Chirality: Nucleophilic substitution reactions of
Optical isomerism occurs due to a chiral C haloalkanes:
(tetrahedral C with 4 different groups attached) Sn1 (tertiary haloalkanes):
= type of stereoisomerism (same structural Middles step = carbocation
formula, atoms/groups have a different Stable because the C’s attached to the central C
arrangement in space) non-superimposable spread out the +ve charge
images of each other, and are object. The carbocation is planar, meaning the
nucleophile can attack from either side. This
means the product is a racemic mixture (no
optical activity)
Enantiomers: Sn2 (primary and secondary haloalkanes):
= mirror images with a chiral centre (isomers) Middle step = transition state
Enantiomers have equal but opposite effects on The nucleophile can only attack from the LHS
plane polarised light because the big halogen is in the way on the
TEST for one enantiomer = will rotate a plane of otherside, and always 4 things bonded (never
polarised light by x° (has optical activity) planar). This means the product is the opposite
Racemic mix enantiomer than the reactant, but there’s no
= an equal mix of two enantiomers. It will have racemic mixture.
no optical activity because the enantiomers
cancel eachother out.
, Carbonyl compounds: Tests:
C=O bond attracts nucleophiles due to δ+ charge on C Ox agent = acidified potassium dichromate (Cr2O7^2-/H+)
Have london forces & dipole-dipole interactions (polar) ∴ Conditions = heat under reflux
some water solubility. Aldehydes → CA
Ketones → x
Ox agent = fehling’s/benedict's solution
Reagents = copper sulphate & NaOH solutions
Conditions = warm gently
Aldehydes: Blue (Cu2+) → Red (Cu1+)
Ketones: No change
Nucleophilic addition 1 - REDUCTION: Ox agent = tollen’s agent
- Reagents = NaBH4 or LiAlH4 (provides H- ions) Reagents = silver nitrate, NaOH & ammonia solutions
- Conditions = aq (H2O provides H+ ion) Conditions = warm gently
Aldehyde + NaBH4 → primary alcohol (+ BH3 + NA salt) Aldehyde: silver mirror on test tube (Ag+ + e-→Ag)
Ketone: no change
Triiodomethane reaction (the iodoform reaction)
Reagents = iodine in KI & NaOH solution
Conditions = cold/room temp
Results: methylketone = yellow ppt & antiseptic smell
- Shows the presence of an aldehyde or ketone in which a
methyl group is one of the groups immediately connected to
Ketone + NaBH4 → secondary alcohol (+ BH3 + NA salt) the carbonyl carbon (only aldehyde = ethanal).
- Or shows the presence of an OH next to a CH3
2,4-DNPH (Brady’s Reagent)
Conditions = room temp
Results = yellow/orange ppt for aldehydes and ketones
- Can then find WHICH aldehyde/ketone, MPT of purified
derivative = characteristic to each
Nucleophilic additions 2 - ADDITION of HCN: (1) (2)
Reagents = KCN and dil. Acid (provides H+ ion)
Conditions = aq
(1)Aldehyde + KCN + acid→ hydroxynitrile
(2)Ketone + KCN + acid → hydroxynitrile
Optical isomerism occurs due to a chiral C haloalkanes:
(tetrahedral C with 4 different groups attached) Sn1 (tertiary haloalkanes):
= type of stereoisomerism (same structural Middles step = carbocation
formula, atoms/groups have a different Stable because the C’s attached to the central C
arrangement in space) non-superimposable spread out the +ve charge
images of each other, and are object. The carbocation is planar, meaning the
nucleophile can attack from either side. This
means the product is a racemic mixture (no
optical activity)
Enantiomers: Sn2 (primary and secondary haloalkanes):
= mirror images with a chiral centre (isomers) Middle step = transition state
Enantiomers have equal but opposite effects on The nucleophile can only attack from the LHS
plane polarised light because the big halogen is in the way on the
TEST for one enantiomer = will rotate a plane of otherside, and always 4 things bonded (never
polarised light by x° (has optical activity) planar). This means the product is the opposite
Racemic mix enantiomer than the reactant, but there’s no
= an equal mix of two enantiomers. It will have racemic mixture.
no optical activity because the enantiomers
cancel eachother out.
, Carbonyl compounds: Tests:
C=O bond attracts nucleophiles due to δ+ charge on C Ox agent = acidified potassium dichromate (Cr2O7^2-/H+)
Have london forces & dipole-dipole interactions (polar) ∴ Conditions = heat under reflux
some water solubility. Aldehydes → CA
Ketones → x
Ox agent = fehling’s/benedict's solution
Reagents = copper sulphate & NaOH solutions
Conditions = warm gently
Aldehydes: Blue (Cu2+) → Red (Cu1+)
Ketones: No change
Nucleophilic addition 1 - REDUCTION: Ox agent = tollen’s agent
- Reagents = NaBH4 or LiAlH4 (provides H- ions) Reagents = silver nitrate, NaOH & ammonia solutions
- Conditions = aq (H2O provides H+ ion) Conditions = warm gently
Aldehyde + NaBH4 → primary alcohol (+ BH3 + NA salt) Aldehyde: silver mirror on test tube (Ag+ + e-→Ag)
Ketone: no change
Triiodomethane reaction (the iodoform reaction)
Reagents = iodine in KI & NaOH solution
Conditions = cold/room temp
Results: methylketone = yellow ppt & antiseptic smell
- Shows the presence of an aldehyde or ketone in which a
methyl group is one of the groups immediately connected to
Ketone + NaBH4 → secondary alcohol (+ BH3 + NA salt) the carbonyl carbon (only aldehyde = ethanal).
- Or shows the presence of an OH next to a CH3
2,4-DNPH (Brady’s Reagent)
Conditions = room temp
Results = yellow/orange ppt for aldehydes and ketones
- Can then find WHICH aldehyde/ketone, MPT of purified
derivative = characteristic to each
Nucleophilic additions 2 - ADDITION of HCN: (1) (2)
Reagents = KCN and dil. Acid (provides H+ ion)
Conditions = aq
(1)Aldehyde + KCN + acid→ hydroxynitrile
(2)Ketone + KCN + acid → hydroxynitrile
