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In para-Toluenesulfonylchloride, there are 5 pi bonds and 1 ring. This suggests a IHD of 6. However, the actual IHD is 4. How come?

Is it because the formula IHD = rings + pi bonds does not apply to compounds containing sulfur? If not, what formula do I use? I have this formula:

$IHD = \frac{2C + N - X - H + 2}{2}$

Using this formula yields a IHD of 4.

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You have to start from the general formula for the degree of unsaturation $$DU = 1 + \frac{1}{2} \sum n_i(v_i-2),$$ where $n_i$ is the number of atoms with a valency of $v_i$.

For para-toluenesulfunylchloride ($\ce{C7H7SO2Cl}$) you will therefore arrive at \begin{align}DU &= 1 + \frac{1}{2}\Big[n_\ce{C}(4-2) + n_\ce{H}(1-2) + n_\ce{S}(6-2) + n_\ce{O}(2-2) + n_\ce{Cl}(1-2)\Big]\\ &=1+\frac12\Big[7(4-2) + 7(1-2) + 1(6-2) + 2(2-2) +1(1-2)\\ &=1 + \frac12\Big[14 -7 +4 +0 -1\Big]\\ &=6, \end{align}

because sulfur uses all six valence electrons in this case.

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  • $\begingroup$ Just to confirm though, you need to know the oxidation state of sulfur before using the equation. $\endgroup$ – jerepierre Sep 8 '14 at 14:08
  • $\begingroup$ @jerepierre Yes of course. I was just making the point that the OP used the wrong formula. $\endgroup$ – Martin - マーチン Sep 8 '14 at 15:15
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There are 5 pi bonds: three C-C in the ring and two S-O pi bonds. Add the ring, and the IHD is 6.

Because of it's ability to exist as higher oxidation states than the second row elements, I don't believe that there is any way to include sulfur in the IHD formula unambiguously. Consider the following two compounds, toluenesulfonyl chloride and an isomer.

enter image description here

The formulas are the same, but the IHD of the left structure is 6 but the right structure is 4.

I think the best way to handle third row (and beyond) is to treat the large atom as if it were it's second row analog (P->N, S->O, etc.), and use the result as the lower limit on the IHD.

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