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Rank the following acids in decreasing order of their acid strength. In each case, explain why the previous compound should be a stronger acid than the one that follows it.

Short Answer

Expert verified

Stronger electron-withdrawing groups stabilize the anion of the conjugate base more than weaker groups do, leading to stronger acids. Chlorine is more electronegative and a stronger withdrawing group than bromine, making chloropentanoic acid a stronger acid than bromopentanoic acid.

Step by step solution

01

     Step 1

Stronger electron-withdrawing groups stabilize the anion of the conjugate base more than weaker groups do, leading to stronger acids. Chlorine is more electronegative and a stronger withdrawing group than bromine, making chloropentanoic acid a stronger acid than bromopentanoic acid.

02

Step 2

Adding a chlorine atom to pentanoic acid increases its acidity, and the effect is larger if the chlorine atom is closer to the acidic group.

03

Step 3

In case of bromopentanoic acid, acidic effect is larger if the bromine atom is closer to the acidic group. Therefore 2-bromopentanoic acid is more acidic than 4-bromopentanoic acid.

04

Step 4

Multiple electron-withdrawing groups stabilize the conjugate base and increase the acidity more than a single group does. Therefore 2,2 dichloropentanoic acid is more acidic than 2-chloropentanoic acid.

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Classify the following hydrocarbons, and draw a lewis structure for each one. A compound may fit into more than one of the following classifications:

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The pKa of ascorbic acid (vitamin C, page 9) is 4.17 , showing that it is lightly more acidic than acetic acid (CH3COOH,pKa =4.74) .

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(b) Compare the stabilities of these four conjugate bases, and predict which OH group of ascorbic acid is the most acidic.

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(a)

(b)

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