$$C{H_3} \equiv CC{H_3}$$   is linear and symmetrical ; thus it has lowest dipole moment.

\[\left( \text{i} \right)\overset{3}{\mathop{C}}\,{{H}_{2}}=\overset{2}{\mathop{C}}\,H\overset{1}{\mathop{C}}\,OOH\to \]      \[\overset{2}{\mathop{C}}\,{{H}_{2}}=\overset{1}{\mathop{C}}\,{{H}_{2}}-COOH\]      loses its preference
\[\left( \text{ii} \right)\]  \[-COOH\]   loses its preference
\[\left( \text{iii} \right)\]  \[\to C{{H}_{3}}C{{H}_{2}}COOH;-COOH\]       does not lose its preference

Only 2 - cylcopropyl butane has a chiral centre.

No explanation is given for this question. Let's discuss the answer together.

Stability of the given species depends upon the hyperconjugation having higher number of a-hydrogens as well as resonance. Hence, the correct stability order will be (I) > (III) > (II) > (IV).

In aromatic acids presence of electron withdrawing substituent e.g. $${\text{ - }}N{O_2}$$  disperses the negative charge of the anion and stablises it and hence increases the acidity of the parent benzoic acid.
Further $$o-$$ isomer will have higher acidity than corresponding $$m$$ and $$p$$ isomers. Since nitro group at $$p-$$ position have more pronounced electron withdrawing than $${\text{ - }}N{O_2}$$  group at $$m-$$ position hence the correct order is the one given above.

$$\eqalign{ & \% \,\,{\text{of}}\,\,S = \frac{{32}}{{233}} \times \frac{{1.02}}{{2.18}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6.42\% \cr} $$