OH | H2PO4 CH3-C-CH2-CH3 ----> CH3-C(double bond)CH-CH3+CH2(double bond)C-CH2-CH
ID: 782965 • Letter: O
Question
OH
| H2PO4
CH3-C-CH2-CH3 ----> CH3-C(double bond)CH-CH3+CH2(double bond)C-CH2-CH3 + H2O
| HEAT | |
CH3 CH3 CH3
(CH3)2COHCH2CH3 ----> CH3CH3CCHCH3+ CH2CCH3CH2CH3+ H20
1)in the equation given for this experiment, only two isomers are shown, what other isomer could also form? was there any GLC evidence of its existence?
2) does GLC analysis allow you to skip the IR as an analytical method? compare the two methods for advantages and disadvantages
3) how can you tell when the starting alcohol is completly mixed with the aqueous acid?
Explanation / Answer
This is the reaction I am seeing:
2-methyl-2-butananol + H3PO4 (acid catalyst) = 2-methyl-1-butene (Hoffman minor product) + 2-methyl-2-butene (Zaitsev major product)
The only other isomer I see possible that could form is 3-methyl-1-butene. This product forms from a tertiary to secondary carbocation rearrangement. This is a minor product since a secondary carbocation is less stable than a tertiary one. Whether there was any GLC evidence of its existence would be determined by it's percent composition. How many peaks did you have on your chromatogram? If 3 besides the air bubble (t0), then there is a chance this product did form. However, the percent composition of the product would be very small.
2.
I don't think the GLC analysis would allow you to skip the IR because the percent composition of the products has to be calculated. The chromatogram doesn't give you the percent composition of the products formed. The IR gives you peaks of the differnent functional groups based on their frequencies.
3.
I actually did this experiment, but my reactants and products were different. Did you isolate your product by distillation? If so, when you heat your reaction mixture (alcohol and acid) in the round bottom flask, first evaporation occurs and the products formed will turn in vapor. When the vaporized products interact with the water-cooled condensor, condensation occurs and the liquid product will collect in the receiving flask. So, once condensation occurs, you know that the alcohol has completely mixed with the aqeous acid.
Hope this helps! :)
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