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CH10136 > Aromatic- Aromacity > Flashcards

Aromatic- Aromacity Flashcards

1
Q

Describe the structure of benzene

A
  1. 6 sp2 hybridised carbon atoms that link together to form a planar ring
  2. Angles of 120 degrees
  3. Bond length 139 pm- in between length of C-C and C=C
  4. Each of 6 sp2 hybridised carbon atoms in benzene has an unhybridized p orbital at right angles to the plane of the ring, containing a single electron
  5. P orbitals are parallel and overlap side on with p-orbitals on adjacent carbon atoms- pi electrons
  6. Create 2 continuos rings of pi electrons above and below the plane of the benzene ring
  7. The electrons are delocalised
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2
Q

What is wrong with kekules structure

A
  1. 3 c=c bonds would be shorter than the 3 C-C bonds- but all are the same length
  2. C=C would be expected to react in electrophilic addition reactions like in alkenes- but actually undergoes substitution
  3. More stable than thought- heat of hydrogenation of benzene is lower than expected- P-electrons are delocalised
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3
Q

How can the stability of benzene be measured

A
  1. C=C can be reduced using H2 and Pd catalyst - hydrogenation
  2. This is exothermic as more thermodynamically stable product is formed
  3. Heat of hydrogenation can be directly measured and gives an indication of stability of unsaturated compounds
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4
Q

How is the heat of hydrogenation of benzene different to expected

A
  1. Despite having 3 double bonds it has a heat of hydrogenation less than cyclohexadienes
  2. More stable than would be expected
  3. Due to delocalisation of the electrons in the ring- resonance
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5
Q

Why does benzene not undergo bromination

A
  1. Activation energy must be overcome
  2. Too large for benzene
  3. Due to aromatic stabilisation energy which must be overcome for aromatics to react with electrophiles
  4. When forced a substituted product is formed instead of addition product
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6
Q

What is Huckel’s Rule

A

Requirements for aromaticity

  1. Molecule must be cyclic
  2. Must be planar- sp2 hybridised
  3. Fully conjugated
  4. Must contain 4n+2 pi-electrons- n is any positive integer e.g. 6, 10, 12, 14 etc
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7
Q

Can heteroatoms be part of the aromatic ring

A
  1. Yes
  2. In some compounds a lone pair of electrons on the heteroatom is part of the ring of pi electrons- pyrrole
  3. Sometimes not- pyridine
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8
Q

What are antiaromatic compounds

A
  1. Cyclic, planar and contain an uninterrupted ring of pi electrons
  2. Less stable than a similar compounds with localised electrons
  3. Contain 4n pi-electrons
  4. Very unstable and reactive
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9
Q

What is a non aromatic compounds

A
  1. Non-planar and have an interrupted or uninterrupted ring of pi electrons
  2. Similar stability to related compounds that are non cyclic
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10
Q

How many MOs are present in benzene

A
  1. 6 carbon p-orbitals in benzene combine to give 6 pi molecular orbitals
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11
Q

Describe bonding MOs in benzene

A
  1. Lowest energy MO occurs due to all 6 p-orbitals combining in phase - no nodes
  2. Next energy level contains 2 degenerate MOs with one node each- one node across bonds and other with a node across atoms
  3. All these three MOs are lower in energy than the contributing AOs- all bonding MOs
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12
Q

Descirbe the anti-bonding MOs in benzene

A
  1. 3rd energy level contains 2 degenerate MOs, with 2 nodes each
  2. Highest energy level MO occurs due to all 6 p-orbitals combining out of phase - 3 nodes
  3. These 3 MOs are higher in energy than the contributing AOs and are therefore all anti-bonding MOs
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13
Q

Where do the electrons stay in benzene

A
  1. The 6 p-electrons are all paired and fill all bonding MOs

2. Contribute to bonding effects across the entire ring- why aromatic compounds are so stable

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14
Q

How can you use a Frost circle

A
  1. Draw a circle and place a horizontal line through its centre
  2. Draw a regular polygon with same number of sides as molecule in question- apex needs to point down
  3. Draw horizontal lines where the polygon touches the circle- vertices- represent the relative energy of MO
  4. Fill MO with number of electrons p-orbitals of the molecule- following correct rules
  5. Bonding= below line, non-bonding= on line, antibonding= above line
  6. If all paired and in bonding or non-bonding MO then the molecule is aromatic
  7. If there are unpaired electrons in non-bonding or anti-bonding orbitals- anti-aromatic
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15
Q

What can aromatic compounds be classed as and what is the symbol

A
  1. Arenes

2. C6R5= aryl (Ar) groups

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16
Q

What are the different aryl groups called

A
  1. C6H5- Phenyl (Ph)
  2. Phenyl group linked through a methylene unit- Benzyl (Bn)
  3. Phenyl group linked through a carbonyl- benzoyl (Bz) group
17
Q

