Degrees of Unsaturation

Correct Answer: D. 11

Degree of Unsaturation (or index of hydrogen deficiency, IHD), or rings plus pi bonds, lets you determine the number of rings, double bonds and triple bonds present in the compound if the molecular formul is known. It does not give the exact number of rings or double or triple bonds, but rather the sum of the number of rings and double bonds plus twice the number of triple bonds.

Degree of Unsaturation = C – (H/2) – (X/2) + (N/2) +1 (C=#carbon, H=#hydrogen, X=#halogen, N=#nitrogen)

= 16 – (13/2) – (1/2) + (2/2) + 1 = 11

Extended Explanation
First use the saturated formula CnH(2n+2) which will get you to C16H34

Now lets deal with the Cl. How does it play a role in this molecule? It connects with a carbon and TAKES the place of a H. So the saturated molecule should have 1 less Hydrogen from what we predicted.

Now we have C16H33

Now the two nitrogens. Lets assume they are amines for the saturated solution. They connect with the carbon and TAKE the place of a hydrogen BUT add on TWO hydrogens (NH2). Net gain of 1 hydrogen per N. So the saturated molecules would have 2 more H of the current formula we predicted.

Now we have C16H35.

All you have to do now is subtract the H difference divided by two.

35-13 = 22/2 = 11 degrees of unsaturation

Credits to extended explanation to DatRav27 on SDN.

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