Which of the following order regarding acidity of the following compounds is correct?

R O H > H C \equiv C H > N H_{3}

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R O H < H C \equiv C H < N H_{3}

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R O H > H C \equiv C H < N H_{3}

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R O H < H C \equiv C H > N H_{3}

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Solution

The correct option is A $R O H > H C \equiv C H > N H_{3}$
Order of acidity is:
Carboxylic acid > Alcohol > Alkyne > Amines > Alkene > Alkane (in general)

As the alcohols contain $- O H$ which contains more electronegative oxygen so easily can lose $H^{+}$

So, alcohol is more acidic and alkynes have $s p$ carbon which is electronegative so more acidic H than $N H_{3}$

Option A is correct.

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Similar questions

H C \equiv C H + 2 N a N H_{2} + 2 E t B r ⟶ (A) C s, N H_{3} ⟶ (B) B r_{2} ⟶ (C)

The correct statement(s) about the following compound
$H_{3} C (O H) H C - C H = C H - C H (O H) C H_{3} (X)$ is/are:

C H_{2} = C H - C H = C H - \oplus N H_{3}

\oplus C H_{2} - C H = C H - ⊖ C H - \oplus N H_{3}

\oplus C H_{2} - C H = C H - C H = N H_{3}

H C \equiv C H

H_{2} C = C H_{2}

H_{3} C - C H_{3}

What is the decreasing order of strengths of the following bases?
$⊖ O H, N H_{2}^{⊖}, H - C \equiv C^{⊖}, a n d C H_{3} - C H_{2}^{⊖}$

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