chemistry 30 – organic chemistry – part 1 to accompany inquiry into chemistry powerpoint...

Chemistry 30 – Organic Chemistry – Part 1

To accompany Inquiry into Chemistry

PowerPoint Presentation prepared by Robert Schultzrobert.schultz@ei.educ.ab.ca

Organic Chemistry – Preparation – VSEPR

• Recall VSEPR Theory (valence shell electron pair repulson theory) from Chemistry 20

• Organic chemistry will involve 3 particular groupings:

• 0 lone pairs, 4 bonding pairs - tetrahedral

H C H

H

H

••

••

•••

•

Organic Chemistry – Preparation - VSEPR

• 0 lone pairs, 3 bonding pairs – trigonal planar

• 0 lone pairs, 2 bonding pairs - linear

O C O••

••••

••

••

••

••

••

H C H

O••••••

••

•• ••

Organic Chemistry - Preparation

• Recall polarity of covalent bonds from Chemistry 20:

2 particular polar bonds important in organic chemistry

• C – H bonds are virtually non-polar

O — H

C = O

Organic Chemistry – Preparation – Intermolecular Forces

• London Dispersion Forces – all moleculars – temporary dipoles – affected by total # of e- and shape

• Dipole-dipole Forces – polar moleculars

• Hydrogen Bonding (H covalently bonded to F, O, or N)

affect melting point, boiling point, and solubility

Organic Chemistry – 14.1 - Introduction

• Organic compounds – originally defined to be compounds from living or once-living organisms

• Wohler, 1828, synthesized urea (an organic compound) from inorganic chemicals

• Today organic compounds defined to be molecular compounds of carbon – exception: oxides of carbon – CO, CO2

Organic Chemistry – 14.1 - Introduction

• Most existing compounds are organic!

• Special things about carbon that allow it to form so many different compounds:• 4 bonding electrons• ability to form single, double, triple bonds with itself• ability to bond with itself in many different configurations

Organic Chemistry – 14.1 - Introduction

• Classification:organic compounds

hydrocarbons

C and H only

hydrocarbon derivativesC and H along with O, N, and/or halogen atoms

aliphatics without

aromatics with

alkynes – 1 triple bond between C’s – CnH2n-2

alkenes – 1 double bond between C’s – CnH2n

alkanes – all single bonds – CnH2n+2

Organic Chemistry – 14.2 - Hydrocarbons

• Alkanes - saturated hydrocarbons

• Term saturated used because alkanes have the maximum number of hydrogens

• General formula: CnH2n+2

butane

first 4 alkanes

methane ethane

propane

Organic Chemistry – 14.2 - Hydrocarbons

• The unbranched alkanes are a homologous series because they differ by the number of CH2 units in each

• Alkanes are tetrahedral around each carbon

Organic Chemistry – 14.2 - Hydrocarbons

• Since carbons and hydrogens can join up in so many ways, structural formulas are used

• Different types of structural formulas:

3

3 32

we won’t use this type

Organic Chemistry – 14.2 – Hydrocarbons: Alkanes

• Nomenclature of alkanes:• You must learn the

following prefixes:# of C’s

prefix

1 meth

2 eth

3 prop

4 but

5 pent

6 hex

7 hept

8 oct

9 non

10 dec

Organic Chemistry – 14.2 – Hydrocarbons: Alkanes

• Start naming by finding the longest continuous chain of carbon atoms. Name the long chain using its prefix with an ane ending.

• Identify branches, and name using their prefix with a yl ending.

• Number the longest continuous chain from the end closest to the branching and use the numbers like addresses for the branches.

