Benzoic acid gives benzene on being heated with X and phenol gives benzene on being heated with Y. Therefore, X and Y are respectively                                                              

1. Sodalime and copper

2. Zn dust and NaOH

3. Cu and sodalime

4. Sodalime and zinc dust

Which one of the following can be oxidised to the corresponding carbonyl compound?          

1. 2-hydroxy propane

2. Ortho-nitro phenol

3. Phenol

4. 2-methyl-2-hydroxy propane

The -OH group of an alcohol or the -COOH group of a carboxylic acid can be replaced by -Cl using     

1. phosphorus pentachloride

2. hypochlorous acid

3. chlorine

4. hydrochloric acid

Acetone reacts with iodine (I₂) to form iodoform in the presence of      

1. CaCO₃   

2. NaOH     

3. KOH     

4. MgCO₃

A and B in the following reactions are  

1. A=RR'CH₂CN,              B=NaOH

2. A=RR'C${<}_{\mathrm{COOH}}^{\mathrm{OH}}$ ,          B=CH₃

3. A=RR'C${<}_{\mathrm{COOH}}^{\mathrm{CN}}$ ,          B=CH₃

4. A=RR"C${<}_{\mathrm{OH}}^{\mathrm{CN}}$,              B=LiAlH₄

Acetaldehyde normally reacts with                

1. only electrophiles

2. only nucleophiles

3. only free radicals

4. both electrophiles and nucleophiles

Nucleophilic addition reaction will be most favoured in                   [2006]

1. $$
$\mathrm{O}$
$\parallel$
${\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2\mathrm{C}-{\mathrm{CH}}_3$

2. $({\mathrm{CH}}_3{)}_2\mathrm{C}=\mathrm{O}$

3. ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CHO}$

4. ${\mathrm{CH}}_3\mathrm{CHO}$

The enolic form of acetone contains:

1. 9 $\mathrm{\sigma }$-bonds, 1$\mathrm{\pi }$-bond and 2 lone pairs

2. 8 $\mathrm{\sigma }$-bonds, 2$\mathrm{\pi }$-bonds and 2 lone pairs

3. 10 $\mathrm{\sigma }$-bonds, 1$\mathrm{\pi }$-bond and 1 lone pair

4. 9 $\mathrm{\sigma }$-bonds, 2$\mathrm{\pi }$-bonds and 1 lone pair

In this reaction,

CH₃CHO + HCN $\to$CH₃CH(OH)CN $\stackrel{\mathrm{H}.\mathrm{OH}}{\to }$CH₃CH(OH)COOH

an asymmetric compound is generated. The acid obtained would be            

1. 50% D + 50% L-isomer

2. 20% D + 80% L-isomer

3. D-isomer

4. L-isomer

Formaldehyde can be distinguished from acetaldehyde by:

1. Fehling's solution

2. Schiff's reagent

3. Ammonia

4. Ammoniacal ${\mathrm{AgNO}}_3$