A) You need to synthesize 1 gram of propoxynaphthalene. Assuming that naphthol i
ID: 988698 • Letter: A
Question
A) You need to synthesize 1 gram of propoxynaphthalene. Assuming that naphthol is the limiting reagent how much naphthol will you need? (Write your answer in the form x.xxx g)
Answer for part A: _______________
B) Assume your goal is to synthesize 1 g of product and naphthol will be your limiting reagent.
You have read that it is best to use 5 equivalents of propyl iodide. How many mL is this? (Write your answer in the form x.xx mL)
Answer for part B: _______________
C) Assume your goal is to synthesize 1 g of product and naphthol will be your limiting reagent.
You have read that this reaction requires 0.03 equivalents of tetrabutylammonium bromide (Bu4NBr). If this is the case, what can you say about the role of tetrabutyl ammonium bromide? Choose from the following:
A)Tetrabutylammoniumbromide must be a solvent in this reaction. The ionic bond will dissolve the reagents well.
B)Tetrabutylammoniumbromide must be a catalyst in this reaction, It is used to speed up the reaction, but it is regenerated so it is not consumed.
C)Tetrabutylammoniumbromide must be a reagent in this reaction.
A)Tetrabutylammoniumbromide must be a solvent in this reaction. The ionic bond will dissolve the reagents well.
B)Tetrabutylammoniumbromide must be a catalyst in this reaction, It is used to speed up the reaction, but it is regenerated so it is not consumed.
C)Tetrabutylammoniumbromide must be a reagent in this reaction.
NaOH, H20 CH2OH OH Bu4NBr mw 144 g/mol mp= 122 C mw = 169.99 g/mol d = 1.743 g/mL mw = 186.25 g/molExplanation / Answer
The reaction is Napthol + iodopropajne ----->propoxynaphthalene,
moles of propoxynaphthalene in 1 g =1/186.25=0.005369
hence moles of napthol= 0.005369 molecular weigght of napthol= 144, mass of napthol= 144*0.005369=0.773154 gms
mole of propyl iodide= 5 *0.005369 =0.0268 moles
Mass of propyl iodide= 0.0268*169.99=4.5024 gm
density - 1.743 g/ml
volume= 4.5024/1.743 ml=2.6 ml
C) Tetrabutyl ammonium chloride must be solvent. The ionic bond will dissolve the reagents.
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