reactions i need to know-c06j-part 1

8/3/2019 Reactions I Need to Know-C06J-Part 1

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Reactions I need to know

ALKANES

Preparation

a) Fractional distillation of crude petroleum

Crude petroleum consists mainly of hydrocarbons.

Mixtures of hydrocarbons can be separated based on theirboiling points.

Process called fractional distillation.

Boiling range of

fraction (C)

No. of carbon atoms Use

Below 20 C1-C4 Natural gas

20 60 C5-C6 Solvents

60 100 C6-C7 Solvents

40 200 C5-C10 Gasoline

175 325 C12-C18 Kerosene, jet fuel

250 400 C12 and higher Gas oil, diesel

Nonvolatile C20 and higher Mineral oil, wax


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Distillation tower at an oil refinery Fractional distillation apparatus in the laboratory

b) Reduction of alkenes

Reduction meaning addition of hydrogen atoms.

Alkenes possess C=C double bond.

Upon reduction the C=Cbond is converted to a C-C single bond.

The reagent used to carry out this reaction is hydrogen.

Reaction requires a catalyst e.g. Pd/C, Ni, Pt

CH3 CH2 CH CH2

H2, Pd/ C

CH3 CH2 CH2 CH3

CH3 CH2 C

CH3

CH CH3 CH3 CH2 CH

CH3

CH2 CHH2, Pt


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c) Reduction of carbonyl compounds.

Carbonyl compounds are compounds with a C=O double bond.

They are aldehydes or ketones.

By completely removing the oxygen atom (reduction) an alkane can be formed.

H2N-NH2 is hydrazine

The reaction is known as the Wolff-Kishner reduction

Example:

H3CC

R

O

R= H or Alkyl or phenyl

H2N NH2+NaOH

CH3 CH2 R

heat

H3C

C

CH2

O

H2N NH2+NaOH

CH3 CH2 CH2CH3

CH3 CH2 C

O

H

H2N NH2+NaOH

heat

heat

CH3 CH2 CH3

CH3

ketones

aldehydes


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If a ketone is treated with hydrochloric containing amalgamated zinc the C=O is

completely removed.

Amalgamated zinc is a zinc-mercury mixture.

Reaction is known as the Clemmensen reduction.

Example:

d) From alkyl halides.

Alkyl halides are hydrocarbons containing at least one halogen atom.

Reaction with Grignard reagents

Grignard reagents are : RMgX, X= halogen, Chlorine, Bromine, Iodine

H3CC

CH3

OZn(Hg)

CH3 CH2 CH3

CH3 CH2 C

O

CH2

CH3 CH2 CH2

HClreflux

CH3

Zn(Hg)

HClreflux

CH3

R Cl + + MgClBrR'MgBr R R'

H3C ClH+ H--

CH3MgI+


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Example:

Note: there is an extension of the chain at the point of

reaction.

CH3 CH2 Br CH3 CH2 CH3+

+ MgBr2

CH 3MgBr

CH3 CH Cl CH3 CH CH2+

+ MgCl2

CH3CH2MgCl

CH2

CH3

CH3

CH2

CH3


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ALKANES

Reactions

a) Combustion Alkanes will burn in oxygen to form carbon dioxide and water.

This reaction is used to determine molecular formulae of alkanes.

Combustion Analysis

Using Avogadros Law:

Equal volumes of all gases, under the same conditions of temperature and

pressure, contain equal numbers ofmolecules

And Gay-Lussacs Law:

When gases react, they do so in volumes that bear a simple ratio to one another

and to the volume of the product.

Example:

Consider combustion of ethane

Ratio ofreacting volumes: C2H6 : O2 is 1 : 3

Ratio of products volumes: CO2

: H2O is 2 : 3

If the relative ratios of volumes ofreactants and products can be determined then the

MolecularFormula can be found.

CxHy (g) + (x + y/4) O2(g) x CO2(g) + y/2 H2O (g)

C2H6(g) + 7/2 O2(g) 2CO2(g) + 3 H2O(g)


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Method:

Fixed volume of hydrocarbon mixed with a fixed volume of oxygen and the

mixture ignited.

The change in volume is determined. Note the new volume is due to steam

(H2O

),CO

2 and excessO

2. The mixture cooled to room temperature (H2O).

The mixture passed through sodium hydroxide solution (absorbs CO2).

Remaining volume due to excess O2.

E.g. 30 mL of a hydrocarbon is mixed with 180 mL ofO2 and the mixture ignited. The resulting

was cooled to room temperature and the new volume determined to be 120 mL. After passing

the mixture through sodium hydroxide the volume was reduced by 75%. Determine the

molecular formula of this compound.

Hint: Determine the volumes of the different reactant and products, and apply Laws.

Solution:

120mL due to excess O2 and CO2

After treating with NaOH volume reduced to 30 mL.

Excess O2 is 120 90 = 30 mL therefore O2used = 180 30 = 150 mL

CO2 produced is 90 mL

Ratio- 30 CxHy : 150 O2 : 90 CO2

1 CxHy : 5O2 : 3 CO2

For liquid and solid hydrocarbons the combustion analysis tends not to be as accurate as

you are unable to calculate the ratio with respect to the hydrocarbon.

Methods are in place to determine the percentage composition of carbon by mass.

