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CHEM1002 2011 Week 2 notes You should now be able to

Understand the basis of drawing organic structures Convert between a condensed molecular formula and a skeletal or line

structure Determine the formula of a molecule from its skeletal representation Know and recognise the functional groups Be able to name an alkane

Unless otherwise stated, all images in this file have been reproduced from:

Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, 2007 (John Wiley)

ISBN: 9 78047081 0866

FIRST YEAR CHEMISTRY CHEM1002

Tutorials

•  Start in week 1

•  Complete the assignment sheet after working through the ‘critical thinking’ problems in the tutorial

Laboratory Work

•  Starts in week 2 – check your timetable for your session Assessment

•  15% laboratory assessment (see first lab session for details)

•  15% tutorial quizzes (3 per semester: weeks 5, 9 and 12)

•  10% spectroscopy assignment (begins week 4 with a deadline of week 7)

•  60% 3 hour exam at the end of semester

CH4CH3CH3CH3CH2CH3CH3CH2CH2CH3CH3CH2CH2CH2CH3CH3CH2CH2CH2CH2CH3CH3CH2CH2CH2CH2CH2CH3CH3CH2CH2CH2CH2CH2CH2CH3CH3CH2CH2CH2CH2CH2CH2CH2CH3CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3

Naming Alkanes: Stems and Substituents· Alkanes have the ending -ane· The stem is given by the number of carbons

Stemanemethaneethanepropanebutanepentanehexaneheptaneoctanenonanedecane

CH4C2H6C3H8C4H10C5H12C6H14C7H16C8H18C9H20 C10H22

· To name substituents just replace the ending -ane with -ylAlkanepropane CH3CH2CH3pentane CH3CH2CH2CH2CH3

Alkyl substituentpropyl CH3CH2CHpentyl CH3CH2CH2CH2CH

Nomenclature

Number(s) Substituent(s) Stem Ending

The number of carbons in the longest chain containing the functional group

The functional group present

The parts of the molecule that are not the stem

Locates the substituent(s) in the molecule

How do you name an organic compound? Need a unique name for every unique compound

Find and name the longest carbon chain containing the fuctional group - this is the stem - and add the endingIdentify substituent(s)Number the longest chain to give the lowest possible numbering for substituent(s)Allocate a number to every substituentList substituents in alphabetical orderIdentical substituents are indicated by prefixes: di (2), tri (3), tetra (4), then penta (5), hexa (6)...then write it all out as one word

The name has several parts:

The Rules1)

2)3)4)5)6)

Branching Out

2-methylpropane

2-methylbutane

2,2-dimethylbutane

3-methyl-pent-2-ene

2,4-dimethylhexane

5-methyl-hex-2-yne

Useful older names

Isopropyl-

tert-butyl

•  Find the longest chain to work out your prefix

•  If multiple bonds are present, your ‘longest chain’ must include them

•  Use the ‘methyl’, ‘ethyl’, ‘propyl’ etc to indicate branches + ‘2-’ etc to show where

…

HOH

H

H

cholesterol

OH

O

O

OH3C

Aspirinbenzene

cyclohexene2-butyne

Functional Groups

•  Functional groups are the “interesting bits” of a molecule …

•  C–C’s and C–H’s form the skeleton

•  multiple bonds and heteroatoms are the action centres

alcohol

alkene

arene

carboxylic acid

ester

ether

alcohol

amide

amine

amine

O

HO

OH

OHO

OO

NH2O

Alcohol

Ketone

Aldehyde

Carboxylic Acid

Ester

Amide

-ol

-one

-al

-oic acid

-oate

-amide

Propanol

Butanone

Ethanal

Hexanoic acid

Methyl butanoate

Butanamide

Functional Groups

Ether

Amine

Alkyl chloride*

Acid chloride*

NH2

O

Cl

Cl

O

-ether

Amino- or –amine

Chloro- or –chloride

-anoyl chloride

Diethyl ether

Ethyl amine

2-Chlorobutane

Propanoyl chloride

Functional Groups

* Similarly for F, Br & I: alkyl / acid fluorides, bromides & iodides

You should now be able to:

1. Understand the basis of drawing organic structures

2.  Convert between a condensed molecular formula and

a skeletal or line structure

3. Determine the formula of a molecule from its skeletal

representation

4.  Know and recognise the functional groups

5.  Be able to name an alkane

Next: Isomerism

What have we done to date?

