Explanation
Nitrobenzene is reduced to phenylammonium ions using a mixture of tin and concentrated hydrochloric acid. The mixture is heated under reflux in a boiling water bath for about half an hour.
Under the acidic conditions, rather than getting phenylamine directly, you instead get phenylammonium ions formed.
(i) Out of two -NH2 groups in semicarbazide, one is a part of amide functional group. In this, the lone pair on nitrogen atom is involved in the resonance with carbonyl group. Hence, this -NH2 group cannot act as a nucleophile. Hence, it is not involved in the formation of semicarbazones.
(ii) In cyclohexanone, the carbonyl carbon is not hindered. Hence, the nucleophile CN-ion can easily attack the carbonyl carbon. However in 2,2,6-trimethyl cyclohexanone, the carbonyl carbon is sterically hindered due to presence of methyl groups. Hence, the nucleophile CN- ion cannot easily attack the carbonyl carbon.
(iii) This is because of simply the difference between the electronegativities of carbon and double bonded oxygen and as you know oxygen the second most electronegative element because of which partial positive and partial negative charges appears and carbon and oxygen respectively because of which dissociation of Alpha hydrogen becomes easier (as we know the acidic character is how easily a particular compound loses H ion). Thus, it show acidic character
Pyrrole is least basic because the lone pair of $$N$$ is involved in the resonance and thus they are not available for formation of a new bond with proton.
Pyridine is more basic than pyrrole as in pyridine the lone pair of $$N$$ is not involved resonance.
Piperidine is most basic because the lone pair is present in $$sp^3$$ hybrid orbital of $$N$$ while In pyridine it is present in $$sp^2$$ hybrid orbital. Greater the $$s$$ character of orbital more strongly the electrons will be bonded with the atom and less will be the basicity.
morpholine is less basic than piperidine but more basic than pyridine. In morphiline the electron withdrawing tendency of ring $$O$$ makes the lone pair of $$N$$ to be less available for the bonding with proton.
$$ Piperidine>\ morpholine>\ pyridine>\ pyrrole$$
$$\mathbf{I > III > II > IV}$$
$$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (D)}$$
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