The compound on hydrolysis of 50% aqueous sodium hydroxide that produces the corresponding alcohol and acid is:

1. $C_6{H}_{5}C{H}_{2}CHO$

2. $C_6{H}_{5}CHO$

3. $C{H}_{3}C{H}_{2}C{H}_{2}CHO$

4. ${\mathrm{CH}}_3-\stackrel{\mathrm{O}}{\stackrel{\left|\right|}{\mathrm{C}}}-{\mathrm{CH}}_3$

The product in the below mentioned reaction is:

1.
2.
3.	C6H5OC6H5
4.	C2H5OC2H5

$\underset{\left(\mathrm{ii}\right)<mspace\; width="1em"></mspace>{\mathrm{H}}_2\mathrm{O}}{\overset{\left(\mathrm{I}\right)<mspace\; width="1em"></mspace>{\mathrm{CH}}_3\mathrm{MgI}}{\to }}X\stackrel{Cu/575<mspace\; width="1em"></mspace>K}{\to }Y$

The compound Y in the above sequence is
1. 2-Methyl-2-phenyl-1-propanol
2. 2-Phenyl-2-propanol
3. Acetophenone
4. 2-Methyl-1-Phenyl-2-propanol

1-propanol and 2-propanol can be best distinguished
by:

1. Oxidation with alkaline KMnO₄ followed by
reaction with Fehling solution.

2. Oxidation with acidic dichromate followed by
reaction with Fehling solution

3. Oxidation by heating with copper followed by
reaction with Fehling solution.

4. Oxidation with conc. H₂SO₄
followed by reaction with Fehling solution

$\stackrel{{\mathrm{H}}^{+}/∆}{\to }A\left(major\right)$

The product ‘A’ is

R–CH₂–CH₂–OH can be converted into
R–CH₂–CH₂–COOH. The correct sequence of
reagents is

1. PBr₃, KCN, H₃O⁺

2. PBr₃, KCN, H₂

3. KCN, H₃O⁺

4. HCN, PBr₃, H₃O⁺

$\underset{\mathrm{heat}}{\overset{{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }}\underset{major}{P}\stackrel{NBS}{\to }\underset{major}{Q}$

The structure of Q is?

$\underset{\left(Predominant\right)}{\left(\mathrm{B}\right)}\stackrel{PB{r}_{\mathit{3}}}{\leftarrow }$$\stackrel{\mathrm{HBr}}{\to }\underset{\left(Predominant\right)}{\left(A\right)}$

(A) and (B) are respectively :

Which of the following is the strongest acid?

1. HC $\equiv$ CH

2. C₆H₆

3. C₂H₆

4. CH₃OH