By a proper choice of reagents, both symmetrical and unsymmetrical ethers can be prepared by Williamson synthesis which involves the reaction between an alkyl halide and an alkoxide ion. The reverse process involving the cleavage of ethers to give back the original alkyl halide and the alcohol can be carried out by heating the ether with HI at 373K.

Allyl phenyl ether can be prepared by heating

CH₂=CHCH₂Br + C₆H₅ONa

The correct answer is option 2. CH₂=CHCH₂Br + C₆H₅ONa.

Allyl phenyl ether (\(\text{C}_6\text{H}_5\text{OCH}_2\text{CH=CH}_2\)) can be prepared by heating an allyl halide with sodium phenoxide. This type of reaction is known as the Williamson ether synthesis, which involves the reaction of an alkoxide ion with a primary alkyl halide to form an ether.

Let us evaluate the given options:

1. \(\text{C}_6\text{H}_5\text{Br} + \text{CH}_2=\text{CH}-\text{CH}_2\text{ONa}\)

This reaction would result in the formation of an ether, but not allyl phenyl ether. It would more likely produce \(\text{C}_6\text{H}_5\text{OCH}_2\text{CH=CH}_2\) and \(\text{Br}^-\).

2. \(\text{CH}_2=\text{CHCH}_2\text{Br} + \text{C}_6\text{H}_5\text{ONa}\)

This is the correct reaction for the formation of allyl phenyl ether. Sodium phenoxide (\(\text{C}_6\text{H}_5\text{ONa}\)) reacts with allyl bromide (\(\text{CH}_2=\text{CHCH}_2\text{Br}\)) to form allyl phenyl ether (\(\text{C}_6\text{H}_5\text{OCH}_2\text{CH=CH}_2\)).

3. \(\text{C}_6\text{H}_5\text{CH=CHBr} + \text{CH}_3\text{ONa}\)

This reaction would not produce allyl phenyl ether. It would likely produce a different ether and not the desired product.

4. \(\text{CH}_2=\text{CHBr} + \text{C}_6\text{H}_5\text{CH}_2\text{ONa}\)

This reaction involves a different set of reactants that would not lead to the formation of allyl phenyl ether.

So, the correct answer is: \(\text{CH}_2=\text{CHCH}_2\text{Br} + \text{C}_6\text{H}_5\text{ONa}\).