FORMATION OF GRIGINARD’S REAGENT

FORMATION OF GRIGINARD’S REAGENT
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]

SN2-REACTIONS

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 solvent Structure 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

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-
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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-reactions SN2-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

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
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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
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:

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 = ?

P₂ = P₁T₂/T₁= (1atm*438K)/ (373K) = 1.17 atm

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: V_aN_a = V_bN_b. 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

e. 1N

DENSITY

DENSITY

Practice Problems and Answers

MGCCC, Perk Learning Lab

TLM

1. What is the density of benzene in grams per milliliter if 5.5L of benzene weigh 4.8kg?

2. The density of air is 1.3 g/L. How many liters of air weigh 2.0 lb?

3. Given a cylinder with diameter of 39 mm and height of 39 mm, and mass of 1.0 Kg, what is it’s density?

4. How many ounces does a 250ml sample of sulfuric acid weigh? (density = 1.30 g/ml)

5. What is the density of the Earth if 1.00 yd³ weighs 22.0 lb? Express your answer in lb/ft³.

6. The density of brass is 8.0 g/ml. A cube of brass 2.0 inches on an edge will weigh how many grams?

7. An unknown substance has a volume of 0.114 L and has a mass of 467g. What is its density in g/mL?

8. The density of Ethanoic Acetic Acid is 1.05 g/cm³. What is the mass in kilograms of a volume of 5L?

ANSWERS

1) D = mass ?D= 4.8Kg x 1000g x 1L = 0.87 g/mL

vol 5.5L 1Kg 1000mL

2) V= m ?g = 2 lb x 453.6 g = 907.2 g

D 1 lb

?V = 907.2g = 907.2g x 1L = 697.8 L

1.3g 1.3g

1L

3)Volume of a cylinder v= P r²h

r = ½ d, r = ½ (39mm) = 19.5 mm

v= P (19.5 mm)²(39mm) = 46589 mm³

? cm³ = 46589 mm³ x (1m)³ x (100cm)³ = 46.5 cm³

(1000mm)³ (1m)³

D = 1 Kg x 1000g = 21.5 g/cm³

46.5 cm³ 1Kg

4) m= vD

?oz = 250ml x 1.3g x 1 oz = 11.5 oz

1 mL 28.35 g

5) ? lb = 22 lb x (1yd)³ = 0.815 lb/ft³

ft³ 1 yd³ (3ft)³

6) Volume of a cube v = s³

? v = (2in)³ x (2.54 cm)³ = 131cm³

(1 in)³

? v = 131 cm³ x 1 mL = 131 ml

1 cm³

m= vD = 131 mL x 8g = 1048g

1 mL

7) D = m = 467g x 1 L = 0.467 g/mL

v 0.114L 1000mL

8) ? v = 5 L x 1000mL x 1 cm³ = 5000cm³

1L 1mL

?m = vD= 5000cm³ x 1.05g x 1 Kg = 5.25 Kg

1 cm³ 1000g

Isotope Abundance and Atomic Weight

Isotope Abundance and Atomic Weight

Ex: The natural abundance for boron isotopes is: 19.9% ¹⁰B (10.013 amu) and 80.1% ¹¹B (11.009amu).

Calculate the atomic weight of boron.

Aw = [(%abundance of isotope) (mass of isotope)] + [(%abundance of isotope) (mass of isotope)] + [….]

Aw = the atomic weight found in the periodic table

% abundance of EACH isotope is converted to decimal ex: 19.9% ÷ 100% = 0.199

The equation continues on[….] based on the number of isotopes in the problem.

Aw = [(0.199)(10.013)] + [(0.801)(11.009)]

Aw = [1.992587] + [8.818209]

Aw = 10.810796 so, the atomic weight of B = 10.811

If you look in the periodic table you will be able to check that our answer is correct!

The atomic mass of lithium is 6.94, the naturally occurring isotopes are ⁶Li = 6.015121 amu, and ⁷Li = 7.016003 amu.

