| ||
| GRIGINARD’S REAGENT | ||
| In the preparation of a griginard’s reagent great care must be taken that the reactants are absolutely dry and pure. The other experimental details are given below: Magnesium: magnesium ribbon is washed with ether and dried in desicater. Ether: it is distilled over sodium metal and P2O5. Ethyl bromide: it is purified and then distilled over P2O5. Take dry ether in a round bottom flask and add pieces of Mg ribbon in it. Fit it with water condenser and add ethyl bromide gradually through the condenser. Heat a little so that magnesium dissolves. Cool if necessary.A clear solution of griginard’s reagent – ethyl magnesium bromide, in ether appears. This ethereal solution is used as griginard’s reagent. | ||
| C2H5--Br + Mg à C2H5--Mg--Br | ||
| BIMOLECULAR REACTION | ||
| A chemical reaction in which two molecules react together in rate determining step is called bimolecular reaction. A bimolecular reaction may be SN2 or E2. Rate of reaction of A bimolecular reaction is expressed as: | ||
| Rate = K[R--X][Nu] | ||
| UNIMOLECULAR REACTION | ||
| A chemical reaction in which only one molecule takes part in rate determining step.is called a unimolecular reaction. A unimolecular reaction may be SN1 or E1. Rate of reaction of a unimolecular reaction is expressed as: | ||
| Rate = K[R--X] | ||
Friday, July 24, 2009
FORMATION OF GRIGINARD’S REAGENT
SN2-REACTIONS
| ||
| SN2-REACTIONS | ||
| In SN2-REACTIONS, the formation of the carbon-nucleophile bond [C--Nu] and the cleavage of the carbon- halogen bond [C--X] occur simultaneously. These are nucleophilic substitution reactions which occur in one step . | ||
| MECHANISM | ||
| The reaction proceeds in one step. The attack of nucleophile and release of halide group take place at the same time.i.e | ||
| Nu- + R-X ® [Nu…R…X] ® R--Nu + X- Transition state | ||
| Reactions of secondary alkyl halide may be SN1 or SN2 depending upon the nature of solvent. | ||
| RATE OF SN1-REACTIONS | ||
| The rate of an SN2 reactions depend on the concentrations of nucleophile and the substrate. i.e Rate of reaction = k[substrate][Nu] Since both substrate and nucleophile are present in rate expression therefore ifst is a bimolecular reaction. It may be defined as "a bimolecular nucleophilic substitution reaction which occur in one step . | ||
| FACTORS AFFECTING THE NATURE Of SN REACTIONS | ||
The mechanism of SN-reactions depend upon : The structure of alkyl group Nature of the solventStructure Of Alkyl Group Reactions of primary alkyl halides take place in SN2 mechanism. Reactions of tertiary alkyl halides take place in SN1 mechanism. Reactions of secondary alkyl halides take place both in SN1 and SN2 mechanism but it depends upon the nature of solvent. The order of stability of carbonium ion is as under: R3C+[tertiary carbonium ion]>R2C+H[secondary carbonium ion]>RC+H2[primary carbonium ion] Nature Of Solvent:A polar solvent favours SN1 mechanism due to capability of ionization where as a non- polar solvent favours SN2 mechanism. | ||
NUCLEOPHILIC SUBSTITUTION REACTIONS
| ||
| DEFINITION | ||
| " A reaction in which a nucleophile displaces another nucleophile and takes its position is called a nucleophilic substitution reaction. " | ||
| Nucleophilic substitution reaction are represented by the symbol "SN". | ||
| EXPLANATION | ||
| In alkyl halides, the C-X bond is polar because C-atom is attached to a highly electro-negative halogen atom.Therefore in alkyl halides carbon atom is electrophilic and halogen has a nucleophilic character. | ||
 | ||
