GOC & Aromaticity - 1


20 Questions MCQ Test GATE Chemistry Mock Test Series | GOC & Aromaticity - 1


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This mock test of GOC & Aromaticity - 1 for Chemistry helps you for every Chemistry entrance exam. This contains 20 Multiple Choice Questions for Chemistry GOC & Aromaticity - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this GOC & Aromaticity - 1 quiz give you a good mix of easy questions and tough questions. Chemistry students definitely take this GOC & Aromaticity - 1 exercise for a better result in the exam. You can find other GOC & Aromaticity - 1 extra questions, long questions & short questions for Chemistry on EduRev as well by searching above.
QUESTION: 1

Which is more nucleophilic than the other of the following compounds?
P. (X) R2N-    (Y) R-O   (Z) F-
Q. (X) R-S-    (Y) Cl-
R. (X) R-S-    (Y) R-O-
S. (X) R-OH  (Y) R-SH
T. (X) OH-     (Y) -O-O-H
U. (X) NH3    (Y) H2N-NH2

Solution:

Nucleophilicity increases with the increase in the size of atom in the same period and decreases with the increase in electronegativity.

QUESTION: 2

The correct order of the pKa values for the conjugate acids of heterocyclic compounds given below is:


Solution:

Basicity: Electron pair donating ability

Delocalised lone pair present on nitrogen which has the lowest basicity.

Pyridine has poor basic character. In pyridine, lone pair present on the nitrogen atom are perpendicular
to the orthogonal orbital, containing sp2 hybrid orbital.

 (Aziridine having more s% character than pyrrolidine. so, pyrrolidine is more basic than Aziridine).

So, the correct order of pka value.

III>II>IV>I

QUESTION: 3

Pick out the incorrect statements:

Solution:
  • Resonance involves the displacement of lone pair and double bonds of the molecules without changing the position of the atoms.
  • Tautomerization involves the change in position of lone pair and double bond with atom.
QUESTION: 4

The correct order of pKa for the following compounds is:

 

Solution:

  • Electron withdrawing group increases the acidic character.
  • Electron donating group decreases the acidic behaviour.
  • Higher the Ka, lower will be pka, higher will be acidity ⇒ acidic behaviour 
  • Nitro Benzoic acid is more acidic as compared to acetic acid, which in turn is more acidic as compared to phenol.
  • Thus the order of acidic behaviour will be:
    III > IV > I > II
    Hence, the correct order of pkwill be:
    III < IV < I < II
QUESTION: 5

Consider the following reaction sequence:

Product (P) and (Q) are made to react with sodium amide to yield R and S respectively. Identify the correct statement.

Solution:

QUESTION: 6

Consider the bonds marked as (A), (B), (C) and (D) in the following structures.

What is the correct order of their homolytic bond dissociation energies?

Solution:
  • If the intermediate free radical which is formed via homolytic cleavage is more stable, the bond breakage is more feasible, and thus bond dissociation energy will be less.
  • Since (C) forms, stabilized conjugated free radical, its homolytic bond dissociation energy will be least. 
  • Followed by the same approach, the order will be  A > B > D > C. Hence, the correct option is (c).
QUESTION: 7

The correct order of dipole moment in pyrrole, thiophene and furan respectively, is:

Solution:

There are three major factors contributing to the overall dipole moments:

(i) The σ bond dipole associated with C-0 and C—N bonds.

(ii) The π bond dipole associated with the delocalisation of π electrons from the heteroatom to the ring.

(iii) The dipole moment associated with the unshared electron pair (for 0) or N—H bond (for N).

  • All these factors have a greater moment towards rather than away from the heteroatom for furan than for pyrrole.
  • For pyrrole, the C-N π dipole should be larger and the N—H moment in the opposite direction from furan. These two factors account for the reversal in the direction of overall dipole moment.
  • In furan and thiophene, the dipole moment is influenced by resonance in the same way as in the pyrrole. However, the effect of resonance is not sufficient enough to exceed the effect of heteroatom electronegativity. Thus, the direction of dipole moment will be: 

QUESTION: 8

The rotational energy barrier along 1-2 bond will be the lowest in:

Solution:
  • As we know that rotation around double bond requires breakage of laterally overlapped bond. Hence, it is not energetically favourable to rotate around double bond as compared to single bond.
  • In structure given in option (b) and (d). There is considerable double bond character around 1-2 bond due to delocalisation.
  • (a) will remain in energetically favourable uncharged form, because there no driving force like aromaticity to push it in charged form.

