A compound A of formula C₃H₆Cl₂ on reaction with alkali can give B of formula C₃H₆O or C of formula C₃H₄. B on oxidation gave a compound of the formula C₃H₇O₂. C with dilute H₂SO₄ containing Hg²⁺ ion gave D of formula C₃H₆O, which with bromine and NaOH gave the sodium salt of C₂H₄O₂. Then A is :

1. CH₃CH₂CHCl₂

2. CH₃CCl₂CH₃

3. CH₂ClCH₂CH₂Cl

4. CH₃CHClCH₂Cl

Identify the product (A) in following reaction series, ${\mathrm{CH}}_3\mathrm{CN}\stackrel{\mathrm{Na}/{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}{\to }$(X)$\stackrel{{\mathrm{HNO}}_2}{\to }$(Y)$\stackrel{\left[\mathrm{O}\right]}{\to }$(Z)$\stackrel{\mathrm{Tollens}<mspace\; width="1em"></mspace>\mathrm{reagent}}{\to }$(A) :

1. CH₃CHO

2. CH₃CONH₂

3. CH₃COOH

4. CH₃ - CH₂-NHOH

Compound (A), C₈H₉Br , gives a white precipitate when warmed with alcholic AgNO₃. Oxidation of (A) gives an acid (B), C₈H₆O₄. (B) easily forms anhydride on heating. Identify the compound (A).

1.
2.
3.
4.

Which of the following will give yellow precipitate with I₂/NaOH?

1.	CH3-CO-O-CO-CH3	2.
3.		4.	Both (2) and (3)

A nitrogeneous substance X is treated with HNO₂ and the product so formed is further treated with NaOH solution, which produces blue colouration. X can be

1. CH₃CH₂NH₂

2 CH₃CH₂NO₂

3. CH₃CH₂ONO

4 (CH₃)₂CHNO₂

The compound which reacts fastest with Lucas reagent at room temperature is:
1. Butan-1-ol
2. Butan-2-ol
3. 2-methylpropan-1-ol
4. 2-methylpropan-2-ol

Identify (X) in the sequence :

${\mathrm{C}}_3{\mathrm{H}}_8\mathrm{O}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\left(\mathrm{X}\right)<mspace\; width="1em"></mspace>\underset{{\mathrm{H}}_2{\mathrm{SO}}_4}{\overset{{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7}{\to }}<mspace\; width="1em"></mspace>{\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}<mspace\; width="1em"></mspace>\underset{\mathrm{Warm}}{\overset{{\mathrm{I}}_2<mspace\; width="1em"></mspace>+\hspace{0.17em}\mathrm{NaOH}}{\to }}<mspace\; width="1em"></mspace>{\mathrm{CHI}}_3$

1. CH₃-CH₂-CH₂OH

2. ${\mathrm{CH}}_3-\mathrm{CH}-{\mathrm{CH}}_3$
$<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>|$
$<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\mathrm{OH}$

3. CH₃-O-CH₂-CH₃

4. CH₃-CH₂-CHO

In order to accomplish the following conversion, what reagent and conditions would be required?

1. Cold sodium hydroxide

2. Hot conc. sodium hydroxide

3. Potassium tertiary butoxide and heat

4. Hot water

An amine is reacted with benzene sulphonyl chloride then a solid compound is formed which is insoluble in alkali. The amine is:

1. CH₃-CH₂-NH₂

2. CH₃-NH-CH₂-CH₃

3. (CH₃)₃N

4. C₆H₅-NH₂

Nitrobenzene on electrolytic reduction in a strongly acidic medium gives:

1. Aniline         
2. p-Aminophenol
3. m-Nitroaniline    
4. Nitrosobenzene