C6H5-CH2Br is an activated halide (kind of allylic position) and thus despite it is a R-X compound it displays reactivities between
secundary and tertiary halides (R2CHBr and R3CBr) .But reactivity isn't stability and if (CH3)3C(+) is stabilised by the three CH3 (I+
effect), in the benzylic position all the aromatic ring participates to the stability and the resonance will help a bit.
Thus so far t-butyl bromide > or = to Ar-CH2-Br on reactivity and stability.
The introduction of a -C#N will make the benzilic position more electronegative (more acidic) and thus renders the formation of a
carbocation less easy...it starts to prefer to allow H(+) ore Br(+) to leave a carbanion.
Actually the more aromatic ring or electron withdrawing groups (-CO2R, -NO2, -C#N, -CO-R, ...) on a Carbon the more it favors radicalar
stabilisation/splitting...radicalar splitting means the carbon keeps one electron.
See (Ar)3C-C(Ar)3, (Ar)3C-Br, (N#C-)3C-C(-C#N)3, Br-C(NO2)3.
Thus to my feeling:
3>or =2 >1
Johannes Babach wrote:
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