Hello there Here is an experiment for orgo lab We had to identify our compound b

Question

Hello there Here is an experiment for orgo lab We had to identify our compound based on the melting point that we discovered. So my question is that is there something about the solubility of each compound deprotinated by HCO3 And NAOH that leads us to one compound over the other? Please ask if you need further explaination could 03-59-230 W18 Experiment 2 ba bon EXPERIMENT 2: Quantitative Separation and Identification of an Unknown Mixture by Extraction, Recrystallization, and Melting Point Determination TESTNGA SAMPLE FOR BERYLLUM THAT? YEAH TERSA ANT YOU SEE THEM?HAT THINGS ARE? BUNCH OF BERYLLUMS OF HETAL IN YOUR EVERYONE THOUGHT I LR6 WERD FACE. LOTS OF Courtesy Randall Munroe, XKCD Grading Scheme Experiment 2 counts for 1/7th of your total lab grade that are subdivided into 30% for experimental outcomes and 70% for the lab report. This lab requires you to determine and report the yields of recovered unknowns (5% each), their melting points (5% each, see grading scheme), and their chemical identity (5% each, no partial marks). Although the chemical identity marks are included in the experimental outcomes marks, your assignment of identity is to be included in the lab report.

Explanation / Answer

The compound which are deprotonated by HCO3 minus would be more acidic than which are deprotonated by NaOH, beacuase NaOH is stronger base than HCO3 - , so compound which are deprotonated by NaOH must be soluable in highly basic medium that is they contain less acidic grouop and very less soluable in water but the compound which can deprotaned by HCO3- must be soluable in slightly basic medium that means that compound would have more soluability in water compare to others

the solubility differnce can help in identification of compound

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