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The decreasing order of the basic character of nitrogens at a, b, c, d in the given compound is:

(A) a > c > b > d
(B) d > c > b > a
(C) c > b > d > a
(D) b > c > d > a

Answer
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Hint: The term ‘basic character’ means which of the compounds has a tendency to donate a lone pair of electrons or to accept a hydrogen ion.

Complete Step by Step Answer:
Among the four nitrogens present in the given compound, the nitrogen (d) is the most basic. The reason for this is that the electron pair can be easily delocalized.

Now, in the other three nitrogens (a), (b), and (c), the basic character is determined by electronic factors and$-C{{H}_{3}}$. In nitrogen (c), the electron donating $-C{{H}_{3}}$ group is in para position to $N{{H}_{2}}$. So, its basicity is higher than nitrogen (a) and (b).

The nitrogen (a) has a methyl group at ortho position to it. In this case, when the ammonium ion accepts the hydrogen, it gets out of the plane. The methyl group causes steric hindrance and thus has less tendency to donate a lone pair of electrons. As a result, nitrogen (a) has the least basic character of the given compound's nitrogens.
Correct option : (B) d > c > b > a.

Note: If an electron withdrawing group has been attached to the compound, the basic character of a compound decreases, that is, lone pair availability decreases. Alternatively, when an electron donating group is attached to a compound, its basic character, i.e. lone pair availability, increases.