I don’t know where you heard about the water molecule interacting with d orbitals before displacement takes place, but the idea is wrong. Phosphorus’ d orbitals are energetically too far removed to take part in bonding. Remember that according to the aufbau principle, 3d orbitals are higher in energy than 4s! They exist and you can excite electrons into them but that’s it (and you need high energies for that excitation).
Rather, the oxygen of water interacts with the $\sigma^*_{\ce{P-Cl}}$ orbital. Since phosphorus uses an orbital with a very high p-content, it has a notable lobe on the far side allowing for interaction. An intermediate tetracoordinated phosphorus species is formed which displaces the better leaving group (chloride). Water can lose a proton at any point in this mechanism.
The mechanism is basically identical to that of a carbon-centred $\mathrm{S_N2}$ reaction except that the intermediate breaks down much faster in carbon-based reactions.
I don’t know where you heard about the water molecule interacting with d orbitals before displacement takes place, but the idea is wrong. Phosphorus’ d orbitals are energetically too far removed to take part in bonding. Remember that according to the aufbau principle, 3d orbitals are higher in energy than 4s! They exist and you can excite electrons into them but that’s it (and you need high energies for that excitation).
Rather, the oxygen of water interacts with the $\sigma^*_{\ce{P-Cl}}$ orbital. Since phosphorus uses an orbital with a very high p-content, it has a notable lobe on the far side allowing for interaction. An intermediate tetracoordinated phosphorus species is formed which displaces the better leaving group (chloride). Water can lose a proton at any point in this mechanism.
The mechanism is basically identical to that of a carbon-centred $\mathrm{S_N2}$ reaction except that the intermediate breaks down much faster in carbon-based reactions.
I don’t know where you heard about the water molecule interacting with d orbitals before displacement takes place, but the idea is wrong. Phosphorus’ d orbitals are energetically too far removed to take part in bonding. Remember that according to the aufbau principle, 3d orbitals are higher in energy than 4s! They exist and you can excite electrons into them but that’s it (and you need high energies for that excitation).
Rather, the oxygen of water interacts with the $\sigma^*_{\ce{P-Cl}}$ orbital. Since phosphorus uses an orbital with a very high p-content, it has a notable lobe on the far side allowing for interaction. An intermediate tetracoordinated phosphorus species is formed which displaces the better leaving group (chloride). Water can lose a proton at any point in this mechanism.
The mechanism is basically identical to that of a carbon-centred $\mathrm{S_N2}$ reaction except that the intermediate breaks down much faster in carbon-based reactions.
I don’t know where you heard about the water molecule interacting with d orbitals before displacement takes place, but the idea is wrong. Phosphorus’ d orbitals are energetically too far removed to take part in bonding. Remember that according to the aufbau principle, 3d orbitals are higher in energy than 4s! They exist and you can excite electrons into them but that’s it (and you need high energies for that excitation).
Rather, the oxygen of water interacts with the $\sigma^*_{\ce{P-Cl}}$ orbital. Since phosphorus uses an orbital with a very high p-content, it has a notable lobe on the far side allowing for interaction. An intermediate tetracoordinated phosphorus species is formed which displaces the better leaving group (chloride). Water can lose a proton at any point in this mechanism.
The mechanism is basically identical to that of a carbon-centred $\mathrm{S_N2}$ reaction except that the intermediate breaks down much faster in carbon-based reactions.
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