Some Basic Concepts of Organic Chemistry

IUPAC name : 3-ethyl-5-methylheptane

In Lassaigne's test, the elements present in an organic compound are converted from covalent form to ionic form.

So, the correct answer is:

Option B: Covalent form to ionic form

Reason:

In organic compounds, elements like nitrogen, sulphur, and halogens are usually present in the covalent state. Since covalent compounds do not give the usual ionic reactions directly, the compound is fused with sodium. This converts these elements into ionic sodium salts, such as:

$ \text{NaCN, Na}_2\text{S, NaX} $

These ionic forms can then be easily tested.

Therefore, the answer is:

$ \boxed{\text{Covalent form to ionic form}} $

Possible monochlorination products:

Total 6 isomers.

For cyclic ethers O should be in ring * carbon here is chiral

Total number of isomers $=2+1+1+1+2+1+2=10$

I. carbon of this bond is sp$^2$ hybridised

II. carbon of this bond is sp hybridised

III. carbon of this bond is sp$^3$ hybridised

Higher the percentage s character, stronger is C-H bond. Correct order of bond dissociation energy of C-H bond:

$II > I > III$

Cis-trans isomers shown by:

Condition: Restricted rotation around double bond

Or

Different group around double bond

$ \text { 4-ethyl-1-fluoro-2-nitrobenzene } $

Steam distillation is a separation technique used for purifying organic compounds that are immiscible with water and have high boiling points. In this method, the organic compound co-distills with water. The key concept here is that the total vapor pressure of the mixture is the sum of the vapor pressures of the organic compound and water. When the total vapor pressure equals atmospheric pressure, the mixture will boil.

For a compound that has a boiling point of $250^{\circ} \mathrm{C}$ and is immiscible with water, during steam distillation, the boiling point of the mixture will be influenced by both the vapor pressure of water and the organic compound. Since water has a boiling point of $100^{\circ} \mathrm{C}$, steam distillation allows the mixture to boil at a temperature lower than the boiling point of the organic compound when in pure form. This is because the partial pressures of both the water and the organic compound add up to reach the atmospheric pressure more quickly than either would alone.

Practically, the mixture will boil at a temperature close to the boiling point of water, which is $100^{\circ} \mathrm{C}$. Hence, the correct answer is:

Option D close to but below $100^{\circ} \mathrm{C}$

$\begin{array}{lll} \text { Lake test } & - & \mathrm{Al}^{3+} \\ \text { Nessler's reagent } & - & \mathrm{NH}_4^{+} \\ \text {Potassium sulphocyanide } & - & \mathrm{Fe}^{3+} \\ \text { Brown ring test } & - & \mathrm{NO}_3^{-} \end{array}$

Methyl group attached to a positively charged carbon atom stabilizes the carbocation due to hyperconjugation and +I effect.

The two statements provided deal with molecular structure and boiling points of isomers of pentane. Let's analyze both statements:

Statement I: This statement lists the boiling point order of the three isomers of pentane as n-pentane > isopentane > neopentane. To verify this statement, let's consider the molecular structure and boiling points of each isomer:

• n-pentane: It is a straight-chain alkane with the formula $C_5H_{12}$, and it has the highest boiling point among the isomers because of its larger surface area which allows for greater van der Waals forces (dispersion forces).
• isopentane (also known as methylbutane): It has one branch in its carbon chain. The branching reduces the surface area slightly, which slightly weakens the van der Waals forces compared to n-pentane.
• neopentane (also known as dimethylpropane): It has a highly branched structure making it almost spherical. This shape minimizes the surface area significantly, thereby reducing the van der Waals forces drastically compared to the other two isomers.

Thus, the boiling point order correctly reflects the influence of molecular structure and intermolecular forces. Hence, Statement I is correct.

Statement II: This statement elaborates on why branching leads to lower boiling points. The assertion is that increased branching gives the molecule a more spherical shape, which then results in a smaller surface area, and thus weaker intermolecular forces (primarily van der Waals forces). Weaker intermolecular forces correspond with a lower energy requirement for the liquid to gas phase transition, thereby lowering the boiling point. This explanation is coherent with concepts in physical chemistry regarding molecular interactions and phase change. Therefore, Statement II is also correct.

In summary, both Statement I and Statement II are correct, reflecting the accurate relationship between molecular structure, intermolecular forces, and physical properties like boiling points. Therefore, the correct answer is:

Option A: Both Statement I and Statement II are correct.