Organic Chemistry – 14.2 – Hydrocarbons: Alkanes

• These rules will be introduced by the following examples

• Several additional rules will be presented with the examples

Organic Chemistry – 14.2 – Hydrocarbons: Alkanes

Example 1:

CH3 – CH – CH – CH2 – CH2 – CH3

CH3

CH2 - CH3

CH3 – CH – CH – CH2 – CH2 – CH3

Root name: hexane

CH3

CH2 - CH3

Organic Chemistry – 14.2 – Hydrocarbons: Alkanes

Example 1:

CH3 – CH – CH – CH2 – CH2 – CH3

CH3

CH2 - CH3

CH3 – CH – CH – CH2 – CH2 – CH3

Root name: hexane

CH3

CH2 - CH3 Identify side

groups

ethyl

methyl

number carbon chain to locate branches

1 2 3 4 5 6

Organic Chemistry – 14.2 – Hydrocarbons: Alkanes

• Compound name:3-ethyl-2-methylhexane

long chain

side group

side group

position on long chain

Additional rule: list side groups in alphabetical order

Organic Chemistry – 14.2 – Hydrocarbons: Alkanes

CH3 – CH – CH – CH – CH3

CH3

CH3 CH3

CH3 – CH – CH – CH – CH3

CH3

CH3 CH3

CH3 – CH – CH – CH – CH3

CH3

CH3 CH3

CH3 – CH – CH – CH – CH3

CH3

CH3 CH3

CH3 – CH – CH – CH – CH3

CH3

CH3 CH3

Example:

No matter how the long chain is selected, the name is the same: 2, 3, 4 - trimethylpentaneNote the tri; use di, tri, tetra, etc, but don’t

use them for alphabetical order

Organic Chemistry – 14.2 – Hydrocarbons: Alkanes

• Example:

CH3 – CH2 – C – CH3

CH2 – CH3

CH – CH3CH2 – CH3

Organic Chemistry – 14.2 – Hydrocarbons: Alkanes

CH3 – CH2 – C – CH3

CH2 – CH3

CH – CH3CH2 – CH3

3 – ethyl – 3, 4 – dimethylhexane or

4 – ethyl – 3, 4 - dimethylhexaneWhich one???

lowest set of numbers

Organic Chemistry – 14.2 – Hydrocarbons: Alkanes

• Doing the reverse process is actually easier – draw your long chain and attach the groups in the addressed spots

• Start by drawing the long chain without any hydrogens – don’t worry about orientation

• Add side groups in their addressed spots• Add hydrogens (each C gets 4 bonds)

• Do alkane nomenclature worksheet

Organic Chemistry – 14.2 – Hydrocarbons: Alkanes

• Physical Properties of Alkanes:• All alkanes are non-polar,

only intermolecular forces = London Dispersion Forces – boiling point and melting point increase with number of carbons (see chart page 551) KNOW

all alkanes are insoluble in water

Organic Chemistry – 14.2 – Hydrocarbons: Alkenes

• Alkenes are hydrocarbons with 1 double bond

• Note dienes and trienes also exist – we’ll focus on compounds with 1 double bond

• Alkenes with 1 double bond have the general formula, CnH2n

• Since they have 2 less hydrogens than corresponding alkanes, they’re called unsaturated hydrocarbons

Organic Chemistry – 14.2 – Hydrocarbons: Alkenes

• Alkene formulas:

• Alkenes are trigonal planar around the doubly bonded C’s and tetrahedral around the others

3

3

3

we won’t use this type

Organic Chemistry – 14.2 – Hydrocarbons: Alkenes

• Nomenclature of alkenes:

find longest continuous chain of carbons that contains the double bond – same prefixes as for alkanes

add ene to the prefix along with a number to indicate the position of the double bond (for ethene and propene a position number is not needed)

number the long chain from the end closest to the double bond (not the branching)

Organic Chemistry – 14.2 – Hydrocarbons: Alkenes

• Example:

CH3 – CH2 – CH2 – C = CH2 CH2

CH3

CH3 – CH2 – CH2 – C = CH2 CH2

CH3

2 – ethylpent-1-enelength of long chain containing double bond

side-group

position of side-group

position of double bond

Organic Chemistry – 14.2 – Hydrocarbons: Alkenes

• Do questions 10 – 14 on pages 554-5

Organic Chemistry – 14.2 – Hydrocarbons: Alkenes

• Physical properties of alkenes:

• Like alkanes, alkenes are non-polar and are insoluble in water

• Boiling points are slightly lower than those for alkanes with the same number of carbonsWhy?