| For example in iso propyl bromide , C-atom has partial positive charge and Br-atom has partial negative charge. The C-atom has capability to accept a pair of electrons to form a new bond. In doing so the C-Br bond breakes and Br- leaves the molecules. The net result is that one nucleophile displaces another nucleophile. Such displacement reactions are called nucleophilic substitution reactions. | ||
 | ||
| Following examples illustrate the nature of "SN" reactions: OH- + CH3+-I- à OH – CH3 + I- In this example hydoxyl ion is an attacking nucleophile while iodide-ion is called leaving group. CH3-CH2-I + OH- àCH3-CH2-OH +I- CH3-I + -OCH3 àCH3-O-CH3 +I- CH3-Br + CH3COO- àCH3COOCH3 + Br- | ||
| For latest information , free computer courses and high impact notes visit : www.citycollegiate.com | ||
| GENERAL EXPRESSION | ||
| A nucleophilic substitution reaction (SN). Nu- + Rd+-Xd- à Nu-R + X- In general it can be written as: Alkyl halide + nucleophile à new compound + halide ion | ||
| MECHANISM OF SN REACTIONS | ||
There are two classes of SN-reactions: SN1-reactionsSN2-reactions | ||
| SN1-REACTIONS | ||
| Nucleophilic substitution reactions which take place in two steps are referred to as "SN1-reactions. | ||
| MECHANISM | ||
| SN1 mechanism occurs in two steps.In the first step, the substrate ionizes slowly and reversibly to produce carbonium ion. R-X àR+ + X- In the second step, nucleophile reacts with carbonium ion to produce a new compound. R+ + Nu- à R-Nu | ||
| RATE OF SN1-REACTIONS | ||
| The rate of an SN1 reaction depends upon the concentration of substrate (R-X) only and not on the nucleophile.i.e. Rate = K[R-X] This means that it is a unimolecular reaction. Therefore, SN1 is a unimolecular nucleophilic substitution reaction. | ||
PREPARATION OF ALKYL HALIDES
| ||
| Mono haloalkanes or Alkyl halides can be prepared by a number of methods.Some important synthetic methods are as follows: | ||
| METHOD No. 1 | ||
| BY USING ALCOHOL AS A MAIN REAGENT | ||
| A number of reagents such as halogen acids, PCl5, PCl3, SOCl2 etc. react with aliphatic alcohols to produce corresponding alkyl halides. | ||
| FROM HALOGEN ACID | ||
| Alcohols react with halogen acids (HCl ,HBr or HI) to produce alkyl halides. | ||
| R-OH + HX ® R-X + H2O CH3-OH + HCl® CH3-Cl + H2O (in the presence of ZnCl2/H2SO4) C2H5OH + HI® C2H5I + H2O C2H5OH + HBr® C2H5Br + H2O CH3-CH2-CH2-OH + HCl ® CH3-CH2-CH2-Cl + H2O | ||
| FROM PHOSPHORUS PENTA CHLORIDE (PCl5) | ||
| When alcohols are treated with phosphorus penta chloride, corresponding alkyl chlorides are obtained. | ||
| CH3-CH2-OH + PCl5 ® CH3-CH2-Cl + POCl3 + HCl | ||
| For latest information , free computer courses and high impact notes visit : www.citycollegiate.com | ||
| FROM THIONYL CHLORIDE(SOCl2) | ||
| CH3-CH2-OH +SOCl2 ® CH3-CH2-Cl + SO2 + HCl | ||
| Above reaction is carried out in the presence of Pyridine. Pyridine absorbs HCl as it is formed. | ||
| FROM PHOSPHORUS TRI BROMIDE (PBr3) | ||
| 3C2H5OH + PBr3 ® 3C2H5Br + H3PO3 | ||
| FROM PHOSPHORUS TRI IODIDE (PI3) | ||
| 3CH3OH + PI3 ® 3CH3I + H3PO3 | ||
| METHOD No. 2 | | |
| BY THE HALOGENATION OF ALKANES | ||
| Alkanes react with chlorine and bromine in presence of sunlight and undergo photo chemical series of substitution reactions to yield alkyl halides. | ||
| CH4 + (Cl2) ® CH3Cl + CH2Cl2 + CHCl3 + CCl4 | ||
| REMEMBER: THIS IS NOT AN APPROPRIATE METHOD FOR THE LABORATORY PREPARATION OF ALKYL HALIDES BECAUSE THE RESULTANT MIXTURE CONTAINS POLYHALOGEN DERIVATIVES WHICH ARE DIFFICULT TO SEPARATE. | ||
| METHOD No. 3 | ||
| FROM ALKENE | ||
