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.	\(\mathrm{I}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}\)

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

1.

2.

3.

4. 

A compound that gives an optically inactive compound on treatment with H₂(2 moles)/ Pt is-

1. 3-Methyl-1-pentyne

2. 4-Methyl-1-hexyne

3. 3-Methyl-1-heptyne

4. None of the above

Compound (x) will be:

(a) 

(b) 

(c) 

(d) 

Which alkyne yields butanoic acid (CH₃CH₂CH₂CO₂H) as the only organic product on treatment with ozone followed by the hydrolysis?

1. 1-Butyne             

2. 4-Octyne               

3. 1-Pentyne               

4. 2-Hexyne

The degree of unsaturation in compound (A) is -

(1) 7                       

(2) 8                       

(3) 9                             

(4) 10

For the given compounds, the correct order of rate of EASR (electrophilic aromatic substitution reaction) is:

1.		2.
3.		4.

1. c > b > a > d

2. c > d > a > b

3. a > b > c > d

4. c> d> b > a

The major product formed in the reaction is:

1. 
2. 
3. 
4.