1. We need to identify which reaction does not give propionic acid.

Propionic acid is: $CH_3CH_2COOH$

So we check each option.

2. Option A: $CH_3CH_2COCH_3 \xrightarrow{OI^- / H_3O^+} ?$

Here $OI^-$ represents hypoiodite/iodoform reaction conditions. The compound is 2-butanone, a methyl ketone: $RCOCH_3 \xrightarrow{I_2/OH^-} RCOO^- + CHI_3$ After acidification: $RCOO^- \xrightarrow{H_3O^+} RCOOH$

For $CH_3CH_2COCH_3$, we have $R = CH_3CH_2$. Hence the product is: $CH_3CH_2COOH$ which is propionic acid.

So, A yields propionic acid.

3. Option B: $CH_3CH_2CH_3 \xrightarrow[OH^- / H_3O^+]{KMnO_4,\ heat} ?$

This is propane. Strong oxidation with hot $KMnO_4$ does not selectively convert an alkane into the corresponding carboxylic acid in this way. Alkanes are generally resistant, and under drastic oxidation they undergo cleavage to mixtures/complete oxidation, not clean formation of propionic acid.

So this reaction will not yield propionic acid.

4. Option C: $CH_3CH_2CCl_3 \xrightarrow{OH^- / H_3O^+} ?$

Compounds of type $R-CCl_3$ undergo hydrolysis to carboxylic acids: $R-CCl_3 \xrightarrow{OH^-} R-COO^- \xrightarrow{H_3O^+} R-COOH$

Here $R = CH_3CH_2$. Thus product is: $CH_3CH_2COOH$ which is propionic acid.

So, C yields propionic acid.

5. Option D: $CH_3CH_2CH_2Br \xrightarrow[\text{dry ether}]{Mg} CH_3CH_2CH_2MgBr$ Then: $CH_3CH_2CH_2MgBr + CO_2 \xrightarrow{H_3O^+} CH_3CH_2CH_2COOH$

This gives a carboxylic acid with one extra carbon compared to the alkyl halide. So the product is: $CH_3CH_2CH_2COOH$ which is butanoic acid, not propionic acid.

Therefore, D does not yield propionic acid.

6. Final conclusion

• A gives propionic acid
• B does not give propionic acid
• C gives propionic acid
• D gives butanoic acid, so does not give propionic acid

Thus, there are actually two options that do not yield propionic acid: B and D.

Since the question is marked as MCQ (single correct), the stored answer $D$ is incomplete/wrong because $B$ also does not yield propionic acid.