Organic Reagents

Front 1

Alkene → Alkane

Back 1

Hydrogen, nickel catalyst, 150°

Front 2

Alkane → haloalkane

Back 2

UV light, Cl2 (Free radical)

Front 3

Alkene → haloalkanes

Back 3

Halogens give di,

Hydrogen halides give you single

Electrophilic addition

Front 4

Alkene → alcohol

Back 4

- Steam, acid catalyst (Hydration)

- [O] in acidic conditions for Diol

Front 5

Haloalkanes → alcohol

Back 5

OH-, reflux (mixture of eth and water) NS

Likewise with Silver Nitrate and eth/water (H2O = nucleophile)

Front 6

Haloalkanes → nitrile

Back 6

Heat reflux, in eth CN group (K/Na)

Adds C to chain

Front 7

Haloalkanes → amines

Back 7

Conc ammonia in XS, eth, sealed

Forms salt (RNH3+Ha-) which reversibly reacts to make RNH2 + NH4+Ha- = 1°

More NH3 means more forward reaction.

The lone pair on RNH2 can attack another RHa to form R2NH (2°) amine

The lone pair on RNH2 can attack another RHa to form R2NH (2°) amineAnd again R3N: (3°)And again R4N+Ha- (4°)

The lone pair on RNH2 can attack another RHa to form R2NH (2°) amineAnd again R3N: (3°)And again R4N+Ha- (4°)

And again R3N: (3°)

And again R4N+Ha- (4°)

Limiting RHa prevents this furthering.

Front 8

Haloalkanes → alkenes

Back 8

OH- in ETHANOL reflux

Elimination

Front 9

Alcohols combustion

Back 9

CO2 + H2O

In calculations, moles of C = moles of CO2, moles of H = 2x moles of H2O

Moles of O = (mass of CO2 - mass of C)/16

Front 10

Alcohol → Haloalkane

Back 10

•ROH + PCl5 →RCl + POCl3 + HCl

•(KBr + 50% conc H2SO4 → HBr)

ROH + HBr → RBr + H2O

•ROH + PI3 → RI + H3PO3 (balance)

Front 11

Alcohol → Aldehyde

Back 11

Dilute Sulf Acid + [O] to oxidise:

1° OH → Aldehyde (distill)

Forms COOH under reflux

2° OH → Ketone (Reflux)

So 1° and 2° Potassium Dichromate = orange → green, but not 3°

Fehling's/Benedict's/Tollens': ketone (2°) no change, aldehyde (1°): F/B: blue→brick/dark red. T: silver mirror

Front 12

Alcohol→ alkene

Back 12

Dehydration/elimination

XS phosphoric acid, 170° (is catalyst)

→ alkene + H2O

Front 13

Reduction of carbonyl

Back 13

Lithiumtetrahydridoaluminate [H] in dry ether

Ketone→ 2° OH

Aldehyde→ 1° OH

Front 14

Aldehyde→ Carboxylic acid

Back 14

[O], acidified PD

Front 15

Aldehyde → nitrile

Back 15

HCN (dangerous and weak acid (little dissociation), so acidified KCN as well)

CN attacks above/below, = opens, H from nearby HCN joins too.

Forms racemic mixture as the C is bonded to 4 different groups and is different depending on which side CN attacker from. Unlike ketones, where regardless where CN attack from it'll be the same product.

Front 16

2,4-DNPH test

Back 16

Aliphatic: yellow (proof of carbonyl)

Aromatic: orange

Figure out the compound by checking melting point data

Front 17

Carbonyl + iodine

Back 17

Iodoform/Triiodomethane:

CH3CRO (ethanal or ketone) +3I2 +3OH- → CI3CRO + 3I- + 3H2O (H from CH3 and OH)

+OH- → CHI3 breaks off, leaving RCOO-

CHI3 is yellow precipitate

Front 18

Racemic mixture note:

Back 18

SN1 does form, SN2 doesn't. This is because SN1 forms a carbocation where we can attack from either side. SN2 form transition state and can only be attacked from one side.

Front 19

Carboxylic acid→ aldehyde→Alcohol

Back 19

Reduction

[H]

Front 20

Nitriles → COOH

Back 20

Acid hydrolysis: reflux with dilute H+ (aq) to create RCOOH + NH4+

Alkaline hydrolysis: RCN + OH- (aq) → RCOO + NH3. Add acid to make RCOOH.

