Hydrocarbon (A) reacts with bromine by substitution reaction to form an alkyl bromide B. B undergoes the Wurtz reaction to give a gaseous hydrocarbon containing less than four carbon atoms.

The formula of (A) is-

1. $\mathrm{CH}\equiv \mathrm{CH}$
2. ${\mathrm{CH}}_2={\mathrm{CH}}_2$ 
3. ${\mathrm{CH}}_3-{\mathrm{CH}}_3$
4. ${\mathrm{CH}}_4$

The compound ${\mathrm{C}}_7{\mathrm{H}}_8$ undergoes the following reactions

${\mathrm{C}}_7{\mathrm{H}}_8\stackrel{3{\mathrm{Cl}}_2/\mathrm{hv}}{\to }\mathrm{A}\stackrel{{\mathrm{Br}}_2/\mathrm{Fe}}{\to }\mathrm{B}\stackrel{\mathrm{Zn}/\mathrm{HCl}}{\to }$ C

The product 'C' is-

1. m–Bromotoluene

2. o–Bromotoluene

3. 3–Bromo–2,4,6–trichlorotoluene

4. p–Bromotoluene

P, Q, and R in the above-mentioned sequence of reactions are respectively -

1. 

 

2. 

 

3. 

 

4. 

Predict the correct intermediate and product in the following reaction.

H₃C-C$\equiv$CH $\underset{{\mathrm{HgSO}}_4}{\overset{{\mathrm{H}}_2\mathrm{O}, {\mathrm{H}}_2{\mathrm{SO}}_4}{\to }}$ Intermediate $\to$Product

                                                    (A)                   (B)

The correct order of acidity is-

1. \(CH \equiv CH > CH_3 - C \equiv CH \\ > CH_2 = CH_2 > CH_3 - CH_3\)
2. \(CH \equiv CH > CH_2 = CH_2 \\ > CH_3 - C \equiv CH > CH_3 - CH_3\)
3. \(CH_3 - CH_3 > CH_2 = CH_2 \\ > CH_3 - C \equiv CH > CH \equiv CH \)
4. \( CH_2 = CH_2 > CH_3 - CH_3 \\ > CH_3 - C \equiv CH > CH \equiv CH\)

In the given reaction 
 
The product P is

1. 

2.  

3.  

4.  

The compound that will react most readily with gaseous bromine has the formula-
1.  ${\mathrm{C}}_3{\mathrm{H}}_6$
2.  ${\mathrm{C}}_2{\mathrm{H}}_2$
3.  ${\mathrm{C}}_4{\mathrm{H}}_{10}$
4.  ${\mathrm{C}}_2{\mathrm{H}}_4$