Which is the stronger acid in each of the following pairs?
(i) $C H_{3} - C O O H$ and $C H_{2} = C H - C O O H$
(ii) $C_{6} H_{5} - C O O H$ and $C_{6} H_{5} - C H_{2} - C O O H$
(iii) $C H_{3} - C H_{2} - C O O H$ and $C H_{3} - C H | O H - C O O H$ .

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Solution

$(i)$ For more acidic character, we need an electron-withdrawing factor which will stabilizing the anion formed after loss of proton. In this case, the second compound will be more acidic than the first as the second compound has an $s p^{2}$ hybridized adjacent carbon and it is more electron withdrawing than an $s p^{3}$ hybridized one.

$(i i)$ In this case, the $1^{s t}$ compound will be more acidic as the anion here is stabilized by the phenyl ring. There is no such factor in $2^{n d}$ compound.

$(i i i)$ The $- O H$ group in second compound shows stronger $- I$ effect due to which it becomes more acidic than first compound, having no such group.

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Similar questions

(A) C H_{3} - C H_{2} - C | C O O H H - - C H - C H_{3}

(B) C H_{3} - C | C O O H H - C H_{2} - - C H - C H_{3}

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

C H_{3} - C H_{2} - C O O H + B r_{2} P B r_{3} - -- \to C H_{3} - C | B r H - C O O H

H_{2} O, C H_{3} N H_{2} - -------- \to

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

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

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

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

C H_{3} C O O H

H C O O H

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