Basics of Organic Chemistry

The IUPAC name of the compound is 4-ethyl-3-methyl-hex-2-ene.

2-Cyclopropen-1-one cation is a cyclic $\pi $-system with two electrons. It has highest dipole moment which can be explained on the basis that the dipolar resonance structure contributes more to the resonance hybrid as it involves stable aromatic cyclopropenyl cation.

Among the given compounds, cyclopentadiene is most reactive to an aqueous solution of sodium carbonate. It forms cyclopentadienyl anion which is aromatic and therefore, highly stable.

The IUPAC name of the compound shown is 2-Ethyl-1,1-dimethylcyclohexane

A water insoluble organic base (amine) can be protonated by reaction with an acidic solution (HCl) and it can thus move from the organic layer into the aqueous layer.

In partition chromatography, which includes paper chromatography, the stationary phase is typically water held by cellulose fibers in the paper. This paper acts as an inert support. When a mixture is applied to the paper, its components partition between this stationary phase and the mobile phase (which can be a liquid or a gas). The varying affinities of the mixture's components for the stationary and mobile phases allow for their separation.

Now, let's address the options :

Option A : Mobile phase can be a gas - This is true. In partition chromatography, the mobile phase can be either a liquid or a gas.

Option B : Stationary phase is a finely divided solid adsorbent - This is typically false for partition chromatography. Here, the stationary phase is a liquid (like water in paper chromatography) that's held by an inert solid support (like cellulose paper).

Option C : Separation depends upon equilibration of solute between a mobile and a stationary phase - This is true. In partition chromatography, the mixture's components equilibrate between the mobile and stationary phases, leading to their separation.

Option D : Paper chromatography is an example of partition chromatography - This is true. As explained above, in paper chromatography, the stationary phase (water) is held by an inert support (cellulose paper), fitting the definition of partition chromatography.

(a) I is a benzylic halide, thus, it undergoes S_N1 reaction easily as benzylic carbocation is resonance stabilized. III also follows S_N1 mechanism as it is 3$^\circ$ alkyl halide.

(b) Compounds I and II are 1$^\circ$ alkyl halides, then they undergo S_N2 mechanism.

(c) Correct. In S_N1 reaction, the nucleophile approaches the substrate carbon from the back side with respect to the leaving group. Nucleophilic displacement of the leaving group in an S_N2 reaction causes inversion of configuration at the substrate carbon.

(d) Stability of carbocations follows the order:

$\mathop {2^\circ \,benzylic}\limits_{(IV)} > \mathop {3^\circ \,alkyl}\limits_{(III)} > \mathop {1^\circ \,benzylic}\limits_{(I)} $

The best technique for separating glycerol from spent-lye in the soap industry is Distillation under Reduced Pressure, also known as Vacuum Distillation (Option D).

The reason for this is that glycerol has a high boiling point (~${290^ \circ }$$C$). If we were to use simple or fractional distillation, which involves boiling at standard atmospheric pressure, the high temperature could degrade or decompose the glycerol.

In contrast, vacuum distillation allows the separation to occur at a much lower temperature ( around ${180^ \circ }C$) by reducing the pressure. This helps to prevent degradation of the glycerol, making it a better method for this particular separation. Steam distillation isn't suitable here, as it's typically used for temperature-sensitive, volatile substances, which isn't the case for glycerol.

As the branching increases, the boiling point decreases. This is because the van der Waals forces of attraction decrease.

(i) The structure of tert-butyl cation $\left[\left(\mathrm{CH}_3\right)_3 \mathrm{C}^{\oplus}\right]$ is as follows :

The carbocation is highly stable due to positive hyperconjugative $(+\mathrm{H})$ and inductive $(+\mathrm{I})$ effect.

(ii) The hyperconjugative effect is due to delocalication of $\sigma$ (sigma) electrons of $\mathrm{C}-\mathrm{H}$ sigma ( $\sigma$ ) bond on carbon adjacent to carbocation with an empty P orbital (on carbcation).

Hence, carbocation is stabilised due to formation of pi bond with adjacent carbon (due to delocalised) electrons of $\mathrm{C}-\mathrm{H}$ sigma bond). Its a s to p (empty) delocalisation.

(iii) There are eight more protons on carbons adjacent to carbocation. This gives eight more hyperconjugative structures. This impart stability to carbocation.

(iv) The structure of 2-butene is as follows :

$ \mathrm{CH}_3-\mathrm{C}=\mathrm{C}-\mathrm{CH}_3 $

The hyperconjugative effect is due to the delocalisation of sigma ( $\sigma$ ) electrons of $\mathrm{C}-\mathrm{H}$ bond with adjacent carbon containing pi bond.

The electrons of $\mathrm{C}-\mathrm{H} $ $\sigma$ bond are donated to anti-bonding $\left(\pi^*\right)$ orbitals of carbon-carbon double bond. Its $\sigma $ $\mathrm{s}$ to $\pi^*$ delocalisation.

