16)

CHCH→X→CH3-CH2-OH→ CH3-CH2-Br 

1) CH3COOH

2) CH3CHO 

3)Br2 

4)CH3 CHBr2 

5)HBr

The reaction sequence is:

CH≡CH + X → CH3-CH2-OH → CH3-CH2-Br

1) First, CH≡CH reacts with X to form an intermediate compound. The type of X is not specified, but it could be a halogen such as Cl or Br. The reaction is likely a halogenation, which involves the addition of a halogen to a carbon-carbon double bond. The intermediate compound is not shown, but it would have a halogen atom attached to one of the carbons in the double bond.

2) Next, the intermediate compound reacts with CH3COOH (acetic acid) to form CH3-CH2-OH (ethanol). This is a substitution reaction, where the halogen atom is replaced by an -OH group from the acetic acid. The reaction is likely catalyzed by an acid, such as sulfuric acid (H2SO4).

3) The ethanol is then oxidized by Br2 (bromine) to form an aldehyde, CH3CHO (acetaldehyde). This is an oxidation reaction, where the -OH group is converted to a carbonyl group (C=O). The bromine is reduced to HBr in the process.

4) The aldehyde is then reacted with CH3CHBr2 (1,2-dibromoethane) to form a new compound, CH3-CHBr-CHBr-CH3. This is a nucleophilic substitution reaction, where the aldehyde is attacked by the CH3CHBr2 molecule, which acts as a nucleophile. The product has two bromine atoms attached to the second carbon in the chain.

5) Finally, the product is treated with HBr (hydrogen bromide) to form CH3-CH2-Br (ethyl bromide). This is a substitution reaction, where the CH3CHBr2 molecule is replaced by a bromine atom from the HBr molecule. The final product has a single bromine atom attached to the second carbon in the chain.