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Amines are basic in… (Class 12 Chemistry SQP 2024-25, Q30)

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ChemistryCBSE Class 12 · Sample paper, session 2024-25 · 4 marks · Case-based

CBSE Class 12 Chemistry sample paper 2024-25: question 30

From the official CBSE sample question paper for 2024-25, section D, question 30, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q30 · 4 marks · Case-basedCompetency-basedAmines

Amines are basic in nature. The pKb value is a measure of the basic strength of an amine. Lower the value of pKb, more basic is the amine. The effect of substituent on the basic strength of amines in aqueous solution was determined using titrations. The substituent “X” replaced “-CH2” group in piperidine (compound 1) and propylamine , (compound 2).
Compound 1: piperidine ring with NH and X. Compound 2:

The experimental data is tabulated below:

Substituent “X”Electronegativity of Xsubstituted piperidine compoundpKaSubstituted propylamine compoundpKa
2.55piperidine11.1310.67
NH3.12piperazine9.8110.08
O3.44morpholine8.369.45
3.6N-acetylpiperazine7.949.28
3.7N-benzoylpiperazine7.78___

(source: Hall Jr, H. K. (1956). Field and inductive effects on the base strengths of amines. Journal of the American Chemical Society, 78(11), 2570-2572.)

Study the above data and answer the following questions:

  1. a

    Plot a graph between the electronegativity of the substituent vs pKb value of the corresponding substituted propyl amine (given that pKa + pKb =14). Is there any relation between the electronegativity of the substituent and its basic strength?

    2 marks
  2. b

    The electronegativity of the substituent “” is 3.7, what is the expected pKa value of compound ?
    (i) 9.9 (ii) 9.5 (iii) 9.3 (iv) 9.1

    1 mark
  3. c

    The pKa value of the substituted piperidine formed with substituent “X” is found to be 8.28. What is the expected electronegativity of “X”
    (i)3.5 (ii)3.4 (iii)3.8 (iv) 3.1

    1 mark

  4. OR

    c

    What is the most suitable pKa value of the substituted propylamine formed with substituent “X” with electronegativity 3.0
    (i)10.67 (ii)10.08 (iii)10.15 (iv)11.10

    1 mark

Figure in the paper: Table includes skeletal structures of piperidine and its substituted analogues (NH, O, N-COCH3, N-COC6H5 in the ring). See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

Open one hint at a time and try again before the next.

Hint 1

Convert each pKa to pKb using pKa + pKb = 14 before plotting.

Hint 2

Look at the trend: as EN increases, what happens to pKa and to pKb?

Hint 3

For b and c, place the given value between the two nearest data points and interpolate or extrapolate.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. a. Use pKb = 14 − pKa for the substituted propylamines:

    XEN of XpKapKb
    CH₂2.5510.673.33
    NH3.1210.083.92
    O3.449.454.55
    CH₃CON3.69.284.72

    Plot EN on the x-axis and pKb on the y-axis. The points rise steadily and lie close to a straight line of best fit with positive slope.

    1½ marks

  2. Relation: pKb increases as the electronegativity of X increases. A more electronegative substituent withdraws electron density from N (–I effect), so basic strength decreases with increasing electronegativity.

    ½ mark

  3. b. pKa falls as EN rises (9.45 at 3.44; 9.28 at 3.6). Extrapolating to EN 3.7 gives a value just below 9.28, about 9.1–9.2. Answer: (iv) 9.1.

    1 mark

  4. c. For piperidines, pKa 8.36 (EN 3.44) and 7.94 (EN 3.6). A pKa of 8.28 lies just past O, so EN ≈ 3.44 + (0.08/0.42)(0.16) ≈ 3.47. Answer: (i) 3.5.

    1 mark

  5. c (OR). For propylamines, EN 2.55 → 10.67 and 3.12 → 10.08. At EN 3.0, pKa ≈ 10.67 − (0.45/0.57)(0.59) ≈ 10.2. Answer: (iii) 10.15.

Answer: a. pKb rises (basic strength falls) with increasing electronegativity of X; b. (iv) 9.1; c. (i) 3.5 / OR (iii) 10.15

Where the marks are

  • Correct pKb values and graph with line of best fit (1½ marks)
  • Higher electronegativity gives higher pKb, i.e. weaker base (½ mark)
  • (iv) 9.1 (1 mark)
  • (i) 3.5, OR (iii) 10.15 (1 mark)

Revise the chapter

  • Chapter: Amines (notes, formulas, where marks are lost and every past question)
Practise this chapterBoard-style questions from Amines, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2024-25: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.