AS Organic Chemistry Notes by Fahad Hameed

Typed version of Fahad Hameed's handwritten AS Organic Chemistry notes for Cambridge International AS Level Chemistry (9701). Every page of the original is typed in the same order, with drawn structures named and written as formulae, and drawn mechanisms written as numbered steps, so the notes can be searched and copied. The handwritten original is on the same page of megalecture.com.

Organic Chemistry

Carbon compounds → living organisms (plants / animals).

Arrows lead from here to:

• food
• fossil fuels
• pharmaceutical
• agriculture

Formulas

1. Molecular formula

Total number of atoms of elements in a compound.

C₄H₈O₂, C₆H₁₂O₆. No information about structure.

2. Structural formula

Provides information about the structure.

(a) Full displayed formula

Every single atom / bond is shown.

• C: 4 bonds
• N: 3 bonds
• O: 2 bonds
• H: 1 bond
• halogens: 1 bond

Example drawn in full, with every C, H and O and every bond shown: 4-hydroxybutan-2-one, CH₃COCH₂CH₂OH.

A displayed formula is not the actual shape. Two small sketches show the real shapes: a carbon with four single bonds (one drawn as a wedge, one as a dashed line) with a bond angle of 109.5°, and a carbon with a double bond (C=) with a bond angle of 120°.

Two displayed formulas joined by a double-headed arrow labelled "same": 2-fluoropropenoic acid, CH₂=CFCOOH, drawn once starting from the COOH end (HOOC–CF=CH₂) and once starting from the CH₂ end with the chain bent downwards (CH₂=CF–COOH). Both drawings are the same molecule.

A ring drawn as a full displayed formula: a six-membered carbon ring containing two C=C double bonds (four CH groups), one CH₂ group and one carbon that is double bonded to N–H outside the ring. This is cyclohexa-2,4-dien-1-imine, C₆H₇N.

(b) Condensed structural formula

Running out of space.

First example, drawn three ways:

• Skeletal: 3-methylpentan-2-ol.
• Displayed: a chain of four carbons drawn in a straight line with an arrow along it. The O–H group on the second carbon (drawn above the chain) and the CH₂CH₃ group on the third carbon (drawn below the chain) are circled.
• Condensed: CH₃CH(OH)CH(CH₂CH₃)CH₃. The carbons of the chain are underlined and the circled groups go in brackets.

Second example, drawn three ways:

• Skeletal: prop-1-en-2-yl 2-oxopropanoate.
• Displayed: H₃C–C(=O)–C(=O)–O–C(CH₃)=CH₂, drawn in a straight line with an arrow along it. The CH₃ group above the chain is circled.
• Condensed: CH₃COCOOC(CH₃)=CH₂. The carbons of the chain are underlined.

R–C(=O)–H, aldehyde.
Condensed: RCHO

Displayed + condensed

4-methylcyclohexene drawn as a ring of condensed groups: CH=CH, CH₂, CH(CH₃), CH₂ and CH₂ joined in a six-membered ring. Next to it is the same molecule as a skeletal formula: a cyclohexene ring with one methyl branch.

(c) Skeletal formula

Only C–C bonds shown and any functional group (no C / H atoms are shown).

CH₃–CH₂–CH₂–CH₂–CH₃ (pentane) is drawn as a zigzag of four lines. Each dot on a carbon in the formula matches a corner or an end of the zigzag. The lines are the C–C bonds.

CH₃–CH₂–CH(CH₃)–CH(CH₂CH₃)–CH=CH₂, drawn with an arrow along the main chain and the two branches (CH₃ above, CH₂CH₃ below) circled.

Condensed: CH₃CH₂CH(CH₃)CH(CH₂CH₃)CH=CH₂

Skeletal: 3-ethyl-4-methylhex-1-ene.

CH₃–CH₂–C(=O)–CH(CH₂CH₃)–C(=O)–OH, with the skeletal formula drawn below it: 2-ethyl-3-oxopentanoic acid. The skeletal formula shows the two C=O groups and the OH.

A skeletal formula showing O, two C=O groups and NH₂, with its full displayed formula drawn below it: ethyl 2-carbamoylbutanoate, CH₃CH₂OCOCH(CH₂CH₃)CONH₂.

Naming

Homologous series

Family of organic molecules having:

1. Similar chemical properties.
2. Show trends in physical properties. As molecule size increases:
   • melting point / boiling point higher
   • more viscosity
   • less flammability
   Reason: bigger molecule / more electrons / stronger temporary / induced dipoles, more van der Waals forces.
   Sketch: two CH₃–CH₃ molecules (18 electrons each) side by side. The electrons shift in one molecule so that one end is δ+ and the other δ-, and this induces a matching δ+ and δ- in the molecule next to it, with a dotted line of attraction between them. A second pair shows the same thing with the dipoles the other way round. Label: instantaneous - induced dipole.
3. Same functional group: group of atoms / structure that determines chemical properties.
   CH₂=CH–CH₂–CH₃, with an arrow to the C=C: all reactions.
4. Same general formula: only applicable for simple molecules.
   C_nH_2n
   CH₃–CH=CH₂ is C₃H₆: general formula works.
   CH₂=CH–CH₂–CH=CH–CH₃ is C₆H₁₀: general formula does not work.
5. Successive members differ by CH₂.
   CH₄ → C₂H₆ → C₃H₈ → C₄H₁₀, each step adding CH₂.

Prefix = number of carbon atoms

Prefix	meth	eth	prop	but	pent	hex	hept	oct	non	dec
Carbons	1C	2C	3C	4C	5C	6C	7C	8C	9C	10C

Suffix

The suffix gives the homologous series.

No.	Homologous series	Suffix	Notes
1	Alkane	-ane	saturated hydrocarbon
2	Alkene	-ene	unsaturated hydrocarbon, C=C
3	Alcohol	-ol	–OH, hydroxy group (hydroxyl)
4	Carboxylic acid	-oic acid	–C(=O)–OH
5	Carbonyl compounds
5 (a)	Aldehyde	-al	–C(=O)–H, at the end of the carbon chain
5 (b)	Ketone	-one	–C(=O)–, in the middle of the carbon chain
6	Nitriles	-nitrile	–C≡N, at the end of a carbon chain
7	Amines	-amine	–NH2

A name has three parts:

Prefix	an / en	Suffix
number of C	saturated or unsaturated	homologous series

Alkanes

CH₃CH₂CH₃ (drawn as a displayed formula): propane

Alkenes

CH₃CH=CH₂ (displayed formula): propene

CH₃CH₂CH=CHCH₃ (displayed formula, carbons numbered 5 to 1 from left to right so the C=C is between carbons 2 and 3): pent-2-ene

CH₂=CH–CH=CH–CH₃ (displayed formula, carbons numbered 1 to 5 from left to right): pent-1,3-diene. The numbers 1,3 give the position of C=C.

