Q 1 :

The set of meta directing functional groups from the following sets is:                 [2024]

  • —CN,—NH2,—NHR,—OCH3

     

  • —CN,—CHO,—NHCOCH3,—COOR

     

  • —NO2,—NH2,—COOH,—COOR

     

  • —NO2,—CHO,—SO3H,—COR

     

(4)

–M groups decrease electron density at ortho and para position of benzene, thereby directing the incoming electrophile to meta position. Hence –M groups are meta directing groups for electrophilic aromatic substitution reactions.

Groups of the type –X = Y or –X ≡ Y act as –M groups. Among given species, —NO2,—C≡N,—COR,—COOH,—COOR and —SO3H are –M groups, hence are meta directing groups.

 



Q 2 :

Among the following, total number of meta directing functional groups is ________ . (Integer based) 

—OCH3,—NO2,—CN,—CH3,—NHCOCH3,—COR,—OH,—COOH,—Cl             [2024]             



(4)

        Among given species, —NO2,—C≡N,—COR and —COOH are -M groups, hence are meta directing groups.

 



Q 3 :

The arenium ion which is not involved in the bromination of Aniline is ____.                        [2024]

  •  

  •  

  •  

  •  

(3)

-NH2 due to its +M effect is ortho and para directing group for electrophilic substitution reactions such as bromination. Hence, bromine would not attack the meta position (low electron density position).



Q 4 :

The correct arrangement for decreasing order of electrophilic substitution for above compounds is:                  [2024]

  • (III) > (I) > (II) > (IV)

     

  • (IV) > (I) > (II) > (III)

     

  • (III) > (IV) > (II) > (I)

     

  • (II) > (IV) > (III) > (I)

     

(1)

More is electron density on the ring, more is the rate of electrophilic substitution reaction.



Q 5 :

Total number of deactivating groups in aromatic electrophilic substitution reaction among the following is ______.            [2024]



(2)

-M groups decrease electron density on aromatic ring hence act as deactivating groups in electrophilic aromatic substitution reactions. Groups of the type –X=Y and –X≡Y act as –M groups. Among given options –COOCH3 and –C≡N are –M groups, thus acts as deactivating groups.

–NHCOCH3, –NHCH3 and –OCH3 are +M groups hence act as activating groups for electrophilic aromatic substitution reactions. +M groups have a lone pair of electrons on first atom.

 



Q 6 :

Identify correct statement/s                   [2025]

(A) –OCH3 and –NHCOCH3 are activating group.

(B) –CN and –OH are meta directing group.

(C) –CN and –SO3H are meta directing group.

(D) Activating groups act as ortho- and para directing groups.

(E) Halides are activating groups.

Choose the correct answer from the options given below:

  • (A), (B) and (E) only

     

  • (A), (C) and (D) only

     

  • (A) only

     

  • (A) and (C) only

     

(2)

(A) For –OCH3 and –NHCOCH3, +M effect is more than –I effect, hence these are activating groups.

(B)(C) –CN and –SO3H are –M groups. –M groups decrease electron density at ortho and para position hence are meta directing for electrophilic substitution reactions. –OH is a +M group. +M groups increase electron density at ortho and para positions, hence are ortho and para directing for electrophilic substitution reactions.

(D) Activating groups are +M groups. +M are ortho and para directing for electrophilic substitution reactions.

(E) For halides –I > +M, hence these are deactivating groups.

 



Q 7 :

Given below are two statements:                                   [2025]

Statement I: On nitration of m-xylene with HNO3, H2SO4 followed by oxidation, 4-nitrobenzene-1,3-dicarboxylic acid is obtained as the major product.

Statement II: –CH3 group is o/p-directing while –NO2 group is m-directing group.

In the light of the above statements, choose the correct answer from the options given below:

  • Both Statement I and Statement II are false

     

  • Statement I is false but Statement II is true

     

  • Both Statement I and Statement II are true

     

  • Statement I is true but Statement II is false

     

(3)

Statement I:

Statement II: Because of hyperconjugation effect of alkyl groups, electron density is increased at ortho and para position of benzene. Hence incoming electrophiles are directed to ortho and para position. Because of –M effect of nitro group, electron density is decreased at ortho and para position of benzene. Hence incoming electrophiles are directed to meta position.

