria would require an isomer.

Moreover, even if I draw this out as described, using the method above:

CH_3(A)CH_2(B)CO₂CH_2(B)CH_3(A)

Noting the symmetry of the molecule, I see two types of ¹H, but my solutions manual claims four.

Where am I going wrong?

 

[Borek]

1 - it is not only oxygen that changes shielding, you have to look at the other end - and clearly -CH₃ is different form -C₅H₁₁

2 - have you ever heard about esters?

 

[2h2o]

Thank you for the hint on looking at esters--we've only just scratched carboxylic/carbonyls. But drawing it in ester form makes sense.

re #1:

It's not clear to me why the "D" hydrogens aren't the same. All of the "D" hydrogens each "see" four hydrogens on adjacent carbons across the C-C sigma bond.

 

[Borek]

They are very similar, but NMR is not about only the closest proximity (although the closest is the most important).

 

[DeeNos]

There are two types of protons in compound 1 and four types in compound 2. The number of types of protons in a molecule is determined by the number of unique environments each proton experiences. In compound 1, there are two types of protons because the two CH3 groups are identical and the five CH2 groups are also identical. In compound 2, there are four types of protons because the two CH3 groups are not identical and the two CH2 groups are also not identical. The oxygen atom in compound 1 does not have a significant effect on the chemical shift of the hydrogens, so it does not create a new type of proton. In compound 2, the presence of the carbonyl group creates a new type of proton due to its strong electron-withdrawing effect, which leads to a different chemical shift for the protons on the adjacent carbon. Your impression about the electron affinity of oxygen is not entirely incorrect, but it is important to consider the overall effect of the atom on the chemical environment of the protons in the molecule. In this case, the oxygen does not significantly alter the chemical shift of the protons, so it does not create a new type of proton.