Rank the following compounds in decreasing order of their reactivity toward the \(S_{N} 2\) reaction with sodium ethoxide \(\left(\mathrm{Na}^{+}-\mathrm{OCH}_{2} \mathrm{CH}_{3}\right)\) in ethanol.
methyl chloride tert-butyl iodide neopentyl bromide
isopropyl bromide methyl iodide ethyl chloride
Problem-solving Hint
Do not write \(S_{N} 2\) reactions occurring on tertiary alkyl halides.
Equation Transcription:
Text Transcription:
S_N2
(Na+-OCH_2CH_3)
SN2
Solution 18P
Factors that affect the rate of reactions are as follows:
1. Nature of leaving group
2. Nature of the substrate
Weak bases are good leaving groups. In halogens, the leaving group order is as follows:
The substrate with less steric hindrance has a high reactivity towards the reaction.
The substrate reactivity order is as follows:
Methyl halides are most reactive towards the reactions. So, among the given compounds, methyl halides are methyl chloride and methyl iodide.
Methyl iodide is most reactive towards the reaction than methyl chloride. This is due to iodine is a good leaving group than chloride.
Among the given compounds, the primary halides are ethyl chloride and neopentyl bromide. Among the two alkyl halides ethyl chloride is most reactive towards reaction.
Neopentyl bromide is also a primary alkyl halide. However, due to bulky nature of this substrate, it undergoes reaction with the same rate of tertiary alkyl halide. Thus the reactivity of tert-butyl iodide and Neopentyl bromide are equal.
The isopropyl bromide is a secondary halide, whereas tert-butyl iodide is a tertiary halide. So, isopropyl bromide is more reactive towards the reaction than tert-butyl iodide.
Thus the reactivity of SN2 reaction with sodium ethoxide in decreasing order is as follows:
