A $1.300 g$ sample of ethyl acetoacetate ( $M = 130 g mo l^{- 1}$ ) is dissolved in methanol and made up to exactly $100.0 mL$ . A $10.0 mL$ aliquot is titrated rapidly with $0.0272 M$ bromine solution, which reacts only with the enol form in a $1 : 1$ mole ratio, and $2.50 mL$ is consumed. Calculate the percent of ethyl acetoacetate present as enol, to one decimal place.

Numerical answer type, answer in %, tolerance 0.05.

Answer: 6.8 %

Moles of ester in the aliquot

= (1.300/130) \times (10.0/100.0) = 1.00 mmol

. Moles of

Br X_{2}

used

= 2.50 mL \times 0.0272 M = 0.068 mmol

, which equals the moles of enol. Percent enol

= 0.068/1.00 \times 100 = 6.8%