To achieve above conversion, the reagents used will be:

(a) O₃/H₂O₂, HO^-/$∆$                               

(b) HBr, alc. KOH, O₃, LiAlH₄, H⁺/$∆$

(c)  HBr, t-BuOK, O₃, KMnO₄, $∆$

(d)  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)

Which alkyne(s) will yield 3-ethylhexane upon catalytic hydrogenation?
 

1.		2.
3.		4.	All of the above

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

2.  
3. 
4. Br-(CH₂)₁₂-CH=CH  

Ph - C $\equiv \mathrm{CH}\underset{\mathrm{MeOH}}{\overset{{\mathrm{MeO}}^{-}}{\to }}$ Major product of the reaction is:

(a) 

(b) 

(c) Ph - C $\equiv$C-OMe

(d) 

To carry out the above conversion, (A) and (B) respectively are:

1. NaNH₂, Cl - CH₂ーCH₂ - CH₂ - Br

2. NaNH₂, CH₂-CH₂-CH₂-Br

3. NaNH₂, l-CH₂ーCH₂-CH₂-Br

4. NaNH₂, I-CH₂-CH₂-CH₂-I

H-C$\equiv$C-Ph   Product; Product obtained in this reaction is:

1.		2.
3.	\(\mathrm{Ph}-\mathrm{C} \equiv \mathrm{C}-\mathrm{I}\)	4.	\(\mathrm{I}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}\)