Topic 17 Organic 2 Notes

Question 1

No. 1 requirement for optical isomerism?

Answer 1

All 4 substituents on the central carbon must be different.

Question 2

What name is given to the 2 stereoisomers of a carbon with 4 different substituents?

Answer 2

Enantiomers.

Question 3

Enantiomers

Answer 3

Non-superimposable mirror images of each other.

Question 4

Chiral centre

Answer 4

The carbon with 4 different groups attached. Often labelled with *. A molecule can have more than one chiral centre.

Question 5

What is special about a chiral molecule?

Answer 5

It can rotate the plane of polarised light.

Question 6

Optically active

Answer 6

Term given to chiral compounds because they rotate the plane of polarised light.

Question 7

Achiral compounds

Answer 7

Optically inactive.

Question 8

Why are racemic mixtures/racemates optically inactive?

Answer 8

For every molecule in the mixture that rotates the plane of polarisation in one direction, there is an enantiomer that rotates the plane in the opposite direction.

Question 9

Why does light emerge from a racemic mixture with its plane of polarisation unchanged?

Answer 9

There is an equal concentration of both enantiomers, so the rotations cancel out.

Question 10

Wedge-&-dash notation

Answer 10

The wedge comes towards you, out of the plane. The dashed line goes away from you, into the plane. The two straight lines are next to each other.

Question 11

Why is SN1 unimolecular?

Answer 11

It has only one species in the rate equation.

Question 12

Why is SN2 bimolecular?

Answer 12

It has 2 species in the rate equation.

Question 13

Chloride nucleophile + butanol (secondary alcohol) by SN1

Answer 13

Cl - can approach the cation from either side as the molecule is planar. Thus, both enantiomers form; the race mate is not optically active.

Question 14

Chloride ion + butanol (secondary alcohol) by SN2

Answer 14

Rear-side attack, the chloride attacks the partially positive chiral carbon. Inversion occurs, and the Cl switches with the OH group. Only one enantiomer is formed, so the product is optically active.

Question 15

Carbonyls

Answer 15

Aldehydes & ketones.

Question 16

-ate

Answer 16

Almost always 3 or 4 oxygens.

Question 17

-ide

Answer 17

Just the element.

Question 18

Break down the name: phosphide

Answer 18

Reduced form of phosphorous.

Question 19

Break down the name: phosphate

Answer 19

Oxidised form of phosphorous.

Question 20

How to find the name from the formula

Answer 20

Use oxidation states, and work backwards from the acid.

Question 21

Why do tertiary molecules react via SN1?

Answer 21

It forms a more stable carbocation by the positive inductive effect.

Question 22

Aldehyde functional group

Answer 22

R-CHO

Question 23

Carbonyl functional group

Answer 23

R-COR’

Question 24

Under carefully-controlled conditions, what are primary alcohols converted into?

Answer 24

Aldehydes.

Question 25

What are primary alcohols converted into, if there is an excess of oxidant?

Answer 25

Carboxylic acids.

Question 26

Reduction

Answer 26

Add H. Add electrons. Lose O.

Question 27

What are secondary alcohols converted into?

Answer 27

Ketones.

Question 28

Why can’t tertiary alcohols be oxidised?

Answer 28

There is no H on the carbon to which the OH is bonded.

Question 29

Oxidising agent

Answer 29

Acidified (H2SO4) potassium dichromate. K2Cr2O7. Cr2O7 -2/ H+.

Question 30

Acidified potassium dichromate colour change

Answer 30

Orange to green.

Question 31

Why do aldehydes have lower boiling points than alcohols?

Answer 31

Alcohols form H bonds, whereas aldehydes have only London dispersion forces. This means aldehydes can be distilled off as they form.

Question 32

Oxidation of a primary alcohol: technique to obtain an aldehyde

Answer 32

Distillation

Question 33

Oxidation of a primary alcohol: technique to obtain a ketone

Answer 33

Heating under reflux.

Question 34

Alternative oxidising agent (not typically used for alcohol oxidation)

Answer 34

KMnO4 /H2SO4.