Number (smaller number should be given to higher precedence group):

1. carboxylic acid
2. nitrile
3. aldehyde
4. ketones
5. alcohols (hydroxy)
6. amines (amino)

Alcohols

CH₃CH₂CH(OH)CH(OH)CH₂OH (displayed formula, the three OH groups circled, numbering starts from the right-hand end): pentan-1,2,3-triol. The "an" shows a saturated carbon chain.

CH₂=CH–CH=CH–CH(OH)–CH₂OH (displayed formula, the two OH groups circled, carbons numbered 6 to 1 from left to right): hex-3,5-dien-1,2-diol

Aldehyde

Suffix -al, group –C(=O)–H

CH₃CH₂CHO (displayed formula): propan-1-al, the same as propanal. It is unnecessary to write the 1.

CH₂=CH–CH₂–CHO (carbons numbered 4 to 1 from left to right, starting from the CHO carbon): but-3-en(-1-)al

Ketones

–C(=O)– in the middle of the chain. Suffix -one.

CH₂=CH–CH₂–CO–CH₃ (carbons numbered 5 to 1 from left to right): pent-4-en-2-one

CH₃–CO–CH₂–CO–CH₃ (numbered from the right): pentan-2,4-dione

Nitriles

Group –C≡N

CH₃–CH₂–CH₂–CH₂–C≡N: pentanenitrile

CH₂=CH–CH₂–C≡N (carbons numbered 4 to 1 from left to right, the C of C≡N is carbon 1): but-3-enenitrile

Amines

CH₃–CH₂–CH₂–NH₂: propanamine

Branched molecules and multiple functional groups

• number, number → comma
• number - letter → dash

Structure drawn with the main chain highlighted and numbered 1 to 6: CH₃CH(NH₂)C(OH)(CH₃)CH=C(CH₃)CHO. Carbon 1 is the CHO carbon. The branches are circled and labelled: methyl on carbon 2, hydroxy and methyl on carbon 4, amino on carbon 5.

Name: 5-amino-4-hydroxy-2,4-dimethylhex-2-enal

Position - branches, alphabetically.

Main chain: longest possible / continuous carbon atom chain, containing:

• suffix functional group
• maximum C=C, C≡C
• maximum groups attached to it

Number (smaller number should be given to higher precedence group). The name used when the group is a branch is in brackets:

1. carboxylic acid
2. nitrile (cyano)
3. aldehyde (formyl)
4. ketones (oxo)
5. alcohols (hydroxy)
6. amines (amino)

Structure drawn with the chain highlighted: OHC–CH(OH)–CH(CH₂CH₃)–COOH. The COOH carbon is carbon 1 (carboxylic acid). The circled groups are labelled formyl (the H–C=O at the left-hand end), hydroxy (the OH on carbon 3) and ethyl (the CH₂CH₃ on carbon 2).

Name: 2-ethyl-3-formyl-3-hydroxypropanoic acid

• Halogen branches: chloro, bromo, iodo
• Carbon branches: alkyl, methyl etc.

CH₃–CO–CHCl–CHO (displayed formula, carbons numbered 4 to 1 from left to right, the Cl circled, an arrow pointing at the CHO carbon as carbon 1): 2-chlorobutan-3-one-1-al

Isomers

C₂H₄O₂. Six different displayed formulas are drawn for this one molecular formula:

• ethanoic acid, CH₃COOH
• hydroxyethanal, HOCH₂CHO
• ethene-1,2-diol, HOCH=CHOH
• methyl methanoate, HCOOCH₃
• ethene-1,1-diol, CH₂=C(OH)₂
• oxiran-2-ol, a three-membered ring of two carbons and one oxygen, with an OH group on one of the carbons

Structural isomers
Molecules having same molecular formula and different structural formulas.

Methyl methanoate drawn twice, once in a straight line as CH₃–O–CHO and once bent round as OHC–O–CH₃, joined by a double-headed arrow labelled "same". These are not isomers, they are the same molecule.

Types of structural isomers

(a) Chain isomer

Four displayed formulas of carboxylic acids with five carbons, with the carbon chain highlighted in each:

• pentanoic acid, CH₃CH₂CH₂CH₂COOH (straight chain of five carbons)
• 2-methylbutanoic acid, CH₃CH₂CH(CH₃)COOH (chain of four with one branch)
• 3-methylbutanoic acid, (CH₃)₂CHCH₂COOH (chain of four with one branch)
• 2,2-dimethylpropanoic acid, (CH₃)₃CCOOH (chain of three with two branches, drawn as a cross)

(b) Positional isomers

Three displayed formulas of a five-carbon chain with the OH group circled in a different position each time:

• pentan-1-ol, CH₃CH₂CH₂CH₂CH₂OH
• pentan-3-ol, CH₃CH₂CH(OH)CH₂CH₃
• pentan-2-ol, CH₃CH₂CH₂CH(OH)CH₃

An arrow leads from pentan-2-ol to a three-dimensional drawing of it: a central carbon marked with a star (C*) bonded to H (up), CH₃ (dashed bond), OH (wedge bond) and CH₃CH₂CH₂. A vertical line is drawn beside it.

3. Functional group isomer

Change the functional group.

Functional group	Example drawn	Isomer with a different functional group	Example drawn
alkenes	propene, CH2=CHCH3	cycloalkanes	cyclopropane, a ring of three CH2 groups
aldehydes	propanal, CH3CH2CHO	ketones	propanone, CH3COCH3
carboxylic acid	ethanoic acid, CH3COOH	ester	methyl methanoate, HCOOCH3

Stereoisomer

Molecules having same molecular formula, same structural formula but different 3-dimensional geometry (shape is different).

Picture: a left hand and a right hand on either side of a mirror. They are mirror images and cannot be superimposed.

1. Geometric isomer

• C=C must be present, each C in C=C must be bonded to two different groups.

Each C of the C=C is sp² hybridised: 120°, trigonal planar. Sketch: a carbon atom with three sp² orbitals pointing outwards in one plane, each holding one electron, and one p orbital drawn above and below the plane.

Sketch: C=C with H and Cl on each carbon and the π bond drawn as two shaded regions, one above and one below the line joining the carbons. Independent rotation is restricted.

Sketch for comparison: two carbons joined by a single bond, one carrying Cl, Cl and H and the other carrying H, Cl and H. Rotation possible around a σ bond.

