[Fe(CN)63-] × 104
(mol dm-3 ) / [CHP] × 103
(mol dm-3 ) / [OH-]
(mol dm-3 ) / [I-] × 108
(mol dm-3) / kobs × 103 (s-1)
Exptl. / Calculated
2.0 / 1.0 / 0.1 / 5.0 / 3.96 / 3.94
2.0 / 1.0 / 0.3 / 5.0 / 7.67 / 7.89
2.0 / 1.0 / 0.5 / 5.0 / 9.82 / 9.86
2.0 / 1.0 / 0.7 / 5.0 / 11.1 / 11.0
2.0
2.0 / 1.0
1.0 / 1.0
0.3 / 5.0
0.0 / 12.6
0.90 / 12.1
-
2.0 / 1.0 / 0.3 / 1.0 / 1.50 / 1.57
2.0 / 1.0 / 0.3 / 2.0 / 3.29 / 3.15
2.0 / 1.0 / 0.3 / 3.0 / 4.54 / 4.73
2.0 / 1.0 / 0.3 / 4.0 / 6.29 / 6.31
2.0 / 1.0 / 0.3 / 5.0 / 7.67 / 7.89
2.0 / 1.0 / 0.3 / 6.0 / 9.32 / 9.47
2.0 / 1.0 / 0.3 / 7.0 / 10.5 / 11.0
2.0 / 1.0 / 0.3 / 8.0 / 12.1 / 12.6
2.0 / 1.0 / 0.3 / 9.0 / 13.8 / 14.2
2.0 / 1.0 / 0.3 / 10.0 / 15.6 / 15.7

TableS1Effect of the variation of OH- and I- concentrations on the iodide mediated oxidation of chloramphenicol by hexacyanoferrate(III) in aqueous alkaline medium at 25C,µ =1.10 mol dm-3

TableS2 Effect of ionic strength and dielectric constant on the rate of reaction.

µ
( mol dm-3) / kobs × 103
(s-1) / % of t-butanol-H2O
(v/v) / D / kobs × 103
(s-1)
0.40 / 3.89 / 0 / 78.50 / 7.67
0.60 / 5.04 / 5 / 75.12 / 5.87
0.80 / 6.25 / 10 / 71.74 / 4.74
1.10 / 7.67 / 15 / 68.36 / 3.97
1.50 / 9.02 / 20 / 64.98 / 3.05
2.00 / 12.82 / 25 / 61.60 / 2.24
- / - / 30 / 58.22 / 1.60

Fig.S1 First order plots of iodide mediated oxidation of chloramphenicol by hexacyanoferrate(III) in aqueous alkaline medium at 25oC [CHP] = 1.0 x 10-3, [OH-] = 0.30; [I -] = 5.0 x 10-8mol dm-3 [Fe(CN)63-] x 104mol dm-3= (i ) 0.20 (ii) 0.50 (iii) 1.0 (iv) 2.0 and (v) 3.0

Fig.S2 FT-IR spectra of p-nitrobenzaldehyde, the product obtained during the oxidation of chloramphenicol by hexacyanoferrate(III)

Fig.S3 GC-MS spectra of the product p-nitrobenzaldehyde showed molecular ion peak and base peak at m/z 151 amu

Fig.S41H NMR spectra of p-nitrobenzaldehyde, the product obtained during the oxidation of chloramphenicol by hexacyanoferrate(III)

Supplementary information(Rate law):

From Scheme 1, the rate law (12) can be derived as follows:

From second step of Scheme 1 we have,

Therefore,

Substituting eq.(3) in eq.(2) we get

From first step of Scheme 1 we have,

Substituting eq.(7) in eq.(6) we get,

The total concentration of [CHP]t is given by,

Where subscripts ‘t’ and ‘f’ refers to total and free concentrations

Therefore,

In view of low concentration of iodide used in the experiment the term,

can be neglected,

Similarly,

In view of low concentration of CHP used, the term in the denominator can be neglected,

Therefore total concentration of OH- is given by,

Similarly,

Substituting eq.(10), (12) and (13) in eq.(8) and omitting the subscripts we get,

OR

eq.(14) can be rearranged to eq.(15), which is suitable for verification