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?          

(a) 2-hydroxy propane

(b) Ortho-nitro phenol

(c) Phenol

(d) 2-methyl-2-hydroxy propane

The product Z in the below mentioned reaction is                      

1. Toluene

2. Benzaldehyde

3. Benzoic acid

4. Benzene

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

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

Benzaldehyde and acetaldehyde can be distinguished by:

1. Iodoform test

2. 2:4 DNP test

3. ${\mathrm{NH}}_3$ reaction

4. Wolff-Kishner's reduction

Acetaldehyde 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) ${\mathrm{CH}}_3\mathrm{CHO}$

What is present in acetone's enolic form?

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