Why it is difficult to remove cl as nucleophile from benzene rather than alkyl halide?
Answers
Answered by
0
CH3-CH2-Cl + OH- —→ CH3-CH2-OH + Cl- ?
As the Cl- leaves, the alkyl group becomes positively charged CH3-CH2+
Methyl groups, ethyl groups etc are considered to be electron-donating groups (+I inductive effect). Therefore in this case, the CH3 group stabilizes the positive charge on the CH2+ by pushing electrons towards it (neutral is stable, charged is unstable; meaning lesser the magnitude of charge on a C atom, more stable it will be). This makes the chlorine leave easily (otherwise the positive charge on the C atom will attract the Cl- very strongly and will not allow its departure). The OH- is attracted by the positive charge on CH2+ and becomes fixed on it (fills the vacancy) to give the alcohol.
[The C atom of CH3-CH2-Cl is sp3-hybridized; It momentarily becomes sp2-hybridized in CH3-CH2+ and again changes back to sp3-hybridized in CH3-CH2-OH]
This argument is supported by the fact that the reaction rate increases if the halogen atom is on a secondary carbon.
CH3-CH(Cl)-CH3 + OH- —→ CH3-CH(OH)-CH3 + Cl-
Now when Cl- leaves, the positive charge is on the middle carbon. It is stabilized by two methyl groups (on both sides) and is more stable than in the first example. So more chance of OH- replacing the Cl-.
But in the case of chlorobenzene (C6H5Cl), the benzene ring is considered to be an electron-withdrawing group (-I inductive effect). Therefore it tends to destabilize the positive charge on the C atom (makes it more positive) when Cl- tries to leave. Also, the C atoms in a benzene ring are already sp2-hybridized and the ring is stable due to resonance.
As the Cl- leaves, the alkyl group becomes positively charged CH3-CH2+
Methyl groups, ethyl groups etc are considered to be electron-donating groups (+I inductive effect). Therefore in this case, the CH3 group stabilizes the positive charge on the CH2+ by pushing electrons towards it (neutral is stable, charged is unstable; meaning lesser the magnitude of charge on a C atom, more stable it will be). This makes the chlorine leave easily (otherwise the positive charge on the C atom will attract the Cl- very strongly and will not allow its departure). The OH- is attracted by the positive charge on CH2+ and becomes fixed on it (fills the vacancy) to give the alcohol.
[The C atom of CH3-CH2-Cl is sp3-hybridized; It momentarily becomes sp2-hybridized in CH3-CH2+ and again changes back to sp3-hybridized in CH3-CH2-OH]
This argument is supported by the fact that the reaction rate increases if the halogen atom is on a secondary carbon.
CH3-CH(Cl)-CH3 + OH- —→ CH3-CH(OH)-CH3 + Cl-
Now when Cl- leaves, the positive charge is on the middle carbon. It is stabilized by two methyl groups (on both sides) and is more stable than in the first example. So more chance of OH- replacing the Cl-.
But in the case of chlorobenzene (C6H5Cl), the benzene ring is considered to be an electron-withdrawing group (-I inductive effect). Therefore it tends to destabilize the positive charge on the C atom (makes it more positive) when Cl- tries to leave. Also, the C atoms in a benzene ring are already sp2-hybridized and the ring is stable due to resonance.
Similar questions