Organic Chemistry

P 267

Hydrocarbon• Hydro: hydrogen• Carbon• Additional element: Nitrogen, Oxygen, SulphurAtomic number of carbon = 6Valence electrons of carbon = 4Each carbon atom has 4 bond in compoundEx: alkanesAlkanes: hydrocarbon that consist of hydrogen and

carbon, (single bond)

CH4, methane

C2H6, ethane

C3H8, propane

C4H10, butane

C5H12, pentane

C6H14, hexane

C7H16, heptane

C8H18, octane

C9H20, nonane

C10H22, decane

Structure Types of Propane

CH3CH2CH3

Skeletal Formula

Stick and ball modelDisplayed Formula

Structural Formula

Molecular Formula

C3H8

3 dimensional Formula

HH

H

HH

HH

H

ALCOHOL (-OH)

PROPANE PROPANOL

ALCANOIC ACID

BUTANE BUTANOIC ACID

ALDEHID/ALCANAL

PENTANE PENTANAL

KETONES

Butan-2-one

Alkylamine R-NH2

CH3CH2CH2NH2

Propylamine

ESTHER (Alkyl Alkanoate)

methyl propanoate

Hydrocarbon with Branch

-Find the longest carbon chain, give the name- Find branch and give the name to the Branch:Carbon:Alkyl;1 C atom: methyl2 C atom: ethyl3 C atom: propylHalogen: F(Floro), Cl(Chloro), Br(Bromo), I(Iodo)- Give the number, branch number is the smallest number

123456

Hexane Methyl 2-

Same branch:2 branchs: di3 branchs: tri4 branchs: tetra

the branches names arranged alphabetically

3-bromo 9-chloro 4-methyldecan-3-ol

6-chloro 4-ethyl octan-3-one

Alcanal, and alcanoic numbering begin from C attached to the function

Pentanal

12345

4- methyl

• Same branch:• 2 branchs: di• 3 branchs: tri• 4 branchs: tetra

1 2 3 4 5 6 7 8 9

nonane4,6 2,4,7 -trimethyl-diethyl

Alcanal, and alcanoic numbering begin from C attached to the function

6-ethyl 3-floro 4,7-dimethyl octanoic acid

Remember that the branches names arranged alphabetically

Task

• Copy the table 15.2 page 270 in to your book

Determination of element mass in compound

Compound: AmBn

Mass of element A =

Mass of element B = Find mass of carbon in 88 gr CO2

Mass of carbon = =

= 24 gr

Determination of empirical formulae of organic compound

100,00 g amino acid(CpHqOrNs) react with oxygen and produced 117,10 g of carbon dioxide and 59,92 g of water. In kjeldahl a second of amino acid produced 22,64 g ammonia, determine the empirical formula of amino acidAnswer- write the information mass of amino acid = 100 gr mass of carbon dioxide (CO2) = 117,10 gr mass of water(H2O)= 59,92 g mass of ammonia (NH3) = 22,64

• Find mass of carbon in carbon dioxide= 31,94 g• Find mass of hydrogen in water

= 6,66 g• Find mass of nitrogen in ammonia

= 18,64 g• Find mass of oxygen in amino acid=mass of amino acid – carbon mass –

hydrogen mass - nitrogen mass= 100 – 31,94 – 6,66 – 18,64= 42,76 gr

• Compare mass of the elementsC : H : O : N

31,94 : 6,66 : 42,7 : 18,64• Compare mol of the elements 31,94/12 : 6,66/1 : 42,76/16 : 18,64/14 2,67 : 6,66 : 2,67 : 1,33• Divide by the smallest mole

2 : 5 : 2 : 1

• Determine the empirical formula

C2H5O2N

Reactions

Subtitution

C

H

H

H

H

C

H

Br + OH C

H

H

H

H

C

H

+ Br-OH-

AdditionDouble bondsingle bondTriple bonddouble or single bond

Eliminationsingle bondDouble bonddouble Triple bond

C

H

H

H

C H + H2O OH C

H

H

H

H

C

H

OHC

H

H

H

H

C

H

C

H

H

H

C H H2O+

HydrolysisReaction involve breaking covalent bonds by

reaction with water

C

H

H

H

H

C

H

Br + OH C

H

H

H

H

C

H

+ HBrH2O

Breaking Bonds in Different Ways

+

Homolytic fissionEach atom break to free radical (atom with unpaired electron

Heterolytic fissionEach atom break to ion (cation and anion)

+ + -

Another example

Homolytic fissionCH3Br CH3 + Br

Heterolytic fissionCH3Br CH3

+ + Br-

+Carbocation

Cycloalkane

cyclopropane cyclobutane cyclopentane

cyclohexane

Methyl cyclohexane

Alkylamine (R-NH2)

CH3NH2 methylamine

CH3CH2NH2 ethylamine

CH3CH2CH2NH2 propylamine

Alkenes C=C

CH2=CH2

CH2=CH-CH3

CH2=CH-CH2-CH3

CH2=CH-CH2-CH2-CH3

CH3-CH=CH-CH2-CH3

Double bond number is the smallest number in the longest carbon chain

ethenepropene

butene

pentene

pent-2-ene

Isomerism

Structural isomerDifferent structural formula same molecular

formula

CH3-CH2-CH2-CH3Butane, C4H10

Methyl propane, C4H10

Functional isomerism- Alcohol and eter

- Ethanol: CH3-CH2-OH, molecular formula: C2H6O

- Ether : CH3-O-CH3, molecular formula: C2H6O

- Aldehid and ketones- pentanal: ,molecular formula:C5H10O

- Penta-2-one C5H10O

- Alkanoic acid and esthers- Propanoic acid- Methyl ethanoate

Prove it !

Stereo isomerGeometric isomer (cis-trans isomerism)But-2ene: CH3-CH=CH-CH3

Cis-but-2-ene trans-but-2-ene

Optical isomerism use 3 dimensional formulaOccurs in the chiral molecules

Butan-2-ol

The star sign indicate chiral center. Chiral center is the carbon atom which carries four different groups

Reactions Routes

Write the reactions routes to convert:a. Propene to propanol– pass propene and steam over heated phosphoric

acid catalyst under pressure

b. Propene to propanal– pass propene and steam over heated phosphoric

acid catalyst under pressure untill propanol forms– Distill on propanol to acidified potassium dichromate

Draw the structure of 5,7 dibromo, 2,6 dimethyl octan-3-oneSolution stepsDraw the longest chainGive the numberPut the functionPut the branches

C-C-C-C-C-C-C-C

1 2 3 4 5 6 7 8O

Br

Br

C

C