The two arrangements of CHCl=CHCl (1,2-dichloroethene), each drawn first with its π bond and again as a plain structure with the Cl atoms circled:

Isomer	Arrangement
cis (same side)	both H on the top, both Cl on the bottom
trans (opposite sides)	one Cl on the bottom left and the other Cl on the top right, H on the top left and bottom right

Examples

(a) CH₃–CH=CH₂, propene. Redrawn around the C=C: the left carbon has H and CH₃ (different groups), the right carbon has H and H (same groups). No geometric isomers.

(b) but-2-ene, CH₃–CH=CH–CH₃

• cis but-2-ene: both CH₃ groups on the same side (both drawn at the bottom), both H at the top
• trans but-2-ene: CH₃ groups on opposite sides (one bottom left, one top right)

Unsymmetric molecules

Multiple cis / trans.

Cl–CH=CH–CH=CH–CH₃. Each of the two C=C can be cis / trans.

First C=C	Second C=C
cis	cis
cis	trans
trans	cis
trans	trans

The cis, trans and trans, cis rows are the same if the molecule is symmetric.

The isomers are drawn out as structures with each C=C labelled. On this page: cis, cis and cis, trans.

Continued at the top of the next page: trans, trans and trans, cis.

Cl–CH=CH–CH=CH–CH=CH–CH₃. With three C=C there are 8 combinations:

First C=C	Second C=C	Third C=C
cis	cis	cis
cis	cis	trans
cis	trans	cis
cis	trans	trans
trans	cis	cis
trans	cis	trans
trans	trans	cis
trans	trans	trans

Arrows pair up rows that are the same if the molecule is symmetric: cis, cis, trans with trans, cis, cis, and cis, trans, trans with trans, trans, cis (each pair is the same order read from the other end).

Number of geometric isomers

2ⁿ

n is the number of C=C that make cis / trans isomers in unsymmetric molecules (also add chiral centres).

Can't form geometric isomers in cyclic structures

Cyclohexene drawn as a ring, –CH=CH–CH₂–CH₂–CH₂–CH₂–, with the two H atoms on the C=C marked. Can't make the trans version as the cyclic structure needs to be broken.

2. Optical isomers

Enantiomers. Mixture of enantiomers → racemic mixture.

Chiral centre needed: C atom that is making 4 single bonds with 4 different groups.

• tetrahedral, 109.5° bond angles, sp³ hybridised

Sketch: a carbon with four H atoms in a tetrahedral arrangement, one bond up, one to the side, one dashed and one wedge.

Two three-dimensional drawings of 2-chlorobutan-2-ol, CH₃CH₂C(OH)(Cl)CH₃, on either side of a vertical mirror line. The chiral carbon is marked C* and carries Cl (up), OH, CH₂CH₃ (dashed bond) and CH₃ (wedge bond). The left drawing is labelled (+) and the right drawing, its reflection, is labelled (-). These are non-superimposable mirror images.

A third drawing below shows the molecule turned upside down (CH₃ on a wedge at the top, Cl at the bottom, HO on the left, CH₂CH₃ on a dashed bond). A curved arrow labelled "same" joins it to one of the two drawings above: turning the molecule over does not make a new isomer.

Enantiomers have almost identical physical / chemical properties.

• Shape specific functionality varies a lot, e.g. biology / enzyme.

Differentiate between enantiomers

Both enantiomers rotate plane polarised monochromatic light in opposite directions, clockwise (+) and anticlockwise (-).

Homolytic bond fission

Bond breaks equally.

Cl–Cl → Cl• + Cl• (free radicals formed)

Two half arrows start from the Cl–Cl bond, one going to each chlorine atom.

Heterolytic bond fission

Bond breaks unequally.

H–Cl → H⁺ + :Cl^-

One full arrow starts from the H–Cl bond and goes to the chlorine atom.

Curly arrows

Represent movements of electrons.

• Half arrow: movement of 1 electron.
• Full arrow: movement of 2 electrons.

Free radical

A species (atom or molecule) with an unpaired electron. Very reactive.

•Cl, •OH, •CH₃ (single dot: unpaired electron)

Dot-and-cross diagrams are drawn for the hydroxyl radical, •OH, and the methyl radical, •CH₃. In each one the central atom has one unpaired electron.

Nucleophile (attracted to +ve charge)

Species which can donate a pair of electrons and is attracted to +ve charge.

Diagram: curly arrows go to X⁺ from the lone pair of each of these nucleophiles, and from the double bond of ethene:

• NC:^-
• :OH^-
• :NH₃
• H₂O:
• ethene, CH₂=CH₂

Electrophile

Species which can accept a pair of electrons and is attracted to -ve charge.

Types of reactions

Type of reaction	Example
(a) Addition	CH2=CH2 + Br2 → CH2BrCH2Br
(b) Substitution	CH3CH3 + Cl2 → CH3CH2Cl + HCl
(c) Elimination	CH3CH2Cl → CH2=CH2 + HCl

Source of organic compounds

Source of organic compounds → fossil fuel.

Printed diagram: fractionating column. Crude oil passes through a heater and enters the bottom of the column.

Fraction	Temperature	Use
Refinery gas	20 °C	Bottled gas
Petrol	70 °C	Petrol for vehicles
Naphtha	120 °C	Chemicals
Kerosene	170 °C	Jet fuel, paraffin for lighting and heating
Diesel	270 °C	Diesel fuels
Lubricating oil	Lubricating oils, waxes, polishes
Fuel oil	Fuel for ships, factories and central heating
Bitumen	340 °C	Roads and roofing

Small molecules (top of the column)	Large molecules (bottom of the column)
Low boiling point	High boiling point
Light in colour	Dark in colour
Easy to light	Hard to light
Runny	Thick

Alkanes

Saturated hydrocarbons. C_nH_2n+2

Structures drawn: 2-methylpropane, (CH₃)₃CH; cyclopropane, C₃H₆ (a ring of three CH₂ groups); methane, CH₄.

Physical properties

• Non-polar, van der Waals (weak).
• Not soluble in water.

When size of molecule increases:

• more van der Waals
• higher MP, BP
• more viscous
• less flammable

Free radical substitution

Halogenation / chlorination / bromination (for I₂ reaction is very slow, F₂ reaction is very uncontrolled).

CH₃CH₃ + Cl₂ → CH₃CH₂Cl + HCl (UV light)

Also drawn below the product: 1,1,2-trichloroethane, CH₂ClCHCl₂. Many random substitution products.

Reaction can happen with any alkyl chain:

propanoic acid, CH₃CH₂COOH + Cl₂ → 2,3,3-trichloropropanoic acid, CHCl₂CHClCOOH + HCl (UV light)

The alkyl chain of the propanoic acid (the CH₃CH₂ part) is underlined.

