Chemistry How to calculate new IHD after finding a molecule in IR spec

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To calculate the new Index of Hydrogen Deficiency (IHD) after identifying a ketone in a molecule with the formula C5H10O, the original IHD is determined to be 1. When subtracting the components of the ketone, it is clarified that only the carbonyl group (CO) should be subtracted, leading to a new IHD of 0. This approach is consistent for other functional groups, where specific atoms or groups are removed based on their structure, such as CO for ketones and aldehydes, CO2 for carboxylic acids, and CON for amides. Understanding these subtractions is crucial for future calculations involving different functional groups. The discussion emphasizes the importance of knowing which components to subtract for accurate IHD calculations.
AMan24
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Homework Statement


Lets say i have C5H10O.. The IHD is (5x2+2-10)/2 = 1 IHD

So now i use my IR chart and i determine i have ketone

Now i want to find the new IHD and also figure out what other atoms i have left. So i need to subtract C5H10O by ketone and figure out the new IHD.

Homework Equations


Given formula: CcHhNnOoXx

IHD = 0.5 * [2c+2-h-x+n]

3. The Attempt at a Solution

So for ketone, my guess would be either subtracting CO from the original formula or subtracting C2H3O then calculating the new IHD which would be either

C5H10O - C2H3O = IHD 0.5 which doesn't make sense
C5H10O - CO = IHD 0, which does make sense, but is it correct

And you might say i don't need to calculate it for this problem, but i just want to know for in the future because for some problems finding the new IHD is useful...

My question is: What am i supposed to subtract exactly? And not just for this example but, let's say i have aldehyde, do i just subtract the CO or do i subtract CHO
 
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AMan24 said:
C2H3O

?
 
Borek said:
?

Isn't that was ketone is? A CO and a CH3 = C2H3O
 
Try to draw a C2H3O molecule.
 
Borek said:
Try to draw a C2H3O molecule.

I think i might have figured it out. So for ketones and aldehydes, its just CO. For carboxylic acids and esters its CO2. For amides its CON.. and so on
 

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