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The acidic character of alcohols is due ...

 The acidic character of alcohols is due the polar nature of O-H bond. An e releasing group `(- CH_(3) -C_(2)H_(5))` increases e density on oxygen thus decreasing the acidity while e withdrawing group increases the acidity.
Which among the following is a better proton acceptor

A

`CH_(3)O^(-)`

B

`OH^(-)`

C

`SO_(4)^(2-)`

D

`NO_(3)^(-)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine which among the given options is a better proton acceptor, we need to analyze the stability of the corresponding anions formed from the alcohols. The more stable the anion, the better it is at accepting a proton (acting as a base). Here’s a step-by-step breakdown of the solution: ### Step 1: Identify the Anions We are given four options, which are likely different anions formed from alcohols. Let's denote them as follows: 1. **Option 1:** CH3O⁻ (Methoxide ion) 2. **Option 2:** HO⁻ (Hydroxide ion) 3. **Option 3:** SO4²⁻ (Sulfate ion) 4. **Option 4:** NO3⁻ (Nitrate ion) ### Step 2: Analyze Stability of Each Anion - **Option 1: CH3O⁻ (Methoxide ion)** - The negative charge is localized on the oxygen atom. The presence of the electron-donating methyl group (CH3) increases electron density on the oxygen, making it less stable and thus a weaker proton acceptor. - **Option 2: HO⁻ (Hydroxide ion)** - Similar to methoxide, the negative charge is localized on oxygen. However, it does not have any electron-donating groups, making it a moderate proton acceptor. - **Option 3: SO4²⁻ (Sulfate ion)** - The negative charges are delocalized over multiple oxygen atoms due to resonance. This resonance stabilization makes the sulfate ion quite stable, thus a weaker proton acceptor. - **Option 4: NO3⁻ (Nitrate ion)** - Like the sulfate ion, the negative charge in the nitrate ion is also delocalized over three oxygen atoms due to resonance. This makes the nitrate ion stable, and it is also a weaker proton acceptor. ### Step 3: Compare the Anions - **Methoxide ion (CH3O⁻)** is the least stable due to the electron-donating effect of the methyl group, making it a better proton acceptor. - **Hydroxide ion (HO⁻)** is more stable than methoxide but less stable than the resonance-stabilized ions. - **Sulfate ion (SO4²⁻)** and **Nitrate ion (NO3⁻)** are both resonance-stabilized and thus are the most stable, making them the least likely to accept protons. ### Conclusion The best proton acceptor among the given options is **Option 1: CH3O⁻ (Methoxide ion)**, as it is the least stable and thus will react more readily to accept a proton.
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Carboxylic acids are distinctly acidic and ionise in water to give H_3 O^+ ions. In general, electron withdrawing groups increase the acidity of carboxylic acids while electron donating groups decrease the acidity of the carboxylic acid. The electron releasing groups such as -CH_3 -OH, -NH_2 and -OCH_3 make benzoic acid weaker while electron withdrawing groups such as -CI, -NO_2 , etc., make benzoic acid stronger. The ortho isomer of every substituted benzoic acid is the strongest among the three isomers due to effect known as ortho effect. Carboxylic acids neutralise alkalies, decompose carbonates and bicarbonates evolving CO_2 with brisk effervescence. Arrange the following in increasing order of acidic strength : FCH_2 COOH ,CH_3 COOH, CICH_2 COOH ,CH_3 CH_2 OH ,C_6 H_5 COOH

Carboxylic acids are distinctly acidic and ionise in water to give H_3 O^+ ions. In general, electron withdrawing groups increase the acidity of carboxylic acids while electron donating groups decrease the acidity of the carboxylic acid. The electron releasing groups such as -CH_3 -OH, -NH_2 and -OCH_3 make benzoic acid weaker while electron withdrawing groups such as -CI, -NO_2 , etc., make benzoic acid stronger. The ortho isomer of every substituted benzoic acid is the strongest among the three isomers due to effect known as ortho effect. Carboxylic acids neutralise alkalies, decompose carbonates and bicarbonates evolving CO_2 with brisk effervescence. among which is the strongest acid .

Knowledge Check

  • The acidic character of alcohols is due the polar nature of O-H bond. An e releasing group (- CH_(3) -C_(2)H_(5)) increases e density on oxygen thus decreasing the acidity while e withdrawing group increases the acidity. Which among the following is most acidic

    A
    B
    C
    D

  • The acidic character of alcohols is due the polar nature of O-H bond. An e releasing group (- CH_(3) -C_(2)H_(5)) increases e density on oxygen thus decreasing the acidity while e withdrawing group increases the acidity. Which is the strongest acid among:

    A
    `CH_(3)OH`
    B
    `CH_(3)CH_(2)OH`
    C
    `CICH_(2)OH`
    D
    `NO_(2) - CH_(2)- OH`

  • The acidic character of 1^@,2^@,3^@ alcohols, H_2O and RC-= CH is of the order

    A
    `H_2O gt 1^@ gt 2^@ gt 3^@ R-= CH`
    B
    `RC-=CH gt 3^@ gt 2^@ gt 1^@ gt H_2O`
    C
    `1^@ gt 2^@ gt 3^@ gt H_2O gt RC-=CH`
    D
    `3^@ gt 2^@ gt 1^@ gt H_2O gt RC-=CH`

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