Mechanism

1. Step 1, Initiation: free radicals formed, homolytic bond fission because of UV light.
   Cl–Cl → Cl• + Cl• (free radicals formed)
   UV light falls on the Cl–Cl bond. Two half arrows start from the Cl–Cl bond, one going to each chlorine atom.
2. Step 2, Propagation: total number of free radicals remains constant.
   CH₃CH₃ + •Cl → CH₃CH₂• + HCl (ethyl radical formed)
   One half arrow goes from the C–H bond to the carbon atom. A second half arrow from the C–H bond and a half arrow from the unpaired electron of •Cl meet between H and Cl to form the H–Cl bond.
   CH₃CH₂• + Cl–Cl → CH₃CH₂Cl + •Cl
   A half arrow from the unpaired electron of the ethyl radical and a half arrow from the Cl–Cl bond meet to form the C–Cl bond. The other half arrow from the Cl–Cl bond goes to the second chlorine atom, which leaves as •Cl.
3. Propagation (continued, further substitution):
   CH₃CH₂Cl + •Cl → •CH₂CH₂Cl + HCl (chloroethyl radical)
   The •Cl takes a hydrogen atom from the carbon that does not carry the chlorine. Half arrows go from the C–H bond and from the unpaired electron of •Cl.
   •CH₂CH₂Cl + Cl–Cl → CH₂ClCH₂Cl + •Cl
   A half arrow from the unpaired electron on carbon and a half arrow from the Cl–Cl bond form the new C–Cl bond, giving 1,2-dichloroethane, and •Cl is regenerated.
4. Step 3, Termination: two free radicals join, no more free radicals → reaction stops.
   •CH₂CH₂Cl + Cl• → CH₂ClCH₂Cl
   CH₃CH₂• + •Cl → CH₃CH₂Cl
   CH₃CH₂• + •CH₂CH₂Cl → CH₃CH₂CH₂CH₂Cl
   In each case a half arrow from the unpaired electron of each radical meets to form the new bond. In the last product the new C–C bond in the middle of the chain is marked.
   Bigger molecules are produced than the original molecule, because two radicals join.

(H₂ gas molecules are never produced as there are no H• radicals.)

Speed up free radical: higher concentration of free radicals.

Pb(CH₃)₄, tetramethyl lead(IV) → •CH₃

Free radical substitution of propane

CH₃CH₂CH₃ + Cl₂ → products (UV light)

Products drawn:

• 1,2,2-trichloropropane, CH₂ClCCl₂CH₃ (ticked)
• 1,1,1,2,2-pentachloropropane, CH₃CCl₂CCl₃ (ticked)
• 2-chlorohexane, CH₃CH₂CH₂CH₂CHClCH₃

How the bigger molecules form (two radicals join):

From	To
propyl radical, CH3CH2CH2• + 2-chloropropyl radical, •CH2CHClCH3	2-chlorohexane, CH3CH2CH2CH2CHClCH3
3-chloropropyl radical, CH2ClCH2CH2• + secondary radical CHCl2–•CH–CH2Cl	CH2ClCH2CH2CH(CHCl2)CH2Cl

Not formed (arrow crossed out): CH₃CHClCH(CH₃)CCl₂CCl₃ (as drawn). Can't split it into two propyl radicals.

All hydrocarbons (or most organic compounds): combustion

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

O₂ in "excess": complete combustion.

In incomplete combustion C (soot) and CO are produced.

CO: deadly poisonous gas, causes choking, combines irreversibly with haemoglobin.

Cracking of long chain hydrocarbons

From	Reagents and conditions	To
Fossil fuel → large molecules (not useful) high MP/BP, more van der Waals solid, semi-solid, wax very viscous less flammable	Catalytic cracking T > 450 °C catalyst: Al2O3 / SiO2 pumice / zeolite	Small molecules (alkanes / alkenes) + hydrogen gas low MP/BP, less van der Waals less viscous very flammable

Drawn example: octane, CH₃(CH₂)₆CH₃, with the chain broken between the sixth and seventh carbon atoms → hex-1-ene, CH₃CH₂CH₂CH₂CH=CH₂ + H₂(g) + ethene, CH₂=CH₂

C₈H₁₈ can crack in different ways:

• C₈H₁₈ → C₂H₆ (ethane) + C₆H₁₂ (hexene)
• C₈H₁₈ → CH₄ (alkane) + C₃H₆ (alkene) + C₄H₈ (alkene)
• C₈H₁₈ → C₃H₆ + C₅H₁₀ + H₂

Alkenes

C_nH_2n

Ethene, CH₂=CH₂. Each carbon atom of the double bond is sp².

Diagram of ethene with the π electron cloud above and below the C–C bond: high charge density π electron cloud.

An arrow points to the cloud from: electrophiles (+ve ions, δ+ charge, electron deficient).

Electrophilic addition reactions

CH₃CH=CH₂ + X–Y (electrophile) → CH₃CHXCH₂Y

A curly arrow goes from the C=C double bond to X–Y.

1. Bromination reaction

CH₃CH=CH₂ + Br–Br (electrophile) → CH₃CHBrCH₂Br (r.t.p., solvent: aqueous or inert (CCl₄))

A curly arrow goes from the C=C double bond to Br–Br.

If aqueous: the product is CH₃CH(OH)CH₂Br.

Test for identification of unsaturated compounds: +ve test → red / orange / brown Br₂ decolourises.

Electrophilic addition mechanism

Drawn first with the π cloud of ethene, then with displayed formulae (boxed):

1. Step 1: a curly arrow goes from the C=C double bond (the π electron cloud) of ethene, CH₂=CH₂, to the Br^δ+ of Br^δ+–Br^δ-. A second curly arrow goes from the Br–Br bond to the Br^δ-. This forms the carbocation intermediate, ⁺CH₂CH₂Br, and :Br^-.
2. Step 2: a curly arrow goes from the lone pair of :Br^- to the C⁺ of the carbocation. The product is 1,2-dibromoethane, CH₂BrCH₂Br.

2. Hydration of alkenes (electrophilic addition)

CH₃CH=CH₂ (unsymmetric) + H₂O → products (300 °C, 60 atm, H₃PO₄ catalyst)

• CH₃CH₂CH₂OH, propan-1-ol, minor product
• or CH₃CH(OH)CH₃, propan-2-ol, major product

Markovnikov's Rule

H likes to be bonded to C in C=C which is already bonded to more H-atoms.

(CH₃)₂C=CH₂ + H₂O → (CH₃)₂C(OH)CH₃ (major product)

In (CH₃)₂C=CH₂ the left carbon of the double bond has 0 × H and the right carbon has 2 × H.