Isomerism

1.  What are isomers?

2.  Constitutional Isomers

3.  Conformational Isomers

4.  Double bond Isomerism

•  How many different compounds are there with the formula C2H2BrClO?*

•  How do we tell them apart?

•  How do we name them?

•  How are they different?

•  Do they have different properties?

*(Excluding those with O–Cl or O–Br bonds)

Isomerism

There are 16 different compounds C2H2BrClO

C C

O

H HCl Br

C C

O

H HBr Cl

C C

O

H BrCl H

C C

O

Br HH Cl

C CH OH

Cl BrC C

H OH

Br Cl

C CCl OH

H BrC C

Br OH

H ClC C

Cl OH

Br H

C CBr OH

Cl H

C C

O

Cl HBr H

C C

O

Br HCl H

H

CC

BrCl

O

H

H

CC

ClBr

O

H

BrCH2 CO

Cl

ClCH2 C Br

O

Ouch!

Isomerism

Isomers same molecular formula

Constitutional Isomers Different nature/sequence of bonds

Stereoisomers Different arrangement of groups in space

Configurational Isomers

Interconversion requires breaking bonds

Conformational Isomers Differ by rotation about a single bond

Enantiomers Non-superposable mirror images

Diastereoisomers Not mirror images

Same molecular formula but different structures

Isomerism










Isomers which differ in nature or sequence of bonding !  Within a homologous sequence of alkanes, the number of

constitutional increases rapidly

n123456:

1020

CnH2n+2CH4C2H6C3H8C4H10C5H12C6H14

:C10H22C20H42

CH4CH3CH3CH3CH2CH3CH3CH2CH2CH3 CH3CHCH3

CH3

No. of Constitutionaliosmers

111235:

75366 319

Constitutional formula

AND

Constitutional Isomers

Draw and name the constitutional isomers of C6H14

hexane2-methylpentane 3-methylpentane

2,3-dimethylbutane 2,2-dimethylbutane

Question

•  The physical and chemical properties of constitutional isomers may be very different, particularly when different functional groups are present

•  For example the molecule with formula C4H8O may be a ketone, aldehyde, alkene/ether or alkene/alcohol Question: which is which?

OH

OO

O

Isomers have Different Properties

ketone

aldehyde

alkene/ether

alkene/alcohol

Isomerism










Isomers which differ in arrangement of groups in space (same nature and sequence of bonding)

•  Two groups: •  Conformational isomers (conformers)

•  differ by rotation about a single (C-C) bond •  not normally separable at room temperature

•  Configurational isomers •  Interconverted only by breaking and remaking bonds. •  This process normally requires considerable energy …

… and does not happen at room temperature

Stereoisomers




Configurational Isomers Interconversion requires breaking bonds

Conformational Isomers

Differ by rotation about a single bond



Back to the Isomer Tree


Use ethane as an example (CH3CH3)

H

H H

H H

H

H

H

H

H H

H

H

H

H

HH

HH

H

H HH

H

Sawhorse representation

Newman projection

eclipsed staggered

rotate back carbon 60°

!  In ‘straight chain’ alkanes, rotation about C-C rapid at R.T. !  Differences in energy arise from steric interaction

