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# What does ihd of 5 mean?

In a hydrocarbon where all the C atoms have only single bonds and no rings are involved, the compound would have the maximum number of H atoms. If any of the bonds are replaced with double or triple bonds, or if rings are involved, there would be a “deficiency” of H atoms. By calculating the index of hydrogen deficiency (IHD), we can tell from the molecular formula whether and how many multiple bonds and rings are involved. IHD is also called the Degree of Unstaturation. This will help cut down the possibilities one has to consider in trying to come up with all the isomers of a given formula.

Here is a summary of how the index of hydrogen deficiency (IHD) works.

Hydrocarbons (\(C_xH_y\)):

\[IHD = \dfrac{2x + 2 - y}{2} \]

(where \(x\) and \(y\) stand for # of C and H respectively.)

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The best thing about the IHD is that it is an easy equation to use that gives you useful information about the structure of an unknown compound. It literally takes a minute to do. Calculating the degrees of unsaturation is THE first task I recommend doing when you are faced with determining the structure of an unknown compound with a known molecular formula. In this post, we see that this calculation was clutch in the determination of the structure of the side chain of deer tarsal pheromone.

Before we get going, see if you can apply the equation in the box above to find the degrees of unsaturation in these extremely well known molecules whose formulae are given below.

The latter three molecules are somewhat “naughty” examples, being illegal in most, if not all states. Hopefully that’s an incentive for you to actually do the calculation!  (answers at the bottom of the post).

Let’s start this discussion by looking for patterns in molecular formulae. We’ll start with hydrocarbons containing no double bonds or rings.

Can you see a relationship between the number of carbons and the number of hydrogens?

More specifically, can you come up with a formula?

You should be able to see that for a hydrocarbon with no rings or double bonds, the number of hydrogens is equal to twice the number of carbons, plus 2.

So for a molecule like dodecane (C12) we’d expect to see (12 x 2) + 2 = 26 hydrogens.

Now: what happens to the molecular formula when we add a double bond to the molecule?

You should be able to see that each multiple bond (π bond ) the number of hydrogens in the formula decreases by two. Ethyne (2 π bonds) has two fewer hydrogens than ethene (1 π bond), which has two fewer hydrogens than ethane (zero π bonds).

Hydrocarbons containing  π bonds are often called “unsaturated” hydrocarbons. They can be treated with hydrogen (Pd/C, H2 ) to give the corresponding alkane with no  π bonds, which is then said to be “saturated” with hydrogen. (Compare “unsaturated fats“, which contain alkenes, and “saturated fats” which do not).

Since every pi bond results in a loss of 2 hydrogens from the molecular formula, we refer to this to as a “degree of unsaturation“.

Let’s turn our attention to cyclic compounds. Do you notice a similar effect?

You should! Every ring in the molecule decreases the number of hydrogens by two.

Therefore, each ring introduces a “degree of unsaturation” into the molecule.

You might ask: what if we have a molecule with rings and multiple bonds? See for yourself.

The effect is additive. That is, the “degrees of unsaturation” is the sum of the number of double bonds and rings.

Note that it doesn’t tell you how many double bonds are present or how many rings are present. It merely tells you their sum.

For instance, thus the molecular formula C6H6 (4 degrees of unsaturation) is satisfied by molecules with

[Historical side note: the formula of benzene was known to be C6H6 from Michael Faraday in 1825, but the correct structure was not proposed until 1865 (Kekule) and not confirmed until 1929, by Catherine Lonsdale’s X-ray studies . Just a reminder that knowing the molecular formula only gets you so far. ]

Helpful hint: if you see 4 degrees of unsaturation in an unknown molecule, thinking “benzene ring” is a good place to start.

So far, so good.  Let’s move on. What about molecules with oxygen?

The same rules apply! We can calculate the number of hydrogens from the number of carbons as if the oxygen didn’t exist.

Note that a π bond which contains oxygen still counts as a “degree of unsaturation” (see formaldehyde, CH2O).

OK. What about halogens like chlorine, fluorine, iodine and bromine?

For the purposes of our formula relating hydrogens to carbons for molecules without rings or double bonds, halogens can be counted as hydrogens. That is, carbon tetrachloride (CCl4) has the same degrees of unsaturation (zero) as CH4.

We can thus modify the equation for a molecule with no double bonds or rings as follows:

We’ve saved nitrogen for last, because it’s a bit weird.  Try to see the pattern of how nitrogen affects our formula.

OK, you might have noticed that we can’t ignore nitrogen like we did oxygen.

The way to make the equation work with nitrogen is to add the nitrogen count to the right hand side of the equation, like this:

Try it with cadaverine (C5H14N2) which contains no double bonds or rings.

So far our equation has only applied to “saturated” compounds with no double bonds or rings.

How can we use this to make a general formula for “degrees of unsaturation”, which is actually… useful?

Math time. If we move the left hand side over to the right, we get this:

Zero equals the degrees of unsaturation in this case.

If we introduce a double bond or ring, the equation will return “2”.

Each successive degree of unsaturation will increase this number by 2. For instance, 2-chloropyridine C5H4NCl gives us 8:

Now 8 is not the degrees of unsaturation. It’s merely the number of hydrogens “missing” from the corresponding non-cyclic alkane of the same carbon count.

In order to make this equation give us the actual degrees of unsaturation, we need to divide everything by two.

We can really condense things by using the symbols H, C, N, and X to represent hydrogens, carbons, nitrogens and halogens, respectively.

This is our final form for the equation for “degrees of unsaturation”, or “index of hydrogen deficiency” if you prefer.

OK, now for those real-life examples I mentioned above.

Let’s plug each of them into our formula. You should get:

Now, let’s look at each of the molecules. You’ll find that the calculated degrees of unsaturation agrees with the sum of [multiple bonds and rings] in the molecule.

Super useful. Super easy.

With the structure in front of you it seems obvious. But if you’re dealing with an unknown it can be very helpful, like we saw with deer tarsal gland hormone in the last post.

In the next post, we’ll start delving into the mysteries of spectroscopy. We’ll start with one of the simplest spectroscopic techniques – UV-Vis spectroscopy – and show how it can be used to deliver clues about the structures of various molecules.

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