Advertisements
Advertisements
प्रश्न
CH3CO2H or CH2FCO2H
Advertisements
उत्तर
|
It is difficult for the proton to escape because the +I effect of −CH3 group increases the electron density in the O–H bond. |
The release of the proton is easy because the −I effect of −F reduces the electron density in the O–H bond. |
|
The +I effect due to the concentration of negative charge destabilises the carboxylate ion. |
The spreading of the negative charge causes the –I effect to stabilize the carboxylate ion. |
Hence, due to lower electron density in the O–H bond and higher stability of FCH2COO– ion, FCH2COOH is a stronger acid than CH3COOH.
APPEARS IN
संबंधित प्रश्न
How is 1-nitropropane prepared from suitable oxime?
Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Give two reasons.
Why pKa of F-CH2-COOH is lower than that of Cl−CH2−COOH?
Which acid of the pair shown here would you expect to be stronger?
CH2FCO2H or CH2ClCO2H
Which acid of the pair shown here would you expect to be stronger?

Write the reactions involved Hell-Volhard Zelinsky reaction?
Account for the following:
Carboxylic acid is a stronger acid than phenol.
What happens when Salicylic acid is treated with (CH3CO)2 O/H+?
Arrange the following compounds in increasing order of their property as indicated:
F - CH2COOH, O2N - CH2 COOH CH3 COOH,HCOOH - acid character.
Complete the following reaction sequence.
\[\begin{array}{cc}
\ce{O}\phantom{...............................................}\\
||\phantom{...............................................}\\
\ce{CH3 - C - CH3 ->[(i) CH3MgBr][H2O] (A) ->[Na metal][Ether] (B) ->[CH3 - Br] (C)}
\end{array}\]
Assertion: Formaldehyde is a planar molecule.
Reason: It contains sp2 hybridised carbon atom.
A mixture of benzaldehyde and formaldehyde on heating with 50% NaOH solution gives
When propionamide reacts with Br2 in the presence of alkali the product is ______.
Describe the action of alcoholic potassium hydroxide (alc. KOH) on isopropyl bromide.
An ester is boiled with KOH. The product is cooled and acidified with conc. HCL. A white crystalline acid separates. The ester is:
The correct order of decreasing acidity of the following aliphatic acids is:




