Organic Chemistry Test I

Question 1

• Toluene
• Benzoic Acid
• Aniline
• Phenol
• Anisole
• Nitrobenzene
• Benzonitrile
• Acetophenone
• acetanilide

Answer 1

• Toluene= CH3
• Benzoic Acid= COOH
• Aniline=NH2
• Phenol= OH
• Anisole= OCH3
• Nitrobenzene= NO3
• Benzonitrile= CN
• Acetophenone= COCH3
• acetanilide= NHOCH3

Question 2

Electrophile

Answer 2

An electron loving species, typically positively charged (e.g. NO2+, Cl+)

Question 3

Benzene—-> bromobenzene

Answer 3

Benzene + Br2/FeBr3 —-> bromobenzene

Question 4

+ I

Answer 4

• Electron donating by induction
• Mainly ortho, para directing

e.g. CH3,

Question 5

• I

Answer 5

• Electron withdrawing by induction
• Mainly meta substitution

e.g. CF3

Question 6

+ R

Answer 6

• Electron donating by resonance

- Mainly ortho, para substitution

Question 7

• R

Answer 7

• Electron withdrawing by resonance
• Mainly meta substitution

e.g. NO2

Question 8

Inductive effects

Answer 8

Operates through pi bonds on account of polarisation in pi bonds

-I for electronegative atoms e.g. O, N
+I electron donating e.g. CH3

Question 9

Resonance effects

Answer 9

Operates where there is a lone pair or double bonds.

Stronger than inductive effects.

Question 10

Reagents for Nitration:

Answer 10

H2SO4/ HNO3

Question 11

Explain why bromination of phenol or aniline forms mainly tribromo products , while bromination of acetaniline affords mainly p-bromoacetanilide.

Answer 11

• NH2 is strongly activating
• Ortho positions are sterically hindered by bulky groups
• Br is deactivating, NH2 is strongly activating

Question 12

Alkylation

Answer 12

• Introducing an R group
• R+ is the ‘active’ reagent

Benzene + R-Cl + AlCl3 —> Benzene-R + HCl

Question 13

Acylation

Answer 13

• Introducing an acyl group (R=O)
• RC=O is the active reagent

Benzene+ RC=OCl + AlCl3/H20 —> Benzene-CR=O + HCl

Question 14

Limitations of Friedel Crafts

Answer 14

• Only aliphatic alkyl halides or alkyl acyl halides react under Friedel Craft conditions
• No reaction for substrates which have an electron withdrawing group (-NO2, -CN, COOH)
• No reaction with aniline or other -NH2 or -NHR groups containing substrates.

Question 15

Reduction of aryl nitro compounds

Answer 15

Benzene + HNO3/ H2SO4 –> nitrobenzene —> Sn/HCl (lab) or H2/Ni —> Aniline

Question 16

Inductive effects

Answer 16

• +I effects are observed when alkyl groups stabilise radicals by donating electron density
  -The greater the number of alkyl groups, the more stable the radical
  Tertiary > secondary > primary

Question 17

Resonance effects

Answer 17

Stabilisation by delocalisation, the radical charge is spread over more than two carbons.

Question 18

Halogenation of benzylic carbons

Answer 18

• Alkyl benzenes are readily halogenated at the benzylic carbon
• If there is a choice of positions to remove protons from, chlorine will yield a mixture of products but bromine is more selective

Question 19

Halogenation of allylic carbons

Answer 19

-Conditions have to be tightly controlled to prevent electrophilic addition to the double bond (a competing reaction)

Question 20

N-Bromosuccinimide

Answer 20

• The reagent of choice for allylic and benzylic halogenations
• NBS does not halogenate, but acts as a carrier for bromine and maintains a low but steady concentration of bromine
• At low concentrations, prevents competing electrophilic addition because the radical reaction occurs more quickly.
• The reaction can be started with an initiator or light which causes NBS to release a bromine radical

Question 21

Anti Markovnikov addition to alkenes

Answer 21

• When HBr adds by the normal mechanism with a carbocation intermediate, the H adds to the carbon with the most protons attached.
• Puts a positive charge on the carbon with the most R-groups attached and hence yields the most stable carbocation.
• The Br will be attached to the most stable carbocation

Question 22

Organoboranes in Anti Markovnikov addition (hydroboration) of alkenes

Answer 22

• In BH3, H is the electronegative part and behaves like an hydride ion (H-)
• BH3 is added to the alkene to form a organoborane which can be then be converted to an alcohol, haloalkane or alkane.
• Steric hindrance is an issue so tends to add to the least substituted carbon.