Test: GOC, Aromaticity, Acidity & Basicity - 1 - Chemistry MCQ

Option 'D' is correct because there in no conjugation of negative charge and without conjugation resonance is not possible.

Single bond contains only a sigma bond, One double bond contains a sigma and a pi bond, whereas a triple bond contains a sigma bond and 2 pi bonds.
So, CH₂ = CH – C ≡ CH
No of σ bonds = 2 + 1 + 1 + 1 +1 + 1 = 7;
No of π bonds = 1 + 2 = 3.

In CH_3​CH=CHC(=O)CH₃​, presence of CH₃​CH= group shows no bond resonance (Baker nathan effect), presence of alternate π bond and electron negative O shows mesomeric effect and CH₃​  positive inductive effect and C=O group shows negative inductive effect. 

CHF₂−COOH. Difluoroacetic acid is strongest because presence of two F atoms increases its acidic nature.

Because in option 'C' all the atoms have their octets fulfilled which is not in the case with the other options.

Resonating structures with all the atoms having complete octate. Therefore, more stable.
It will contribute to the resonance hybrid.

• 1st is uncharged so it is the most stable among the following options.
• 2nd one is charged but has more no of covalent bonds than 3rd and 4th so it is more stable than 3rd and 4th.
• Between 3rd and 4th follow the electronegativity rule which means more electronegative atom bears negative charge. In 4th the positive charge on oxygen atom is against the electronegative prinicple . So it will like to have negative charge rather than positive chargee. So 3rd is more stable tha 4th.

Therefore the order will be I > II > III > IV

Hyperconjugation is the stabilizing interaction that results from the interaction of the electrons in a σ-bond (usually C-H or C-C) with an adjacent empty or partially filled p-orbital or a π-orbital to give an extended molecular orbital that increases the stability of the system.

Carbocation is resonance stabilized when it is adjacent to an oxygen atom.
 
Therefore, 1 and 3 are more stable than 2 and 4 and among 1 and 3 1 is more stable due to +I effect of -CH₃ also.

Stability of hydrocarbons increase in the order of primary,secondary and then tertiary. Therefore 2 is more stable than 4.

As the number of the rings increases, the resonance energy per ring decreases.

• The triple bonds are the strongest and hence the shortest. Then comes double bonds which are of intermediate strength between the triple and single bonds. And finally the single bonds are weaker than the other two. This way, Triple bonds are the shortest. Option (I) is the shortest.
• In III the lone pair over the Oxygen atom is involved in conjugation so it has partial double bond character, means bond length is lesser than single bond but greater than double bond. so bond length of III is greater than II.

Option A is the contributor of the radical as it is the only structure going in delocalization.

Hydrogens must have two electrons and elements in the second row cannot have more than 8 electrons. If so, the resonance structure is not valid. Always look at the placement of arrows to make sure they agree.

CH₃NH₂ is more basic than C₆H₅NH₂ due to +I effect in methylamine, electron density in methylamine at nitrogen increases whereas in aniline resonance takes place and electron density on nitrogen decreases.

The correct option is D
IV > III > I > II
Electron - donation groups decrease while electron withdrawing groups increase the acidity of phenols. Therefore, the overall order of acidity is IV > III > I > II, i.e. option (d) is correct.

Nitro and chlorine act as electro withdrawing group which increases the acidic character in compound where as NH₂ acts as electron donating which decreases the acidic character. Further nitro is strong withdrawing group so first priority is given to para nitro phenol ,afterwards para chloro phenol. So correct answer is (a).

In case of molecule 1 nitrogen captures the negative charge after protonation i.e. delocalization of the -ve charge in aromatic polycyclic ring will be less but in case of molecule 2, P have less electronegativity and hence delocalization of -ve charge can occur more easily within the aromatic rings. and in case of 3 the +I effect of CH₃ disturbs the stability of -ve charge form after protonation.

In 1, the ortho effect is highly effective and hence is most acidic among II AND III . In option II, the tertiary butoxy group increases the electrohillic nature of benezene ring and hence the negative charge formed on phenoxide ion after the removal of proton gets more dispersed in the ring and thus the anion gets stable and becomes more acidic then III.( benzoic acid).
Thus Correct option is A

Carboxylic acids, similar to alcohols, can form hydrogen bonds with each other as well as van der Waals dispersion forces and dipole-dipole interactions. So Max. boiling point. is of carboxylic acid.
The boiling point of aldehydes and ketones are lower than alcohols due to absence of intermolecula hydogen bonding. Alcohol Contains an OH Group Allowing For Hydrogen Bonding Which Is Stronger Than Other Intermolecular Forces.