The major product obtained in the given reaction is:

1.		2.
3.		4.

$\mathrm{Benzene}+{\mathrm{CH}}_3\mathrm{Cl}\stackrel{{\mathrm{AlCl}}_3}{\to } {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_3$

The mechanism & intermediate involved in the above reaction are:

1. Aromatic electrophilic substitution & carbocation

2. Aromatic Nucleophilic substitution & carbanion

3. Aromatic free radical substitution & Free radical

4. Carbene based substitution reaction & Carbene

A compound among the following that does not undergo Friedel-Craft's reaction easily is-

1. Cumene               

2.  Xylene

3. Nitrobenzene         

4. Toluene

 The compounds among the following that can be generated by Friedel craft acylation is/are -

I.
II.
III.
IV.

Cyclopentadiene is much more acidic than cyclopentane, beacuse -

1. Cyclopentadiene has conjugated double bonds.

2. Cyclopentadiene has both sp² and sp³ hybridized carbon atoms.

3. Cyclopentadiene is a strain-free cyclic system.

4. Cyclopentadienyl anion ion, the conjugate base of cyclopentadiene, is an aromatic species and hence has higher stability.

A compound among the following that gives the maximum % of meta product on nitration

(using HNO₃/H₂SO₄) is :

1. Toluene

2. Aniline

3. Benzene

4. Isopropyl benzene

The major product in the following reaction is-

1.		2.
3.		4.

The D in the above-mentioned reaction is -

1.		2.
3.		4.

     
The product B in the above mentioned reaction is -

1.		2.
3.		4.

Consider the following reaction:-

A and B are isomeric compounds. Additionally, A is a monochloro derivative. These two can be differentiated easily by using-

1. AgN$O_3$(aq)

2. ${\mathrm{Br}}_2/{\mathrm{FeCl}}_3$

3. Conc.HN${\mathrm{O}}_3$/$\mathrm{Conc}.{\mathrm{H}}_2{\mathrm{SO}}_4$

4. NaCl(aq)