What products do you get when you react salicyladehyde with 1-bromobutane?
Rxn 1, 0.25g salicyladehyde and 0.56g of 1-bromobutane and 0.13g of tetrabutylammonia bromide and 2ml acetone was mixed and to that 0.40g of K2CO3 …
Rxn 2, same procedure is done as above except instead of K2CO3 , KOH was used.
I would like to know the products been produced and the reaction mechanism?
Well, this may be okay, but without knowing what is being discussed in your class (intended products to form), it can be tricky. The first part is easier. See the following (these examples were taken from "A Guide to Organic Chemistry Mechanisms©" at http://www.curvedarrowpress.com):
Your exact example was not represented in the book, but you should be able to piece together the parts. K2CO3 is not as strong a base as KOH. It will form the anion of the phenol. The phenoxide anion will react with the 1-bromobutane to give the ether product in Rx 1 and 2.
Here is where the problem occurs. K2CO3 is a weaker base than KOH. When K2CO3 is consumed, KHCO3 is formed, a weaker base. If KOH is used, the consumption of KOH give KBr, a neutral salt. No problem there. If K2CO3 were replaced with the same weight of KOH, now we face another reaction which I don't know is part of the problem or not.
This is how it works, approximately, 2 mmol salicylaldehyde is reacted with 4 mmol K2CO3 and 4+ mmol 1-bromobutane. The excess reagents won't interfere with forming the product.
Replacing K2CO3 gives about 8 mmol KOH. After the first reaction is complete, there is a considerable amount of base present (KOH) which can catalyze an aldol condensation with acetone. This is very close to a common student prep of dibenzalacetone. The mechanism for that reaction is Example 3, Chapter 9.
Learn reaction mechanisms with "A Guide to Organic Chemistry Mechanisms©" at http://www.curvedarrowpress.com .