Question

1. Write a balanced chemical reaction

2. identify and label (or name) the base/nucleophile, the electrophile, and the leaving group.

3. Decide whether the reaction is a substitution or an elimination,, label the reaction as E or S.

2-chloro-3-methylbutane (15.0 mmol) was added to a solution of sodium ethoxide (15.0 mmol) in ethanol (50.0 mL) and heated to reflux for 1 h. The mixture was then cooled to room temperature, neutralized with saturated aqueous sodium bicarbonate (a weak base). 2-methyl-2-butene was isolated by fractional distillation (Tvapor = 35-40 °C)

Answer 1

In this since sodium ethoxide is used as base which is a bulkier and stronger base so the reaction will undergo E2 elimination reaction in which Cl^- will be leaving group .

Sodium ethoxide extracts a proton from alpha carbon and simultaneously Cl^- leaves thereby forming an alkene which is major and more stable

so 2-methyl -2-butene is more stable and major product formed in this reaction

The detailed mechanism of this reaction is given below :-

# 2-chloro-3-methyl butane CH3 - CH2 - CH3 di CH3 this is reacted with fodim Ethoxide (NG OCH2 CH3 ) will undergo Elimination (82) Reaction because OCH2 Ch being a strong and bulby bage @viту Яшир = с а се Base = NO CH2 CH3 ç 3 'CH₃ -C- & - CH₂ - G-CH3 HOCH2CH3 + CH3 CH2OH \ CH3 + 11 c olo overall deaction - 2-methyl-2-butere CH3 CH3-CH-CH-CH2 CH3 NaOCH2CH3 Ethanol H3CH=C-CH2 + CH3 CH2OH 2-chloro-3 methyl butane NOCI

In the final step the product is treated with aqueous Sodium bicarbonate which is a weak base . This is done to ensure no acid is left as side product and final make up is done with this weak base .

So we get a stable and pure alkene as product.