Home
Class 12
CHEMISTRY
PK(b) of aniline is less / more than p-C...

`PK_(b)` of aniline is less / more than `p-C_(6)H_(4)(NH_(2))NH_(2)`

Text Solution

AI Generated Solution

To determine whether the \( pK_b \) of aniline is less than or more than that of para-C6H4(NH2)2 (para-benzene-1,4-diamine), we need to analyze the basicity of both compounds. ### Step-by-Step Solution: 1. **Understanding the Compounds**: - Aniline is \( C_6H_5NH_2 \), which has one amino group (-NH2) attached to a benzene ring. - Para-benzene-1,4-diamine (or para-phenylenediamine) is \( C_6H_4(NH_2)_2 \), which has two amino groups attached to the benzene ring at the para position. ...
Promotional Banner

Topper's Solved these Questions

  • ORGANIC COMPOUNDS CONTAINING NITROGEN

    MODERN PUBLICATION|Exercise REVISION EXERCISES (OBJECTIVE TYPE QUESTIONS) (MULTIPLE CHOICE QUESTIONS)|25 Videos

  • ORGANIC COMPOUNDS CONTAINING NITROGEN

    MODERN PUBLICATION|Exercise REVISION EXERCISES PASSAGE BASED QUESTIONS|10 Videos

  • ORGANIC COMPOUNDS CONTAINING NITROGEN

    MODERN PUBLICATION|Exercise QUICK MEMORY TEST B. COMPLETE THE MISSING LINKS|10 Videos

  • HALOALKANES AND HALOARENES

    MODERN PUBLICATION|Exercise UNIT PRACTICE TEST|11 Videos

  • POLYMERS

    MODERN PUBLICATION|Exercise Competition file (OBJECTIVE TYPE QUESTIONS) (C. MULTIPLE CHOICE QUESTIONS)(Integer Type Questions)|6 Videos

Similar Questions

Explore conceptually related problems

Give reasons for the following (a) Acetylation of aniline reduces its activation effect. (b) CH_(3)NH_(2) is more basic than C_(6)H_(5)NH_(2) . (c) Although -NH_(2) is o/p directing group, yet aniline on nitration gives a significant amount of m-nitroaniline.

The number of amines having pK_(b) less than C_(6)H_(5)NH_(2) among the following is P-CH_(3) C_(6)H_(5)NH_(2), o-CH_(3)C_(6)H_(4)NH_(2), m-CH_(3)C_(6)H_(4)NH_(2), C_(6)H_(5)N (CH_(3)) _(2), C_(6)H_(5)NHCH_(3), p-NO_(2)C_(6)H_(4)NH_(2), p-CIC_(6)H_(4)NH_(2), C_(6)H_(5)CH_(2)NH_(2)

pK_(b) of aniline is more than that of methylamine. Why?

Arrange the following in decreasing order of their basic strength: C_(6)H_(5)NH_(2), C_(2)H_(5)NH_(2), (C_(2)H_(5))_(2)NH, NH_(3) in aqueous solution.

Iodobenzene (C_(6)H_(5)l) is prepared from aniline (C_(6)H_(5)NH_(2)) in a two step process as shown below C_(6)H_(5)NH_(2) + HNO_(2) + HCl rarr C_(6)H_(5)N_(2) .^(+)Cl^(-) + 2H_(2)O " "C_(6)H_(5)N_(2) .^(+)Cl^(-) + KI rarr C_(6)H_(5)I + N_(2) + KCl In an actual preparation 9.30 g of aniline was coverted to 16.32 g of iodobenzene. The percentage yield of iodobenzene is :