recit_exp23_sn12
4
1. The reaction between NaBr and H 2 SO 4 is an acid-base reaction. Bromide ion, which results from dissociation of NaBr in water, reacts with sulfuric acid to capture one of its acidic protons. Equilibrium in acid-base reactions favors formation of the weaker acid (HBr) and the weaker base (bisulfate ion). Br + HO S OH O O HBr + O S OH O O The first inorganic salt that forms as a byproduct is therefore NaHSO 4. Remember, sodium is a expectator ion and serves as a counterion for any negative charges present in the medium. hydrogen sulfate, or bisulfate ion stronger acid weaker acid 2. Bromide can react with bisulfate ion in another acid base reaction to yield sulfate ion as another byproduct. Therefore Na 2 SO 4 is the second inorganic salt that forms. Br + HO S O O O HBr + O S O O O sulfate ion 3. The total number of inorganic salts that can be present in the system is therefore three, namely NaBr, NaHSO 4 , and Na 2 SO 4. Although they are soluble in water at room temperature, they may be less soluble in ice water and form crystalline precipitates. EXP. 23 – SIDE REACTIONS
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Transcript of recit_exp23_sn12
EXPERIMENT 21 SIDE REACTIONS
1. The reaction between NaBr and H2SO4 is an acid-base reaction. Bromide ion, whichresults from dissociation of NaBr in water, reacts with sulfuric acid to capture one of itsacidic protons. Equilibrium in acid-base reactions favors formation of the weaker acid(HBr) and the weaker base (bisulfate ion).
Br + HO S OH
O
O
HBr + O S OH
O
O
The first inorganic salt that forms as a byproduct is therefore NaHSO4. Remember,sodium is a expectator ion and serves as a counterion for any negative charges presentin the medium.
hydrogen sulfate,or bisulfate ionstronger acid weaker acid
2. Bromide can react with bisulfate ion in another acid base reaction to yield sulfate ionas another byproduct. Therefore Na2SO4 is the second inorganic salt that forms.
Br + HO S O
O
O
HBr + O S O
O
O
sulfate ion
3. The total number of inorganic salts that can be present in the system is thereforethree, namely NaBr, NaHSO4, and Na2SO4. Although they are soluble in water at roomtemperature, they may be less soluble in ice water and form crystalline precipitates.
EXP. 23 – SIDE REACTIONS
4. In part A (preparation of n-butyl bromide), another Sn2 side reaction leading to theformation of n-butyl ether is possible if the alcohol acts as the nucleophile.
OH2HO+
Sn2
OH
H2OO
H3O +
n-butyl ether
5. Finally, a less probable side reaction -an E2 reaction- can happen with water acting asa weak base. This reaction is less probable because E2 reactions normally require strongbases. However the reaction is still possible.
OH2
H
+ H2O
EXP. 23 – DISTILLATION SETUP MODIFICATION
LEAVE OUT THE HIGHLIGHTED YELLOW PORTION. DO NOT CAP THE CONDENSER, OR YOU’LL BE HEATING A CLOSED SYSTEM.
EXP. 23A FLOWCHART
Carefully mix the ingredients in ice and reflux for 60 min. In this step you use concentrated sulfuric acid (NOT 9M).
Perform part C during this reflux period.
Cool to room temperature. The organic product forms a separate layer, which should be the top layer.
Isolate the organic layer and wash it with 9M sulfuric acid (DO NOT use concentrated acid in this step)
Isolate the organic layer and wash it with water.
Isolate the organic layer and wash it with saturated sodium bicarbonate (aqueous solution).
H3O+ + HCO3- H2O + CO2 (g)
Isolate the organic layer and dry it with anhydrous sodium sulfate.
Record percent yield and save a sample for IR.
NOTE: The flowchart for part C is very similar, except that the first step requires no reflux, and no washing with 9M sulfuric acid is performed.
