Question

Green Oxidation Lab

1. Why did you synthesize a derivative, vs. directly testing your carbonyl product? Hint: In order for a derivative to be useful in identification of an unknown, it must possess certain physical properties. Answer in terms of physical properties. (2 points)

2. Explain why a Tollen's test is selective for aldehydes vs. ketones. (1 point)

3. Using a chemical test or tests, how can you distinguish between 3-pentanone and 2-pentanone. (1 point)

4. Draw two structural isomers for C₅H₁₀O that would both give positive iodoform tests. ( 2 pts.)

5. Draw a possible structure for C₄H₈O that would NOT give a positive 2,4-DNP test. ( 2 pts.)

Answer 1

1) Melting point of the prepared derivative is used to help identify our unknown. For example, let’s assume that unknown sample react with 2,4-Dinitrophenylhydrazine and form reddish-orange precipitates. Formation of a precipitate therefore indicates the presence of an aldehyde or ketone. It was identified to be a ketone with a boiling point of 129 ⁰C. There are two ketones that have identical boiling points. The two possible candidates are 2-hexanone and 5-Hexan-3 –one. When a derivative is prepared, the melting points of each derivative are different (122⁰C and 102⁰ respectively). Therefore the melting point of the derivative confirms the identity of the unknown.

2) A ketone has two alkyl groups bonded to the carbonyl carbon while an aldehyde has one alkyl group and one hydrogen atom bonded to the carbonyl carbon. The presence of that hydrogen atom makes aldehydes very easy to oxidize. Because ketones don't have that particular hydrogen atom, they are resistant to oxidation.

Tollen's reagent is added to an aldehyde; it is oxidized and forms a Silver Mirror. Ketones don’t give Tollens test

R-CHO + 2Ag (NH₃)²⁺ + 3OH^- --------> 2Ag + RCO₂^- + 4NH₃ + 2H₂O

Aldehyde reduces the Ag^+ to Ag, and the Ag precipitates.

3) Iodoform test could distinguish between 3-pentanone and 2-pentanone.

2-pentanone is a methyl ketone when treated with iodoform gives yellow precipitate of iodoform but 3-pentanone does not react with it.

CH₃CH₂CHCOCH₃ + 3NaOI ----------> CH₃CH₂CH₂COONa + CHI₃ (iodoform) + 2NaOH

4) Two structural isomers for C5H10O that would both give positive iodoform tests

i) CH₃- CO−CH₂−CH₂−CH₃    (pent-2-one )

ii) CH₃−CO−CH (CH₃)−CH₃   (3-methyl but-2-one )