 



Q 8 :

In the following series of reactions identify the major products A & B respectively.                       [2025]

  •  

  •  

  •  

  •  

(2)

 



Q 9 :

Consider the following compounds.

Arrange these compounds in the increasing order of reactivity with nitrating mixture.             [2026]

  • b < c < a

     

  • b < a < c

     

  • c < b < a

     

  • c < a < b

     

(2)

In Ph-OMe, -OMe is an electron donor group (+M).

Ph-NO2, -NO2 is a strong withdrawing group (-M).

Ph-Cl, -Cl is an electron withdrawing group.



Q 10 :

Given below are two statements:                                           [2026]

Statement I: Benzene is nitrated to give nitrobenzene, which on further treatment with CH3COCl/AlCl3 will give 

Statement II: –NO2 group is a m-directing and deactivating group.

In the light of the above statements, choose the most appropriate answer from the options given below.

  • Statement I is correct but Statement II is incorrect

     

  • Both Statement I and Statement II are incorrect

     

  • Both Statement I and Statement II are correct

     

  • Statement I is incorrect but Statement II is correct

     

(4)

Nitrobenzene does not give Friedel–Craft acylation since it is highly deactivated ring.



Q 11 :

Method used for separation of mixture of products (B and C) obtained in the following reaction is:    [2026]

  • Simple distillation

     

  • Fractional distillation

     

  • Sublimation

     

  • Steam distillation

     

(2)

B & C separate by

Fractional Distillation method

Due to their different boiling point.



Q 12 :

How many groups o/p director in the electrophilic aromatic substitution?



(4)

I, V, Vi, VII are O,P Directing groups while others are meta directing



Q 13 :

Consider the three aromatic molecules (P, Q and R) whose structures have been given below:

[IMAGE 186]

The correct order regarding the reactivity of these compounds with Ph-N(+)≡NCl(-) under optimum but slightly acidic medium is:        [2026]

  • P > Q > R

     

  • R > P > Q

     

  • R > Q > P

     

  • P > R > Q

     

(1)

This is an electrophilic aromatic substitution reaction. More is the electron density on aromatic ring, more is the reactivity towards electrophiles. Presence of bulky groups at ortho position of -NMe2 group pushes -NMe2 out of plane of benzene, thereby reducing its +M effect. This reduces the electron density in ring and hence the reactivity towards electrophilic aromatic substitution.

 



Q 14 :

For the following Friedel Craft's alkylation reaction, which of the statements are correct?

[IMAGE 187]

A. Major product is n-propyl benzene.

B. iso-propyl carbocation intermediate is also generated.

C. Multiple substitution is inevitable.

D. Introducing electron-donating substituent on benzene will not produce any alkyl benzene.

Choose the correct answer from the options given below:                    [2026]

  • A and D only 

     

  • B and C only

     

  • A and C only

     

  • B and D only

     

(2)

[IMAGE 188]

Poly substitution occurs in Friedel Craft reaction as alkyl benzene is more reactive than benzene towards electrophilic aromatic substitution reactions. Higher reactivity of alkyl benzene is because of electron donating +I and hyperconjugation effect of alkyl group.



Q 15 :

Consider the following sequence of reactions to give the major product (X)

[IMAGE 199]

P g of the major product (X) formed is reacted with NaHCO3 solution to liberate a gas which occupied 11.2 dm3 at STP.

P = ______ g.

(Given molar mass in g mol-1 H : 1, C : 12, O : 16, Cl : 35.5)                   [2026]



(78)

[IMAGE 200]

[IMAGE 201]

(M=7×12+5×1+2×16+35.5=156.5 g)

Moles of X=Moles of CO2

=Volume of gas at STP (in dm3)22.4

=11.222.4=0.5 mol

Mass of X(P)=moles of X×molar mass=0.5×156.5 g

                          =78.25 g≃78 g