Could you explain how to do this with great detail? **Synthesis of (E)-1,2-dibromohex-1-ene from Hex-1-yne and Bromine**To synthesize (E)-1,2-dibromohex-1-ene efficiently from hex-1-yne and bromine (Br₂), follow the chemical reaction shown below:1. **Starting Material:** - Hex-1-yne (depicted with a triple bond between the first two carbon atoms).2. **Reagent:** - Bromine (Br₂, 1 mole).3. **Product:** - (E)-1,2-dibromohex-1-ene (showing the addition of bromine across the double bond to form the dibromo compound with an E-configuration).**Options for Synthesis:**- **Choice 1:** 1 mole of bromine should be added to 1 mole of hex-1-yne.- **Choice 2:** 1 mole of hex-1-yne should be added to 1 mole of bromine.- **Both choice 1 and choice 2 will yield the product in equally good yield.**- **Neither choice 1 nor choice 2 will give a good yield of the product.**Select the appropriate synthesis choice based on your understanding of the reaction dynamics between hex-1-yne and bromine.

The conversion of hex-1-yne to (E)1,2-dibromohex-1-ene is the halogenation reaction of alkyne. When Alkyne reacts with 1 mole of halogen then disubstituted (with halogen) trans haloalkene is formed. When alkyne reacts with 2 moles of halogen then tetrasubstituted haloalkane is formed. Alkyne is nucleophilic in nature. When bromine approached alkyne a dipole is generated on br2. And a cyclic halonium ion will formed in the intermediate The reaction and mechanism is shown in an image below.