Topic 16: benzene Flashcards
(25 cards)
Define properties of benzene
- Unsaturated = doesn’t undergo electrophilic addition
- Undergoes substitution reactions
- Black smoke = high C:H ratio = carbon unburnt due to not enough oxygen to complete combustion
Describe benzene stability
- More stable than alkenes
- Reaction doesn’t cause permenent breaking double bonds
- Unsaturation = stability
Describe how stability is measured
- Via heat of hydrogenation
- High = unstable
- Low = stable
Explain bond lengths for stability
- Typical C-C = 1.54
- Benzene C-C = 1.39 = all C-C in benzene identical
- Typical C=C = 1.34
- Benzene C-C closer to typical double bond
- Shorter the bond - stronger
Explain why benzenes are stable
- All carbon atoms = sp2 hybridized
- All perpendicular 2p e- join = solid ring above + below plane of C atoms
- Solid ring = stability from delocalized e-
Define aromaticity
- Unsaturated compounds behave chemically like benzene
- No typical reactions = undergo substitution
- Aromatic compound
- Special stability = due to delocalized e-
Give the requirements for aromaticity
- Cyclic
- Planar
- Conjugated
- 4n+2 π e- delocalized
- Huckel rule
- Magic number = 2/6/10/14
Describe Huckel rule
- Gives magic number of π e- to make compound aromatic
- 1 π bond = 2 π e-
- n = variable
Define conjugated
- Alternative double-single-double C-C bonds
Describe electrophilic substitution
- H-atom substitutes by incoming atom
- Addition/elimination mechanism
How is electrophile formed?
- Br-Br + FeBr3 → Br+ = electrophile
Describe the mechanism for benzene bromination
- Benzene attacks electrophile = 1 double bond breaks = attachs to Br = Be-C-H
- To get rid of H = use base = HB+
- Bromobenzene formed
- Same for chlorination
Describe iodination
- I2 = not as reactive
- Add oxidizing agent = H2O2 + HNO3 + CuCl2 salt = produce electrophile
Describe nitration
- Add NO2
- Via HNO3 + H2SO4
Describe reduction of nitro groups
- NO2 →NH2
- Use reducing agent OR HCl/OH- aq base
Describe benzene alkylation
- Add CH3/methyl
- When adding propyl group = n-propyl minor + isopropyl major
- Use AlCl3
- Al not octet = unstable + reactive
Explain carbocation rearrangement of propyl carbocation to isopropyl carbocation
- n-propyl = primary carbocation
- Isopropyl = secondary carbocation = more stable
- Due to hydride shift = H atom + its e- in n-propyl shift to C+ atom
Explain methyl shift
- CH3 in primary carbocation shifts from
Give the limitations of Friedel-crafts alkylation
1) Only alkyl halides can be used
2) e- withdrawing group = cannot be on benzene = no C=O/-X/NO2
3) Difficult to stop alkylation at 1 substituation = more than 1 alkyl group can go
4) Skeletal rearrangements possible
Describe Friedel crafts acylation
- Makes benzyl ketones
- Acyl = R-C=O
- Reagent = acid chlorine
- Catalyst = AlCl3
- AlCl3 takes Cl from reagent → R-C=O→ C attaches to benzene
- No 2nd substitution possible = carbonyl group takes e- density from benzene = less reactive to substitution
How does acylation fix the problem of rearrangement?
- No 2nd substitution possible after propylation = carbonyl group takes e- density from benzene = less reactive to substitution
- Therefore if we do acylation of benzene 1st + then C=O reduction = carbonyl deactivates benzene = n-propyl is major product
- This way no rearrangement + 2nd product
Describe reduction of carbonyl group
- With H2 = if carbonyl in position 1
- If in other position = cannot used H2 = use LiAlH4
Describe alkyl side chain oxidation
- Alkyl group → COOH
- Use KMnO4 in any chain length OR O2/Co3+
- For oxidation 1st C must have H attached
Describe alkyl side chain oxidation
- Use NBS = Br in benzylic position
- Benzylic position = on C1 = replaces H
- Post bromination = KOH = alkene