Determine the percent abundance of each isotope.

Aw = [(%abundance of isotope) (mass of isotope)] + [(%abundance of isotope) (mass of isotope)] + [….]

6.94 =[(% ⁶Li)(6.015121)] + [(%⁷Li)(7.016003)]

Since I don’t know what the percentage are, I will have to use variables.

100% of Lithium is determined by these two naturally occurring isotopes.

We will let ⁶Li = x and ⁷ Li = 1-x; we use 1 – x instead of 100 – x because the small number is easier to work with.

(in other words we reduced 100% to decimal form 1.00)

Now let’s plug our variables in: 6.94 =[(% ⁶Li)(6.015121)] + [(%⁷Li)(7.016003)]

6.94 = [(x)(6.015121)] +[(1-x)(7.016003)]

6.94 = 6.015121x + 7.016003 – 7.016003x

Combine like terms: 6.94 -7.016003 = (6.015121x - 7.016003x)

-0.076003 = -1.000882 x

Solve for x: -0.076003 = x

-1.000882

X = 0.075936, therefore ⁶Li = 0.075936 x 100% = 7.59%

1-X = 1 -0.075936 = 0.924064, therefore ⁷Li = 0.924064 x 100% = 92.41%

PROBLEMS (SOLUTIONS ON BACK)

1. Calculate the weight of silicon using the following data for the percent natural abundance and mass of each isotope:

92.23% ²⁸Si (27.9769 amu); 4.67% ²⁹ Si (28.9765); 3.10% ³⁰Si (29.9738 amu).

2. Thallium has two stable isotopes, ²⁰³Tl and ²⁰⁵Tl. Knowing that the atomic weight of thallium is 204.4,

which isotope is the more abundant of the two?

3. Verify that the atomic mass of magnesium is 24.31, given the following:

24Mg= 23.985042amu, 78.99%

25Mg= 24.985837 amu, 10.00%

26Mg= 25.982593, 11.01%

4. Copper exists as two isotopes: ⁶³Cu (62.9298 amu) and ⁶⁵Cu (64.9278 amu).

What are the percent abundances of the isotopes?

SOLUTIONS

Aw = [(%abundance of isotope) (mass of isotope)] + [(%abundance of isotope) (mass of isotope)] + [….]

1. Aw= [(0.9223)(27.9769)] + [(0.0467)(28.97650)] + [(0.031) (29.9738)]

Aw = 25.803 + 1.353 + 0.929

Aw for Silicon = 28.085

Hint: double check the periodic table to see if the answer is the same or close

2. ²⁰³Tl % = x

²⁰⁵Tl % = 1 – x

204.4 = [(x)(203)] + [(1- x)(205)]

204.4 = 203x + 205 – 205x

-0.6 = -2x

-0.6 = x

-2

X= 0.3 ²⁰³Tl = 0.3 x 100% = 30 %

²⁰⁵Tl = 1 –x = 1 – 0.3 = 0.7 x 100% = 70%

Therefore ²⁰⁵Tl = 70% is more abundant.

3. Aw= [(0.7899)(23.985042)] + [(0.1)(24.985837)] + [(0.1101)(25.982593)]

Aw = 18.946 + 2.499 + 2.861

Aw for Mg = 24.31, this checks with the Aw that was given in problem #3.

4. Since the overall atomic weight for copper is not given in the problem, you must look it up in the

periodic table to work this solution. Aw for Cu = 63.546

⁶³Cu % = x

⁶⁵Cu % = 1 – x

63.546 = [(x)(62.9298)] + [(1-x)(64.9278)]

63.546 = 62.9298x + 64.9278 – 64.9278x

-1.3818 = -1.998x

-1.3818 = x

-1.998

X = 0.6916 ⁶³Cu = 0.6916 x 100% = 69.16%

⁶⁵Cu = 1 – x = 1 – 0.6916 = 0.3084 x 100% = 30.84%