Test: General Organic Chemistry Level - 3 - Chemistry MCQ

A and D have 4n+2 electrons but are not aromatic as they are non-planar.

Two different molecules are said to be enantiomers if they are non-superimposable mirror images of each other. They have same refractive index, density, boiling and melting points but different direction of rotation in the polarimeter. They interact in different manner with other chiral molecules and are difficult to separate. Enantiomers differ in their configuration such as R or S at the stereogenic centre. We already know that these configurations are assigned by the Cahn-Ingold-Prelog (CIP) rules.
Diastereomers are the non-superimposable non-mirror images but have two or more chiral centres. When four different groups are attached to an asymmetric carbon atom, it is a chiral molecule. They are easy to separate as they have different melting and boiling points. For example, geometric isomers such as cis/trans.
To convert into enantiomers, both chiral carbon atoms must change their configuration. All stereocenters must be different for molecules to be enantiomers. And to convert into diastereomers, only one chiral carbon atom must change configuration. This is an important condition that at least one stereocenter must be different and at least one stereocenter must be same for molecules so as to be diastereomers.
Identical molecules are those two molecules that can be superimposed on each other, through rotation of bonds i.e. conformational changes or via rotation of the molecule itself. These are considered to be the same molecule or identical twins. They differ from each other only in the spatial orientation of groups in the molecule.
Let us write all the given molecules in Fischer projection from sawhorse projections and write the configurations of each chiral carbon as per CIP rules.

From these configurations we can easily see that M and N are diastereomers as they have different configuration at one chiral carbon and same at another chiral centre. therefore, M and N are non-mirror image stereoisomers. So, option (A) is correct.
Let us look at option (B), M and O are identical. Yes, we can clearly see that both the molecules look identical. The position of every atom and configuration of each chiral carbon is same in both. So it is a correct option.
In option (C), it is stated that M and P are enantiomers. We can see that the configurations are opposite at each chiral carbon in the two molecules. This is the case of enantiomers in which both chiral carbon atoms must change their configuration. So, this option is also correct.
In the last option, it is mentioned that M and Q are identical but they are not because the configurations are changed and not same. So, this option is incorrect.
Hence, the correct options are (A), (B) and (C).

Correct Answer :- c

Explanation : Aromatic π-sextets are defined as six π-electrons localized in a single benzene-like ring separated from adjacent rings by formal CC single bonds.Therefore, Aromatic sextet compound does not shows aromaticity.

Corect Answer : a,d

For aromatic-compound should be cyclic, sp² hybridised, planar, (4n+2)πe^- for anti-aromatic-compound should be cyclic, sp² hybridised, planar, (4n)πe^-
Hence compound A and D has anti-aromatic character.

Group which increases electron density on benzene ring help the compound to undergo faster electrophilic aromatic substitution.

Correct Answer :- a,b

Explanation : The term aromaticity is used to describe a cyclic, planar molecule with a ring of resonance bonds that exhibits more stability than other geometric or connective arrangements with the same set of atoms.

Only the peripheral pie electrons are considered! doesn't matter whether that central carbon is sp2 or not.

A,B have all sp2 carbon atoms and follow (4n+2)pie electron rule so are aromatic.

C and D are non-aromatic because C is nonplanar and D is having one carbon without hybridization. 

Correct Answer :- b,c

Explanation : (B) Cyclobutadiene follows Huckel's criteria for anti-aromaticity [(4n)−π electrons],  hence unstable.

(C) Follows Huckel's criteria for anti-aromaticity [(4n)−π electrons], hence, unstable.

Correct Answer :- a

Explanation : In 3rd compound delocalisation of electron is not possible so hybridization = 2sigma + 2loan pair of electron

= sp3.

In 1st,2nd,4th compound one loan pair of oxygen can delocalise so these electron are count as pi electron there for hybridization of oxygen = 2sigma + 1loan pair of electron

= sp2

he reaction depicted is a nucleophilic addition-elimination reaction between a cyclic ketone and methylamine (CH₃NH₂) at pH 4.5. Given the mildly acidic conditions (pH 4.5), the reaction will likely proceed through the formation of an imine.

Here's the step-by-step mechanism:

1. The lone pair of electrons on the nitrogen of methylamine attacks the carbonyl carbon of the ketone, forming a tetrahedral intermediate.
2. Protonation of the oxygen occurs due to the acidic conditions, making it a better leaving group.
3. The intermediate collapses, releasing a water molecule and forming a double bond between the nitrogen and the carbon, thus creating an imine.

The product of this reaction would be a cyclic imine, where the oxygen of the ketone has been replaced by a double-bonded nitrogen from the methylamine, creating a C=N bond.

The correct option is the one that shows a cyclic structure with a double bond between a nitrogen and a carbon atom, where the nitrogen has a methyl group attached to it (CH₃). This corresponds to Option B, as it shows the correct cyclic imine structure resulting from the reaction of cyclohexanone with methylamine under mildly acidic conditions.

(1) CH₃CH₂CH₂CHO - 2.720 Debye
(2) CH₃CH = CHCHO - 3.720 Debye
(3) CH₃CH₂CH = CH₂  - 0.359 Debye

Correct Answer :- c

Explanation : According to Hucckel’s rule

We take pi bonds only on the periphery and do not count one not on boundaries or periphery.

4n + 2 = 14

4n = 12

n = 3

Pyrrolidine is strongest base, thus having the weakest conjugate acid so, option B correct.

Correct Answer :- b

Explanation : In R negative charge delocalised into two rings make them aromatic which are further stabilized by 3 benzene rings. while in P and Q negative charge delocalised to one ring make it aromatic which are further stabilized by two benzene rings. but Q have extra resonance in compare to P

Correct answer is. Q and S.

By considering Huckel’s Rule P and S are correct.

Hence D is the correct answer.

H₂C=CH−C≡N
H₂C⇒sp²
CH⇒sp²
C⇒sp
N⇒sp
Hence (a) is the correct answer.

Option c is the right answer.

Here 3rd compound is more stable than 2nd due to hyperconjugation effect here 9 alpha hydrogens are present. but in 2nd it has only resonance effect no alpha hydrogen is there. in 1st element it has only small resonance effect

Correct Answer :- a

Explanation : Option A is correct because more opposite charges are closer, more is the stability of the azulene is.