Front 21

COOH → esters (C=OOC)

Back 21

Conc Sulf acid catalyst

React with alcohol

RCOOH + 'ROH →RCOO'R +H2O

Front 22

COOH → Salt

Back 22

Bases, eg sodium ethanoate (CH3COONa, where Na is ionic bond)

Front 23

COOH → Acyl Chloride

Back 23

RC=OCl (eg ethanoyl chloride CH3COCl)

PCl → POCl

RCOOH + PCl5 → RCOCl + POCl3 +HCl

Front 24

Acyl Chloride + OH

Back 24

RCOCl + XOH → RCOOX + HCl

So with H2O: RCOCl + H2O → RCOOH + HCl

With alcohol: RCOCl + 'ROH → RCOO'R (ester) + HCl

Front 25

Acyl Chloride + NH2

Back 25

Conc Ammonia:

RCOCl + NH3 → RCONH2 + HCl but then XS Ammonia reacts with HCl: NH3 + HCl → NH4+Cl-

Therefore: RCOCl + 2NH3 → RCONH2 + NH4+Cl-

(AMIDE = CONH2)

Amine:

RCOCl + 'RNH2 → RCONH'R (N Substituted amide, since it's amide where a H at the N has been substituted by 'R) + HCl

Then step 2: 'RNH2 + HCl →'RNH3+Cl-

So: RCOCl + 2'RNH2 →RCONH'R + 'RNH3+Cl-

Front 26

Hydrolysis of esters

Back 26

Acidic conditions:

RCOO'R + H2O → RCOOH + 'ROH

Alkaline:

RCOO'R + XOH → RCOOX + 'ROH

Both reversible so use XS

Front 27

Polyesters

Back 27

Condensation

Dicarboxylic acids and diols

H from diol and OH from dicarboxylic acid forms water. COOC ester bonds form between molecules.

Front 28

Benzene combustion

Back 28

Smoky flame due to C:H ratio

Front 29

Halogenation of benzene

Back 29

Only occurs with catalyst (AlCl3/FeBr3) due to stability of shared double bonds

X2 dipole, so splits. X- + AlCl3 → AlCl3X-, and X+ is electrophile so joins.

Forms halobenzene +HX. (X leaves catalyst at this stage)

Front 30

Nitration of benzene

Back 30

Mixture of H2SO4 (catalyst) and HNO3 at 50°

Step 1: HNO3 + 2H2SO4 → NO2+ +H3O+ + 2HSO4-

Step 2: NO2+ is electrophile so can substitute H on benzene

Forms nitrobenzene or dinitrobenzene at higher temp and H2O. Sulf acid is regenerated using H.

Hint 30

Arrow from ring, forms + smiley face with both attached then H leaves

Front 31

Friedel-Crafts reactions

Back 31

Halogenoalkane or acyl chloride refluxed with benzene AlCl3 catalyst which becomes negative in formation of electrophile.

Front 32

Amine vs Amide

Back 32

Amine: NR3

Amide: CONR2

Front 33

Making amines:

Back 33

Nitrobenzene reflux with tin and HCl at 100°

CN + 4[H] or 2H2 with nickel catalyst

NH3 + RHa

Front 34

Amine + H2O

Back 34

Formation of base (N: acts as base and accepts H)

Forms ammonium ion and OH-

Front 35

Amine + Acid

Back 35

RNH2 accepts H+ → RNH3+ and bonds with A- to form RNH3+A- salt

(Similar to NH3)

Front 36

Amine + Cu2+

Back 36

Ligand exchange:

[Cu(H2O)6]2+ + 2RNH2 reversibly forms [Cu(OH)2(H2O)4] (s) as RNH2 deprotonates water on hexaaqua ion

XS RNH2 forms [Cu(RNH2)4(H2O)2]2+, but if R is bigger, fewer RNH2 will fit so formula can change

Front 37

Amine + RHa

Back 37

Much like NH3 reaction:

Forms 2° salt (R2NH2+Ha-) in XS amine will form salt (RNH3+Ha-) and R2NH

R2NH can then react to make R3NH+Ha- (3°) but XS will form R3N and salt

4° formed by R3N: + RHa →R4N+Ha-

Front 38

Amine + COCl

Back 38

RCOCl + 'RNH2 → HCl + RCONH'R

Then HCl + 'RNH2 → 'RNH3+Cl-

So:

RCOCl + 2'RNH2 → RCONH'R + 'RNH3Cl

Front 39

Grignard reagents

Back 39

Halogenoalkane + Mg → RMgHa (Grignard)

+ CO2 and HCl → RCOOH + MgHaCl

Or

+ CHO/COC (any carbonyl) +HCl → ROH + MgHaCl

Both in dry ether