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JEE Main 202429 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Evening Shift
Subject
Chemistry
Chapter
General Organic Chemistry

JEE Main 2024 · 29 January · Evening Shift · Chemistry · General Organic Chemistry

Match List I with List II Choose the correct answer from the...

Match List I with List II
Match List I with List II Choose the correct answer from the options given below:
Choose the correct answer from the options given below:

Options

  1. AA-I, B-II, C-III, D-IV
  2. BA-IV, B-I, C-II, D-III
  3. CA-III, B-IV, C-I, D-II
  4. DA-II, B-I, C-IV, D-III

Correct answer

D. A-II, B-I, C-IV, D-III

Step-by-step solution

pKa=−logKa{pK}_{a} = - {logK}_{a}, here Ka{K}_{a} acid dissociation constant. More the Ka{K}_{a} value, stronger is the acidic nature of the respective acid.

Hence, the lower pKa{pK}_{a} value denotes a more powerful acid.  Phenols are more acidic than alcohols. Hence, ethanol has more pKa{pK}_{a} value. Nitro group is the electron withdrawing group, hence it increases the acidic nature of phenols.  Among the given phenols para nitrophenol  is stronger acid. The pKa{pK}_{a} values are matched below.

Ethanol →\to15.915 . 9

Phenol→10\to 10

m−m -Nitrophenol →8.3\to 8 . 3

p−p -Nitrophenol→7.1\to 7 . 1

Hence, the answer is option D.

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