Smaller # of electrons, weaker LDF lower boiling point

Organic Chemistry – 14.2 – Hydrocarbons: Alkynes

• Alkynes are unsaturated hydrocarbons with 1 triple bond

• General formula CnH2n-2

• Alkynes are linear around the triply bonded carbons and tetrahedral around other carbons

Organic Chemistry – 14.2 – Hydrocarbons: Alkynes

• Alkynes are non-polar aliphatic hydrocarbons like alkanes and alkenes

• They are insoluble in water

Organic Chemistry – 14.2 – Hydrocarbons: Alkynes

• Note that alkynes have higher boiling points than alkanes or alkenes

• Obviously the explanation used for alkenes being lower than alkanes doesn’t apply here Table 14.5, page 557

Organic Chemistry – 14.2 – Hydrocarbons: Alkynes

• Accepted explanation is that for short chain alkynes, the linear structure around triple bond allows them to come closer together than alkanes or alkenes with same number of carbons, causing stronger London Dispersion Forces

Organic Chemistry – 14.2 – Hydrocarbons: Alkynes

• Nomenclature of alkynes is identical to that of alkenes, the only exception is the ending:yne, not ene

• Do Practice Problems 16 and 17 on pages 556 and 557

Organic Chemistry – 14.2 – Hydrocarbons: Cyclics

• Cyclic analogues exist for alkanes, alkenes, and alkynes

• General formulas will contain 2 less hydrogens than the open chain hydrocarbons:cycloalkanes CnH2n, cycloalkenes CnH2n-2, cycloalkynes CnH2n-4

• Small cycloalkynes don’t exist because of the large bond strain that would exist around the linear triple bond

Organic Chemistry – 14.2 – Hydrocarbons: Cyclics

• Line structures are commonly used for the ring part of cyclic hydrocarbons

• Always draw them this way

• Examples:

CH2

CH2 CH2

cyclopropane:

notCH2

CH2

CH

CHnot

cyclobutene:

Organic Chemistry – 14.2 – Hydrocarbons: Cyclics

• Cyclics will always have names ending with cyclo_____ane or cyclo_____ene

• Don’t worry about cyclo_____ynes, you will not encounter them, except my favourite one,

• Consider the following examples to learn how to do the nomenclature for substituted cyclics

Name? stopsyne!

STOP

Organic Chemistry – 14.2 – Hydrocarbons: Cyclics

CH2 – CH3

ethylcyclopentane

No numbers needed. Why?

CH2 – CH3

3-ethylcyclopenteneAlways start at far side of double bond and number clockwise or counter-clockwise towards group

CH2 – CH3

CH3

4-ethyl-3-methlycyclopentene

As above. This one must be numbered counter-clockwise to give lowest set of numbers, even though 1st group gets a higher number

Organic Chemistry – 14.2 – Hydrocarbons: Cyclics

CH2 – CH3

CH3 1-ethyl-2-methylcyclopentane

This time the numbering is clockwise since double bond isn’t a factor and when possible lowest number goes on first group

Do Practice Problems 18 – 23 page 559 and 560

Do Aliphatics Review WS

Quiz coming up!

Organic Chemistry – 14.1 - Introduction

• Classification:organic compounds

hydrocarbons

C and H only

hydrocarbon derivativesC and H along with O, N, and/or halogen atoms

aliphatics without

aromatics with

alkynes – 1 triple bond between C’s – CnH2n-2

alkenes – 1 double bond between C’s – CnH2n

alkanes – all single bonds – CnH2n+2

finished with aliphatics; aromatics today

Organic Chemistry – 14.2 – Hydrocarbons: Aromatics

• Aromatics: all contain the grouping

• Originally this grouping thought to be:

• Problems: • all bonds found to be equal length • this compound should be very reactive but is actually very stable

