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

Predict the correct intermediate and product in the following reaction.

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

                                                    (A)                   (B)

The correct order of acidity among the following is:

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:

X and Y in the above-mentioned reaction are respectively:
1. X = 2–Butyne; Y = 3–Hexyne
2. X = 2-Butyne; Y = 2-Hexyne
3. X = 1-Butyne; Y = 2-Hexyne
4. X = 1-Butyne; Y = 3–Hexyne