Basics of Organic Chemistry

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.

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

(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 $K_a$.

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.

Stability of the following species depends upon the no. of $\alpha$ hydrogen which can undergoes hyperconjugation as well as resonance. Higher the no. of $\alpha$ hydrogen, higher will be the stability of the carbocation.

Stabilizes by resonance and have six $\alpha$-hydrogen atoms (hyperconjugation).

Stabilizes by resonance and have three $\alpha$-hydrogen atoms.

It have five $\alpha$-hydrogen atoms.

It have only two $\alpha$-hydrogen atoms.

$\therefore$ (I) > (III) > (II) > (IV)

Hyperconjugation effect is one of the type of permanent effect in which delocalisation of $\sigma$ electrons of C-H bond of an alkyl group directly attached to an atom of the unsaturated system or to an atom with an unshared p orbital. The main role of hyperconjugation is to stabilise the carbocation because it helps in the dispersal of positive charge. Higher the number of alkyl groups attached to a positively charged carbon atom, greater is the hyperconjugative effect and stability of carbonation. It generally involves the interaction of electrons in a C-H sigma orbital with nearby non-bonding p or anti-bonding $\sigma^*$ or $\pi^*$ orbitals to provide extended conjugation which further increases the stability of the carbocation. Sometimes, low lying anti-bonding $\sigma^*$ also interact with filled orbitals of lone pair character. This type of conjugation is termed as negative hyperconjugation.

If we add halogens on alkenes it is mostly anti addition. Anti addition of $\mathrm{Br_2}$ on $trans$ alkene gives meso compound.

So, here only 1 stereoisomer is formed.

$trans$-2-butene is

Bromination of $trans$-2-butene

Among the given resonance structures, the least stable resonance structure is given by option (A) as the two positive charges lie adjacent to each other and cause maximum repulsion.

(A)

Same charges are present at nearest position making the molecule less stable.

Structural Isomers

(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)} $

(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.

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 possible stereoisomers of the given compounds are as follows:

The correct statements concerning the structures E, F and G are:

(B) E, F and G are tautomers.

E is in keto form (C=O) and F and G are in enol form (C=C$-$OH)

(C) F and G are geometrical isomers. F is Z isomer and E is E isomer.

(D) F and G are diastereomers. They are not mirror images of each other and non-superimposable.

When 2-methyl butane is treated with $\mathrm{Cl}_2$, the chlorination of the compound occur (via free radical mechanism) and an isomeric mixture of products is obtained with chemical formula, $\mathrm{C}_5 \mathrm{H}_{11} \mathrm{Cl}$.

The products formed are shown below.

The chlorination yields total 6 different compounds. The compounds I, II, III and IV have chiral centres present in it. The compounds I and II, III and IV are a pair of enantiomers due to presence of chiral centres. Hence, value of N will be 6 .

The enantiomers have identical boiling points. The method of fractional distillation cannot be used for separating enantiomers.

Hence, only four compounds will be separated which are shown below. The compounds I and II, III and IV will remain as racemic mixture as fractional distillation will unable to separate each enantiomer.

Hence, value of $M$ will be 4 .