What is the benzylic position

A
  1. The CH2 carbon in benzene groups is known as the benzylic position
18
Q

Describe the substitution positions

A
  1. Aromatic carbon atom directly bonded to substituent= ipso
  2. Adjacent to this - ortho (o)
  3. Next position around - meta (m)
  4. Position on opposite side of the ring - para (p)
19
Q

What are the basics of naming aromatics

A
  1. Find highest priority root name

2. Find lowest numbering system

20
Q

What is the closest heterocycle relative to benzene

A
  1. pyridine
  2. 6-membered nitrogen-containing compounds
  3. C-H unit has been replaced with a N: unit
  4. It conforms to Huckel’s rules and so is aromatic
21
Q

Describe the position of the lone pair of nitrogen in pyridine

A
  1. Nitrogen atom is sp2 hybridised
  2. Its lone pair of electrons is in an sp2 orbital perpendicular to the aromatic ring
  3. The lone pair is not delocalised around the aromatic ring and is isolated to the sp2 orbital
22
Q

What do the properties of pyridine mean it is used for

A
  1. Commonly-used organic base

2. Nucleophilic organocatalyst- DMAP (derivative)

23
Q

Can 5-membered heterocycles be aromatic and describe an example

A
  1. Yes, if an entire C=C unit is replaced with an N-H unit - pyrrole
  2. The nitrogen atom is sp2 hybridised but the lone pair is in the unhybridised p-orbital not sp2
  3. Therefore it is delocalised around the aromatic ring
  4. 6 pi-electrons delocalised across 5 cyclic, planar, fully conjugated sp2 hybridised orbitals
  5. Also electron rich
24
Q

What happens if C-H atoms in pyrrole are replaced with further N atoms

A
  1. The extra nitrogen atom is said to be pyridine-like - lone pairs are in sp2 hybridised orbitals and are perpendicular to the aromatic ring
25
Q

What can Nitrogen in pyrrole be replaced with and describe what happens

A
  1. Other heteroatoms- sulfur or oxygen
  2. Heteroatoms are also sp2 hybridised with one lone pair in a p-orbital delocalised amongst the aromatic ring and one lone pair in an sp2 hybridised orbital perpendicular to the ring
26
Q

What happens if a benzene ring is fused to an aromatic heterocycle e.g. pyrrole

A
  1. Give extended aromatic heterocyclic molecules
  2. All the heavy atoms are sp2 hybridised with 10 pi-electrons delocalised across them - aromaticity is observed over the entire molecule
27
Q

What is a PAH and how is it formed

A
  1. polycyclic aromatic hydrocarbons

2. Caused by fusion of benzene to itself multiple times

28
Q

Why are PAHs important for scientific research

A
  1. Suspected carcinogens
  2. Constituents of tobacco smoke
  3. PAH detection and monitoring- estimate more than 20% of all carbon in universe can be found in PAHs
29
Q

Why are helicenes interesting

A
  1. 6 benzene rings
  2. Display axial chirality
  3. Uses in asymmetric synthesis
30
Q

Why are fullerenes useful

A
  1. Can take form of spheres or tubes

2. Used within various applications

31
Q

Why is graphene useful

A
  1. Worlds first 2D material
  2. 200 times stronger than steel but extremely thin diameter
  3. Lightweight, flexible and electrically and thermally-conductive
  4. Application in electronics, energy and biomedical industries
32
Q

What is the simplest PAH and describe its properties

A
  1. Napthalene- 2 benzene rings fused
  2. 10 pi-electrons
  3. Less aromatic than benzene
  4. Only one ring is fully aromatic at a given time
  5. Can be easily reduced under mild conditions
  6. Used in OLED and solar cells
33
Q

What is Huckel’s rule limited to

A
  1. Monocyclic systems
34
Q

What rule can be used to describe aromaticity in PAHs

A
  1. Clar’s sextet theory
35
Q

What does Clar’s sextet theory state

A
  1. Resonance structure with the largest number of disjoint aromatic pi-sextets is the most important for a PAH’s characterisation
  2. PAHs with more pi-sextets are kinetically more stable than those with less
  3. Rings which have the most pi-sextets in resonance structures are more aromatic than other rings and therefore more stable/less reactive
36
Q

What are acenes and which ones are more aromatic

A
  1. Linear structure of benzene rings fused
  2. Only 1 pi-sextet can exist in a resonance structure at 1 time due to linear structure
  3. Large acenes are more stable as more resonance structures present
37
Q

What does it mean if something is fully benzenoid

A
  1. Alternate rings contain pi-sextets with no isolated or conjugated double bonds in other rings
  2. Similar stability and reactivity of benzene

CH10136 (30 decks)

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