| we know that alkenes are unsaturated hydrocarbons.Therefore when alkenes are treated with halogen acid, addition reaction takes place and alkyl halides are formed. Additionof HX takes place according to MARKONIKOV'S PRINCIPLE. | ||
| CH2 = CH2 + HCl ® CH3 – CH2 – Cl | ||
| CH3-CH=CH2 + HBr ® CH3-CHBr-CH3 | ||
| The order of reactivity of halogen acid is as follows: | ||
| HI > HBr > HCl | ||
CHEMISTRY OF ALKYL HALIDES
| ||
| ALKYL HALIDES | ||
| Organic compound containing halogen atom as a functional group are called alkyl halides. Alkyl halides are very reactive organic compounds . The general formula of alkyl halide is CnH2n+1-- X, where X= Cl, Br, I. Example: CH3-Cl (Methyl Chloride),CH3-Br (Methyl bromide) ,C2H5-Br (Ethyl Bromide) ,C3H7-Br (Propyl bromide) etc. etc. | ||
| CLASSIFICATION OF ALKYL HALIDES | ||
| Alkyl halides are classified into three classes: PRIMARY ALKYL HALIDE: Alkyl halides containing primary C-atom are called primary alkyl halides. | ||
| For example: CH3-CH2-I (Ethyl iodide) ,  | ||
| SECONDARY ALKYL HALIDE: Alkyl halides containing secondary c-atom are called secondary alkyl halides. For example: | ||
 | ||
| TERTIARY ALKYL HALIDE: Alkyl halides containing tertiary c-atom are called tertiary alkyl halide. For example: | ||
 | ||
Thursday, April 16, 2009
GAS LAWS WORKSHEET
GAS LAWS WORKSHEET
1. A gas occupies 3.5L at 2.5 mm Hg pressure. What is the volume at 10 mm Hg at the same temperature?
2. A constant volume of oxygen is heated from 100°C to 185°C. The initial pressure is 4.1 atm. What is the final pressure?
3. A sample of 25L of NH3 gas at 10°C is heated at constant pressure until it fills a volume of 50L. What is the new temperature in °C?
4. A certain quantity of argon gas is under 16 torr pressure at 253K in a 12L vessel. How many moles of argon are present?
5. An unknown gas weighs 34g and occupies 6.7L at 2 atm and 245K. What is its molecular weight?
6. An ideal gas occupies 400ml at 270 mm Hg and 65°C. If the pressure is changed to 1.4 atm and the temperature is increased to 100°C, what is the new volume?
7. What is the volume of 23g of neon gas at 1°C and a pressure of 2 atm?
8. If 11 moles of HCl gas occupies 15L at 300°C, what is the pressure in torr?
9. The pressure is 6.5 atm, 2.3 mole of Br2 gas occupies 9.3 L . What is the temperature in °C?
10. A 600mL balloon is filled with helium at 700mm Hg barometric pressure. The balloon is released and climbs to an altitude where the barometric pressure is 400mm Hg. What will the volume of the balloon be if, during the ascent, the temperature drops from 24 to 5°C?
11. An unknown gas has a volume of 200L at 5 atm and -140°C. What is its volume at STP?
12. In an autoclave, a constant amount of steam is generated at a constant volume. Under 1.00 atm pressure the steam temperature is 100°C. What pressure setting should be used to obtain a 165°C steam temperature for the sterilization of surgical instruments?
ANSWERS
1. V1= 3.5L, P1= 2.5mm Hg, V2 = ?, P2= 10 mm Hg
V2 = P1V1/P2 = (2.5mm Hg * 3.5L)/ 10mm Hg = 0.875 L
2. T1 = 100°C + 273K = 373 K and T2 = 185°C + 273K = 458K
P1= 4.1 atm, P2 = ?
P2= P1T2/T1 = (4.1atm*458K)/ 373K = 5 atm
3. T1 = 10°C + 273K = 283K
V1 =25L, V2= 50L, T2 = ?
T2= V2T1/V1 = (50L*283K)/ 25L = 566K, 566K – 273K = 293°C
4. P=16torr * (1atm/760torr)= 0.02 atm
T= 253K, V= 12L, n=?
n=PV/RT= (0.02atm*12L)/ [(0.0821 L*atm/mol*K) (253K)] = 0.012 mol
5. m= 34g, V= 6.7L, P = 2 atm, T= 245 K, MW=?
MW= mRT/PV= [(34g)(0.0821L*atm/mol*K)(245K)]/(2atm*6.7L)= 51g/mol
6. P1= 270mm Hg(1atm/760 mm Hg) = 0.35 atm T2 = 100°C + 273K = 373K T1 = 65°C + 273K = 338K
V1= 400mL, P2 = 1.4 atm, V2 = ?