Mechanism for the major product:

1. A curly arrow goes from the C=C double bond of (CH₃)₂C=CH₂ to the H^δ+ of H^δ+–OH^δ-. A second curly arrow goes from the H–O bond to the O^δ-. The H joins the CH₂ carbon, forming the 3° carbocation (CH₃)₂C⁺CH₃ and :OH^-.
2. A curly arrow goes from the lone pair of :OH^- to the C⁺. The product is (CH₃)₂C(OH)CH₃, the major product.

Mechanism for the minor product:

1. A curly arrow goes from the C=C double bond of (CH₃)₂C=CH₂ to the H^δ+ of H^δ+–OH^δ-. A second curly arrow goes from the H–O bond to the O^δ-. This time the H joins the carbon carrying the two CH₃ groups, forming the 1° carbocation (CH₃)₂CHCH₂⁺ and :OH^-.
2. A curly arrow goes from the lone pair of :OH^- to the C⁺. The product is (CH₃)₂CHCH₂OH, the minor product.

Carbocations with more alkyl chains, which are electron donating, are more stable, as they have lower charge density.

1° carbocation	2° carbocation	3° carbocation
CH3CH2+	(CH3)2CH+	(CH3)3C+
CH3 (9 electrons) and two H (1 electron each) around C+ high charge density	three CH3 (9 electrons each) around C+ low charge density
unstable	→	stable

3. Hydrogenation

CH₃CH=CH₂ (unsaturated) + H₂ → CH₃CH₂CH₃ (saturated) (Ni / Pt catalyst, 150 °C)

Molecule size slightly increases, van der Waals increase, higher MP/BP.

Printed diagram: unsaturated vegetable oil (a triglyceride with linoleic, linoleic and oleic chains) → margarine (a triglyceride with stearic, oleic and oleic chains) (H₂, Ni).

Heterogeneous catalyst → drawn as ethene, CH₂=CH₂, and H–H above a nickel surface, in three stages:

1. Ethene and H–H approach the Ni surface.
2. Adsorption: catalyst and reactants form very weak bonds.
3. Desorption: weak bonds between catalyst / reactants break, now new bonds can form. The H atoms join the carbon atoms and the product leaves the Ni surface.

4. Hydrogen halides (electrophilic addition)

CH₃CH=CH₂ → CH₃CHClCH₃ (major) or CH₃CH₂CH₂Cl (minor) (HCl; also HBr, HI)

Mechanism (boxed):

1. A curly arrow goes from the C=C double bond of CH₃CH=CH₂ to the H^δ+ of H^δ+–Cl^δ-. A second curly arrow goes from the H–Cl bond to the Cl^δ-. This forms the carbocation CH₃C⁺HCH₃ and :Cl^-.
2. A curly arrow goes from the lone pair of :Cl^- to the C⁺. The product is CH₃CHClCH₃.

Oxidation of alkenes

1. Mild oxidation

Cold / dilute KMnO₄

CH₃CH=CH₂ + [O] + H₂O → CH₃CH(OH)CH₂OH

Diols formed. Propane-1,2-diol.

Skeletal example: 4-(prop-1-enyl)cyclohexene (a cyclohexene ring with a –CH=CHCH₃ side chain) → 4-(1,2-dihydroxypropyl)cyclohexane-1,2-diol. An OH group is added to each of the four carbon atoms that were in the two double bonds.

2. Strong oxidation

Hot, concentrated, acidified KMnO₄ (potassium manganate(VII)).

* Double bond breaks completely.

From (each end of C=C)	To
1. =CH2	CO2 + H2O
2. =CHR	RCHO (aldehyde) → RCOOH (carboxylic acid)
3. =CR2	R2C=O (ketone)

Example: 2-methylpent-1-ene, CH₃CH₂CH₂C(CH₃)=CH₂. The double bond is broken:

• =CH₂ → CO₂ + H₂O
• CH₃CH₂CH₂C(CH₃)= → CH₃CH₂CH₂C(CH₃)=O (ketone)

Skeletal example: 2-methylbut-2-ene, (CH₃)₂C=CHCH₃, with the double bond broken → ethanoic acid, CH₃COOH (from the =CHCH₃ end) + propanone, (CH₃)₂C=O (from the =C(CH₃)₂ end).

Assignment

Draw products for both mild and strong oxidation.

1. 2,3-dimethylpent-2-ene
2. 2-methylbut-1-ene
3. Skeletal structure: 4-methylcyclohexene
4. Skeletal structure: 1-methylcyclohexa-1,4-diene (as read from the sketch)
5. Skeletal structure: 2,3,5-trimethylhexa-2,4-diene, (CH₃)₂C=C(CH₃)CH=C(CH₃)₂ (as read from the sketch)
6. penta-1,3-diene

Alkenes, addition polymers

Polymer: a very large macromolecule formed when hundreds of monomers join together.

Monomer: ethene, CH₂=CH₂ (six ethene molecules drawn side by side)

→ polyethene: –CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂– with continuation bonds at each end.

Polyethene: plastic bags etc.

Repeat unit: –(CH₂–CH₂)–_n

2-methylpropene

2-methylpropene, (CH₃)₂C=CH₂. The C=C is highlighted. The CH₃ group and the H drawn in line with the double bond are circled, with arrows showing them being moved to above and below the carbon atoms, so the monomer is redrawn with the C=C horizontal and the four groups above and below it.

Five 2-methylpropene monomers drawn side by side → polymer chain –C(CH₃)₂–CH₂–C(CH₃)₂–CH₂–C(CH₃)₂–CH₂–C(CH₃)₂–CH₂–C(CH₃)₂–CH₂– with continuation bonds at each end.

Repeat unit: –(C(CH₃)₂–CH₂)–_n

Two unlabelled sketches at the side: many long polymer chains tangled across each other, and two long chains joined by short links between them.

PVC

Polyvinyl chloride: pipes, tiles, doors, windows.

Monomer: chloroethene, CH₂=CHCl (five drawn side by side) → polymer chain –CH₂–CHCl–CH₂–CHCl–CH₂–CHCl–CH₂–CHCl–CH₂–CHCl– with continuation bonds at each end. The Cl atoms are marked δ- and an H atom is marked δ+.

Dipoles, stronger attraction between polymer chains.

PTFE

Teflon.

Monomer: tetrafluoroethene, CF₂=CF₂. Repeat unit: –(CF₂–CF₂)–_n

• No dipoles, less intermolecular, non-stick in cooking.
• C–F strong, heat resistant.

* Addition polymers: non-biodegradable. C–C bonds are strong and unreactive / non-polar.

Halogenoalkanes

Structure drawn: 2-bromo-1-chlorobutane, CH₃CH₂CHBrCH₂Cl, with the Br and the Cl circled.