Lone pair not available for delocalization into benzene ring.

QUESTION: 9

The compound with lowest heat of combustion will be:

Solution:

More stable olefin will have lesser heat of hydrogenation and lesser heat of combustion.

*Answer can only contain numeric values
QUESTION: 10

Consider the following reactions,

A boy is said to choose one entity from product P, Q, R and S. What is the probability that the chosen compound will be antiaromatic? (Answer should be in decimals)


Solution:

It is found that among P, Q, R and S, P and S are antiaromatic i.e.,

QUESTION: 11

Given compounds A,B,C and D are subjected to IR spectrum analysis. compounds which will show minimum and maximum vco streching frequency in IR respectively will be:

Solution:
  • vco, stretching frequency ∝ strength of bond. Thus, the species, in which there is more delocalisation, C-0 bond have considerable single bond character and thus have a lower value of stretching frequency.
  • Hence, among A, B, C and D, B will have a maximum value of vca, Since, its — C=0 bond is intact.
  • On the other hand, C is an aromatic compound thus, it must have charge separation, which leads to &localisation. Which in turn makes C = 0 bond, considerable single bond character.
  • Thus, C will have a minimum value stretching frequency. 
QUESTION: 12

If polarity of solvent is sufficiently increased, then what will happen?

Solution:
  • Rate of sololysis in aqueous ethanol (weak nucleophile) will proceed preferentially by SN 1 mechanism. since, rate of SN 1 reaction ∝ stability of carbocation.
  • Here, stability order of carbocation will be in the order of :

QUESTION: 13

Consider the following sequence of reaction,

B on reaction with Na in presence of FeCl2 gives a well known organometallic compound C. Identify the correct statement about above reaction sequence.

Solution:

From the above reaction sequence, it is clear that A and B not an aromatic compound, whicle C which is ferrocene is a well know aromatic compound

QUESTION: 14

Among the pairs given below, identify the most acidic compound.

Solution:

Generally, cis acid are more acidic than trans acid, if group are small.

 

(Y) due to the presence of methyl group ortho to carboxylate ion. which are perpendicular to phenyl ring. So. carboxylate group not in conjugate with phenyl ring. Hence. carboxylate negative ion having more conjugation. So. (Y) is more acidic in nature 

Presence of electron releasing group, decreases the stability of negative charged conjugate base. So, less acidic

*Answer can only contain numeric values
QUESTION: 15

The number of anti-aromatic systems present among the following compounds is/are _____________(Answer should be an integer)


Solution:

B, E, F, G and I are anti-aromatic in nature

*Answer can only contain numeric values
QUESTION: 16

Among the following, how many are aromatic in nature? (Answer should be an integer)


Solution:

B, D, H aromatic in nature

*Answer can only contain numeric values
QUESTION: 17

The number of aromatic compound among the following will be: (Answer should be an integer)



Solution:

No such driving force for delocalization present in (d) and (e)

QUESTION: 18

What is the sequence of reagents that will accomplish the synthesis of the following aromatic amine from benzene?

Solution:

Electrophilic Aromatic Substitution of the methyl group with lewis acid leads to toluene. Through nitration, the nitro group is added then the reduction of nitro group leads to the amine.  

*Answer can only contain numeric values
QUESTION: 19

In the following compounds, how many pairs of compounds having correct order w.r.t mentioned properties? (Answer should be an integer)


Solution:

*Answer can only contain numeric values
QUESTION: 20

In the following compounds, how many compounds having more pKa value relative to the H2O? (Answer should be an integer)


Solution:

H2O = pKa value 15.74 (more pka value than H20 means less acidic than H20)

Compounds C,E,F,G having more pKa value than H20

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