The alkene among the following that has the smallest heat of hydrogenation is-

1. ${\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_3$

2. ${\left({\mathrm{CH}}_3\right)}_2\mathrm{C}=\mathrm{CH}-{\mathrm{CH}}_3$

3. ${\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_2-{\mathrm{CH}}_3$

4. ${\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}{\left({\mathrm{CH}}_3\right)}_2$

The correct order of relative rates of hydrogenation of alkenes is:

1. Ethylene > propene > 2-butene > 2-methyl-2-butene

2. 2-methyl-2-butene > 2-butene > Propene > Ethylene

3. 2-butene > propene > ethylene > 2-methyl-2-butene

4. Propene > 2-butene > ethylene > 2-methyl-2-butene

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

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

$\stackrel{\stackrel{\mathrm{H}}{|}}{\mathrm{C}}{\mathrm{H}}_2-\stackrel{\stackrel{\mathrm{Cl}}{|}}{\mathrm{C}}{\mathrm{H}}_2\underset{\mathrm{alc}.}{\overset{\mathrm{NaOH}}{\to }}{\mathrm{CH}}_2={\mathrm{CH}}_2$

The most probable mechanism for the above reaction is-

1. E1                                        

2. E2

3. $E$ $1_{CB}$                                    

4. $\mathrm{\alpha -}$elimination

The main product of the following reaction will be-

1.		2.
3.		4.

3-Phenylpropene on reaction with HBr gives (as a major product)-

1. $C_6{H}_{5}CH\left(Br\right)C{H}_2-C{H}_3$

2. ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{CH}\left(\mathrm{Br}\right){\mathrm{CH}}_3$

3. ${\mathrm{C}}_6{\mathrm{H}}_4\left(\mathrm{Br}\right)<mspace\; width="1em"></mspace>{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_3$

4. ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Br}$

Anti-Markovnikov’s addition of HBr is not observed in-

The reaction of propene with HOCl proceeds via the addition of: