Final answer:
The reaction between 1-bromopropane and various reagents can result in different products. In some cases, an elimination reaction occurs where the bromine atom is eliminated, resulting in the formation of propene. In other cases, a substitution reaction occurs where the bromine atom is replaced by a different functional group. Additionally, a reduction reaction can occur where the bromine atom is replaced by hydrogen.
Step-by-step explanation:
a. The reaction between 1-bromopropane and NaNH₂ would result in the formation of propene (CH₂=CHCH₃) and sodium bromide (NaBr). This is an elimination reaction where the bromine atom is eliminated from 1-bromopropane.
b. The reaction between 1-bromopropane and KOC(CH₃)₃ would result in the formation of propene (CH₂=CHCH₃) and potassium bromide (KBr). This is also an elimination reaction.
c. The reaction between 1-bromopropane and NaI would result in the formation of propene (CH₂=CHCH₃) and sodium bromide (NaBr). This is another elimination reaction.
d. The reaction between 1-bromopropane and NaCN would result in the formation of nitrile (CH₃CH₂CN) and sodium bromide (NaBr). This is a substitution reaction where the bromine atom is replaced by a cyano group (CN).
e. The reaction between 1-bromopropane and NaC=CH would result in the formation of propene (CH₂=CHCH₃) and sodium bromide (NaBr). This is also an elimination reaction.
f. The reaction between 1-bromopropane and Mg, followed by H₂O, would result in the formation of propane (CH₃CH₂CH₃), magnesium bromide (MgBr₂), and hydrogen gas (H₂). This is a reduction reaction where the bromine atom is replaced by hydrogen.