Supplementary material

(a)  Crystallographic data for the structure reported in this paper have been deposited in the Cambridge Crystallographic Data Centre as supplementary materials (CCDC No. 609935). These data can be obtained free of charge via www.ccdc.cam.ac.uk/data_request/cif, or by emailing , or by contacting the Cambridge Crystallographic Data Centre, 12, Union Road, Cambridge CB2 1EZ, UK; fax: +44 1223 336033.

(b)  The detailed calculation of GC yield and determination of isolated yield are given.

Method of determination of GC yield:

Firstly response factor needs to be determined.

Let the response factor be F for the olefins with respect to internal standard.

Now, area of olefin signal / moles of olefin = F ´ area of standard signal / moles of standard.

That is, ao / mo = F ´ as / ms.

Therefore, F = ao ´ ms / as ´ mo…………..(1) where mo and ms are the moles of olefin and that of standard respectively. ao and as are the area of olefin and that of standard respectively at zero time of the mixing.

This equation presumes a linear response of the detector to both olefin and the standard.

Constancy of F has been checked by varying randomly the moles of olefins and recording the chromatogram at zero time.

Similarly, a known quantity of epoxide is mixed with the same quantity of the standard as in equation (1) and is chromatographed to find the response factor of this system.

Therefore, F¢ = ae ´ ms / as¢ ´ me……………(2) where ae and as¢ are the area of epoxide and standard, and me and ms are the moles of epoxide and standard, respectively. The F´ = response factor for the epoxides with respect to standard.

Constancy of F¢ has been checked by varying randomly the moles of olefins. The constancy of both F and F¢ having been checked, varying moles of standard randomly without changing the moles of olefin and epoxide respectively, it can be concluded that equation (1) and (2) should always be valid.

In case of olefin after reaction is over,

ao¢ / mo¢ =F ´ as¢¢ / ms¢

Therefore, mo¢ = ao¢ ´ ms¢ / as¢¢ ´ F……………(3), where ao¢ and as¢¢ are the area of olefin and standard respectively. mo¢ is the unknown moles of olefin and ms¢ is the moles of standard.

From this equation mo¢ can be determined by applying known values of ao¢, as¢¢ and ms¢.

In the case of epoxide after the reaction is over,

ae¢¢ / me¢¢ =F¢ ´ as¢¢¢ / ms¢¢

Therefore, me¢¢ = ae¢¢ ´ ms¢¢ / as¢¢¢ ´ F¢………… (4), where ae¢¢ and as¢¢¢ are the area of epoxide and standard respectively. ms¢¢ is the moles of standard and me¢¢ is the unknown moles of epoxide.

Thus, from the equation (4) me¢¢ have been calculated, as all other terms are known.

Therefore, % GC yield = [moles of epoxide (me¢¢) / {moles of olefin (mo¢) + moles of epoxide (me¢¢)}]´100.

Determination of isolated yield:

The isolated yield in a few cases (Table 1) is obtained by multiple ether extraction of the reaction solution after the reaction is over and then evaporating the ether and acetonitrile by distilling at a mildly reduced pressure (using water aspirator) and kept over P2O5 in a desiccator and weighed (when the GC yield was 98-99%) in a micro-balance and then confirmed the identity of the products by IR and GC analysis. For lower yield % the liquid products were subjected to preparative TLC and the highly intense spot was cut out and plunged in CH2Cl2 which serves as an eluant and then the resulting solution was dried over MgSO4, filtered through a short silica gel pad and finally the filtrate was evaporated by distillation of the solvent to yield only the epoxide. The residue was then kept over P2O5 for 15 minutes and then weighed. For solid epoxides obtained from liquid olefins, the former are simply dried and weighed.

(c). The probable reaction routes for the synthesis of the title complex with or without H3PO4 are given in the supplementary material as equations 1-8 and 9-15 respectively.

12 MoO3 + 2 H2O2 → 2 H2[Mo6O19] + O2 …………………………………………(1)

2 H2[Mo6O19] + CH3COCH2COCH3 + 2 H2O2 → H2[Mo12O36] + 2 CH3CO2H +HCO2H + 2 H2O ……………………………………………...(2)

H2[Mo12O36] + 0.5 H2O2 → H[Mo12O36] + H2O ……………………………….…..(3)

H[Mo12O36] + H3PO4 → H4[PMo12O40] ……………….……………………….….(4)

Adding (1) to (4) we get,

12 MoO3 + 4.5 H2O2 + CH3COCH2COCH3 + H3PO4→ H4[PMo12O40] +

2 CH3CO2H +HCO2H + 3 H2O + O2 …………………………(5)

Multiplying the equation (5) by 2 we get,

24 MoO3 + 9 H2O2 + 2 CH3COCH2COCH3 + 2 H3PO4→ 2H4[PMo12O40] +

+4 CH3CO2H +2 HCO2H + 6 H2O +2 O2 ………………………(6)

Using PPh4Cl for precipitation,

2 H4[PMo12O40] + 8 PPh4Cl → 2 (PPh4)4[PMo12O40] + 8HCl ………..……………(7)

Adding (6) to (7) we get,

24 MoO3 + 9 H2O2 + 2 CH3COCH2COCH3 + 2 H3PO4 + 8 PPh4Cl →

2 (PPh4)4 [PMo12O40] + 4 CH3CO2H +2 HCO2H + 6 H2O +2 O2 +

8 HCl …………………………………………………………………………..(8)

12 MoO3 + 2 H2O2 → 2 H2[Mo6O19] + O2 ………………………………………..(9)

2 H2[Mo6O19] + CH3COCH2COCH3 + 2 H2O2 → H2[Mo12O36] + 2CH3CO2H + HCO2H + 2 H2O ………………….…………………………..(10)

H2[Mo12O36] + 0.5 H2O2 → H[Mo12O36] + H2O ………………………………….(11)

PPh4Cl + 4 H2O2 → H3PO4 + 4 PhOH + HCl …………………………….……… (12)

Adding (1) to (4) we get,

12 MoO3 + 8.5 H2O2 + CH3COCH2COCH3 + PPh4Cl → H[Mo12O36] + H3PO4 +

4 PhOH +2CH3CO2H +HCO2H + 3 H2O + O2 + HCl

= H4[PMo12O40] + 4 PhOH +2 CH3CO2H +HCO2H +

3 H2O + O2 + HCl …….…………….……………………….(13)

Multiplying the equation (5) by 2 we get,

24 MoO3 + 17 H2O2 + 2 CH3COCH2COCH3 + 2 PPh4Cl → 2H4[PMo12O40] +

8 PhOH +4 CH3CO2H +2 HCO2H + 6 H2O +2 O2 + 2 HCl ……………(14)

Using excess PPh4Cl for precipitation,

24 MoO3 + 17 H2O2 + 2 CH3COCH2COCH3 + 10 PPh4Cl → 2 (PPh4)4[PMo12O40] + 8 PhOH +4 CH3CO2H +2 HCO2H + 6 H2O +2 O2 + 10 HCl…(15)