Basics of Organic Chemistry

Metamers are compounds which have different alkyl groups present along both side of polyvalent functional group.

No polyvalent functional group in option (a).

The two functional groups are different in option (b).

Alcohol is not a polyvalent functional group.

Alkyl groups are different along both sides of polyvalent functional group.

For resonance positive charged carbon atom ( Name it carbon 1) should be connected with the carbon atom( Name it carbon 2) using only sigma bond. Then carbon 2 should attach with another carbon with double bond. Here you can see (A) and (B) satisfies this situation but (C) and (D) don’t satisfy the situation.

The purification of a compound using chromatography is dependent on several factors, including the solubility of the compound, the mobility or flow of the solvent system, and the length of the column or TLC plate. However, it is not typically dependent on the physical state of the pure compound. Here's why :

Option A : The solubility of the compound in the mobile phase and its interaction with the stationary phase are key factors in determining how the compound will migrate in a chromatographic system. Hence, the solubility of the compound does affect its purification.

Option B : The mobility or flow of the solvent system, often referred to as the mobile phase in chromatography, also plays a significant role in the separation of components in the mixture. The rate at which the mobile phase moves can influence how far and how quickly the components of the mixture will travel.

Option C : The physical state (solid, liquid, gas) of the pure compound is usually not a factor in the purification process in chromatography. Whether a compound is solid, liquid, or gas does not typically affect its ability to be separated by chromatography, as it is the interactions between the compound and the stationary/mobile phases that are key.

Option D : The length of the column or TLC plate can impact the degree of separation achieved in chromatography. A longer column or TLC plate can provide more opportunities for the interactions that drive the separation process, often leading to better separation of the components.

So, Option C is correct : The physical state of the pure compound is usually not a factor in its purification by chromatography.

Molecules are aromatic if they are

(1) Cyclic

(2) Planar

(3) Conjugated

(4) Molecules who have (4n + 2) $\pi$ electrons [This is called Huckel's Rule]

Aromatic compounds are highly stable.

(a)

(1) Molecule is cyclic.

(2) All atoms which are completing the cyclic ring have sp² hybridisation. So it is shape is planar. If any one atom is sp³ hybridised then molecule shape will be non planar.

(3) It is conjugated means all $\pi$ bonds are showing resonance with each other.

So all atoms which are forming cyclic ring are perticipating in resonance. That is called conjugated.

(b)

(1) This compounds is cyclic

(2) All atoms are sp² hybridised. So molecule shape is planar.

(3) It is conjugated as all $\pi$ bonds are showing resonance.

(4) It has 6$\pi$e^$-$. So obey Huckel Rule.

$\therefore$ Compound is aromatic.

(c)

(1) This compound is cyclic

(2) All atoms are sp² hybridised. So molecule shape is planar.

(3) It is conjugated as $\pi$ bond is showing resonance.

(4) It has 2$\pi$e^$-$. So obey Huckel Rule.

$\therefore$ Compound is aromatic.

Anti-Aromatic :-

It is highly unstable.

Anti-aromatic compounds are -

(1) Cyclic

(2) All atoms which is forming cyclic ring are planar.

(3) it is conjugated.

(4) It has 4n$\pi$e^$-$.

Here no anti-aromatic compounds presents.

Non-Aromatic :-

Those compounds which are not either aromatic or anti-aromatic.

Definition : All cyclic compounds which are comparatively stable with its open chain analogous system, compound is known as non aromatic compound.

(a)

Here one carbon is sp³ hybridised that is why it's structure is non planar. So this cyclic compound will be neither aromatic nor anti-aromatic. It will be non aromatic.

Multiple cyclic rings :-

To find nature of this ring, make maximum possible $\pi$ bonds in the outer ring. This is a hypothetical structure. Here outer ring should be largest conjugated ring means outer ring should have these $\pi$ bonds which are in resonance with each other.

And the nature of the outer ring will be same as the original molecule.

(d)

From this structure we can't decide the nature as all $\pi$ bonds are not present in the outer ring that is why we make right resonating structure where all $\pi$ bonds present in the outer ring.

Outer ring :

As outer ring is aromatic so original compound will also be aromatic.

Note :-

Even though from left resonating structure we can't decide the nature of the molecule, it does not mean it's nature is not aromatic. As different resonating structures don't have different nature. All resonating structure have same nature.

A more stable radical means lesser the bond dissociation energy.

Radicals formed are

Radical formed from Q (PhCH₂$-$H) is most stable due to resonance.

Stability of free radical decreases with increase in % s-character.

P radical $\to$ sp³ $\to$ % s-character 25%

R radical $\to$ sp² $\to$ % s-character 33%

S radical $\to$ sp $\to$ % s-character 50%

Thus, order of stability of free radical is

Q > P > R > S

Order of bond dissociation energy is

S > R > P > Q

The reaction is exothermic with $\Delta$H$^\circ$ = $-$25 kcal mol^$-$1

CH₄(g) + Cl₂(g) $\buildrel {hv} \over \longrightarrow $ CH₃Cl(g) + HCl(g)

(a) Initiation step is breaking of Cl$-$Cl bond which requires energy and hence, this step is endothermic.

(b) Propagation step involving formation of CH₃ is also endothermic as bond between C$-$H is breaking.

(c) Propagation step involving formation of CH₃Cl is exothermic as bond is formed between C$-$Cl

(d) $\mathop H\limits^ \bullet $ + $\mathop {Cl}\limits^ \bullet $ $\to$ HCl, $\Delta$H₄$^\circ$ = $-$103 kcal/mol

Enthalpy of reaction, $\Delta$H$^\circ$ = $\Delta$H$_1^o$ + $\Delta$H$_2^o$ + $\Delta$H$_3^o$ + $\Delta$H$_4^o$

= 58 + 105 $-$ 85 $-$ 103

= $-$25 kcal mol^$-$1

Hence, the reaction is exothermic with $\Delta$H$^o$ = $-$25 kcal/mol

Due to the intramolecular H-bonding, the gauche conformation of butane-2,3-diol is most stable one.

Only 4-methylpent-2-ene shows geometrical isomerism.