• Higher (slightly) melting and boiling points. Polar molecules: chloromethane, CH₃Cl, drawn with C^δ+–Cl^δ-. Strength of dipoles decreases down the group, electronegativity of halogen decreases (bigger molecules have more van der Waals).
• Very slightly soluble in water.

Formation of halogenoalkanes

1. Free radical substitution (see alkanes)

CH₃CH₃ + 2Cl₂ → CH₂ClCH₂Cl + 2HCl (UV light)

2. Electrophilic addition of alkenes (see alkenes)

CH₂=CH₂ → CH₂BrCH₂Br (Br₂ / HCl / HBr)

3. Alcohols (nucleophilic substitution)

From	Reagents and conditions	To
CH3CH2OH	1. PCl3 + heat 2. PCl5, vigorous at r.t.p. 3. SOCl2 + heat (thionyl chloride) 4. HCl + heat (NaCl + conc. H2SO4)	CH3CH2Cl
CH3CH2OH	1. HBr + heat 2. PBr3 + heat (P + Br2)	CH3CH2Br
CH3CH2OH	1. HI + heat 2. PI3 + heat (P + I2)	CH3CH2I

Printed diagram: Heating under Reflux. Labels, from the top: stopper, condenser (water outlet at the top, water inlet at the bottom), round bottom flask, Bunsen burner.

Primary halogenoalkane	RCH2Cl
Secondary halogenoalkane	R2CHCl
Tertiary halogenoalkane	R3CCl

Skeletal example with each halogen labelled: CH₂Cl–CHBr–CCl(CH₃)–CH₂Cl (as read from the sketch). The Cl at each end is primary, the Br is secondary and the Cl on the carbon with the methyl branch is tertiary.

Nucleophilic substitution, S_N

From	Reagents and conditions	Nucleophile	To
CH3CH2Cl	NaOH(aq) + reflux	:OH-	CH3CH2OH
CH3CH2Cl	NaCN (ethanolic), reflux	:CN-	CH3CH2CN
CH3CH2Cl	NH3 (ethanolic) + reflux or NH3(g) in a sealed container + heat	":NH3"	CH3CH2NH2

Nucleophile: species which is attracted to +ve charge, donates a pair of electrons.

S_N2 Reaction Mechanism: Primary Halogenoalkanes

1-chloropropane, CH₃CH₂CH₂Cl, reacting with the hydroxide ion, HO^-. The carbon bonded to chlorine is δ+ and the chlorine is δ-.

Other nucleophiles (boxed): NC^- and H₃N, each with a lone pair.

1. A curly arrow starts from the lone pair on the oxygen of HO^- and goes to the δ+ carbon. At the same time a curly arrow starts from the C–Cl bond and goes to the chlorine.
2. Transition state: [HO···C(H)(H)(CH₂CH₃)···Cl]^-, with partial bonds from carbon to both HO and Cl. C is partially bonded to 5 groups.
3. The C–Cl bond breaks completely, giving propan-1-ol, CH₃CH₂CH₂OH, and a chloride ion, Cl^-.

CH₃CH₂CH₂Cl + OH^- → CH₃CH₂CH₂OH + Cl^-

Rate depends on both reactants. Rate ∝ [R–Cl][OH^-]

S_N2: One Step Mechanism.

S_N1 Reaction Mechanism: Tertiary Halogenoalkanes

2-chloro-2-methylbutane, CH₃C(CH₃)(Cl)CH₂CH₃. The carbon bonded to chlorine is δ+ and the chlorine is δ-.

1. Slow step: a curly arrow starts from the C–Cl bond and goes to the chlorine. The bond breaks to give an intermediate carbocation, CH₃C⁺(CH₃)CH₂CH₃, and a chloride ion, Cl^-. Intermediate carbocation formed.
2. Fast step: a curly arrow starts from the lone pair on OH^- (or CN^-, NH₃) and goes to the positive carbon, giving 2-methylbutan-2-ol, CH₃C(CH₃)(OH)CH₂CH₃.

Rate ∝ [R–Cl] ([OH^-] is crossed out). Rate depends on only one reactant.

2 step reaction.

Why do primary halogenoalkanes undergo S_N2 and tertiary halogenoalkanes undergo S_N1?

Primary	Tertiary
Chloroethane, CH3CH2Cl. If the C–Cl bond broke (curly arrow from the C–Cl bond to Cl) it would give CH3CH2+ and Cl-. The positive charge on the carbon is large.	2-chloro-2-methylpropane, (CH3)3CCl. The C–Cl bond breaks (curly arrow from the C–Cl bond to Cl) to give (CH3)3C+ and Cl-. Each of the three CH3 groups donates electrons to the positive carbon, so the positive charge is small.
Primary carbocations have less electron donating alkyl chains, which increases charge density on the carbocation, making it very unstable, hence SN2 mechanism.	Tertiary carbocations have more electron donating alkyl chains, which decreases charge density on the carbocation, making it very stable, hence SN1 mechanism.

Secondary: both S_N1 / S_N2, depends on conditions.

Reactivity of Halogenoalkanes

R–I > R–Br > R–Cl

R–I: very reactive.

R–Cl: least reactive.

• smaller atoms, make stronger bonds
• shorter bond length
• more orbital overlap
• less shielded nuclei

Elimination Reactions

2-chlorobutane, CH₃CH₂CHClCH₃. The Cl is underlined, and a hydrogen on each of the two carbons next to the C–Cl carbon is underlined with a question mark (either one can be lost with the Cl).

From	Reagents and conditions	To
2-chlorobutane, CH3CH2CHClCH3	NaOH (ethanolic), reflux	but-1-ene, CH3CH2CH=CH2 + HCl
2-chlorobutane, CH3CH2CHClCH3	NaOH (ethanolic), reflux	but-2-ene, CH3CH=CHCH3 + HCl

Identification of halogen in halogenoalkane

R–X ⇌ R⁺ + X^-

X = Cl, Br, I

Ion	Add aqueous AgNO3	Dissolve in NH3
Cl-	AgCl(s), white ppt	dissolves in aqueous NH3
Br-	AgBr(s), cream ppt	dissolves in conc. NH3
I-	AgI(s), yellow ppt	does not dissolve

Uses of Halogenoalkanes

1. CFCs, chlorofluorocarbons. Example drawn: 1,1,2-trichloro-1,2,2-trifluoroethane, CF₂ClCFCl₂. Inert solvents, volatile.
   • solvents, coolants, freons, anaesthetics, aerosols
2. Fire retardants: bromine related compounds.
3. Polymers → PVC: pipes, tiles. Teflon, PTFE: non stick in cooking utensils.

