Some Basic Concepts of Organic Chemistry

Acidic nature of aromatic compounds depend on the ability to donate $\mathrm{H}^{+}$ ions. Presence of electron withdarawing group (i.e. $-\mathrm{NO}_2$) increases the acidic nature.

$-\mathrm{NO}_2$ at ortho-position [i.e. structure $(B)$ ] has ability to form hydrogen bond with $-\mathrm{OH}$ group. Thus, structure $(B)$ is less acidic than structure $(A)$. On the other hand, electron releasing group bonded with the ring (i.e. $-\mathrm{CH}_3$ ) decreases the acidic nature thus is less acidic than phenol. Hence, correct order of acidic nature is

and option (b) is the correct.

Among the given free radicals,

is most stable due to resonance. In the remaining free radicals, more the alkyl groups bonded to the electron deficient carbon, the more stable is the radical. Thus, the correct order of stability of given free radicals is

Key idea Higher be the $\mathrm{p} K_a$ value or smaller be the $K_a$ value weaker is the acid.

$\therefore \quad \mathrm{p} K_a=\log \left[K_a\right]$

In structure (i) fluorine (F) show $(-I)$ effect, thus is most acidic.

In structure (ii) $\mathrm{CH}_3 \mathrm{COOH},-\mathrm{CH}_3$ show $(+I)$ effect, i.e. hyperconjugation, so is weaker acid than (i).

In structure (iii) i.e. $\mathrm{C}_6 \mathrm{H}_5-\mathrm{COOH}$, the $-\mathrm{COOH}$ group is in resonance with the benzene ring. Thus, its carbocation is stable, therefore it is more acidic than (ii) and (iv) but is less acidic than (i).

In structure (iv) larger alkyl group bonded with $-\mathrm{COOH}$ group, which show stronger $(+I)$ effect, thus is the weakest acid.

Therefore, order of acidic nature is

$\text { (i) }>\text { (iii) }>\text { (ii) }>\text { (iv). }$

Hence, correct order of $\mathrm{p} K_a$ values

$\text { (iv) }>\text { (ii) }>\text { (iii) }>\text { (i) }$

Thus, option (b) is the correct.

While numbering in IUPAC system, double bonds get top there after $-\mathrm{NO}_2$ group is placed and $-\mathrm{OCH}_3$ comes in last (as per lowest sum rule).

While writing the name in IUPAC system, substituents are written first in alphabetical order followed by functional group.

Here, methoxy $\left(-\mathrm{OCH}_3\right)$ is at number (5) position, nitro $\left(-\mathrm{NO}_2\right)$ is at number (3) position, both are secondary prefix and 'hex' is the word-root and ene primary and cyclo is primary prefix.

Thus, the correct name is

5-methoxy-3-nitrocyclohexene.

Hence, option (b) is the correct.

The basic nature depends on the ability of atoms in the species to donate lone pairs of electrons. Structure (iii) is the most basic because the $-\mathrm{NH}_2$ group is not directly bonded to the benzene ring, allowing it to donate electrons more freely.

Structure (ii) is the second most basic due to the presence of both $-\mathrm{NH}_2$ and $-\mathrm{OCH}_3$ groups, which enhance its electron-donating ability.

Structure (i) is next in basicity. It is less basic than (iii) and (ii) but more basic than (iv) because of the electron-releasing $\left(-\mathrm{CH}_3\right)$ group.

Structure (iv) is the least basic. This is because the $-\mathrm{NH}_2$ group is directly bonded to the benzene ring, causing the lone pairs on the nitrogen atom to be involved in resonance, reducing their availability for donation.

Hence, the correct order is (iii) > (ii) > (i) > (iv), and option (C) is the correct one.

Among the given structures, structure $(A)$ and $(B)$ are staggered forms of ethan-1, 2 diol. On the other hand, structure $(C)$ and $(D)$ are eclipsed forms of ethan-1, 2- diol eclipsed-forms are less stable than staggered-forms due to electronic repulsion.

Between $(A)$ and $(B)$, structure $(B)$ show H-bonding due to which it is more stable than structure $(A)$. Hence, option (b) is the correct.

Stability of the given cations can be understood by the following structures

Thus, the stability of carbocation decreases in the order,

II > I > III

The correct IUPAC name of the given compound is 5 - hydroxy cyclohex -3 - en - 1 - one. The structure is shown below