Basics of Organic Chemistry

and are not benzenoid compounds, since benzenoid compound contains benzene ring.

(a) (Acenaphthene)


$\to$ 10 $\pi$e^$-$ in cyclic conjugation

$\Rightarrow$ Aromatic

"Note: For a compound with multiple cyclic rings to exhibit aromaticity, it must conform to Hückel's rule, which stipulates that the molecule should have (4n+2) π-electrons in the largest periphery of continuous conjugation. Rings with sp³ hybridized carbon atoms are not considered in this context. Despite their potential π-bonds, these rings cannot participate in resonance with π-bonds of the other rings due to the sp³ hybridization, and thus do not contribute to the overall aromaticity of the compound."

(b)
$\to$ 4 $\pi$e^$-$ in ring conjugation $\Rightarrow$ Anti Aromatic

(c)
$\Rightarrow$ 4 $\pi$e^$-$ in ring conjugation $\Rightarrow$ Antiaromatic

(d)
6$\pi$e^$-$ in ring conjugation $\Rightarrow$ Aromatic

The correct basic strength order is

(IV) > (II) > (I) > (III);

(IV) is strongest base due to steric inhibition to resonance effect.

(III) is weakest base due to $-$M group of three nitro groups present at ortho and para positions.

(II) is stronger than (I) since (III) is tertiary and (I) primary aromatic amine.

So, option (a) is incorrect.

(b) pK_b different between I and II is 0.53 and that of III and IV is 4.6. So, option (b) is incorrect.

(c) and (d) In 2, 4, 6-trinitro aniline (III) due to strong $-$R effect of $-$NO₂ groups, the lone pair of $-$NH₂ is more involved with benzene ring hence it has least basic strength. Whereas (IV) N, N-dimethyl 2, 4, 6-trinitro aniline, due to steric inhibition to resonance (SIR) effect; the lone pair of nitrogen is not in the plane of benzene, hence makes it lone pair more free to protonate.

Triphenylmethane (I) is acidic because its conjugate base is stabilised by resonance.

${(Ph)_3}CH\buildrel {} \over \longrightarrow \mathop {{{(Ph)}_3}C}\limits_{Triphenylmethyl\,carbanion} + {H^ + }$

Cyclopentadiene (IV) is acidic because its conjugate base is aromatic.



Nitrobenzene is more acidic than benzene because nitro group is electron withdrawing. It will stabilise the conjugate base of benzene by $-$R and $-$I effect.

The acidic strength order on the basis of pK_a data is

IV > V > I > II > III.

Hence, the correct options are (a), (b) and (c) only.

(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)} $

Bromination of alkenes always proceeds via trans addition.



' $O$ ' and ' $P$ ' are enantiomers.

$(M$ and $O)$ and ( $N$ and $P)$ are two pairs of diastereomers.

The IUPAC name of the compound:


1. When two or more substituents such as alkane, halogen, nitro groups, etc., are attached to the aromatic ring (e.g., benzene), these substituents are named in the alphabetical order with substituents of lowest alphabet named first followed by the other.

2. These substituents are numbered in a way with lowest number given to substituent whose name starts with the alphabet that appear first in the series. The direction of numbering is chosen such that next substituent gets the lowest number.

3. After the numbering is complete, the substituents are numbered in alphabetical order.

Hence, the IUPAC name of the compound is 1-Chloro-4-methylbenzene.

4. If a substituent attached to benzene ring together has a common name that has been accepted as an IUPAC name, the compound is called by that common name with other group attached to it as its substituent.

Hence, methyl attached to benzene is commonly called toluene and chlorine is substituent at the fourth position.


The compound is named as 4-chlorotoluene.

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.