Decreasing order of reactivity in Williamson’s ether synthesis of the following. I. Me3CCH2Br II. CH3CH2CH2Br III. CH2=CHCH2Cl IV. CH3CH2CH2CH2Cl

Question

Decreasing order of reactivity in Williamson’s ether synthesis of the following. I. Me3CCH2Br II. CH3CH2CH2Br III. CH2=CHCH2Cl IV. CH3CH2CH2CH2Cl (A) I > II > IV > III (B) III > II > IV > I (C) I >III > II > IV (D) II > III > IV > I

Tardigrade Admin · Accepted Answer

The reactivity of alkyl halides is in the order,

C H_{3} > 1^{o} > 2^{o} > 3^{o}

as tendency of alkyl halides to undergo elimination is

3^{o} > 2^{o} > 1^{o} .

C - Br bond length is longer than C - Cl bond. So, C - Br bond is easier to break than C - Cl bond.

CH_{2} = CH C \oplus H_{2}

carbocation is resonance stabilized
i.e.,

[C H_{2} = C H - C \oplus H_{2} \leftrightarrow C \oplus H_{2} - C H = C H_{2}]

So, it reacts faster than n- butyl chloride

Q. Decreasing order of reactivity in Williamson’s ether synthesis of the following. I. Me3​CCH2​Br II. CH3​CH2​CH2​Br III. CH2​=CHCH2​Cl IV. CH3​CH2​CH2​CH2​Cl

I > II > IV > III

III > II > IV > I

I >III > II > IV

II > III > IV > I

Q. Decreasing order of reactivity in Williamson’s ether synthesis of the following.
I. $M e_{3} CC H_{2} B r$
II. $C H_{3} C H_{2} C H_{2} B r$
III. $C H_{2} = C H C H_{2} Cl$
IV. $C H_{3} C H_{2} C H_{2} C H_{2} Cl$