The cylindrical shape of an alkyne is due to 

1. three sigma $\mathrm{C}-\mathrm{C}$ bonds

2. three $\mathrm{\pi }$ $\mathrm{C}-\mathrm{C}$ bonds

3. two sigma $\mathrm{C}-\mathrm{C}$ and one $\mathrm{\pi }$ $\mathrm{C}-\mathrm{C}$ bonds

4. one sigma $\mathrm{C}-\mathrm{C}$ and two$\mathrm{\pi }$ $\mathrm{C}-\mathrm{C}$ bonds

Octane no. of a fuel can be increased by:

1. isomerism 

2. alkylation

3. reforming 

4. all of these

The shortest $\mathrm{C}-\mathrm{C}$ bond distance is found in:

1. Diamond 

2. Ethane

3. Benzene

4. Acetylene

Benzene reacts with n-propyl chloride in the presence of anhydrous ${\mathrm{AlCl}}_3$ to give

1. 3-propyl-1-chlorobenzene

2. n-propyl benzene

3. no reaction

4. iso-propyl benzene

The distance between two adjacent cabron atoms is largest in

1. benzene 

2. ethene

3. butane

4. ethyne

In benzene, $\mathrm{C}-\mathrm{C}$ bond length is 1.39 $\stackrel{°}{\mathrm{A}}$; the $\mathrm{C}-\mathrm{H}$ bond length is:

1. 1.39 

2. 1.08

3. 1.54

4. 1.46

Which reaction produces acrylonitrile (${\mathrm{CH}}_2=\mathrm{CHCN}$)?

1. Ethyne $\underset{{\mathrm{Ba}}^{2+}}{\overset{\mathrm{HCN}}{\to }}$                           

2. Acrylic acid $\stackrel{\mathrm{KCN}}{\to }$

3. Ethyne $\stackrel{\mathrm{KCN}}{\to }$                           

4. Ethyne $\stackrel{\mathrm{HOCl}}{\to }$

 is not converted to  by:

1. Wolff-Kishner reaction

2. Clemmensen reduction

3. Red P + HI at $200°\mathrm{C}$

4. Wurtz's reaction

Products of the following reaction

${\mathrm{CH}}_3\mathrm{C}\equiv \mathrm{C}.{\mathrm{CH}}_2{\mathrm{CH}}_3\underset{\left(\mathrm{ii}\right)\hspace{0.17em}\mathrm{Hydrolysis}}{\overset{\left(\mathrm{i}\right) {\mathrm{O}}_3}{\to }}$... are

1. ${\mathrm{CH}}_3\mathrm{CHO}$ + ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CHO}$

2. ${\mathrm{CH}}_3\mathrm{COOH}$ + ${\mathrm{CH}}_3{\mathrm{COCH}}_3$

3. ${\mathrm{CH}}_3\mathrm{COOH}$ + $\mathrm{HOOC}.{\mathrm{CH}}_2{\mathrm{CH}}_3$

4. ${\mathrm{CH}}_3\mathrm{COOH}$ + ${\mathrm{CO}}_2$

Acetylene reacts with 42% ${\mathrm{H}}_2{\mathrm{SO}}_4$ containing 1% ${\mathrm{HgSO}}_4$ to give:

1. ${\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{HSO}}_4$                               

2. ${\mathrm{CH}}_3\mathrm{CHO}$

3. HCHO                                       

4. ${\mathrm{CH}}_2={\mathrm{CH}}_2$