0° 60° 120° 180° 240° 300° 360°

Ener

gy

Dihedral angle

12 kJ/mol

H

H

H HH

H

H

H

H HH

H

H

H

H HH

H

H

H

HH

HH

H

H

HH

HH

H

H

HH

HH

H

H

H HH

H

Barriers to Rotation

Barriers to Rotation Conformational Isomers

0° 60° 120° 180° 240° 300° 360°

Ener

gy

Dihedral angle

H

H3C

H HH

CH3

H

H

CH3HH

CH3

CH3

H

H HH

CH3

H3C

H

HCH3

HH

H

H3C

HCH3

HH

H

H

CH3

CH3

HH

0 kJ/mol

+4 kJ/mol

+16 kJ/mol

+19 kJ/mol

H

H3C

H HH

CH3

Example: butane







Differ by rotation about a single bond



Back to the Isomer Tree

Straight chain alkanes !  Rotation around each C-C bond readily occurs !  Conformational isomers result

HC

CC

H

H

CH

H

H

H H

H H

C

C

H

H

H

H

Cyclic alkanes !  Rotation is restricted within a ring !  Since rotation would require atoms to pass

through the ring – very high energy barrier


Various isomerism possible !  Constitutional:

Cl

Cl Cl

Cl

ClCl

Cl

Cl

Cl

Cl!  Configurational:

Cycloalkanes

Don’t need to have double bonds to have configurational isomerism!

cis and trans !  These structures are diastereoisomers or

diastereomers !  They have different physical and chemical properties !  Called cis and trans to distinguish them

Cl

Cl

Cl

Cl

Cl Cl

Cl

Cl

cis-1,2-dichlorocyclopentane

trans-1,2-dichlorocyclopentane

Cycloalkanes









One More Look …

Identify the isomerism:

constitutional

conformational

diastereomeric

Question

Will continue with diastereomers and enantiomers in a later lecture

C

C

H

H

H

H

Cyclic alkanes !  Rotation around the C-C bond within a

cycloalkane is restricted compared to that of a hydrocarbon chain

Consequently disubstitute cycloalkanes occur as:

Isomers Resulting from Structural Rigidity

Constitutional isomers:

Diastereoisomers:

Cl

Cl Cl

Cl

ClCl

Cl

Cl

Cl

Cl

Different sequence of bonding

Differ only in stereochemistry

! - bond " - bond

A double bond is contains both a !- and a "-bond •  Pi-bonds result from p-orbital overlap •  Electron density concentrated above and below plane •  Rotation around the C-C axis requires breaking the "-bond

(~128 kJ mol-1) does not occur at room temperature

Alkene ‘diastereoisomers’

H

H H

H

A "-electron overlap requires a fixed geometry around the bonded carbon atoms

No rotation about the two ends

Consequence of restricted rotation about the double bond

(Z)- 2-butene

H3CC

CH

H

CH3

(E)- 2-butene

H3CC

CCH3

H

H

Different compounds!

melting point = -139 ºCboiling point = 4 ºC

melting point = -106 ºCboiling point = 1 ºC

!  If one end of the C=C bond has same two groups not a diastereoisomer

Alkene ‘diastereoisomers’

HC C

HCl

H HCC

H Cl

H

HCC

H Cl

HHC C

HCl

H

All examples of the same molecule, flipping whole molecule gives superimposable structures

!  If both ends of the C=C bond bear two different groups

A

C C

YB

X

A ! B X ! Y(Although A or B can be the same as X or Y)

!  Then TWO stereoisomers are possible called diastereomers and labeled Z or E

HC C

COOHHOOC

H COOHCC

HOOC H

HDifferent

compounds!

Double Bond Isomers

!  Z/E determined by assigning a priority to each of the pairs of groups on each carbon of the double bond

!  The higher the atomic number of the atom attached, the higher the priority

!  If identical atoms attached, work along the chain until the first point of difference, then go by atomic number

!  If high priority groups on same side of C=C plane # Z double bond (German: zusammen, together)

!  If high priority groups on opposite sides of C=C plane # E double bond (German: entgegen, opposite)

Nomenclature – Z and E

HC C

COOHHOOC

H COOHCC

HOOC H

HDifferent

compounds!

