question_answer

The decreasing order of nucleophilicity of \[{{S}_{N}}2\] reaction of the following alkoxide nucleophile is

(I) \[Me{{O}^{\bigcirc -}}\]

(II)

(III) \[M{{e}_{2}}CH{{O}^{\bigcirc -}}\]

(IV) \[M{{e}_{3}}C-{{O}^{\bigcirc -}}\]

A) \[IV>III>II>I\]   

B) \[I>II>III>IV\]  

C) \[I>III>II>IV\]   

D) \[IV>II>III>I\]

Correct Answer: B

Solution :

[b] Steric hindrance decrease the nucleophilicity, e.g.,

I. \[Me{{O}^{\bigcirc -}}\]

II.

III. \[M{{e}_{2}}CH{{O}^{\bigcirc -}}\]

IV. \[M{{e}_{3}}C-{{O}^{\bigcirc -}}\]

Stronger is the conjugate base, stronger is the nucleophile (when nucleophilic centres are same). Acidic order:  \[1{}^\circ alcohols>2{}^\circ \text{ }alcohols>3{}^\circ \text{ }alcohols\] Basic order and nucleophilic order: \[3{}^\circ alcohols>2{}^\circ \text{ }alcohols>1{}^\circ \text{ }alcohols\] But the actual order of nucleophilicity is reversed, i.e., This is because of its bulkiness \[M{{e}_{3}}C{{O}^{\bigcirc -}}\] a poorer nucleophile because of steric hindrance, and basicity and nucleophilicity may diverge. (both \[2{}^\circ \] alkoxide), but R's (alkyi group) in (II) are inside the ring and are 'tied back' (i.e., fixed) away from \[-{{O}^{\bigcirc -}}\]. Such an arrangement permits an easier approach than by \[M{{e}_{2}}CH{{O}^{\bigcirc -}}\]. Hence, (II) is a stronger nucleophile than (III).