Alcools et Phénols Flashcards
(40 cards)
Give general formula and functional groups of alcohols
formula: CnH2n+1OH or ROH
*MUST CLASSIFY
func grp: primary/secondary/tertiary alcohol
Describe nomenclature of alcohols
- For one alcohol, replace ‘e’ with -ol (and relevant number)
- For two alcohol grp, leave ‘e’ with -ol (and number)
What are germinal diols
- Compounds w 2 -OH grp bonded to same C atom
- unstable, readily eliminate H2O molecule to form aldehyde or ketone
Extra case: if got three OH attached to 1 C, carboxylic acid is formed
What bonds form btw alcohol molecules? Why?
id-id and hydrogen bonds
bcos alcohol hv non-polar alkyl grp & polar -OH grp
Compare and explain differences in boiling points of alcohols and other similar Mr organic compounds
Alcohols hv much higher bp than hydrocarbon & other organic compounds of comparable size e- cloud due ti stronger intermolecular H bonding needing more energy to overcome
Why cannot use hydrogen bonding to explain difference in boiling point between alcohols with same no of -OH?
Bcos similar extent of intermolecular hydrogen bonding by alcohols
So, must use other factors eg
- surface area of contact (similar Mr: chain > branched)
- e- cloud size (diff Mr)
Explain solubility of alcohols
- completely miscible w water
bcos
- energy released by form n H bond btw alcohol & H2O enough to overcome H bond btw water molecules & H bond btw alcohol molecules - solubility in water decrease as no of C atom in alcohol increase
bcos
- larger non polar R grp make alcohol more hydrophobic. Main interact n btw large alcohol molecules & H2O become id-id
- energy released fr id-id attract n btw alcohol & H2O molecules inadequate to overcome H bond btw H2O molecules & H bond + id-id btw alcohol molecules
Strength of acid depends on…, which depends on …
stability of conjugate base (the more stable, stronger acid),
intensity of -ve charge (the more intense, the less stable)
Why are alcohols generally very weak acids, even less acidic than water?
e- donating alkyl (R) grp intensify -ve charge on conj base (alkoxide ion, RO-), thus destabilising conj base
=> lose proton less readily and acid strength decreases
ROH ⇌ RO- + H+
The higher the Ka, the … the acid
The lower the pKa, the … the acid
stronger
stronger (pKa=-lgKa)
What reactions do alcohols undergo?
- Acid-metal displacement
- Condensation/ nuc acyl sub n
- Nuc Sub n
- Elimination
- Oxidat n
Name reagent, condit n & type of rxn of ROH + sodium
R&C: Na (or other reactive metal eg K)
Type of rxn: acid-metal displacemt
NOTE:
- Mole ratio is ROH = 0.5 H2
- alcohol no react w base bcos much weaker acid than water
Name reagent, condit n & type of rxn of ROH and carboxylic acids
R&C: RCOOH (specify specific acid if specific ester is stated), conc H2SO4 as cata, heat
type of rxn: condensat n/ nuc acyl sub n
Which part of an ester comes from acid and alcohol?
RC=O comes from acid,
OR’ comes fr alcohol
Name reagent, condit n & type of rxn of ROH + acyl chloride
R&C: RCOCl (specify acyl chloride if specific ester is stated)
Type of rxn: condensat n/ nuc acyl sub n
*Advantage: no heating needed, good yield of ester since rxn is irreversible
Name reagent, condit n & type of rxn of ROH to form RX
type: Nu Sub n
R&C:
- HCl (g), heat –> R-Cl + H2O
- anhydrous PCl5 –> R-Cl + POCl3 + HCl
- anhydrous PCl3 –> R-Cl + H3PO4
- anhydrous SOCl2 –> R-Cl + SO2 (g) + HCl (g)
- NaBr, conc H2SO4, heat/HBr (g), heat –> R-Br + H2O
- anhydrous PBr3 –> R-Br + H3PO4
- red P + I2 (PI3 is unstable) –> R-I + H3PO3
Why should water be absent (anhydrous) in reaction of alcohol with PX3, PCl5 and SOCl2?
- PX3, PCl5 and SOCl2 undergo complete hydrolysis w water to give other side products, so yield of RX pdt decreases
Why is SOCl2 preferred over PCl5 (or PCl3 or HCl)?
using SOCl2 means side pdt (SO2 & HCl) are gases, easier to separate fr rxn mix to obtain pure pdt R-Cl
What is Saytzeff’s rule?
(for elimination rxn)
major pdt is alkene w more no R grp (more substituted alkene) attached to the C=C
Name reagent, conditions, product obtained, observat n & general eqn of oxidation of primary alcohol (HINT: two types of oxidation)
Mild oxidation:
R: K2Cr2O7 (aq), H2SO4 (aq)
C: Heat w immediate distillat n
Pdt: aldehyde (RCHO)
Observ n: orange sol n turn green
eqn: RCH2OH + [O] –> RCHO + H2O
Strong oxidation
R: K2Cr2O7(aq) / KMnO4(aq), H2SO4 (aq)
C: Heat
Pdt: carboxylic acid (RCOOH)
Observ n: orange K2Cr2O7 sol n turn green/ purple KMnO4 sol n decolourise
Eqn: RCH2OH + 2[O] –> RCOOH + H2O
Name reagent, conditions, type of reaction and general equation of oxidation of secondary alcohols
R&C: KMnO4(aq)/K2Cr2O7(aq), H2SO4(aq), heat
type: oxidation
general eqn:
RCH(OH)R’ + [O] –> RC=O R’ + H2O
Describe oxidation of tertiary alcohols
Tertiary ROH cnt b oxidised
- oxidat n of alcohol involve cleavage of α C-H bond, absent in tertiary alcohols (α C-H is C bonded to OH)
Describe special oxidation cases of carbonyl compounds
(methanol & ethane-1,2-diol)
- CH3OH + 2[O] – K2Cr2O7, H2SO4 (aq), heat –> HCOOH + H2O
- HCOOH + [O] –K2Cr2O7, H2SO4 (aq), prolonged heating –> CO2 + H2O
- CH3OH + 3[O] –KMnO4, H2SO4(aq), heat–> CO2 + H2O
CH2(OH)CH2(OH) + 2[O] – K2Cr2O7, H2SO4 (aq), heat –> O=C(OH)C(OH)=O (decompose to 2CO2, H2O w KMnO4) + 2H2O
CH2(OH)CH2(OH) + 5[O] – KMnO4, H2SO4 (aq), heat –> 2CO2 + 3H2O
Describe chemical tests for alcohols
1.
test: add anhydrous PCl5 to compound
observ n: white fumes of HCl will b observed
general eqn: ROH + PCl5 –> RCl + POCl3 + HCl
type: Nu sub n
2.
test: heat compound w KMnO4 in H2SO4 (aq)
observ n: if 1°/2° alcohol present, decolour n of purple KMnO4 sol n is observed (3° ROH resistant to oxidat n)
General eqn:
1°: RCH2OH + 2[O] –> RCOOH + H2O
2°: RCH(OH)R’ +[O] –> RC=O R’ + H2O
type: oxidat n
- (Iodoform test)
test add I2 in NaOH(aq) to compound, heat mix
observ n: Pale yellow ppt of CHI3 will b observed
general eqn: RCH(OH)-CH3/CH2I/CHI2/CI3 + 4I2 + 6NaOH –> RCOO-Na+ + CHI3 + 5NaI + 5H2O