CxHy (g) + (x + y/4) O2(g) x CO2(g) + y/2 H2O (g)


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Once the percentage composition is known then the empirical formula can be

determined.

E.g. % carbon 84.26 : % hydrogen 15.74

Divide by atomic mass:

84.26/12=7.02 ; 15.74/1=15.74

Divide by the smallest value:

7.02/7.02=1 ; 15.74/7.02=2.24 (CH2.24)

Multiply by an integer to give M.F. closest an actual value. E.g. x 8 gives C8H17.92 (C8H18)

or x 9 gives C9H20.14 (C9H20)

b) Free radical halogenations

o Alkanes react with the first three members of the halogen family:

fluorine, chlorine and bromine.

o Multiproducts are obtained from this reaction.

Example: reaction ofmethane with chlorine

The reaction is a substitutionreaction.

One group replaces another. This continues until the alkane is fully substituted (CCl4).

H

C

H

H H + Cl2

Cl

C

Cl

Cl Cl

Cl

C

H

Cl Cl

H

C

H

H Cl

Cl

C

H

H Cl+

+

+ HCl

heat or light


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Mechanism free radical

Reaction is promoted by light or heat.

Reactivity of halogens increases in the order Br < Cl < F.

A

s withm

ost fr

eer

adicalm

echanism

s ther

e ar

e thr

eem

ain steps: Initiation

Propagation

Termination

Initiation

Propagation

Cl Cl Cl2

Cl2 2 Cl

heat or light

heat or light

Cl +

H

C H

H

H

H

C H

H

HCl +

Step 2

CH4 + Cl HCl + CH3

Step 3

H

CH

H

+ Cl Cl

H

C Cl

H

H

Cl2 + CH3 CH3Cl + Cl


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Termination (any two radicals come together)

Multi halogenated products are formed in a similarmanner.

Halogenation of higher alkanes (let us consider ethane and propane)

Halogenation of higher alkanes is also a free radical reaction.

Mono halogenation occurs but

Like methane halogenation (especially chlorination) can lead to multiple products.

H

CH

H

+ Cl

H

C Cl

H

H

Cl + CH3 CH3Cl

H

CH

H

H

C H

H

+ C C

H

H

H

H

H

H

2 CH3 C2H6

Cl +

Cl

C H

H

H

Cl

C H

H

HCl +

Step 2

CH3Cl + Cl HCl + ClCH2

Step 3

Cl

CH

H

+ Cl Cl

Cl

C Cl

H

H

Cl2 + ClCH2 CH2Cl2 + Cl

Cl+


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Mechanism free radical

Initiation: (as before)

Propagation:

Steps 2 and 3 are repeated giving multiple products such as:

Termination (any two radicals come together)

Cl Cl Cl2

Cl 2 2 Cl

heat or light

heat or light

Cl + HCl +

Step 2

C2H6 + Cl HCl + C2H5

Step 3 + Cl Cl

Cl2 + CH3CH2 CH3CH2Cl + Cl

H : CH2CH3 CH2CH3

CH3CH2 CH3CH2Cl + Cl

CH3CH2Cl

CCl 3CCl 3

CHCl2CCl 3CH3CCl 3

CH2ClCCl 3CH3CHCl 2

+ Cl

Cl + CH3CH2 CH3CH2Cl

2 CH3CH2C4H10

CH3CH2

1.

2.

3. 2 Cl Cl2

CH3CH2 Cl


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Halogenation of alkanes with more than 2 carbon atoms gives a mixture ofmono

halogenated alkyl halides.

Hydrogen atoms of higher alkanes are not identical. (1, 2, 3)

Order ofreactivity -1< 2< 3

T

hem

ajor

pr

odu

cts canb

e pr

edicted.

Initiation same as above

Propagation: Looking at propane.

A toss up between abstracting a 2 hydrogen and a 1 hydrogen.

The 2 will be abstracted first.

WHY? The radical formed is more stable.

Stability ofradicals - 3>2>1

Chain propagation Step

Expect Two Major Products

C C C

H

H

H

H

H

H

H

H + X C C C

H

H

H

H

H

H

H

C C C

H

H

H

H

H

H

H

H + X C C C

H

H

H

H

H

H

H

2o

radical

1o

radical

X = halogen

+ HX

+ HX

C C C

H

H

H

H

H

H

H

H + X2

C C C

H

H

H

H

H

H

H

C C C

H

H

H

H

H

H

H

2o

alkyl halide (major)

1o

alkyl halide

X = halogen

heat

X

X


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Due to the high reactivity of chlorine.

There is not much discrimination between 1, 2 and 3 hydrogen.

Different mono chloroalkanes have similar yields.

Bromine is much more selective.

Example:

Exercise:

Predict the products of:

Answer:

+ Br2

H3C C CH3

CH3

Br

H3C C CH2

H

1o

alkyl halide -trace

3o

alkyl haide >99%

heat

CH3

Br

H3C C

CH3

H

CH3

+ Br2 heatH3C C

CH3

H

CH2 CH3

+ Br2

H3C C CH2

CH3

Br

H3

C C CH

H

2o alkyl halide

3o

alkyl halide

heat

CH3

BrH3C C

CH3

H

CH2 CH3

CH3

CH3

H3C C CH2

CH3

H

CH2

Br

1o

alkyl halide

H2C C CH2

CH3

H

CH3

Br

or

1o alkyl halide

reactions i need to know-c06j-part 1

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