Explanation
As the steric hindrance increases from CH3CH2CH2CH2NH2(I) CH3—CHNH2—CH2CH3 (II) (CH3)2CHCH2NH2 (III) to (CH3)3CNH2(IV),the stability of the corresponding conjugated acids decreases and hence the basicity decreases accordingly i.e., I > II > III > IV or increases in the opposite order, i.e., IV < III < II < I. Slight increases in the +I-effect of these groups from I to IV, however, does not affect the basicity appreciably.
In compound II (pyridine), the lone pair of electrons on N is present in a sp2-orbital while in compounds I (piperidine) and III (morpholine), the lone pair of electrons on N is present in a sp3-orbital. Since a sp2-orbital has more s-character (33.33 %) than a sp3–orbital (25%), therefore, the lone pair of electrons on N is more easily available for protonation in I and III than in II. In other words, compound (II) is less basic that I and II than III. Among I and III, III contains an oxygen atom which has—I-effect. As a result, it will attract the lone pair of electrons on N towards itself. Consequently, the lone pair of electrons on N in III is less readily available for protonation than in I. In other words, compound I is more basic than III.
Compound IV (pyrrole) is aromatic in character. Therefore, in accordance with ‘Huckel’s rule’ it has a cyclic cloud of six π-electrons. Out of these, four are contributed by two double bonds while the remaining two are the lone pair of electrons on the N-atom. In other words, the lone pair of electrons on N-atom is contributed towards the aromatic sextet formation and hence is not available for protonation. Therefore, compound IV is the least basic. In fact, it is such a weak base that it is weakly acidic in character and thus reacts with K metal when heated to form the corresponding potassium salt. Thus the overall order of basicity is : I >III > II > IV, i.e. option (4) is correct.
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