6. Draw the hydrolysis reaction of 2-O-linolenoyl-O-oleoylphosphatic dylcholine in the presence of sodium hydroxide. Arrange the stoichiometric coefficients,name the products. 7.

6. The hydrolysis reaction of 2-0 -linolenoyl-1-0 -oleoylcholine in the presence of sodium hydroxide can be represented as follows:<br /><br />2-0 -linolenoyl-1-0 -oleoylcholine + 2NaOH → 2-0 -linolenol + 1-0 -oleoylcholine + 2NaCl<br /><br />In this reaction, the sodium hydroxide (NaOH) acts as a base and reacts with the ester compound, 2-0 -linolenoyl-1-0 -oleoylcholine, to produce the corresponding alcohol, 2-0 -linolenol, and the sodium salt of the other alcohol, 1-0 -oleoylcholine. The stoichiometric coefficients are 1 for 2-0 -linolenoyl-1-0 -oleoylcholine, 2 for NaOH, 1 for 2-0 -linolenol, 1 for 1-0 -oleoylcholine, and 2 for NaCl. The products of this reaction are 2-0 -linolenol, 1-0 -oleoylcholine, and sodium chloride (NaCl).