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The molecular formula fixes the degrees of unsaturation, (2C + 2 + N − H − X) ÷ 2, the rings plus pi bonds a structure must have; infrared says which they are, and fragments choose among what remains, until only one structure fits every constraint.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will calculate degrees of unsaturation, say what features account for them, order the steps of solving an unknown, decide what one degree allows, identify an unknown from combined evidence, and correct the count for nitrogen and halogens.
You can match each spectral observation to the feature it supports. The last step is to combine them, with the molecular formula, until only one structure fits.
The degrees of unsaturation (DU) of a formula count its rings plus pi bonds. A constraint is a fact every candidate must satisfy. Elemental analysis gives the ratio of elements, which with the molecular ion fixes the formula.
A saturated open-chain compound with C carbons has 2C + 2 hydrogens. Every ring or pi bond removes two. So
$$\text{DU} = \frac{2C + 2 + N - H - X}{2}$$
where N counts nitrogens (each adds a bond's worth of hydrogen) and X halogens (each replaces a hydrogen). Oxygen is left out: it does not change the count.
A DU of 4 or more with only six or seven carbons strongly suggests a benzene ring.
Another way: steps
To solve an unknown:
M+ at 72 and C4H8O: DU = (8 + 2 − 8) ÷ 2 = 1. IR: strong at 1715 cm−1, so the degree is a C=O; no peaks near 2720 and 2820 cm−1, so not an aldehyde. The only four-carbon ketone is butanone, CH3COCH2CH3, and its mass spectrum fits: m/z 43 is CH3CO+, from losing an ethyl group. Each constraint alone left several structures; together they left one. That is how structure is established in practice — with NMR added as a further, powerful constraint in later study.
One matching peak proves a complete proposed structure. All constraints must hold together.
Oxygen changes the degrees of unsaturation. It does not.
A halogen is ignored in the count. It counts like a hydrogen.
One degree means a double bond. It could equally be a ring.
DU = (10 + 2 − 8) ÷ 2 = 2.
Two rings or pi bonds.
A peak near 2120 cm−1 is a C≡C, which accounts for both degrees.
IR says which.
So the compound is a pentyne; the position of the triple bond needs further evidence.
Say what is left.
Apply (2C + 2 + N − H − X) ÷ 2.
(8 + 2 − 9 − 1) ÷ 2.
How many degrees of unsaturation?
Zero: a saturated, open-chain bromobutane.
How many degrees of unsaturation — rings plus pi bonds — does a compound C4H8 have?
Answer:
Match each structural feature to the degrees of unsaturation it accounts for.
| 1 | 2 | 4 | |
|---|---|---|---|
| a C=O | |||
| a C≡N | |||
| a benzene ring | |||
| a cyclohexane ring |
Put the steps of working out an unknown's structure in order.
Number the steps in order (write the number in the box):
A compound C4H8O has one degree of unsaturation, and its IR spectrum shows no peak near 1700 cm−1 and no broad O–H band. What must the one degree of unsaturation be?
An unknown has its molecular ion at m/z 72, and elemental analysis gives C4H8O. Its IR shows a strong, sharp peak at 1715 cm−1 and no peaks near 2720 or 2820 cm−1. The mass spectrum has a large fragment at m/z 43. Which compound is it?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
How many degrees of unsaturation does C3H6Cl2 have?
The compound has a degrees of unsaturation.
You can combine formula, infrared and mass spectrum to settle a structure. Tell someone why C6H6 suggests a benzene ring. This completes the organic chemistry course.
8. Your turn: C4H9Br, step 3