$S_{N} 2$ reaction readily occurs in:

C H_{3} C H_{2} - O - C H_{3}

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C H_{3} - C H_{3} | C - | C H_{3} O - C H_{3}

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C H_{2} = C H - C H_{2} - O - C H_{3}

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P h - C H_{2} - O - C H_{2} - C H_{3}

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Solution

The correct option is A $C H_{3} C H_{2} - O - C H_{3}$
SN2 reactions occur more readily at carbons that are less substituted. It is easier for the nucleophile to attack if there is little substitution around the C-leaving group bond. Therefore, the order of reactivity for alkyl halides is methyl>primary>secondary>>>tertiary.

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C H_{3} - C H_{3} | C = O

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C H_{3} - C H_{2} - O | | C - C H_{2} - C H_{3}

{C H}_{3} - {C H}_{2} - O ∥ C - {C H}_{2} - {C H}_{2} - {C H}_{3}

{C H}_{3} - O ∥ C - {C H}_{3} | C | {C H}_{3} - {C H}_{3}

i) C H_{3} - C H_{2} - C H_{3} | C H - C H_{2} - O = C H_{2} - C H_{3}

i i) C H_{3} - C H_{2} - C H_{2} - O - C H_{3} | C | C H_{3} - C H_{3}

C H_{2} = O H | C - C H_{2} - O | | C - C H_{3} ⇌ C H_{3} - O | | C - C H_{2} - O | | C - C H_{3} ⇌ C H_{3} - O H | C = C H - O | | C - C H_{3}

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