Which among the following reagent can be used to distinguish between CH₂BrCH=CH₂ from CH3CH=CHBr?

The correct answer is AgNO₃/C₂H₅OH.

To understand why ethanolic silver nitrate (${\text{AgNO}}_3/{\text{C}}_2{\text{H}}_5\text{OH}$) can be used to distinguish between allyl bromide (${\text{CH}}_2\text{BrCH}={\text{CH}}_2$) and 1-bromoprop-1-ene (${\text{CH}}_3\text{CH}=\text{CHBr}$), let us analyze the reactivity of the bromine atom in each compound towards nucleophilic substitution reactions.

1. Allyl bromide (${\text{CH}}_2\text{BrCH}={\text{CH}}_2$):
In allyl bromide, the bromine atom is attached to an ${\text{sp}}^3$ hybridized carbon atom adjacent to a carbon-carbon double bond. When treated with ethanolic silver nitrate, the weak carbon-bromine bond easily cleaves to form a carbocation:
${\text{CH}}_2\text{Br-CH}={\text{CH}}_2+{\text{Ag}}^{+}\to {[{\text{CH}}_2-\text{CH}={\text{CH}}_2]}^{+}+\text{AgBr}\downarrow$
The resulting allylic carbocation (${[{\text{CH}}_2-\text{CH}={\text{CH}}_2]}^{+}$) is highly stable due to resonance. Because of this high stability, allyl bromide readily undergoes nucleophilic substitution and precipitates silver bromide ($\text{AgBr}$) as a light yellow precipitate rapidly at room temperature.

2. 1-Bromoprop-1-ene (${\text{CH}}_3\text{CH}=\text{CHBr}$):
In 1-bromoprop-1-ene, the bromine atom is directly attached to an ${\text{sp}}^2$ hybridized carbon atom (a vinyl halide). The lone pairs of electrons on the bromine atom are in conjugation with the $\pi$-bond of the double bond, which gives the carbon-bromine bond partial double-bond character due to resonance:
${\text{CH}}_3-\text{CH}=\text{CH}-\ddot{\text{Br}}\leftrightarrow {\text{CH}}_3-{\text{CH}}^{-}-\text{CH}={\text{Br}}^{+}$
This partial double-bond character makes the C-Br bond extremely strong and difficult to break. Consequently, vinylic halides do not react with ethanolic ${\text{AgNO}}_3$ under ordinary conditions, and no precipitate is observed.

Why other options are incorrect:
- Br₂/CCl₄: Both compounds contain a carbon-carbon double bond, so both will decolorize the reddish-brown bromine solution.
- cold KMnO₄ (Baeyer's reagent): Both compounds are unsaturated and will decolorize cold, alkaline potassium permanganate, turning it from purple to a brown precipitate of ${\text{MnO}}_2$.
- Ag(NH₃)₂OH (Tollens' reagent): This reagent is used for detecting terminal alkynes or aldehydes, and neither of these compounds reacts with it.