The rules:

(Z) (E)

Higher priority groups on the same sideof double bond

alkene is denoted (Z)

Higher priority groups on opposite sides

of double bond

alkene is denoted (E)

HC C

CH3H3C

H CH3

CCH3C H

H

(Z)- 2-butene (E)- 2-butene

ClCC

Br CH3

H

(E)-1-bromo-2-chloropropene

higher priority groups on same side

higher priority groups on opposite sides

higher priority groups on opposite sides

Double Bond Isomers

Nomenclature of AlkenesFind longest chain containing double bond - this gives the stemFor alkenes the ending is -ene (instead of -ane for alkanes) Give the double bond the lowest number possibleName the substituents as usualDetermine if the double bond is E or Z

Name the following alkenes:

CH3

H3C

H

CH3

Note: Alkynes have no diastereomersGeometry of both carbon atoms is linearThere is only one way to attach two substituents in a straight line!

Name these molecules … fully:

A....................................... B.......................................

C....................................... D.......................................

What is the isomeric relationship between: A and B…………………..…… C and D………………………..

Pop Quiz

Name these molecules:

What is the isomeric relationship between: A and B…Constitutional isomers… C and D…Diastereoisomers.……..

A.....1-hexene.......... B.......(E)-3-hexene.....................

C....(E)-1,3-hexadiene... D.........(Z)-1,3-hexadiene......

Pop Quiz

You should now be able to 1.  Describe alkane conformational isomers 2.  Understand the difference between constitutional

isomers and stereoisomers 3.  Recognise constitutional, conformational and

diastereomeric (cis/trans) isomers 4.  Name isomeric structures correctly

Next !  Organic Reactions!

Summary

Organic Reactions

Four general types of organic reaction

Acid – base reactions (See later: Carboxylic acids) - loss and gain of protons (H+)

Substitution reactions - one group replaces another

Addition – elimination reactions - group(s) added or taken away

Oxidation – reduction reactions - loss and gain of electrons

Two key types of reactant:

Electrophiles: ‘electron loving’, chase negative things (– or $–)

# positive (+ or $+) themselves

eg. H+, HCl

..

Two types of reagent

Nucleophiles: ‘nucleus loving’, chase positive things (+ or $+)

# negative (– or $–) themselves

eg. -OH, Br-, -CN, NH3

Pop Quiz

Classify these reagents as nucleophile or electrophile

Br

O

OCH3

C N

NH(CH3)2

Cl

CN

O

N(CH3)2

Br+ +

+ + HOCH3

+ HCl

nucleophile

nucleophile

electrophile

Curly Arrows

Curly arrow indicates electron pair movement •  A curly arrow starts at the electron pair that moves and ends

at the atom to which the electron pair has moved

•  Arrow from a bond # bond breaks •  Arrow between two species # new bond formed between them

OO H O H O HH

OC

O

R O

H

H HOC

O

R H + O HH

quiz

Add curly arrows to complete this mechanism …

BrC N

CNBr+ +

Remember: •  Curly arrow starts at the electron pair that moves •  Curly arrow ends at atom to which the electron pair has moved •  Arrow from a bond # bond breaks •  Arrow between two atoms # new bond formed between them

…

quiz

Remember: •  Curly arrow starts at the electron pair that moves •  Curly arrow ends at atom to which the electron pair has moved •  Arrow from a bond # bond breaks •  Arrow between two atoms # new bond formed between them

…

Br

C N

CNBr+ +

Reminder: Alkene Structure

A double bond has two potions: • One ! bond (sp2 + sp2 overlap and one " bond (p + p

overlap)

• The electron density of the " bond lies above and below the plane of the double bond

! - bond " - bond

C CH H

H H

all six atoms lie in one plane

Addition to an Alkene

General reaction: •  Reactions of alkenes occur at the weak and

accessible "-bond •  Addition occurs with replacement of a " bond and

a A-B ! bond with two ! bonds: energetically favourable

•  A-B could be H2, Cl2, Br2, HCl, HBr, HI, H2O (H-OH)