The stability of carbocation can be described by the hyperconjugation. Greater the extent of hyperconjugation, more is the stability of carbocation.

Stability order of carbocations $=(4) > (2) > (1) > (3)$

The technique described in the question is known as sublimation. Sublimation is a process where a solid turns directly into a gas without passing through the intermediate liquid state. This physical change occurs under specific conditions of temperature and pressure and is characteristic of certain substances.

Option B: Sublimation is the correct answer because it directly describes the process of solid substances transforming into vapour without becoming liquid first. For example, substances like iodine, naphthalene, and dry ice (solid carbon dioxide) sublimate when heated under normal atmospheric conditions. Sublimation is not only a fascinating chemical process but is also exploited in various scientific and industrial applications, including purification of substances, where impurities do not sublimate and can thus be easily separated from the vaporized material.

The other options provided do not describe this process:

• Crystallization (Option A) involves the formation of solid crystals from a homogeneous solution. It is typically used to purify solids wherein the impurities remain dissolved in the solvent.
• Distillation (Option C) is a process used to separate mixtures based on differences in the conditions required to change the phase (liquid phase) of the components of the mixture. It does not involve direct transition from solid to gas.
• Chromatography (Option D) is a method for separating components of a mixture based on differences in their movement through a stationary medium under the influence of a solvent or carrier gas. It does not involve the phase transition of solid to gas directly.

Mohr's salt is the ammonium iron(II) sulfate, with the chemical formula $\mathrm{(NH_4)_2Fe(SO_4)_2 \cdot 6H_2O}$. It is known for its stability compared to the other iron(II) salts which tend to readily oxidize to iron(III) salts when exposed to the air. To prevent the oxidation and hydrolysis of the $\mathrm{Fe}^{2+}$ ion during the preparation of Mohr's salt, an acid is added. This acid serves several purposes: it maintains the acidic environment necessary to prevent hydrolysis, aids in the solubilization of iron(II) sulfate, and minimizes the oxidation of $\mathrm{Fe}^{2+}$ ions to $\mathrm{Fe}^{3+}$.

The options provided give four different types of acids, from which one needs to be selected for the preparation of this salt. We can evaluate them based on their appropriateness:

• Dilute Hydrochloric Acid: While capable of maintaining a low pH to prevent oxidation, HCl can potentially introduce chloride ions ($\mathrm{Cl^-}$), which can form complexes with iron, changing the composition of the solution.
• Concentrated Sulphuric Acid: This choice is too strong and can lead to excessive acidity as well as potential safety issues during handling and dilution. It is unnecessary for this application.
• Dilute Nitric Acid: Nitric acid is an oxidizing agent and can promote the oxidation of $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$, which is undesirable in the preparation of Mohr's salt.
• Dilute Sulphuric Acid: This is a very common choice since it helps maintain the stability of the $\mathrm{Fe}^{2+}$ ion without contributing extraneous ions that could form undesired complexes or products. Moreover, it keeps the solution acidic, helping prevent oxidation and hydrolysis, while not introducing any oxidizing characteristics.

Given these considerations, the best choice for preventing hydrolysis of the $\mathrm{Fe}^{2+}$ ion during the preparation of Mohr's salt is Dilute Sulphuric Acid (Option D). This is because it maintains the suitable acidic conditions needed for stabilizing the ferrous ion and does not interfere with the redox stability of the iron by providing an oxidative chemical environment.

Step 1: Identify the main factor influencing basicity

• The basicity of amines and related compounds depends on the availability of the lone pair on the nitrogen atom for donation.

• Anything that decreases the availability of this lone pair decreases basicity.

Step 2: Examine all options

Let us write the names and analyse:

Option A: Aniline ($\ce{C6H5NH2}$)

Option B: p-Nitroaniline ($\ce{p-NO2-C6H4NH2}$)

Option C: Benzylamine ($\ce{C6H5CH2NH2}$)

Option D: Ammonia ($\ce{NH3}$)

Step 3: Analyse effects

1. Aniline ($\ce{C6H5NH2}$):

• The lone pair on nitrogen in aniline is partially delocalized into the aromatic ring (resonance), reducing its availability for protonation $\rightarrow$ decreases basicity.

