$C H_{3} - C H_{2} - O H$ $C H_{3} - C H_{2} - \oplus O H_{2}$ $C H_{3} - O - C H_{3}$
$I$ $I I$ $I I I$
The order that has the following compounds correctly arranged with respect to increasing acidity is:

i < ii < iii

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i < iii < ii

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ii < i < iii

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iii < i < ii

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Solution

The correct option is D iii < i < ii
The correct order is
$C H_{3} - C H_{2} -^{\oplus} O H_{2} > C H_{3} - C H_{2} - O H > C H_{3} - O - C H_{3}$
Because the conjugate bases kept after removal of $H^{+}$ ions ae $C H_{3}^{\oplus} C H_{2}, C H_{3} C H_{2} - O^{⊖}$ since 1 is more stable than 2 because of $+ I$ effect of $C_{2} H_{5}$ group in II increase $⊖$ charge on $0$ and make it unstable.
so, 1 is more acidic.
Ether are lewis bases

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Similar questions

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}

I

I I

I I I

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

⇌ C H_{3} - O H | C = C H (I I I) - O | | C - C H_{3}

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

H C O O H

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

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

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

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

C H_{3} - C H_{2} - N H_{2}

C H_{3} - N H - C H_{3}

C H_{3} - N H - P h

C H_{3} - N H - C H O

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