HC C

HH

H HC C

BA

H+ A B H H

Alkene Reactions

HC C

CH3H

CH3 HC C

HH

CH3+ H H H CH3

1. Hydrogenation (addition of hydrogen)

HC C

HH

CH3 HC C

BrBr

CH3+ Br Br H H

2. Halogenation (addition of a halogen)

Pd/C catalyst

Alkene Reactions

H3CC C

HH

CH3 H3CC C

BrH

CH3+ H Br H H

3. Hydrohalogenation (addition of a hydrogen halide)

H3CC C

HH

CH3 H3CC C

OHH

CH3+ H OH H H

4. Hydration (addition of water)

H2SO4 catalyst

Hydrohalogenation

Two steps !  Step 1: H+ adds

An unstable intermediate carbocation is formed

!  Step 2: Br- adds The carbocation is intercepted by bromide anion

HCC

H3C H

H3C+ H Br

step 1 HCC

H

H3CH

H3C+

a carbocation

Br! !

HCC

H3C H

BrHH3C

HCC

H

H3CH

H3C+

step 2

a carbocation

Br

Halogenation

As the halogen approaches, it is polarised by the alkene

!  Step 1: “Cl+“ adds One end of the polarised halogen reacts with alkene

!  Step 2: Cl- adds The carbocation is intercepted by chloride anion

HCC

H3C H

H3C+Cl Cl

HCCH

H3CCl

H3C+ Cl

! !

HCC

H3C H

ClClH3C

HCCH

H3CCl

H3C+ Cl

Cl Cl.

.

…

Hydration

Addition of water: H+ catalyst required (eg., dilute H2SO4) Again two main steps … but also a third mini-step Step 1: H+ adds to give a carbocation

Step 2: H2O intercepts (using spare electrons on O)

Step 3: H+ is removed (by conjugate base of H2SO4)

HCC

H3C H

H3C+ H

HCC

H

H3CH

H3Ccarbocation

step 1

HCCH

H3CH

H3C+

carbocation

HOH

step 2 HCC

H3C H

OHH3C

HH

water molecule

HCC

H3C H

OHH3C

Hstep 3

HSO4-

+ H2SO4

Question

Draw the products of these electrophilic addition reactions

+ HI

+ Br2

+ dil H2SO4

+ HI

+ Br2

+ dil H2SO4

I

Br Br

OH

+ HI

+ Br2

+ dil H2SO4

I

Br Br

OH

+ HI

+ Br2

+ dil H2SO4

I

Br Br

OH

H

H

Product Isomers?

•  When HX adds to a symmetrical alkene: only one product

•  But HX addition to unsymmetrical alkene … two possibilities

+ HCl

H Cl Cl H

(Z)-2-butene2-chlorobutane

+ HCl

H Cl Cl H+

2-methyl-2-butene 3-chloro-2-methylbutane2-chloro-2-methylbutane

Markovnikov’s Rule

One product dominates

Markovnikov’s Rule: The hydrogen of an unsymmetrical reagent adds to the end of the double bond that has the greater number of hydrogen atoms already directly attached

H

CCH3C H

H3CTwo H'sattached

No H'sattached

… but why?

+ HCl

H Cl Cl H+

Minor product Major product

Carbocations

The intermediate in the reaction is a carbocation !  Carbocations have 6 electrons and a positive charge !  All carbocations are unstable !  But some are more unstable than others …

Carbocation stability

HCH3C

H

CH3CH3C

H3C

HCH3C

H3C

a tertiary alkyl carbocation

a secondary alkylcarbocation

a primary alkylcarbocation

HCH

H

a methylcarbocation

leaststable

morestable

Markovnikov’s Rule

Major product results from more stable carbocation intermediate

HCC

H Cl

H3CCH3H3C

HCC

Cl H

H3CCH3H3C

HCC

CH3

H3C

H3C

H

HCC

CH3

H3CH

H3C

ClCl

HCC

CH3

H3C

H3C

H Cl

step 1

less stable carbocation

more stable carbocation

majorproduct

minorproduct

Pop Quiz

Predict the major products of these reactions:

HCC

CH3

H3C

H3C

+ H2O / H(dil H2SO4)