1. p-Nitroaniline ($\ce{p-NO2-C6H4NH2}$):

• Same as aniline, but now a highly electron-withdrawing nitro group ($\ce{NO2}$) is attached at the para position.

• The nitro group (–M effect and –I effect) pulls electrons even more strongly from the ring, further reducing the electron density on the nitrogen.

• This makes the lone pair least available for donation, so very low basicity.

1. Benzylamine ($\ce{C6H5CH2NH2}$):

• The $\ce{NH2}$ group is not directly attached to the aromatic ring.

• Thus, no resonance with the aromatic ring — lone pair is fully available.

• More basic than both aniline and p-nitroaniline.

1. Ammonia ($\ce{NH3}$):

• A simple molecule; only weakly basic, but nitrogen lone pair is completely available.

• Comparable to benzylamine but more basic than aniline and p-nitroaniline.

Step 4: Order of Basicity

Based on above:

$ \text{Benzylamine} > \text{Ammonia} > \text{Aniline} > \text{p-Nitroaniline} $

Final Step: Answer

The least basic compound is Option B: p-Nitroaniline.

Reason: The electron-withdrawing nitro group at the para position, along with resonance of the aniline system, greatly decreases the electron density on the nitrogen and makes its lone pair least available for bonding with a proton.

Rate of Dehydrohalogenation : II < III < I

To address these statements, let's analyze each one :

Statement I : "In an organic compound, when inductive and electromeric effects operate in opposite directions, the inductive effect predominates."

• The inductive effect refers to the transmission of charge through a chain of atoms in a molecule by electrostatic induction. It's a permanent effect.


• The electromeric effect is a temporary effect, occurring when a reagent approaches a double bond or a polar molecule and causes electrons to shift entirely from one atom to another.


• Generally, in organic chemistry, the inductive effect is considered to be a weaker effect compared to the electromeric effect. When both effects are present and oppose each other, the electromeric effect, being a more dominant electronic effect, usually prevails.

Statement II : "Hyperconjugation is observed in o-xylene."

• Hyperconjugation is a phenomenon where electrons in a sigma bond (usually C-H or C-C) are delocalized into an adjacent empty or partially filled p-orbital or pi-orbital, stabilizing the system.


• o-Xylene is an aromatic compound with two methyl groups attached to the benzene ring in ortho positions. However, hyperconjugation typically occurs in alkenes or carbocations where there is an adjacent p-orbital. In o-xylene, the aromatic system is already stabilized by resonance, and the hyperconjugation of the methyl groups does not significantly contribute to this stability.

Based on these explanations :

• Statement I is false because the electromeric effect is generally stronger than the inductive effect.
• Statement II is true as hyperconjugation can be observed in o-xylene, although its impact on the molecule's stability is minimal compared to the aromatic resonance.

Thus, the correct option is :

Option B : Statement-I is false but Statement-II is true.

Sol. In case nitrogen and sulphur both are present in an organic compound, sodium thiocyanate is formed, it give blood red colour and no prussian blue since there are no free cyanide Ions

$\mathrm{Na}+\mathrm{C}+\mathrm{N}+\mathrm{S} \rightarrow \mathrm{NaSCN}$

$\mathrm{Fe}^{+3}+\mathrm{SCN}^{\ominus} \longrightarrow[\mathrm{Fe}(\mathrm{SCN})]^{2+}$ Blood red

can not act as lewis acid because they does not contain vacant orbital

BF$_3$ $\to$ Contains vacant orbital on central atom (Boron).

Huckle's rule = (4n + 2)$\pi$ electrons

Comp (i), (ii), (v), (vii) obey Huckle's rule

Cl has pronounced $-$I effect than +R effect due to large size difference between carbon and chlorine, also due to high electronegativity of chlorine.

$\bullet$ Lesser the steric hinderance on halide carbon, more will be the reactivity of alkyl halide towards S_N2 reaction.

$\bullet$ So correct order towards S_N2 reactivity is :

$-$COOCH₃ has higher priority than $-$ C = C $-$ and $-$ CHO in IUPAC nomenclature.

C₁ = sp²

C₂ = sp³

Planar, cyclic, conjugated species containing (4n + 2)$\pi$ electrons will be aromatic in nature (n is an integer)

The stereoisomers related to each other as non-superimposable mirror image are called enantiomers.

Kjeldahl method is not applicable to compounds containing nitrogen in nitro group, azo groups and nitrogen present in the ring (e.g., pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.

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