Archive for October 17th, 2008
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What kind of reaction happens between 1-bromobutane and sodium hydroxide?
Is the reaction SN1 or SN2? Can it be used in the same way that you use anhydrous sodium sulfate to dry something? If not, how would you explain why it can’t?
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The Reaction:
CH3CH2CH2CH2—Br + NaOH —–> CH3CH2CH2CH2—OH + NaBr
It can be understood by considering the preparation of 1-bromobutane :
Preparation of 1-Bromobutane: An SN2 Reaction
A common technique for converting a primary alcohol to an alkyl halide involves treating the alcohol with a hydrogen halide H-X (H-X = H-Cl, H-Br, or H-I).. This reaction is reversible, and displacing the equilibrium to the right normally involves using a large excess of the acid, a strategy in accord with the LeChatlier principle.
H—X + R—OH—–> R—X + H2O
where X = Cl, Br, I
1-Bromobutane may be prepared by heating 1-butanol with hydrobromic acid, H-Br, in the presence of sulfuric acid . The mechanism for this reaction has been shown to occur in two steps . The alcohol is protonated in the first H2SO4
CH3CH2CH2CH2—OH + HBr—-> CH3CH2CH2CH2—Br + H2O
step via a Lewis acid-base reaction to give the oxonium ion 1. This oxonium ion then undergoes displacement by the bromide ion to form 1-bromobutane and water. This process is an SN2 reaction in which water is the leaving group and bromide ion is the nucleophile. The sulfuric acid serves two important purposes: (1) It is a dehydrating agent that reduces the activity of water and shifts the position of equilibrium to the right, and (2) it provides an added source of hydrogen ions to increase the concentration of oxonium ion 1. The use of concentrated hydrobromic acid also helps to establish a favorable equilibrium.
No reaction occurs between 1-butanol and NaBr in the absence of strong acid, because leaving groups in nucleophilic substitution reactions must be weakly basic, as is water in Equation . If the reaction of 1-butanol and NaBr were to occur ), the leaving group would necessarily be the strongly basic hydroxide ion; thus the forward reaction ..does not occur. On the other hand, the reverse reaction between 1-bromobutane and hydroxide proceeds readily, since the leaving group in this reaction is the weakly basic bromide ion.
CH3CH2CH2CH2—OH + NaBr —–> CH3CH2CH2CH2—Br + NaOH
October 17th, 2008
Chemistry Tutor Site - Chemistry Homework Help
What is the effect of increasing the concentration of water in an equilibrium reaction? In an equilibrium reaction, if the products are aq ions and water, apparently when water is added the reaction will favour the left to decrease the conc of water. But since the conc of ions is decreasing, won’t the equation favour the right to increase the conc of ions?
Can it be answered qualitatively, or does it depend on the individual concentrations? Does it have something to do with the fact that the system is closed?
Chemistry Tutor Site - Chemistry Homework Answer
According to Le Chatelier’s Law, if a system at equilibrium encounters any stress such as a change in concentration, temperature, or total pressure, it will adjust itself to counteract that change. So in your equilibrium system, there is water and aqueous ions in the products. If you add water to the system, it will shift to the left to produce more reactants and use up the extra water.
This is because any time you add a substance to an equilibrium system, it shifts accordingly to use up the added substance and maintain the equilibrium. The system will shift to the right to produce more ions in addition to water only when ions are removed from the system. Then it will shift toward the products to replace the lost ions and maintain the equilibrium
October 17th, 2008
Chemistry Tutor Site - Chemistry Homework Help
Choose all of the condition the apply to what changes in conditions would favor the products in the following equilibrium?
PCl5(g) <–> Cl2(g) + PCl3(g) ΔH = 88 kJ
(choose all of the things below that apply to the question)
- Increase the phosphorus pentachloride concentration.
- Increase the pressure.
- Increase the temperature.
- Increase either of the product concentrations.
- Decrease the temperature.
- Decrease either of the product concentrations.
- Decrease the pressure.
- Decrease the phosphorus pentachloride concentration.
Chemistry Tutor Site - Chemistry Homework Answer
Read up on Le Chatelier’s principle*
PCl5(g) <–> Cl2(g) + PCl3(g) ΔH = 88 kJ
Which will favor products? AKA: Shift the equilibrium to the right?
Inc. PCl5 - shift right to reduce PCl5 concentration
inc pressure - shift left, fewer molecules, lower P
Inc T - shift right, to reduce heat (heat is a “reactant” since it is endothermic.
inc products - shift left to use up product
dec T - shift left to make more heat
dec product - shift right to make more product
dec P - shift right to make more molecuels and raise P
dec PC5 - shift left to make more PCl5
Le Chatleier’s principle says for a system in equilibrium, if a stress is applied, then the equilibrium will shift in a direction that will relieve the stress.
The “stress” can be adding or removing a reactant or product, Increasing or decreasing the volume of a container of gases, thereby changing the pressure of the gases, or by adding or removing heat.
Chemists will say that the equilibrium “shifts to the right”, or “shifts to the left”. If it shifts to the right more product is produced and more of the reactants are used up. If the reaction shifts to the left, more reactants are made and some of the products are used up.
2A(g) + B(g) <==> C(g) + heat
Increasing the amounts of A or B will shift the equilibrium to the right to use up some A or B and get the concentration back closer to what it was originally. Removing C will shift the equilibrium to the right in order to make more C and bring the concentration of C back closer to what it was.
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Removing A or B will shift the equilibrium to the left to make more A or B and use up C. Adding C will shift the equilibrium to the left to use up some of the added C.
Reducing the volume of the container will increase the pressure and shift the equilibrium in a direction that results in fewer molecules and lowered pressure. That would be to the right. There is one molecule of gas on the right and three on the left.
If we heat the system up, that is adding heat, which will shift the equilibrium to the left to use up some of the added heat. It will also use up some C and make more A and B
October 17th, 2008
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