6 6C H

or

Organic Chemistry – 14.2 – Hydrocarbons: Aromatics

• Today we believe it to be made up of bonds that are neither single nor double but a hybrid of both

• We draw the structure

• Its name is benzene

• Benzene is the root common to all aromatics

Organic Chemistry – 14.2 – Hydrocarbons: Aromatics

• Nomenclature of Aromatics:page 561 Where numbering

starts

Organic Chemistry – 14.2 – Hydrocarbons: Aromatics

• Examples:

CH3

CH2 – CH3

CH2 – CH2 – CH3

CH

3 –

CH

–

CH

3

propylbenzene

1-ethyl-3-methylbenzene

2-phenylpropane

Organic Chemistry – 14.2 – Hydrocarbons: Aromatics

• Do Practice Problems 24 – 27, page 562

• Aromatics WS

Organic Chemistry – 14.2 – Hydrocarbons: Aromatics

Organic Chemistry – 14.3 – Hydrocarbon Derivatives

hydrocarbon derivativesC and H along with O, N, and/or halogen atoms

alkanes – all single bonds – CnH2n+2

organic compounds

hydrocarbons C and H only

aliphatics without

aromatics with

alkynes – 1 triple bond between C’s – CnH2n-2

alkenes – 1 double bond between C’s – CnH2n

alcoholsR-OH

akyl halidesR-X

carboxylic acids

R-C-OH

=

O

esters

R1 – C – O – R2

=

O

Organic Chemistry – 14.3 – Hydrocarbon Derivatives

• Hydrocarbon derivatives contain other elements besides C and H; most commonly O, N, or halogen atom

• Functional group: group of atoms that gives the compound its characteristic properties

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Alcohols

• Alcohols – functional group:“-OH” hydroxyl group

• Common alcohols: table 14.7, page 566

3

3

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Alcohols

• Nomenclature of alcohols

• Key points – long chain must have “–OH” attached to it

• Numbering of the long chain starts from the end closest to “-OH”

• Ending of root is ol

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Alcohols

CH3 – CH2 – CH2 CH3 – CH – CH2 – CH2 –

OH

CH3 – CH2 – CH2 CH3 – CH – CH2 – CH2 –

OH 3-methylhexan-1-ol

side group

position of side group

position of

OH

length of longchain containing OH*

* don’t count OH in length of chain

Example

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Alcohols

• Example

CH2 – CH – CH2 OH

OH

OH

CH2 – CH – CH2 OH

OH

OH

propane - 1, 2, 3 - triollength of long chain containing OH’s

position of OH’s

number of OH’s

common name of this compound: glycerol

Advantages to above name??

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Alcohols

• Do Practice Problems 28 – 30 on page 567

• Omit 28d, 29c, 30a

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Alcohols

• Physical properties of alcohols

• Because of the hydrogen bonding between OH groups in adjacent molecules,

• alcohols have much higher boiling points than hydrocarbons (1-12 C’s are liquids at SATP)

• small alcohols are totally miscible with water, but ……………

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Alkyl Halides

• Alkyl halides contain at least 1 halogen atom, (F, Cl, Br, I)

• Alkyl halides are all synthetic compounds

• CFC’s (chlorofluorocarbons) are alkyl halides

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Alkyl Halides

• Nomenclature of alkyl halides:

long chain must be attached to halogen atom(s)

identical to nomenclature of hydrocarbons

side groups end in o, not yl – fluoro, chloro, bromo, iodo

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Alkyl Halides

• Example:

CH3 – CH2 – CH – CH – CH – CH3 Br

Cl

Br

CH3 – CH2 – CH – CH – CH – CH3 Br

Cl

Br

2, 4 – dibromo – 3 - chlorohexane

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Alkyl Halides

• Do Practice Problems 31, 32, page 569

• Do Alcohols/Alkyl Halides Nomenclature WS

Br

Br

Cl 1,4 – dibromo – 2 - chlorocyclohexane

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Carboxylic Acids

• Carboxylic acids are weak organic acids containing the carboxyl functional group,

often written –COOH

• When carboxylic acids, ionize, the process is:

