True. When 1-bromobutane reacts with sodium methoxide (NaOCH3), it undergoes an elimination reaction, specifically an E2 (bimolecular elimination) reaction.
This reaction typically favors the formation of the alkene product over substitution products.
Here's the step-wise mechanism for the E2 reaction between 1-bromobutane and sodium methoxide:
1. Deprotonation: Sodium methoxide (NaOCH3) acts as a strong base, and it abstracts a proton from the carbon adjacent to the bromine atom in 1-bromobutane. This carbon is called the alpha carbon.
1-bromobutane:
CH3-CH2-CH2-CH2-Br
After deprotonation, you get a carbanion intermediate:
CH3-CH2-CH=CH2 + NaBr + CH3OH
2. Elimination: Simultaneously with deprotonation, the bromine atom leaves as a bromide ion (Br-) in a single step, creating a double bond between the alpha and beta carbons. This is the elimination step, and it leads to the formation of the alkene product:
CH3-CH2-CH=CH2 + NaBr + CH3OH
the reaction between 1-bromobutane and sodium methoxide predominantly leads to the formation of the alkene product, and this is a true statement.