+ HBr

+ HI

HCC

CH3

H3C

H3C


+ HBr

+ HI

CH2CH3C

H3C

OH

Br

CH3

I

HCC

CH3

H3C

H3C


+ HBr

+ HI

CH2CH3C

H3C

OH

Br

CH3

I

HCC

CH3

H3C

H3C


+ HBr

+ HI

CH2CH3C

H3C

OH

Br

CH3

I

Hydrogenation

Catalyst required (eg Pd on charcoal) to break strong H-H bond

•  Both H’s added to same face of C=C •  Mechanism more complex (H2 adsorbed onto Pd surface)

H3C CH3

+ H2

Pd/CH3C CH3

+ H2

Pd/C H H

Quiz

dil H2SO4

H2 / Pd catalystHBr

Br2

dil H2SO4

H2 / Pd catalystHBr

Br2

OHBr

Br

Br

Answers

Benzene

CC

CCC

CH

H

HH

H

H

Benzene

Benzene does not behave like 1,3,5-cyclohexatrieneIt is about 140 kJ mol-1 more stable than predictedIt is less reactive than expected

carbon-carbon bondC-C (alkane)C=C (alkene)

benzene

Bond length154 pm133 pm139 pm

Bond strength356 kJ/mol636 kJ/mol518 kJ/mol

Further evidence for the unusual structure of benzene comes from the bond lengths:

Br2

Br2

Br

Br

without catalystno reaction

Structure of BenzeneThere are 2 bonding models used to describe the structure of benzene

1) Valence bond modelEach of the C atoms in benzene is sp2 hybridised, forming ! bonds to two neighbouring C atoms and a ! bond to one H atom.Each carbon also has a p orbital which can form a " bond.This gives alternating double and single bondsThere are 2 possible representations with the " bonds in different positions

HHH H

H H

HHH H

H H

These structures are related by movement of electron pairs and are resonance forms.

Structure of Benzene1) Valence bond model continued

The bonding in benzene is best described as an average or 'resonance hybrid' of the two bonding arrangements

The more resonance structures possible for a given molecule, the more stable the molecule.

The resonance hybrid will have :Half a !-bond between each adjacent pair of carbonsA bond strength and bond length between that of a C-C single bond (no !-bond) and a C=C double bond (one !-bond)

Resonance - recap from semester 1Resonance structures differ only in the position of their !-electronsAtoms do not move in resonance structuresResonance forms must be valid Lewis structures (i.e. only 8 electrons for C, N, O; only 2 electrons for H)

Structure of Benzene2) Molecular Orbital Model

The 6 p-orbitals (one from each carbon atom) form part of a delocalised !-bond system characteristic of aromatic compounds. The electron delocalisation results in a stabilisation of 150 kJ/mol compared to the hypothetical cyclohexatriene structure:

Delocalisation of electrons leads to greater stability

This results in clouds of electron density above and below the plane of the benzene ring.

Localised electrons = constrained to one atom or shared between two atoms (e.g. in a "-bond)

Delocalised electrons = shared between three or more atoms

Blackman Figure 15.16

Benzene

Benzene might look like 3 x C=C’s but … !  The reactions of benzene are not like an alkene at all

!  Substitution rather than addition is observed !  And a catalyst is required

!  Benzene has special properties: it is aromatic !  6 x " electrons (i.e., 3 x C=C) in a ring

# compound is aromatic

Br2

FeBr3(catalyst)

Br+ HBr

Fe or AlBr3 can also be used as the catalyst

Requirements for Aromaticity

•  Cyclic system; ring of atoms •  (nearly) planar •  Conjugated (sideways overlap of pz orbitals

•  4n+2 "-electrons in cyclic, conjugated system n=0 or integer 1, 2, 3 etc. Hence 6, 10,14 "-electrons

Recap

You should now be able to: "  Distinguish between conformational and

configurational isomers of alkanes "  Identify alkene diastereoisomers as E or Z "  Identify electrophiles and nucleophiles "  Give the major products obtained from the

reaction of alkenes with electrophiles (remember Markovnikov’s rule!)

"  Understand why benzene does not react like an alkene

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