There are five more hydrogens from methyl groups attached to carbon-carbon double bond. Hence, five more hyperconjugative structures.

The relation between the given compounds can be determined assigning them R and S configuration. The given structures can be represented as

M and N are diastereomers.

M and O are identical.

M and P are enantiomers (non-superimposable mirror images).

M and Q are diastereomers.

(a) The compound is monocyclic but non-planar due to presence of two $s p^3$ hybridised carbon atoms. As a result, delocalisation of pi $(\pi)$ electrons (or a conjugate system) is not possible. The compound is non-aromatic.

(b) The compound is monocyclic and planar. Carbon atoms are $s p^2$ hybridised and the pi $(\pi)$ electrons are conjugated, i.e., there is delocalisation of pi $(\pi)$ electrons. It follows $4 n$\pi$ electron system with $4 \pi$ electrons. This makes the compound anti-aromatic and least stable.

(c) The compound is monocyclic and planar. The carbons including carbonyl carbon are $s p^2$ hybridised. There is no extended delocalisation of pi ( $\pi$ ) electrons (due to exocyclic carbonyl double bond). It also follows $4 n \pi$ electron system with $4 \pi$ electrons inside the ring. This makes compound anti-aromatic and least stable.

(d) The compound is monocyclic and planar. The all carbons including carbonyl carbons are $s p^2$ hybridised. Though, there is no extended delocalisation of electron inside the ring, but ring has $(4 n+2) \pi$ electrons, i.e., $6 \pi$ electrons. This makes compound aromatic and most stable.

(i) The structures of the different organic compounds are as follows :

(A) p-Nitrophenol

(B) p-Hydroxybenzoic Acid

(C) o-Hydroxybenzoic Acid

(D) p-Toluic Acid

(ii) According to the $pK_a$ values, the order of acidity of these compounds is as follows :

(iii) Regardless of the substitution, phenols are generally less acidic than carboxylic acids. Therefore, p-nitrophenol is the least acidic, despite the presence of an electron-withdrawing nitro group.

(iv) The "ortho effect," i.e., the presence of an electron-withdrawing or releasing substituent at the ortho position relative to the carboxylic acid group, increases the acidity of substituted benzoic acids.

Therefore, o-hydroxybenzoic acid is more acidic than p-hydroxybenzoic acid.

(v) The presence of an -OH group at the para position decreases the acidity of carboxylic acids due to its electron-donating nature (via the positive mesomeric effect, +M). This effect is stronger compared to the electron-donating nature of a methyl group via the positive inductive effect (+I) or hyperconjugative effect (+H).

Hence, the order of acidity of the compounds is :

(C) o-Hydroxybenzoic Acid

(D) p-Toluic Acid

(B) p-Hydroxybenzoic Acid

(A) p-Nitrophenol

Option (C) has the lowest $pK_a$ or the highest $K_a$.

For compound in option (A) : Only two of the conformers (cisiod and transoid) have all the atom in the same plane.

For compound in option (B) : The terminal hydrogen of allene will be perpendicular to each other plane.

For compound in option (C) : All the atoms are in one plane in all the possible conformations. There is no atom on oxygen.

In the structure of 2,2-dimethylbutane:

On C$_2$$-$C$_3$ bond axis X = CH$_3$, Y = CH$_3$.

On C$_1$$-$C$_2$ bond axis X = H, Y = C$_2$H$_5$.

The hydride shift from C-2 will lead to a resonance stabilized secondary carbocation (conjugated with oxygen).

Among the resonance structures, the more stable ones are those with:

(i) more number of covalent bonds

(ii) all the atoms having octet of electrons complete,

(iii) less separation of opposite charges and more dispersal of charge.

Based on these, the order of stability of the resonance structures is (I) > (III) > (II) > (IV).

We know that, carboxylic acids are more acidic than phenols. Further, presence of electron withdrawing groups on the ring increases the acidic nature and electron releasing group decreases the acidic strength. So, structure III, due to the presence of carboxylic acid is the strongest acid $(p{K_a} = 4.17)$.

In structure IV, the +I effect of alkyl group reduces the acidic strength $(p{K_a} = 4.37)$ of the carboxylic acid. Structure III, due to the presence of electron withdrawing substituent (Cl), is more acidic than phenol $(p{K_a} = 9.38)$. Phenol (structure I) is the weakest acid amongst all $(p{K} = 9.98)$.

The priority order of the substituents on the ring is

$-$CN > $-$Br > $-$OH

Cyanide group is given the highest priority.

Hence, it is written as suffix benzonitrile and other functional groups behave as substituents. The numbering of substituents is done in such a way that sum of the locants is least.

The names of substituents are written in alphabetical order.