V2 = P1V1T2/T1P2 = (0.35 atm*400mL*373K)/ (338K * 1.4atm)= 110 mL
7. T= 1°C + 273K = 274K
m= 23g of Neon, P= 2 atm, V=?
***look up the molecular weight of Neon on the periodic table
MW= 20.2g/mol
V= mRT/MWP= [(23g)(0.0821 L*atm/mol*K)(274K)]/ [(20.2g/1mol)(2atm)]= 12.8L
8. T= 300°C + 273K = 573 K
n=11 mol, V= 15L, P = ? torr
P= nRT/V = [(11mol)(0.0821 L*atm/mol*K)(573K)]/(15L)= 34 atm
34 atm * (760torr/1atm) = 26,218 torr
9. P= 6.5 atm, n = 2.3 mol, V= 9.3L, T = ? °C
T = PV/nR = (6.5atm*9.3L)/ [(0.0821 L*atm/mol*K)(2.3 mol)]= 320 K
320K –273K = 47°C
10. T1= 24°C + 273K = 297 K T2 = 5°C + 273K= 278K
V1= 600mL, P1=700mm Hg, P2= 400mm Hg, V2 = ?
V2= P1V1T2/T1P2 = (700mm Hg* 600mL * 278K) / (297K * 400 mm Hg) = 982 mL
11. T1 = -140°C + 273K = 133K
V1 = 200 L, P1= 5 atm, P2= 1 atm, T2=273K, V2 = ?
V2= P1V1T2/T1P2= (5atm*200L*273K)/ (133K*1atm) = 2,052 L
12. V1= V2, T1 = 100°C + 273 = 373 K T2 = 165°C + 273K = 438K
P1 = 1 atm, P2 = ?
Molarity, Molality and Normailty
Molarity, Molality and Normailty
Molarity = Moles of solute/Liters of Solution (abbreviation = M)
Molality = Moles of solute/Kg of Solvent (abbreviation = m)
Normality = equivalent solute/Liters of Solution (abbreviation = N)
1. How many moles of ethyl alcohol, C2H5OH, are present in 65ml of a 1.5M solution?
2. How many liters of a 6.0M solution of acetic acid CH3COOH, contain 0.0030 mol acetic acid?
3. You want 85 g of KOH. How many of 3.0 m solution of KOH will provide it?
4. If you dissolve 0.70 moles of HCl in enough water to prepare 250ml of solution, what is the molarity of the solution you have prepared?
5. A solution is prepared by adding 2.0L of 6.0 M HCl to 500 ml of a 9.0 M HCl solution. What is the molarity of the new solution? (Remember, the volumes are additive)
6. What is the molality of a 0.10M solution of ethylene glycol C2H6O2? The solutions density is 0.90 g/ml.
7. Convert the following Molarities to Normalities.
a. 2.5M HCl = _________ N
b. 1.4M H2SO4 = _________ N
c. 2.0M PbCl3 = _________N
d. 1.0M NaOH = ________N
e. 0.5M Ca(OH)2 = __________N
8. A commonly purchased disinfectant is a 3.0% (by mass) solution of hydrogen peroxide (H2O2) in water. Assuming the density of the solution is 1.0g/cm3 , calculate the molarity, molality and mole fraction of H2O2.
9. A solution is made by dissolving 25g of NaCl in enough water to make 1.0L of solution. Assume the density of the solution is 1.0 g/cm3. Calculate the molarity, normality and molality of the solution.
10. When two solutions react and you have their concentrations expressed in normality, you can write: VaNa = VbNb. How many liters of a 0.30 N solution of KMnO4 can react with 5.0 L of a 0.10 N solution of H2C2O4 ?
11. What is the concentration (M) of NaOH if 75.6 ml of NaOH is required to react with 270.0ml of a 0.3M solution of H2SO4?
ANSWERS
1. 0.098 mol alcohol 8. 0.88M, 0.91m, mole fraction 1.6 x 10-2
2. 5.0 x 10-4 L solution 9. 0.43M, 0.43N, 0.44m
3. 5.9 x 10 2 g solution 10. 1.7L
4. 2.8M 11. 2.1M
5. 6.6M
6. 0.11m
7. a. 2.5N
b. 2.8N
c. 6.0N
d. 1.0 N
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