Distinguishing Formaldehyde from Acetaldehyde

Formaldehyde (HCHO) and acetaldehyde (CH₃CHO) are both aldehydes, but they can be distinguished based on their different chemical behaviors with specific reagents. The key difference is that formaldehyde is a reducing aldehyde that lacks an alpha-hydrogen (hydrogen attached to the carbon adjacent to the carbonyl group), while acetaldehyde has alpha-hydrogens. This affects their reactions with certain tests.

Step-by-Step Analysis of Each Reagent:

Step 1: Evaluate Tollen's Reagent

Tollen's reagent (ammoniacal silver nitrate) is a test for aldehydes. Both formaldehyde and acetaldehyde reduce Tollen's reagent to give a silver mirror. The reaction is:

$\mathrm{RCHO}+2{\left[\mathrm{Ag}\right({\mathrm{NH}}_3\left)2\right]}^{+}+3{\mathrm{OH}}^{-}\to 2\mathrm{Ag}+\mathrm{RCOO}{\mathrm{NH}}^4+2H_{2}O$

Both give a positive test (silver mirror), so Tollen's reagent cannot distinguish them.

Step 2: Evaluate Fehling's Solution

Fehling's solution (alkaline copper tartrate) is also a test for aldehydes. Both formaldehyde and acetaldehyde reduce Fehling's solution to give a red precipitate of cuprous oxide (Cu₂O). The reaction is:

$\mathrm{RCHO}+2{\mathrm{Cu}}^{\mathrm{2+}}+5{\mathrm{OH}}^{-}\to \mathrm{RCOO}-+{\mathrm{Cu}}_{2}O+3H_{2}O$

Both give a positive test (red precipitate), so Fehling's solution cannot distinguish them.

Step 3: Evaluate Schiff's Reagent

Schiff's reagent is used to detect aldehydes. It gives a magenta color with aldehydes. However, both formaldehyde and acetaldehyde produce a pink/magenta color with Schiff's reagent. Although formaldehyde gives a more intense color, it is not a reliable distinction. Thus, Schiff's reagent cannot clearly distinguish them.

Step 4: Evaluate I₂/Alkali (Iodoform Test)

The iodoform test (I₂ with NaOH) is used to detect methyl ketones (CH₃COR) or alcohols that can be oxidized to methyl ketones. Acetaldehyde (CH₃CHO) has a methyl group attached to the carbonyl and undergoes the iodoform reaction to give a yellow precipitate of iodoform (CHI₃). The reaction is:

$\mathrm{CH}3\mathrm{CHO}+3I_2+4\mathrm{NaOH}\to \mathrm{CHI}3+\mathrm{HCOONa}+3\mathrm{NaI}+3H_{2}O$

Formaldehyde (HCHO) does not have a methyl group attached to the carbonyl and does not give the iodoform test. Thus, only acetaldehyde gives a positive iodoform test (yellow precipitate), while formaldehyde does not react.

Final Answer: I₂/Alkali (iodoform test) can distinguish formaldehyde from acetaldehyde.

Related Topics:

• Carbonyl Compounds: Aldehydes and Ketones
• Tests for Aldehydes: Tollen's, Fehling's, Schiff's
• Iodoform Test for Methyl Ketones and Acetaldehyde
• Alpha-Hydrogen in Aldehydes/Ketones

Important Formulae and Theory:

Tollen's Test: RCHO + 2[Ag(NH₃)₂]⁺ + 3OH^- → 2Ag + RCOONH₄ + 2H₂O (Positive for aldehydes)

Fehling's Test: RCHO + 2Cu²⁺ + 5OH^- → RCOO^- + Cu₂O + 3H₂O (Positive for aldehydes)

Iodoform Test: CH₃COR (or CH₃CH(OH)R) + 3I₂ + 4NaOH → CHI₃ + RCOONa + 3NaI + 3H₂O (Positive for acetaldehyde, methyl ketones, ethanol, etc.)

Key Theory: The iodoform test is positive for compounds with CH₃C=O group or those oxidizable to it (e.g., CH₃CH(OH)-). Formaldehyde lacks this group.