Which one of the following reactions will not yield propionic acid? 1. $\ce{CH3CH2COCH3 + OH- / H3O+}$ 2. $\ce{CH3CH2CH3 + KMnO4 (Heat), OH- / H3O+}$ 3. $\ce{CH3CH2CCl3 + OH- / H3O+}$ 4. $\ce{CH3CH2CH2Br + Mg, CO2 \text{ dry ether} / H3O+}$ what is the first reaction called. ?

Key concepts you must know

1. Propionic (propanoic) acid: $\ce{CH3CH2COOH}$ – it owns 3 carbon atoms.

2. Haloform reaction

   • Substrate: any compound containing the $\ce{-COCH3}$ group (methyl-ketone)

   • Reagents: $\ce{X2/\, OH^-}$ (X = Cl, Br, I)

   • Outcome:

     $\ce{RCOCH3 \xrightarrow{X2, OH^-} RCOO^- + CHX3}$

   • The acid product has one carbon fewer than the original ketone.

3. Side-chain oxidation with $\ce{KMnO4}$

   • Very effective on benzylic positions. On straight alkanes, vigorous conditions may break C–C bonds and generally give mixtures of shorter acids.

4. Hydrolysis of trichloromethyl derivatives ($\ce{-CCl3}$)

   • When the $\ce{-CCl3}$ group sits on the same carbon that bears the rest of the chain ($\ce{R-CCl3}$), hot aqueous alkali attacks it to give the acid of the same carbon length:

     $\ce{R-CCl3 \xrightarrow{OH^-} R-COOH + 3Cl^-}$

5. Carboxylation of a Grignard reagent

   • Sequence: $\ce{RBr \xrightarrow{Mg} RMgBr \xrightarrow{CO2} RCO2MgBr \xrightarrow{H3O^+} RCOOH}$
   • The acid product has one extra carbon than the starting alkyl halide.

Logical chain you would follow

1. Translate each starting molecule into the number of carbons it brings to the future acid.
2. Apply the rule for the specific reaction type (lose 1C, keep same C-count, add 1C, possible cleavage, etc.).
3. Compare the projected carbon count with the target 3.
4. Pick the option whose count is NOT 3.
5. Lastly, recognise the classical name of reaction 1 – the haloform reaction.

What is the question asking?

We have four different laboratory recipes that each start from a different molecule. The recipes try to make a new molecule: propionic acid (also called propanoic acid, a 3-carbon acid).

The question is simply: which recipe fails – which one does NOT give propionic acid?

How could you check?

1. Count how many carbons each starting molecule can finally give to the acid.
2. Recall the key reaction ideas:

• A haloform reaction chops a methyl-ketone (…COCH3) into a smaller acid with one carbon fewer.
• Strong oxidiser KMnO4 turns an alkane side chain on an aromatic ring into an acid, but for a simple alkane it often breaks chains apart.
• A molecule that owns a CCl3 group beside a carbon can be converted into an acid of the same carbon length.
• A Grignard + CO2 always adds one new carbon (from CO2) before making the acid.

3. Use those ideas to see which route cannot land at a 3-carbon acid.

And what is the name of the first reaction?

When a methyl-ketone reacts with base (and a halogen such as I2/Br2/Cl2) and ends up as a smaller acid + CHX3 (haloform), we call it the Haloform reaction.

Step-by-step analysis

1. Option 1: $\ce{CH3CH2COCH3}$ (2-butanone) + $\ce{OH^-}$ then $\ce{H3O^+}$
   Type: Haloform reaction (needs X2 but assumed present).
   Carbon bookkeeping: Start 4C; product acid has 3C → propionic acid possible.

2. Option 2: $\ce{CH3CH2CH3}$ (propane) + hot $\ce{KMnO4}$
   Type: Strong oxidation; entire chain may cleave into smaller fragments such as acetic acid + CO2. Formation of a neat 3-carbon acid is unlikely. (Many syllabi still mention that straight alkanes do not give a simple acid in one step.)

3. Option 3: $\ce{CH3CH2CCl3}$ + $\ce{OH^-}$ then $\ce{H3O^+}$
   Type: Base hydrolysis of $\ce{-CCl3}$ to $\ce{-COOH}$.
   Carbon bookkeeping: Same 3C chain survives → propionic acid forms.

4. Option 4: $\ce{CH3CH2CH2Br}$ (1-bromopropane) → $\ce{RMgBr}$ → $\ce{CO2}$ → acid
   Type: Grignard carboxylation; adds one carbon.
   Carbon bookkeeping: Start 3C, end 4C → butanoic acid (4C), not propionic acid.

Hence, option 4 does NOT yield propionic acid.

Name of the first reaction

The transformation of a methyl-ketone into a carboxylate ion with loss of one carbon and formation of a haloform ($\ce{CHX3}$) is called the Haloform reaction.