6.1 halogen derivatives Flashcards
why do halogenoalkanes have a higher boiling point than alkanes?
- alkanes and halogenoalkanes have simple molecular structure
- alkanes are non-polar with id-id interactions while halogenoalkanes are polar with pd-pd interactions
- moreover, size of electron cloud is greater for halogenoalkane
- energy required to overcome interactions larger
why does boiliing point of halogenoalkanes increase from -F < -Cl < Br < I ?
- size of electron cloud and hence extent of distortion cloud increases
- strength of id-id interactions increases
- energy required to overcome interactions increases
(cannot use pd-pd to explain)
briefly describe why halogenoalkanes can undergo nucleophilic substitution
- the C-X bond is polarised by the electronegative halogen atom which attracts the shared pair of electrons away from C
- the carbon atom in the C-X bond is associated with a partial positive charge δ+, making it susceptible to attack by nucleophiles
- the halogen atom is displaced as a halide ion
describe the Sn2 mechanism of nucleophilic substitution
- in a one-step process, the nucleophile attacks the δ+ carbon atom from the opposite side of the halogen atom
- a pentavalent transition state is formed with nucleophile and halogen bonded to C
- bond breaking and bond forming are taking place simultaneously
why does the nucleophile attack from the opposite side of the halogen?
the large bulky halogen atom sterically hinders the nucleophile from attacking from the same side, since there will be repulsion between electron clouds
what are the methods of preparation for halogenoalkanes (3)?
- free-radical substitution of alkanes (limited X2, UV light)
- electrophilic addition of alkenes (HX(g) or X2(aq) in CCl4)
- nucleophilic subtitution of alcohols (HX(g), heat OR PX3 (s/l), heat OR PCl5(s) OR SOCl2(s) in pyridine, heat)
describe the Sn1 mechanism of nucleophilic substitution
step 1: formation of carbocation (slow)
- with X atom δ- and C atom δ+, the C-X bond breaks heterolytically to form a carbocation intermediate and a halide ion
step 2: attack of nucleophile (fast)
why will Sn1 mechanism with a chiral reactant result in an optically inactive mixture?
- carbocation is trigonal planar with respect to the carbon atom, hence the nucleophile attacks from the top and bottom of the plane with equal probability
- if the carbon is chiral, an optically inactive racemic mixtrure is formed as both enantiomers are formed in equal proportion
when is Sn2 favoured?
for methyl or primary halogenoalkanes
- has no or one alkyl group bonded to the δ+ carbon, allowing easy approach of the nucleophile to form pentavalent transition state (as opposed to steric hindrance by tertiary)
when is Sn1 favoured?
for tertiary halogenoalkanes
- has 3 electron-donating methyl groups to disperse the positive charge on the carbocation intermediate, giving a stable tertiary carbocation intermediate
also for benzene
- resonance stabilisation of benzyl carbocation by delocalisation of electrons from benzene rring
how does the rate equation show Sn1 or Sn2?
Sn1: k [halogenoalkane]
Sn2: k [nucleophile][halogenoalkane]
what reactions do halogenoalkanes undergo (2 mechanisms, 4 reactions)?
- nucleophilic substitution
(OH- to alcohols, -CN to nitriles, -NH3 to amines) - elimination to alkenes (KOH in ethanol, heat under reflux)
state the reagents and conditions for nucleo sub to alcohols
NaOH or KOH, heat under reflux
state the reagents and conditions for nucleo sub to nitriles and its subsequent reactions
KCN or NaCN in ethanol, heat under reflux
acidic hydrolysis to -COOH: HCl, heat under reflux
basic hydrolysis to -COO-: NaOH, heat under reflux
reduction to CH2NH2: LiAlH4 in dry ether
state the reagents and conditions for nucleo sub to primary amines
excess NH3 in ethanol, heat in a sealed tube
why will quarternary ammonium salts be formed with limited ammonia?
alkyl groups on alkylamines arre electron-donating, hence increasing the electron density on alkylamine and making it a stronger nucleophile than NH3, competing with NH3 for RX
what is the trend of reactivity from F to I?
reactivity depends on the strength of C-X bond
- from C-F to C-I, bond length increases, hence bond strength decreases and bond dissociation energy decreases
- hence reactivity F < Cl < Br < I
why do halogenoarenes not undergo reaction unless subjected to drastic conditions?
- the p orbital of the halogen atom overlaps with the p orbitals of the carbon atoms in the benzene ring
- the lone pair of electrons from the halogen can delocalise into the delocalised electron cloud of the benzene ring
- this forms a partial double bond character in C-X bond
- more energy is required to break C-X bond
also, sterically hindered by the benzene ring; electron cloud of benzene ring will repel attacking nucleophiles
describe the chemical test for halogen derivatives and the expected observations
add an equal volume of NaOH to RX and heat in a water bath. cool the mixture and add excess HNO3. finally, add AgNO3.
I: yellow ppt
Br: cream ppt
Cl: white ppt
F: no ppt
chlorobenzene: no ppt
