Question

Which of the following acids has the smallest dissociation constant?
A. $C{{H}_{3}}CHFCOOH$
B. $FC{{H}_{2}}C{{H}_{2}}COOH$
C. $BrC{{H}_{2}}C{{H}_{2}}COOH$
D. $C{{H}_{3}}CHBrCOOH$

Answer 1

Due to the presence of electron withdrawing or electron donating groups in a molecule there is an effect on electron density and this phenomenon is called Inductive effect. The inductive effect is going to be indicated with ‘I’.

Complete answer:
- In the question it is given to find the acid which has the lowest or smallest dissociation constant.
- Coming to given options, Option A, $C{{H}_{3}}CHFCOOH$ . Due to the presence of high electronegative atoms (fluorine, -I effect) that are too very near to carboxylic functional group increases the dissociation constant of the acid.
- Coming to option B, $FC{{H}_{2}}C{{H}_{2}}COOH$ . Due to the presence of a high electronegative atom (fluorine, -I effect) in the molecule results in the enhancement dissociation constant of the given acid.
- Coming to the option C, $BrC{{H}_{2}}C{{H}_{2}}COOH$ . The electronegativity of the bromine is little bit less when compared to fluorine and that too bromine atom is little bit far away from the functional group then the dissociation constant of the option C is very less when compared to option A, and B.
- Coming to option D, $C{{H}_{3}}CHBrCOOH$ . The electronegativity of the bromine is little bit less when compared to fluorine and bromine atoms are nearer to the functional group then the dissociation constant of the option D is high when compared to option C.
- Therefore $BrC{{H}_{2}}C{{H}_{2}}COOH$ is the acid which has very less or smallest dissociation constant.

So, the correct option is C.

Note:
If an electron withdrawing group is present in a molecule then the molecule will experience –I effect and if the electron donating group is present in a molecule then the molecule will experience +I effect to the respective molecule.