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Test: GOC, Aromaticity, Acidity & Basicity - 1


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20 Questions MCQ Test Chemistry for GRE Paper II | Test: GOC, Aromaticity, Acidity & Basicity - 1

Test: GOC, Aromaticity, Acidity & Basicity - 1 for Chemistry 2022 is part of Chemistry for GRE Paper II preparation. The Test: GOC, Aromaticity, Acidity & Basicity - 1 questions and answers have been prepared according to the Chemistry exam syllabus.The Test: GOC, Aromaticity, Acidity & Basicity - 1 MCQs are made for Chemistry 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: GOC, Aromaticity, Acidity & Basicity - 1 below.
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Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 1

Which of the following is not a resonance structure of the others?

Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 1

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

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 2

The number of sigma and pi-bonds in 1-butene 3-yne are:

Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 2


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.

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 3

Which of the following molecules has all the effects : inductive, mesomeric and Baker Nathan effect ?

Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 3

In CH3​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. 

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 4

Which one is strongest acid among following options?

Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 4

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

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 5

Most stable resonating structure of the given carbocation is: 

Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 5

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

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 6

Which of the following resonance structures contributes the most to the resonance hybrid?

Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 6

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

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 7

Stability order of the following resonating structure will be:




Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 7
  • 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

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 8

Hyperconjugation is best described as:

Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 8

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.

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 9

The correct stability order for the following species is:




Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 9

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.

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 10

Increasing order of pKa values (pKa = –log Ka) of H2O, CH3OH and C6H5OH is:

Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 10

pKa values are -logKa.

Ka is the dissociation constant of the acid.

Smaller the Ka value, weaker is the acid.

Therefore, pka values will go on decreasing from strong acid to weak acid.

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 11

Identify order of per ring resonance energies of each of the following:



Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 11

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

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 12

Identify the correct C-O bond length order:



Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 12
  • 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.

 

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 13

Which of the following systems are resonance contributors of the radical shown below?

Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 13

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

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 14

Which of the following statements about the resonance structures is false?

Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 14

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.

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 15

Which comparison is not correct as indicated?

Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 15

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.

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 16

The acidity order for the following compound follows the order:

Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 16

Acidity of a compound depends upon the stability of the conjugate base.Decreasing order of the stability is  aromatic, homoaromatic, non aromatic, anti aromatic.

  • Ist conjugate base is aromatic.
  • 2nd conjugate base is non aromatic.
  • 3rd conjugate base is anti aromatic.
Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 17

In the following compounds the order of acid strength is:

Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 17

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).

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 18

The acidity of protons H in each of the following is:



Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 18

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.

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 19

Identify the correct acidic strength order in the following compounds?



Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 19

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

Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 20

 Identify the correct order of boiling points of the following compounds:

Detailed Solution for Test: GOC, Aromaticity, Acidity & Basicity - 1 - Question 20

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.

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