- C – OH ,

=

O

R - C – OH ,

=

O

R - C – OH(aq)

=

O

R - C – O-

(aq)

=

O

H+(aq) +

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Carboxylic Acids

• Common carboxylic acids, acetic acid (active ingredient of vinegar) and citric acid

• Nomenclature of carboxylic acids:

In all carboxylic acids the carboxyl group is at one end of the molecule

It is always carbon #1 in the chain

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Carboxylic Acids

• Example:

CH3 – C – CH2 – CH2 – C – OH

= O

CH2

CH3

CH3

CH3 – C – CH2 – CH2 – C – OH

= O

CH2

CH3

CH3

4, 4 – dimethylhexanoic acid

note that the carboxyl carbon does get counted in the long chain – it is carbon #1

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Carboxylic Acids

• Do Practice Problems 33 – 35, page 570

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Carboxylic Acids

• Physical properties of carboxylic acids:

• Like alcohols they have hydrogen bonding, but hydrogen bonding at 2 sites, -C=O and –OH

• This leads to higher boiling points and greater solubility than alcohols with same number of C’s

• Carboxylic acids with 1-4 C’s are completely miscible in water

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Esters

• Esters have the general formula:

often written RCOOR′

• Esters are formed from the reaction of an alcohol and a carboxylic acid; the formation or esterification reaction is the key to naming them

R(or H) - C – O – R′

=

O

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Esters

R - C – O - H

=

O

+ H - O - R′

R - C – O - R′

=

O

+ HOHcarboxyli

cacid

alcohol

ester

water

It’s important that when you look at ester, that you’re able to recognize part that came from alcohol and part that came from acid

Acid part contains C; alcohol part is bonded directly to O

O

=

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Esters

• General form of name:_______yl _________oate

fromalcohol

fromacid

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Esters

• Examples:

CH3 – CH2 – C – O – CH3

= O

alcohol part: methyl

acid part: propanoate

methyl propanoate

CH3 – CH2 – CH2 – CH2 – O – C – H

= O

alcohol part:butyl

acid part:methanoate

butyl methanoate

Organic Chemistry – 14.4 – Refining and Using Organic Compounds

• Do questions 37 and 38 page 572

Organic Chemistry – 14.3 – Hydrocarbon Derivatives: Esters

• Physical properties of esters:

• fruity odour in some cases• polar but lack of OH bond means no

hydrogen bonding, so lower boiling points than alcohols and carboxylic acids

• esters with few carbons are polar enough to be soluble in water

Organic Chemistry – 14.3 – Hydrocarbon Derivatives

• Structural isomers: compounds with same molecular formula but different structural formulas

Organic Chemistry – 14.4 – Refining and Using Organic Compounds

• Petroleum: mixture of hydrocarbons (primarily alkanes and alkenes) found in natural gas, crude oil, and bitumen (from tar sands)

• Petrochemicals: hydrocarbon materials from petroleum used to produce plastics and other synthetic materials

Organic Chemistry – 14.4 – Refining and Using Organic Compounds

• Fractional distillation: a means of separating petroleum components based on differing boiling points

Organic Chemistry – 14.4 – Refining and Using Organic Compounds

• Read and discuss page 578 regardingfractional distillation

• Fractional distillation is a physicalprocess; mixture is separated intofragments with a small range of boilingpoints – there is no chemical change in thefractions

Organic Chemistry – 14.4 – Refining and Using Organic Compounds

• Next stages of petroleum refining are chemical processes:

• cracking – breaks carbon-carbon bonds• reforming – forms carbon-carbon bonds alkylation (special case of reforming) forms 2,2,4-trimethylpentane from smaller

hydrocarbons

• Both of these can be divided into many subgroups

• Read page 579-80 and page 581

Organic Chemistry – 14.4 – Refining and Using Organic Compounds

Organic Chemistry – 14.4 – Refining and Using Organic Compounds