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Chemistry 212 - Lab Exam
- Why is extraction of biological compounds important ANS For isolation and
iden- tification
- What does a mild base during extraction allow for? ANS Deprotonate
functional groups, increasing their water solubility, simplifying the extraction
- Define extraction ANS Preferentially isolating a compound of interest from a
complex mixture by using selective solubility characteristics
- How can an emulsion be broken ANS a long glass rod
agitating it gently re-agitating the mix with brine
- Define brine ANS A saturated NaCl solution that pulls water into itself
through its hypertonic nature
- Define emulsion ANS A turbid mixture of 2 immiscible liquids. In the case of
this lab, it is usually a suspension of water in an organic phase caused by the presence of precipitates and/or material which is soluble in both phases 1 / 2
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- Why is brine pre-drying done before a drying agent is added ANS to
minimize the amount of drying agent required
- If dichloromethane and water were placed simultaneously in a sep. funnel,
which would be the top layer ANS water as it is less dense than dichloromethane
- Why is an erlenmeyer flask used to contain organic solvents instead of a
beaker ANS Angled sides prevent spilling Shape towards top prevents vapours from escaping as much and can also be stoppered
- What is the structure of (+)catechin ANS
- If the pKa of the (+) catechin is 10.1 and is mixed with a mild base (such as
- Define alkaloid ANS any of a class of nitrogenous organic compounds of plant
sodium carbonate) with a pH of 12, would you expect catechins solubility to be improved, decreased or unaltered? Calculate the ratio of the deprotonated catechin (a base) to the protonated form (the acid) at pH 12.0 using the Henderson-Hassellbach expression ANS
origin that have pronounced physiological actions on humans
- What is TLC used for? ANS identifying individual components of a mixture by
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examining how the analyte moves by capillary action up a stationary phase at