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

Compound (C) in above sequence of reaction is:

1.    

2. 

3.  

4. 

$\underset{{\mathrm{R}}_2{\mathrm{O}}_2/\mathrm{hv}}{\overset{\mathrm{HBr}}{\to }}$(A);   

Major product (A) is:

(a) 

(b) 

(c) 

(d) 

What type of product will be formed in the following reaction?


1. A mixture of diastereomers  
2. Optically active
3. A pure enantiomer with 100% optical purity
4. A racemic mixture

 $\underset{∆}{\overset{{\mathrm{H}}^{+}}{\to }}$Possible products $\stackrel{{\mathrm{Br}}_2/{\mathrm{CCl}}_4}{\to }$(y) products

The number of possible products for x and y is:

(a) 2,4

(b) 3,5

(c) 3,6

(d) 3,4

Product (C) of the above reaction is :

1. 1,3-Hexadiene
2. 1,4-Pentadiene
3. 1,3-Butadiene
4. 1,3-Heptadiene

Which reagents are suitable to accomplish the mentioned conversion?

1. O₃/H₂O₂, HO^-/$∆$                               

2. HBr, alc. KOH, O₃, LiAlH₄, H⁺/$∆$

3. HBr, t-BuOK, O₃, KMnO₄, $∆$

4. HBr, alc. KOH, O₃, KMnO₄, $∆$

$\underset{0°\mathrm{C}}{\overset{{\mathrm{PCl}}_5}{\to }}\underset{70-80\%}{\left(\mathit{A}\right)}\underset{\left(2\right) {\mathrm{H}}^{\oplus }}{\overset{\left(1\right) 3{\mathrm{NaNH}}_2, \mathrm{mineral} \mathrm{oil}}{\to }}\underset{46\%}{\left(B\right)}$; Product (B) is:

Product (K) of the above reaction is:

1. 

2. 
3. 

4. 

CH₃-CH₂-CH₂-C$\equiv$CH+LiHN₂$\to \left(\mathrm{A}\right)\stackrel{({\mathrm{CH}}_3{)}_2{\mathrm{SO}}_4}{\to }$(B)

Give the structural formula of compound (B):

(a) CH₃-(CH₂)₂-C$\equiv$C-SO₃H

(b) CH₃-(CH₂)₂-C$\equiv$C-CH₃

(c)  CH₃-(CH₂)₂-C$\equiv$C-CH₂-O-$\begin{array}{c}\stackrel{\mathrm{O}}{\parallel }\\ \mathrm{s}\\ \underset{\mathrm{O}}{\parallel }\end{array}$-H

(d) CH₃-CH₂-C$\equiv$C-CH₂

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.	(i) KOH, (ii)