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3 answer choices.
1) 2) 3) question 3 answer choices. You add 0.58 mL of n-propanol (0.803 g/mL, 60.1...
You add 0.58 mL of n-propanol (0.803 g/mL, 60.1 g/mol) to an excess of acetic acid to create the pear ester propyl acetate (102.1 g/mol). If you obtain 0.53 grams of the product, what was your percent yield? You react acetic acid with an unknown alcohol, and obtain the following NMR of the product. What is the unknown alcohol? LLLLLLLL LES LL 20 1-butanol (CH3CH2CH2CH2OH) 2-propanol (CH3)2CHOH) 1-propanol (CH3CH2CH20H) 2-methyl-1-propanol (CH3)2CHCH2OH) 2 butanol (CH3CHOHCH2OHCH) 3 Mi Consider the below general...
Question 2 2 pts You add 0.58 mL of n-propanol (0.803 g/mL, 60.1 g/mol) to an excess of acetic acid to create the pear ester propyl acetate (102.1 g/mol). If you obtain 0.59 grams of the product, what was your percent yield?