[grapesugar]

Aldehydes and Ketones

Reduction
Reagents: NaBH4
Reaction: Nucleophilic Addition


Aldehyes are reduced to primary alcohols
CH3CH2CHO + 2[H] gives CH3CH2CH2OH

Ketones are reduced to secondary alcohols
CH3COCH3 + 2[H] gives CH3CH(0H)CH3

Aldehyde or Ketone???

[/b]Reaction with 2,4-dinitrophenylhydrazine
Add this to mixture, get bright orange precipitate if aldehyde or ketone present.


THEN,
Using the product from this reaction

• 
  recystallise
  Measure Melting Point
  Use data book to find out structure

Test for aldehydes:


Oxidation tests because only aldehydes undergo oxidation! RCH0 + [0] gives RCOOH.

1st method
refluxing with acidified dichromate(VI)(aq) ions
Results: Orange to green shows aldehyde has been oxidised to carboxylic acid.
2nd method
use tollens reagent (made by ammonia & silver nitrate) warm this mixture.
Results: Silver mirror forms if aldehyde has been oxidised to a carboxylic acid.
Aldehyde gets oxidised to carboxylic acid, while Ag+ ions are reduced to silver metal.

Another reaction we must know:

Formation of 2-hydroxypropanoic acid using ethanal
Stage 1
Reagents: HCN
Conditions KCN
Reaction Nucleophilic Addition
Equation: CH3CHO + HCN ® CH3CH(OH)CN

Stage 2
Reagents: H+ (aq)
Conditions : Warm, Reflux
Reaction Hydrolysis
Equation: CH3CH(OH)CN + 2H20 + H+ gives CH3CH(OH)COOH + NH4

--------------

If these are any help to anyone, then I can put on all the other stuff we need to know for this unit?!?!

[Shehna]

http://lanther.co.uk/notes/CHM4.pdf

Hope it helps, thanxs for the notes cordelia

[Kooldood]

Nice one cordelia. 7 days to go. You got anything on NMR spec.

[grapesugar]

NMR Checklist
Count the number of peaks
Explain the positions of the peak:
Say what type of proton (e.g. the protons in the CH3 group)
Say where the chemical shift is
Say ‘this is consistent with the data sheet value’
How many protons are in that environment
Spitting (use n + 1 ) rule, where n is the number of hydrogen’s on the adjacent carbon


Use of D2O in N.M.R.
Protons in -OH are labile.
If an organic molecule contains –OH groups & is mixed with deuterium oxide then the protons are replaced with deuterium atoms . Since the deuterium atom has an even number of particles in its nucleus (a proton and a neutron) it does not show up in proton N.M.R.
So, if spectra are taken of a molecule before and after the use of D2O, a comparison of the two spectra can reveal any labile hydrogens in the molecule.

For good nmr practice go to: http://www.chem.uic.edu/web1/OCOL-II...NMR/FRAMES.HTM

[grapesugar]

Arenes

Structure of Bezene

• Planar Molecule
• All C-C bonds same length
• P orbitals above and below the ring
• Overlapping of p-orbitals forms pi bonds
• Electrons are delocalised

Resistance to electrophilic addition
· Doesn't polarise electrophiles well
· Electrophiles less attracted
· Stable delocalised system would need to be disrupted

Electrophilic Substitution
Nitration of bezene
Reagents : HNO3 and H2SO4 (catalyst)
Conditions: approx. 60°C
Equation C6H6 + HNO3 gives C6H5NO2 + H20
Mechanism:
Step 1 = generation of electrophile: 2H2SO4 + HNO3 2HSO4- + NO2+ + H3O+
Step 2 = Electrophilic Substitution
Step 3 = regeneration of catalyst: H+ + HS04- gives H2SO4

Halogenation = Friedal-Crafts (can do this with Br2 as well)
Reagents : Cl2 and ALCL3 (catalyst)
Conditions: anhydrous
Equation C6H6 + Cl2 gives C6H5Cl+ HCl
Mechanism:
Step 1 = generation of electrophile: Cl2 + ALCL3 gives Cl+ + AlCl4
Step 2 = Electrophilic Substitution
Step 3 = regeneration of electrophile: H+ + AlCl4 gives AlCl3 + HCl

Phenols
Form salts by its reactions with NaOH and Na
C6H5OH + NaOH gives C6H5O-Na+ + H20
C6H5OH + Na gives C6H5O-Na+ + 1/2 H2

Phenols react with bromine. the bromine is decolorised and white cystals of 2,4,6-tribromophenol are formed.

Phenol Vs Bezene

• Phenol does not require halogen carrier & reacts instantly with bromine
• OH- group activates benzene ring
• This increases electron density around ring (especially at 2,4, 6)
• This polarises the halogen & increases the attraction for the halogen

uses of phenols
TCP is used as antiseptics & disinfectants!

Oh ya, and Shehna thanks for that link, tis good

[Day Dream]

thanks for these! they are saving my life lol! dyou think u could get them all in like a document or summin tho instead of havin dem in diff posts? pwease?

[Aired]

Nice notes! Just one thing...

Test for aldehydes:


Oxidation tests because only aldehydes undergo oxidation! RCH0 + [0] gives RCOOH.

1st method
refluxing with acidified dichromate(VI)(aq) ions
Results: Orange to green shows aldehyde has been oxidised to carboxylic acid.

That could be a bit confusing. That isn't the test for an aldehyde as a positive result can also show that a primary alcohol was present. If you knew that the substance you were testing was either an aldehyde or a ketone though, it would show that an aldehyde was present.