Harmful effect of CFC

Destroy ozone. Free radicals in upper atmosphere / strong UV rays.

Trichlorofluoromethane, CFCl₃: the C–Cl bond breaks, with one electron of the bond going to each atom.

CFCl₃ → •CFCl₂ + •Cl

Cl• radicals attack O₃, ozone.

Cl• + O₃ → ClO• + O₂

ClO• + O₃ → Cl• + O₂ + O₂

Ozone reflects harmful UV rays.

Alcohols

Ethanol, CH₃–CH₂–OH. The –OH group is the hydroxy group.

1. high melting / boiling points
2. soluble in water

→ form hydrogen bonds.

Diagram: methanol, CH₃–O–H, with two lone pairs on the oxygen (δ-) and δ+ on the hydrogen. A dotted line (hydrogen bond) runs from the δ+ hydrogen of the alcohol to a lone pair on the δ- oxygen of a water molecule, H–O–H, whose hydrogens are δ+.

Primary alcohol	Secondary alcohol	Tertiary alcohol
propan-1-ol, CH3CH2CH2OH	pentan-3-ol, CH3CH2CH(OH)CH2CH3	3-ethylpentan-3-ol, (CH3CH2)3COH

Formation of Alcohols

1. Hydration of alkenes (see Alkenes)

CH₂=CH₂ + H₂O → CH₃CH₂OH (300 °C, 60 atm, H₃PO₄ catalyst)

Dehydration of alcohols

Butan-2-ol, CH₃CH₂CH(OH)CH₃. The OH is crossed out, and a hydrogen on each of the two carbons next to the C–OH carbon is underlined (either one can be lost with the OH).

From	Reagents and conditions	To
butan-2-ol, CH3CH2CH(OH)CH3	concentrated H2SO4 (dehydrating agent), or heat with Al2O3 / SiO2 (pumice, zeolite)	but-2-ene, CH3CH=CHCH3
butan-2-ol, CH3CH2CH(OH)CH3	concentrated H2SO4 (dehydrating agent), or heat with Al2O3 / SiO2 (pumice, zeolite)	but-1-ene, CH3CH2CH=CH2

2. S_N nucleophilic substitution of halogenoalkanes (see Halogenoalkanes)

CH₃CH₂Cl → CH₃CH₂OH (NaOH(aq), reflux)

Alcohols are extremely weak acids (not acids, no acid-base reactions)

Ionization is even weaker compared to H₂O.

CH₃–CH₂–O–H ⇌ CH₃–CH₂–O^- + H⁺

An arrow from the CH₃–CH₂ group points towards the O^-, and the charge on the oxygen is labelled large. Almost no ionization.

Alcohols do react with Group 1 reactive metals, very slow.

CH₃CH₂–O–H + Na → CH₃CH₂–O^- Na⁺ + H₂

The H of the O–H gains an electron. Na loses an electron. Slight effervescence.

cyclohexane-1,3-diol, C₆H₁₀(OH)₂ + 2Na → disodium salt of cyclohexane-1,3-diol, C₆H₁₀(O^-Na⁺)₂ + H₂

Oxidation of Alcohols

Primary alcohols → aldehydes → carboxylic acids

From	Reagents and conditions	To
primary alcohol, R–CH2–OH	1. K2Cr2O7 / H+, heat with reflux (orange to green) 2. KMnO4 / H+, heat with reflux (purple to colourless)	aldehyde, R–CHO
aldehyde, R–CHO	1. K2Cr2O7 / H+, heat with reflux (orange to green) 2. KMnO4 / H+, heat with reflux (purple to colourless) 3. Tollens reagent, [Ag(NH3)2]+ (mix NH3(aq) + AgNO3(aq)). Silver mirror formed, black ppt. Ag+ + 1e- → Ag 4. Fehling solution + heat. Cu2+ gets reduced to a brick red ppt of Cu2O	carboxylic acid, R–COOH

Stopping the reaction at the aldehyde: use distillation + heat instead of heat with reflux.

Aldehydes have a lower boiling point compared to alcohols, carboxylic acid, so they will immediately evaporate once they are formed.

Secondary alcohols → ketone (no further oxidation)

From	Reagents and conditions	To
secondary alcohol, R–CH(OH)–R	1. K2Cr2O7 / H+, heat with reflux (orange to green) 2. KMnO4 / H+, heat with reflux (purple to colourless)	ketone, R–CO–R

* Tertiary alcohols are not oxidized.

Examples: Oxidation of alcohols

Skeletal structure of the starting compound: a cyclohexane ring carrying four alcohol groups.

• One ring carbon carries a methyl group and an –OH group, labelled tertiary.
• The next ring carbon carries a –CH₂OH group, labelled primary.
• The ring carbon opposite the tertiary one carries a methyl group and a side chain –CH(OH)–CH(OH)–CH₃. Both –OH groups on the side chain are labelled secondary.

Reagents and conditions: K₂Cr₂O₇ / H⁺, reflux.

Product: the tertiary –OH is unchanged, the primary –CH₂OH becomes –COOH, and the two secondary –CH(OH)– groups become C=O, giving the side chain –CO–CO–CH₃.

Reduction

Gain of H / loss of O.

From	Reagents and conditions	To
carboxylic acid: ethanoic acid, CH3COOH	LiAlH4	aldehyde: ethanal, CH3CHO
aldehyde: ethanal, CH3CHO	LiAlH4 or NaBH4	primary alcohol: ethanol, CH3CH2OH
ketone: propanone, CH3COCH3	LiAlH4 or NaBH4	secondary alcohol: propan-2-ol, CH3CH(OH)CH3

LiAlH₄, reducing agent. Lithium aluminium hydride. AlH₄^- is drawn as AlH₃ with a fourth H^- (carrying a lone pair) bonded to the Al. It is a source of H^- ions.

In R–CO–R the carbonyl carbon is δ+. The lone pair on H^- attacks the δ+ carbon and the C=O bond pair moves to the oxygen. H^- ions are nucleophiles.

Example: the product of the oxidation example above (cyclohexane ring with a tertiary –OH and methyl on one carbon, –COOH on the next carbon, and a methyl and –CO–CO–CH₃ side chain on the opposite carbon) + LiAlH₄ → the –COOH becomes –CH₂OH and the two C=O groups become –CH(OH)–, giving the side chain –CH(OH)–CH(OH)–CH₃. The tertiary –OH is unchanged.

Uses of Alcohols

• Alcoholic drinks.
• Industrial methylated spirits.
• Use of ethanol as a fuel.
• Ethanol as a solvent.
• Methanol as a fuel.
• Methanol as an industrial feedstock.

Aldehyde and Ketones

Aldehyde: R–CHO (R–C(=O)–H). Ketone: R–CO–R (R–C(=O)–R).

