Chem 353 W 2011 Final : Laboratory

Part 7: LABORATORY

Part A:

Experimental yields.... First you need to balance the reaction equation and then work out the moles of each reagent used to determine the limiting reagent:

	3 CH₃CH=CH₂ + BH₃ -- (then work up) -> 3 CH₃CH₂CH₂OH
MW (g/mol)	42	15	60
amount	4.2g	20mL of 1M	3.00g
moles	0.1	0.02	0.05
moles / coefficient	0.1 / 3 = 0.033	0.02 / 1 = 0.02

Therefore, in the last row of the table where the stoichiometric coefficient from the balanced equation is divided into the number of moles, we can see that the limiting reagent is the borane. This means that the maximum amount of product that can be formed is 0.06 mol (based on the reaction stoichiometry, 1 mole borane gives 3 moles of the alcohol product). Hence the % yield, based on obtained / max. possible = 0.05 / 0.06 = 83%.

Common general errors: (1) not balancing the reaction equation, (2) not determining the limiting reagent correctly (3) not knowing how the calculate an experimental yield and (4) not knowing what borane was (note other information in the question should have helped trigger the memory). This section was done poorly considering the number of times yields are calculated in the laboratory during the semester and that there was a similar question on the midterm and last years final. Note that it doesn't matter whether you work with grams or moles of product the answer should be the same.

Part B:

The molecular formula of C₁₀H₁₀O implies and IHD = 6.
The insolubility data suggests that we have a neutral functional group.

From the chemical tests we get:

2,4-DNP test is positive and therefore indicates the presence of an aldehyde or ketone.
The Tollen's test is negative confirms it's not an aldehyde.
The ferric chloride test is negative indicating that it's not a phenol.
The iodoform test is negative indicating that it's not a methyl ketone.

The H NMR data tells us:

7.25 to 8.0 ppm suggest 4 aromatic H and therefore probably a disubstituted benzene system.
3.0 ppm shows a 2H triplet, i.e. a -CH₂- coupled to 2H.
2.65 ppm shows a 2H triplet, i.e. a -CH₂- coupled to 2H.
2.15 ppm shows a 2H pentet, i.e. a -CH₂- coupled to 4H.

Summary....
Molecular formula of C₁₀H₁₀O implies and IHD = 6.
Functional groups test suggest a ketone, C=O.
H NMR suggests a disubstituted benzene, C₆H₄ and -CH₂CH₂CH₂-

Summing these pieces matches to the molecular formula and we have the IHD = 5 so far....(C=O, plus 4 for the benzene system)... which helps reveal another ring

Altogether...

The fragments we have are :C₆H₄ , -CH₂CH₂CH₂- , and C=O

(i.e. three middle pieces)

Assembling the pieces....the ortho aromatic substitution pattern is supported by the aromatic region peaks (4 types ArH, no "singlet").

solution

The final step should always be to check what you have drawn. The easiest thing to check is usually the coupling patterns you would expect to see, and the chemical shifts of each unit. You should be asking yourself : "Does my answer give me what the H-nmr shows ?"