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Chemistry revision > 20.1 Types of organic reactions > Flashcards

20.1 Types of organic reactions Flashcards

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1
Q

Explain nucleophiles

A

Nucleophiles - electron rich - attack areas with electron deficiency - act as Lewis bases - donate e pair - new covalent bond

Ex: OH-, H2O, NH3, CN-

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2
Q

Explain electrophiles

A

Electrophiles are electron deficient - accept e pairs from nucleophiles - act as Lewis bases

Ex: H+, NO2+

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3
Q

Explain nucleophilic substitution

A

Sn1/Sn2 - depends on what kind of halogenalkane (1/2/3)

Halogenalkanes - subsitution involves heterolytic fission (bond breaks, e to one of the two)

Halogen - leaving group

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4
Q

Reaction of primary halogenalkane

A

Sn2

  1. Nucleophile attacks as e are attracted to halogen
  2. Transition state
  3. LG leaves

Bimolecular reaction - rate depends on both the conc of halogenalkane and nucleophile:

rate = k [halogenalkane] [nucleophile]

Sn2 stereospecific because 3D arrangement of reactants influences arrangement of products

Favoured by polar aprotic solvents - leave nucleophile not solvated (ex propanone)

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5
Q

Explain structural inversion of molecule

A

When nucleophile attacks, attacks on opposite side of LG - after the reaction molecule is inversed - inversion

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6
Q

Reaction of tertiary halogenalkane

A

Sn1

  1. LG leaves by breaking the bond heterolytically
  2. Carbocation forms (stabilised by three alkyl groups)
  3. Nucleophile attacks

Due to steric hindrance a nucleophile cannot attack - LG leaves (slow, so rate determining step) but carbocation is symetrical - nucleophile can attack from any side - Sn1 not stereospecific

Unimolecular - rate determined only by conc of halogenalkane

rate = k [halogenlakane]

Polar protic (H bonds) solvents favour Sn1 - solvate the carbocation (ex water, alcohols, carboxylic)

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7
Q

Reaction of secondary halogenalkanes

A

Sn1/Sn2 - mixture of both

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8
Q

Reaction rates in halogenlakanes

A
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9
Q

Rate of LG leaving by halogenalkane halogen

A

Determined by two opposing:

1) polarity of halogen - C bond (the least polar break off faster)
2) strength of halogen - C bond (the weakset break off more easly)

But bond strength more important - LG according to 2)

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10
Q

Sn1 vs Sn2

A
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11
Q

Reaction of alkenes

A

Electrophilic addition

Alkenes - unsaturated - double bonds - 120 degrees - open structure for C attack - pi bonds electron dense - electrophiles attracted - electrophiles produced in heterolytic fission

Addition of halogens or hydrogen halides

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12
Q

Electrophilic addition reaction mechanism

A
  1. Halogen/halide molecule is polarised by an alkene - electron rich region
  2. Halogen/halide splits in heterolytic fission - as poles have been formed (ex: Br+ and Br- / H+ and Br-)
  3. Electrophilic halogen (ex Br+) atom breaks pi bond - attaches to now alkane - left carbocation
  4. Carbocation reacts withe the other halogen atom (ex Br-)
  5. Halogenalkane formed
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13
Q

Explain unsymmetric electrophilic addition

A

Unsymetrical alkene reacts with halide - two possible products - Markinkov’s rule: H will attach to that C that is already bonded to the greater number of H

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14
Q

Reaction of benzene

A

Electrophilic substitution

Delocalised pi e - attract electrophiles - high activation E - slow

  1. Electrophile attacks pi bond
  2. Non aromatic carbocation forms
  3. H leaves and the electrophile stays
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15
Q

Explan nitration of benzene

A

Electrophilic substitution

  1. Electrophile - nitronium oxide NO2+ - generated by nitrating mixture
  2. NO2+ attacks pi bond (temp. H2SO4)
  3. Carbocation forms
  4. H leaves
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16
Q

Explain production of NO2+

A

By nitrating mixture

17
Q

Explain reduction of carbonyl compounds

A

Reduction (reverse of oxidation) - involves reducing agents: NaBH4, LiAlH4

18
Q

Explain reduction of nitrobenzene

A

Nitrobenzene can be converted into phenylamine

  1. Nitrobenzene interacted with Sn and conc HCl - phenylammonium ion
  2. Phenylammonium ion reacts with NaOH - removes one H - phenylamine
19
Q

Summary of all reaction mechanisms

A

Chemistry revision (31 decks)

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