Structural Determination of Organic Compounds

25min Part 4 / Ch8 / Lesson 3

Prerequisites: 4-7-2 , 4-8-1

Objectives

Determine molecular formulas from elemental analysis
Calculate degree of unsaturation and use it as a structural clue
Identify functional groups from reaction results

Elemental Analysis

Completely combust the organic compound → measure CO₂ and H₂O masses to find C and H.

CO₂ mass × $\frac{12}{44}$ = C mass
H₂O mass × $\frac{2}{18}$ = H mass
O mass = sample − C − H (− N if present)

→ Mass ratio of each element → empirical formula → use molecular weight to find molecular formula

A 6.0 mg compound of C, H, O produced 8.8 mg CO₂ and 3.6 mg H₂O on combustion. MW = 60. Find the molecular formula.

C: $8.8 \times \frac{12}{44} = 2.4\,\mathrm{mg}$

H: $3.6 \times \frac{2}{18} = 0.40\,\mathrm{mg}$

O: $6.0 - 2.4 - 0.40 = 3.2\,\mathrm{mg}$

$\mathrm{C : H : O} = \frac{2.4}{12} : \frac{0.40}{1} : \frac{3.2}{16} = 0.20 : 0.40 : 0.20 = 1 : 2 : 1$

Empirical formula: CH₂O (formula weight = 30)

MW 60 ÷ 30 = 2 → Molecular formula: C₂H₄O₂

Degree of Unsaturation

Degree of unsaturation (index of hydrogen deficiency) $U$ :

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

(Halogens count as H; O is not included)

Structural feature	Unsaturation
C=C double bond	1
C≡C triple bond	2
C=O double bond	1
Ring	1
Benzene ring	4 (3 double bonds + 1 ring)

Example: C₂H₄O₂ → $U = \frac{2(2) + 2 - 4}{2} = 1$ → one double bond or ring

Identifying Functional Groups

Test	Positive result means	Notes
Reacts with Na → H₂	-OH or -COOH
NaHCO₃ → CO₂	-COOH (carboxylic acid)	Phenolic -OH does not react
Silver mirror / Fehling’s	-CHO (aldehyde)	Formic acid HCOOH also positive
FeCl₃ → purple	Phenolic -OH	Alcoholic -OH does not react
I₂ + NaOH → yellow ppt	CH₃CO- (iodoform test)	CH₃CH(OH)- also positive
Br₂ water decolorized	C=C (unsaturated)	Consumed by addition

Functional group identification by reaction

Practical Structural Determination

Compound A (C₂H₄O₂): reacts with Na (H₂ evolved), reacts with NaHCO₃ (CO₂ evolved), negative silver mirror test. Determine A's structure.

$U = \frac{2(2)+2-4}{2} = 1$ → one double bond or ring

Na reaction → has -OH or -COOH
NaHCO₃ reaction → has -COOH
No silver mirror → no -CHO

C₂H₄O₂ with -COOH → acetic acid CH₃COOH

(The unsaturation of 1 corresponds to the C=O in -COOH)

Check Your Understanding

Q1 What is the degree of unsaturation of benzene C₆H₆?

Q2 Which functional group reacts with NaHCO₃ to produce CO₂?

Q3 In elemental analysis, what does the mass of CO₂ tell you?

Exercises

Q1. Compound B (C₃H₆O) gives a positive silver mirror test and a positive iodoform test. Determine B’s structure.

Solution

$U = \frac{2(3)+2-6}{2} = 1$ → one double bond or ring

Positive silver mirror test → has -CHO (aldehyde)
Positive iodoform test → contains CH₃CO- structure

C₃H₆O with -CHO → R must be C₂H₅ → propanal CH₃CH₂CHO

(Note: the iodoform test is also positive for CH₃CH(OH)- groups, but the positive silver mirror test confirms -CHO. In actual problems, conditions must be examined carefully.)

Answer: CH₃CH₂CHO (propanal)