SME
Organic Chemistry I · Problem type
Rings and π bonds from a formula
Use this for “A compound is C₅H₇N. How many rings and π bonds can it have?”
Diagram
IHD
=
2 × 6 + 2 + 0 − 10 − 0
2
=
2.
One structure with these counts: cyclohexanone, C₆H₁₀O: 1 ring + 1 π bond = 2.
O and S add no H, so they leave the IHD as it is; a triple bond is 2 π bonds.
This is an example. Type your own values to start a new problem.
IHD
=
2
×
+
2
+
−
−
2
Formulas
IHD
=
2 ×
C
+ 2 +
N
−
H
−
X
2
2
=
2 × 6 + 2 + 0 − 10 − 0
2
Assumptions
•
An open chain of single bonds with C carbons holds 2C + 2 hydrogens; each ring or π bond takes 2 of them away.
•
A double bond counts 1, a triple bond 2, and a benzene ring 4 (one ring and three π bonds).
•
Oxygen and sulfur leave the count alone; each nitrogen adds room for one H, and each halogen stands in for one H.
•
The picture draws one known structure that fits the typed counts, when it has one.
Step-by-step
WE KNOW
C = 6, H = 10, N = 0, X = 0
FIND
index of hydrogen deficiency (IHD)
Step 1 · Find index of hydrogen deficiency (IHD)
Use IHD = (2 × C + 2 + N − H − X) ÷ 2
Count
the
H
an
open
chain
would
hold,
2
C
+
2,
plus
one
for
each
N
.
Take
away
the
H
and
halogens
there
are,
then
halve.
IHD
=
2 × 6 + 2 + 0 − 10 − 0
2
IHD
=
12 + 2 − 10
2
IHD
=
4
2
IHD = 2 → IHD = 2: rings and π bonds add up to 2; a triple bond counts 2.
Check
Put the numbers back in. Both sides should match.
2
=
2 × 6 + 2 + 0 − 10 − 0
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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