[Sasuke!]

safe

[stratomaster]

Here's the revision notes we use in college:

http://www.jeffrowell.co.uk/mod4.html

[grapesugar]

oOo those notes are really good, far better than mine =) thankyou!

[Sasuke!]

your a re better

[Kooldood]

Excellent Notes Stratomaster And Cordelia!!! Keep It Up

[Jasdeep]

great notes there mate, im just checking they are ocr right? seems the same as what im doing.

[Bravery]

Safe hustler

[grapesugar]

Polymers

Addition Polymerisation
Sigma bond breaks
Many molecules join on

Addition Polymers can have different structures because:

For any polymer other than poly(ethene), the carbon atom attached to the R group in the alkene unit becomes a chiral centre when a polymer chain is formed: (with 4 different groups attached)

Side groups can be arranged in either:
Alternating chirality = Syndiotactic
Random chirality = Atactic
Same chirality = Isotactic

Properties

Isotactic = chains closely packed = Strong intermolecular forces = High Melting Point

Syndiotactic = chains closely packed = Strong intermolecular forces = High Melting point

Atactic = lack of regularity makes it impossible for the chains to lie closely together = Weaker intermolecular forces = Low Melting point

Condensation Polymerisation
A molecule of water is eliminated
Ester or peptide link will be formed
Polyamides = di-carboxylic acid + diamine = amide link
Polyesters = di-carboxylic acid + diol = ester link
Polypeptides/Proteins = many amino acids joined = amide link

Polymers we MUST know (for OCR)

1. Terylene
Monomers = benzene-1,4-dicarboxylic acid and ethane-1,2-diol
Uses = fibres, clothing
2. Kelvar
Monomers = benzene-1,4-dicarboxylic acid and benzene-1,4-diamine
Uses = bullet proof clothing, tennis rackets
3. Nylon-6,6
Monomers = hexane-1,6-dicarboxlic acid and 1,6-diaminohexane
Uses = fibres, clothing

Dipeptides

Two amino acids join together.
Via condensation reaction
COOH reacts with NH2 group.
Amide/Peptide link formed.
Can be two different structures because you can switch the two R groups around in position.

Polypeptides
Many amino acid monomers join in condensation reaction, forming amide links

[stratomaster]

Cool, thanks for typing that up. I've gotta do a lot of revision for this module!

[Kooldood]

Yeh me too after looking at all that, although I know the stuff i just have to revise the proper way of saying it. Any notes on the AS stuff which we need to know about for this Chains, rings and spec module. There is a bit we need to know from AS according to the specification, so please post if you have anything. Cheers

[grapesugar]

I always miss out on really stupid marks which all build up and ocr exam board are really strict, its so annoying!

[Kooldood]

Yes I know me too, and these stupid marks come down to the lack of practice. PRACTICE PRACTICE PRACTICE!!!

[grapesugar]

yup, i bet the night before i wont be able to sleep and so i will just be dead in the exam =(

Here is some as stuff, i know we need to know it, but im not sure if my notes are all that great.

AS Stuff To Know

Structural Isomerism = same molecular formula, different structural arrangement

Optical Isomerism = Same structural formula, different spatial arrangement of atoms

Cis-Trans

• This is due to non rotation about the C=C double bond
• Needs two different groups on each Carbon atom of the C=C double bond

Cis = similar groups on same sides
Trans = similar groups on opposite sides

Terms to know

Free radical = contains a single unpaired electron (e.g Cl’)
Nucleophile = an electron pair donor (e.g OH-)
Electrophile = an electron pair acceptor (e.g H+)

Electrophilic addition occurs in reactions between alkenes and halogens
Nucleophilic substitution occurs in the hydrolysis of halogenoalkanes (use e.g. water)

Must know the basic ideas of addition polymerisation

Must know about oxidation of alcohols

Alcohols can be oxidised using acidified potassium dichromate , (K2Cr207/H2SO4)
To show oxidation has occurred: colour change from orange to green
Choice of apparatus = important when oxidising a primary alcohol
Refluxing gives carboxylic acid
Distillation gives aldehyde

Must be able to do equations for this:

Oxidation of Primary alcohols to aldehydes
E.g. CH3OH + [0] gives HCHO + H20

Oxidation of Primary alcohols to carboxylic acids: (or you can just oxidise the aldehyde straight to a c.acid)
E.g. CH3OH + 2[0] gives HCOOH + H20

Oxidation of secondary alcohols gives Ketones
e.g. CH3CH0HCH3 + [0] gives CH3COCH3 + H20


More stuff to know: this is only some of it, i think we have to know all about the mole and calculations etc *sigh* (never really got the mole)

Moles = Mass/Mr
Liquid = Moles = concentration x volume (dm3)
Liquid = Concentration = Moles /volume(dm3)
OR moles = concentration x volume/1000 (volume in cm3)
Gas volume = moles = volume in dm3/24 or moles = volume in cm3/24000

Molar mass = ·add up the masses of the atoms in the formula

Empirical and Molecular Formulae

· Write down mass or % of each element
· Divide each one by the atomic mass of that element
· Find out the ratio of the numbers (divide them all by the smallest one)


Percentage Composition by Mass
e.g C7H8O

For Carbon: do 7 (no.of carbon atoms in compound * 12 (Mr of Carbon) / 108 (Mr of whole compound) Then * this figure by 100
Then, do the same for the other atoms



T