Post Lab 2 DEHYDRATION AND SUBSTITUTION OF 2-METHYL-2-PROPANOL PDF

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Summary

DEHYDRATION AND SUBSTITUTION OF 2-METHYL-2-PROPANOl
part of this laboratory experiment you will follow the lower reaction pathway and prepare
2-chloro-2-methylpropane. You will verify the formation of this product by boiling point, determine the yield and
calculate the percent yi...

Description

Dolores Apreda Post-Lab 2 Questions 1. Draw the expected products when 2-methylpropene reacts with Br2 and with KMnO4 solutions.

2. Will unreacted HCl in the aqueous or organic layer? Explain. Unreacted HCl will be in the aqueous layer because of its polarity. 3. Look in a reference book to find the boiling points of 2-methyl-2-propanol and 2-chloro-2methylpropane. From the data collected in this experiment, verify that your product was actually 2chloro-2-methylpropane. 2-methyl-2-propanol 82°C 2-chloro-2-methylpropane 51°C The boiling point found in lab for the product collected was 51°C which is the boiling point of 2-chloro-2methylpropane listed in the reference book. 4. Calculate the theoretical yield and percent yield of 2-chloro-2-methylpropane prepared, assuming the 2-methyl-2-propanol is the limiting reagent. Empty container 147.69g Container w/ decanted product 150.24g Mass of product  2.55g Theoretical yield 9.67g (CH3)3CCl C4H10O + HCl (CH3)3CCl + H2O C4H10O 10mL * (0.775g/mL) = 7.75g HCl 25mL * (1.18g/mL) = 29.5g %yield 26.34% (CH3)3CCl 7.75g C4H10O * (1mol/ 74.12g C4H10O) * (92.57g (CH3)3CCl /mol)=9.67g (CH3)3CCl

%yield=

actual yield 2.55 ∗100 % → ∗100 → 26.34 % 9.68 theoretical yield

5. Explain why you didn't get 100% yield (other than sloppy procedure) for the 2-chloro-2methylpropane. A 100% yield for 2-chloro-2-methylpropane cannot be obtained due to not all of the HCl acid reacting with 2-methyl-2-propanol. The carbocation formed is relatively stable therefore it may not react completely....