Benzene reacts with CH₃Cl in the presence of anhydrous AlCl₃ to form.

1. toluene

2. chlorobenzene

3. benzylchloride

4. xylene

In the following reaction, What will be the major product?

1. 

2.  

3. 

4. 

The compound X in the above reaction is:

1.		2.
3.		4.

1. ${\mathrm{Cl}}^{\oplus }$

2. Cl^-

3. Cl

4. Cl²⁺

 Product (A) in this reaction is:

1. 

2. 

3. 

4. 

$\underset{\left(ii\right) C{H}_{3}Br}{\overset{\left(i\right) NaN{H}_2.N{H}_2}{\to }} \left(A\right) \underset{Lindiar catalyst}{\overset{H_2}{\to }} \left(B\right) ; Product \left(B\right) is :$

1.

2.

3.

4. 

$\mathrm{Br}-{\left({\mathrm{CH}}_2\right)}_{12}-\mathrm{C}\equiv \mathrm{CH} \stackrel{{\mathrm{NaNH}}_2}{\to } \left(\mathrm{A}\right) \underset{\mathrm{Catalyst}}{\overset{\mathrm{Lindlar}}{\to }} \left(\mathrm{B}\right)$; Product (B) is :
1.

2.  
3. 
4. Br-(CH₂)₁₂-CH=CH  

The reagents used for the following conversion is/are:

1.	(i) NaNH2 , (ii) CH3CHO
2.	(i) NaNH2 , (ii) CH3-CH2-CH2-Br
3.	(i) KOH, (ii) CH3-CH2-CH2-Br
4.

$\left(\mathrm{A}\right)\stackrel{Na/N{H}_3\left(l\right)}{\to }\left(B\right)\stackrel{MCBPA}{\to }\left(C\right)$

Compound (C) in above sequence of reaction is:

1.    

2. 

3.  

4. 

The major product (A) of the above reaction is:   

1.		2.
3.		4.