Final answer:
The reaction of 1-bromopropane with sodium amide (NaNH2) would result in an elimination reaction to produce propene, sodium bromide (NaBr), and ammonia (NH3).
Step-by-step explanation:
The student is asking about the reaction products of 1-bromopropane with sodium amide (NaNH2). Sodium amide is a strong base and a nucleophile that can deprotonate an alkane to form an alkene. The reaction between 1-bromopropane and sodium amide would likely result in an elimination reaction, where the bromide leaving group is replaced by hydrogen, giving rise to propene.
The reaction can be represented as:
CH3CH2CH2Br (1-bromopropane) + NaNH2 -> CH3CH=CH2 (propene) + NaBr + NH3
The NaNH2 removes a hydrogen atom from the 1-bromopropane, forming a double bond between carbon atoms and producing propene, bromide anion (NaBr), and ammonia (NH3).