Question

After years of study, you have recently finished your graduate degree, and have just accepted a position at a small liberal arts college. You show up the first day and look into your new lab, and see that there is one unlabeled container. The previous occupant has retired at the beginning of the summer and is apparently somewhere in the middle of the Atlantic Ocean now enjoying retirement (and coincidentally cannot be reached). You’re unsure whether or not the compound is useful, but thankfully the department you’re in has a GC-MS, NMR and IR spectrometer, and a number of Meltemps. Between these instruments and online resources, you’re confident that you can identify this unlabeled compound. Although there is no label, you know based on the previous professor’s research that the compounds are likely to be from one of the following classes of compounds: Alcohols and phenols, aldehydes and ketones, amines, or carboxylic acids, amides, and esters. You’ll be provided with MS, IR, melting point, boiling point and NMR data on Blackboard.

1 2 2 4 3 10 PPM

Answer 1

Solution:

Step-1: Identification of functional group

IR spectrum has the following major frequencies.

Sl.No	Frequency	Band	Assignment
1	About 1725cm-1	Strong, Sharp band	C=O Stretch of aliphatic aldehyde.
2	About 2860 - 2700cm-1	Two weak bands. First band is obscured by sp3 C-H absorbances.	C-H Stretch of CHO group.
3	About 2970 - 2950cm-1	Strong, sharp band	C-H Stretch of sp3 hybridized carbon.

Hence the compound is having one O atom and is an aliphatic aldehyde.

Step-2: Finding out the empirical formula / molecular formula

We know that the molecular ion peak, m/e = 100.1. That means the molecular weight of the compound is 100.1g/mol.

Use the rule of 13, find the CnH(n+r)

100.1 / 13 = n is 7; r is 9. Therefore, the formula is C7H16

We know from the IR spectrum that the compound has one O. Replace CH4 from the formula to accommodate O atom.

Empirical formula = [(C7H16 - CH4) + O] = C6H12O

The empirical formula C6H12O accounts for the molecular weight of the compound, 100.1g/mol. Hence the molecular formula is same as empirical formula, C6H12O.

Step-3: Identification of double bond equivalence (DBE)

DBE = [C - (H/2)] + 1 where C is number of C atom and H is number of H atom.

DBE = [6 - (12/2)] + 1 = 1.

Hence the compound has one double bond which is C=O.

Step-4: Identification of the compound

We know that the molecular formula of the unlabeled compound is C6H12O and does not have any ring. Further, six peaks in 13C NMR confirms that there are six C atoms and all the C atoms are in different magnetic environment. Also the 1H NMR confirms that there are 12 protons with different magnetic environment.

Hence the unlabeled compound is 1- Hexanal and the structure is CH3CH2CH2CH2CH2CHO.