Carbonyl compound: C=O, with O δ- and C δ+. Polar molecule. Permanent dipole - permanent dipole attraction.

Example

• CH₃CH₂CHO, propanal
• CH₃COCH₂CH₃, butan-2-one

Identification of Carbonyl Compounds

Add 2,4-DNPH (2,4-dinitrophenylhydrazine), H₂NNHC₆H₃(NO₂)₂ → orange precipitate.

R–CO–R + H₂N–NH–C₆H₃(NO₂)₂ → H₂O + R–C(R)=N–NH–C₆H₃(NO₂)₂ (condensation)

The O of the C=O and the two H atoms of the –NH₂ group are circled together. They leave as H₂O. The product, with a C=N bond, is the orange precipitate.

How to differentiate between aldehyde / ketone

Aldehydes get oxidized, ketones don't.

aldehydes → carboxylic acids, R–COOH ("oxidation")

1. K₂Cr₂O₇ / H⁺, heat with reflux (orange to green)
2. KMnO₄ / H⁺, heat with reflux (purple to colourless)
3. Tollens reagent, [Ag(NH₃)₂]⁺ (mix NH₃(aq) + AgNO₃(aq)). Silver mirror formed, black ppt. Ag⁺ + 1e^- → Ag
4. Fehling solution + heat. Cu²⁺ gets reduced to a brick red ppt of Cu₂O

Carbonyl Compounds, nucleophilic addition (mechanism, 2 step reaction)

R–CO–R, with the carbonyl carbon δ+ and the oxygen δ-.

1. A curly arrow starts from the lone pair on CN^- (from catalyst) and goes to the δ+ carbon. A second curly arrow starts from the C=O double bond and goes to the oxygen. Intermediate formed: R–C(O^-)(CN)–R, with a lone pair and a negative charge on the oxygen.
2. A curly arrow starts from the lone pair on the O^- of the intermediate and goes to H⁺. Product: cyanohydrin, R–C(OH)(CN)–R.

R–CO–R → R–C(OH)(CN)–R (HCN, NaCN catalyst)

Iodoform Test (NaOH(aq) + I₂)

Identifies a specific structure.

• H/R–CO–CH₃ (a C=O with a CH₃ group attached)
• H/R–CH(OH)–CH₃ (a CH(OH) with a CH₃ group attached)

Not carboxylic acid or esters. R represents an alkyl chain.

The bond between the carbon and the CH₃ group is marked as the bond that breaks. Two products:

• CHI₃, triiodomethane, yellow ppt
• H/R–COO^- Na⁺

Alkaline condition: RCOOH → RCOO^-Na⁺

Further acidification can form RCOOH.

Carboxylic Acids

R–COOH (R–C(=O)–OH)

• hydrogen bonds
• high melting / boiling points
• very soluble in water

Weak acids

pH = 3 - 7

Partially dissociate and produce fewer H⁺ ions.

R–COOH ⇌ R–COO^- + H⁺

Acid + Metal → Salt + H₂

RCOOH + Na → RCOO^-Na⁺ + ½H₂

Redox reaction. Na loses an electron, H⁺ gains an electron.

Acid + Metal oxide / hydroxide → Salt + H₂O

RCOOH + NaOH → RCOO^-Na⁺ + H₂O

Acid + Metal carbonate → Salt + H₂O + CO₂

2RCOOH + Na₂CO₃ → 2RCOO^-Na⁺ + H₂O + CO₂

Acid + NH₃ → Ammonium salt

RCOOH + NH₃ → RCOO^-NH₄⁺

Esterification

Formation of esters

R–CO–OH + H–O–R → R–CO–O–R + H₂O (condensation, conc. H₂SO₄ catalyst, heat with reflux)

The OH of the acid and the H of the alcohol's O–H are crossed out. They leave as H₂O. The carbonyl carbon of the acid is δ+ and the alcohol oxygen has lone pairs.

Hydrolysis of esters

R–CO–O–R + H₂O → R–CO–OH + H–O–R (hydrolysis, dil. NaOH(aq) or dil. H₂SO₄(aq), reflux)

An arrow from H₂O points to the C–O single bond of the ester, the bond that breaks.

In alkaline hydrolysis the acid reacts with the alkali: RCOO^-Na⁺

Naming of esters

CH₃–CH₂–CH₂–CO–O–CH₃, split by a line between the C=O carbon and the –O–.

• CH₃–CH₂–CH₂–CO: 4C, butanoate
• O–CH₃: 1C, methyl

Hydrolysis of Nitriles

R–C≡N + H₂O → R–COOH + NH₄⁺ (dilute H₂SO₄, reflux, or dilute NaOH, reflux)

In alkaline conditions: RCOO^-Na⁺

RCOO^-Na⁺ → RCOOH (add acid, H⁺)

Infrared Spectroscopy

Diagram: propanone, CH₃COCH₃, drawn in three dimensions. Infrared radiation (labelled infrared spectrum) is shown arriving at the C=O bond. Double-headed arrows on the C=O bond and on a C–H bond are labelled vibrate.

Each bond will vibrate / stretch at a very specific frequency.

Table 4.2 (1/λ ∝ frequency)

bond	functional group containing the bond	characteristic infrared absorption range (in wavenumbers) / cm-1
C–O	hydroxy, ester	1040–1300
C=C	aromatic compound, alkene	1500–1680
C=O (circled)	amide carbonyl, carboxyl ester	1640–1690 1670–1740 1710–1750 (all three circled)
C≡N	nitrile	2200–2250
C–H	alkane	2850–3100
N–H	amine, amide	3300–3500
O–H (circled)	carboxyl hydroxy (alcohol)	2500–3000 3200–3650

40 Which diagram shows the infrared spectrum of a compound that contains both a C=O and an O–H group?

Four infrared spectra, A to D, each plotted as transmittance / % (0 to 100) against wavenumber / cm^-1 (from above 3000 down to below 1000). The teacher's markings:

Spectrum	What the spectrum shows	Teacher's marking
A	A sharp absorption just below 3000 and a strong sharp absorption near 1700.	The absorption near 3000 is marked with a cross. The absorption near 1700 is marked C=O. The region below about 1500 is marked hard to identify.
B	A broad absorption around 3000 to 3500 and a strong sharp absorption near 1700.	The broad absorption is marked alcohol. The absorption near 1700 is marked C=O. B is circled as the answer.
C	A strong sharp absorption just below 3000 and a sharp absorption near 1400.	The absorption near 1400 is marked probably not C=O.
D	A broad, strong absorption around 3300 and absorptions below about 1500.	The broad absorption is marked alcohol. Marked no C=O.