This question is about distinguishing between acetaldehyde (MeCHO) and benzaldehyde (PhCHO) based on their reactions with different reagents. The key difference lies in the absence of alpha-hydrogen in benzaldehyde, which makes it unable to undergo certain reactions that acetaldehyde can.

Step-by-Step Analysis of Each Reagent:

Option 1: NaOI (Sodium hypoiodite, Iodoform test)

This is the correct answer. The iodoform test is positive for aldehydes and ketones that have a methyl group attached to the carbonyl (i.e., have at least one alpha-hydrogen and that alpha-carbon must be a methyl group, CH₃-).

Acetaldehyde (CH₃CHO) has three alpha-hydrogens. It gives a positive iodoform test, producing a yellow precipitate of iodoform (CHI₃). $CH 3_{} CHO + 3 NaOI \to CHI 3_{} ↓ + HCOONa + 2 NaOH$
Benzaldehyde (C₆H₅CHO) has no alpha-hydrogens. It does not give the iodoform test and no precipitate is formed.

Thus, NaOI can distinguish between them.

Option 2: KCN (Cyanide ion, Cyanohydrin formation)

Both aldehydes react with KCN to form cyanohydrins. This reaction is common to most aldehydes and ketones.

Acetaldehyde: $CH 3_{} CHO + HCN \to CH 3_{} CH (OH) CN$
Benzaldehyde: $C 6_{} H 5_{} CHO + HCN \to C 6_{} H 5_{} CH (OH) CN$

Since both react similarly, KCN cannot distinguish between them.

Option 3: KOH, Δ (Potassium hydroxide, heat - Cannizzaro reaction)

Concentrated alkali like KOH causes the Cannizzaro reaction, which is specific to aldehydes lacking an alpha-hydrogen.

Benzaldehyde (no alpha-H) undergoes disproportionation (Cannizzaro reaction). $2 C 6_{} H 5_{} CHO \overset{KOH}{\to} C 6_{} H 5_{} CH 2_{} OH + C 6_{} H 5_{} COOK$
Acetaldehyde (has alpha-H) undergoes the Aldol condensation reaction instead of Cannizzaro. $CH 3_{} CHO + CH 3_{} CHO \overset{KOH}{\to} CH 3_{} CH (OH) CH 2_{} CHO$

They give different products, so this reagent can also distinguish them. However, the question asks for the reagent that gives a "different type of reaction", and NaOI is a more specific and classic test for this distinction.

Option 4: MeMgBr (Methyl magnesium bromide, Grignard reagent)

Both aldehydes react with Grignard reagents in the same way (nucleophilic addition) to form secondary alcohols after hydrolysis.

Acetaldehyde gives 2-propanol. $CH 3_{} CHO + CH 3_{} MgBr \to CH 3_{} CH (OMgBr) CH 3_{} \overset{H_{2} O}{\to} CH 3_{} CH (OH) CH 3_{}$
Benzaldehyde gives 1-phenylethanol. $C 6_{} H 5_{} CHO + CH 3_{} MgBr \to C 6_{} H 5_{} CH (OMgBr) CH 3_{} \overset{H_{2} O}{\to} C 6_{} H 5_{} CH (OH) CH 3_{}$

Both reactions are of the same type (nucleophilic addition), so MeMgBr cannot distinguish between them based on reaction type.

Final Answer:

NaOI is the reagent that gives a different type of reaction with MeCHO (positive iodoform test) and PhCHO (negative test).

Related Topics & Formulae:

Iodoform Test: A chemical test used to detect the presence of a methyl ketone (R-CO-CH₃) or an aldehyde with the structure R-CH(OH)CH₃ (which can be oxidized to a methyl ketone). A positive result is indicated by the formation of a yellow precipitate of CHI₃.

Alpha-Hydrogen: A hydrogen atom attached to the carbon adjacent to the carbonyl carbon. Its presence or absence dictates the type of reactions an aldehyde/ketone can undergo (e.g., Aldol condensation vs. Cannizzaro reaction).