Question 35

Potassium permanganate: colour change

Answer 35

Purple to pale pink.

Question 36

Reducing agents

Answer 36

NaBH4 sodium borohydride, LiAlH4 Lithium aluminium hydride IN DRY ETHER.

Question 37

What is the colour change of Fehling’s solution when warmed with an aldehyde?

Answer 37

Blue to red. The solution contains blue Cu (II) (aq), which is reduced to a red precipitate of Cu (I) oxide.

Question 38

What happens to the aldehydes when warmed with Fehling’s solution?

Answer 38

It is oxidised to a. A Carboxylic acid.

Question 39

Why do Tollen’s reagent and Fehling’s solution have no effect on ketones?

Answer 39

Ketones cannot be oxidised.

Question 40

Tollen’s reagent

Answer 40

Alkaline ammoniacal silver nitrate solution. Silver nitrate is dissolved in aqueous ammonia to form Tollen’s reagent, incorporating [Ag(NH3)2]+ diamine silver (I) ion.

Question 41

What happens to the aldehyde when warmed with Tollen’s reagent?

Answer 41

The aldehyde is oxidised to a carboxylic acid, and the silver (I) ions are reduced to silver metal. 2Ag+ + RCHO + H2O —> 2Ag + RCOOH + H+. A silver mirror is formed.

Question 42

Reduction of aldehydes with sodium tetrahydridoborate (III) NaBH4

Answer 42

When dissolved in ethanol, sodium tetrahydridoborate reduces aldehydes to primary alcohols, e.g., ethanal + 2[H] — NaBH4 in ethanol—> ethanol. Occurs by Nucleophilic addition.

Question 43

Reaction of ketones with NaBH4

Answer 43

When dissolved in ethanol, NaBH4 reduces ketones to secondary alcohols by Nucleophilic addition.

Question 44

Reduction of carbonyls using lithium aluminium hydride LiAlH4 (a more powerful reducing agent)

Answer 44

LiAlH4 must be used dry/dissolved in ether. Aldehydes are reduced to primary alcohols. Ketones are reduced to secondary alcohols. Occurs by Nucleophilic addition, e.g., propanone + 2[H] —> propan-2-ol.

Question 45

Another name for the iodoform test

Answer 45

Triiodomethane reaction.

Question 46

Iodoform

Answer 46

Carbonyl + I2 in the presence of an alkali. A positive test indicates a carbonyl + a methyl group, forming an insoluble yellow precipitate and an antiseptic smell. E.g., ethanal + iodine —> triodoethanal. Triodoethanal + alkali (NaOH) —> triiodomethane.

Question 47

Which compounds give a positive result for the iodoform test?

Answer 47

Organic compounds with a methyl group next to a carbonyl group, or secondary alcohols which will oxidise under these conditions to form a methyl group next to a carbonyl group.

Question 48

Nucleophilic addition of hydrogen cyanide to aldehydes & ketones

Answer 48

HCN is dissolved in KCN (aq). This provides CN - ions which can attack from either side of the planar carbonyl molecule, forming a racemate. An extra carbon is added to the organic molecule, an alcohol is formed, and the CN- nucleophile regenerated.

Question 49

Reaction of carbonyls with 2,4-DNP in an alcoholic solution

Answer 49

Forms a red/orange precipitate. Positive for both ketones and aldehydes. E.g., aldehyde + 2,4-DNP in alcoholic solution —> …-anal 2,4-dinitrophenylhydrazine by Nucleophilic addition. The product reacts with H2SO4, and a water molecule is eliminated.

Question 50

What is the significance of the products of the reaction of 2,4-DNP being mostly crystalline solids?

Answer 50

Easy to handle. The melting points are different enough to be able to distinguish between the various carbonyl compounds. Several of the pure carbonyl compounds have very similar melting points.

Question 51

Why is the carbon bonded to the oxygen in a carbonyl group susceptible to Nucleophilic attack?

Answer 51

The oxygen is more electronegative than the carbon meaning there is a partial positive charge on the carbon.p and a partial negative charge on the oxygen.