General chemistry in the organic course

JOHN J.

exom quef tion exchange General Chemistry in the Organic Course Parls Svoronos Qmnsbnough C m h y College Bayside, NY 11364

Students have the tendency t o forget General Chemistry once they register in an Organic Chemistry course. There are several problems though that can combine the rationale of General and Organic Chemistry. Here is a way to identify an organic compound using titrimetric methods. This question requires skills a t the applications level of the Bloom taxonomy. Ouaatlon 3.61 g of a lipid (molecular weight 722) is saponified with excess NaOH, and the unreacted base is hack-titrated with HCl. Thus, 50.0 mL of 0.4 M NaOH is added, and the excess base is neutralized with 20.0 mL of 0.25 M HC1. The only acid found in the hydrolysate is myristic acid. What is the structure of the lipid? Acceptable Solutlon The overall procedure can be summarized as follow.

Lipid

-

+ NaOH\

Myristate + Alcohol

University of Cincinnati Cincinnati. Ohio 45221

Number of moles of NaOH added = M X V = 0.4 rnol L-' X 0.050 L = 0.020 rnol Number of HCI added = Number of molesof NaOH (excess) = M X V = 0.25 mol L-'X 0.020 L = 0.005 mol Number of moles of NaOH needed to hydrolyze the lipid = 0.020 0.005 = 0.015 mol

Thus, 1rnol of the lipid is saponified completely by 3 rnol NaOH. The lipid, therefore, should have three ester groups with myriatie (CHs(CHzhaC00)as the acid part.

Three myristate ions have a total mass of 681 amu. This leaves (122 681 =) 41 amu for the mass of the alcohol part. The only possibility is CsHs, the alcohol being glycerol and the lipid glycerol trimyristate

-

I

CHOH

I Number of moles of lipid = (Mw/MW) = (3.611722) = 0.005 rnol

edited by: ALEXANMR

CHaOH glycerol

Volume 66

I

CHOOC(CHa)&Ha

I CHaOOC(CHz)&Hs :glycerol myristate

Number 5

May 1989

429