MCQ
In the following compounds

The order of basicity is

  • A
    $IV > I > III > II$
  • B
    $III > I > IV > II$
  • C
    $II > I >III > IV$
  • $I > III > II > IV$

Answer

Correct option: D.
$I > III > II > IV$
d
$(I)$ Piperidine is the most basic. The $N$ lone pair is in an $s p^3$ hybrid orbital and there is no resonance.

$(II)$ In pyridine the $N$ lone pair is in an $s p^2$ hybrid orbital but is not part of the electron aromatic system of ring nor it is involved in any resonance (perpendicular). The $s p^2$ hybrid is smaller than the $s p^3$ hybrid, so there is a stronger attraction to the nucleus, so less basic.

$(III)$ In morpholine, the $N$ Ione pair is in an $s p^3$ hybrid orbital and there is no resonance. But the electron withdrawing effect of $O$ from ring makes it less available for donation.

$(IV)$ Pyrrole is a very weak base. The $N$ lone pair is involved in the electron aromatic system. Protonation will destroy the aromaticity of the electron aromatic system and this makes it an unfavourable process.

Therefore the correct order of basicity is $I\,>\,I I I\,>\,I I\,>\,I V$

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