Question

(a) Give chemical tests to distinguish between the following pairs of compounds:

(i) Acetophenone and Benzophenone

(ii) Propanal and Propanone

(iii) Pentan-2-one and Pentan-3-one

(b) (i) Which of the following acids is stronger and why?

\( \mathrm{CH_2FCOOH} \), \( \mathrm{CH_3COOH} \)

(ii) Arrange the following compounds in the increasing order of their boiling points:

\( \mathrm{CH_3OCH_3} \), \( \mathrm{CH_3CH_2CH_3} \), \( \mathrm{CH_3CH_2OH} \), \( \mathrm{CH_3CHO} \)

Hint:

Use the iodoform test for methyl ketones and Tollens' or Fehling's test for aldehydes. The -I effect of F makes CH2FCOOH stronger. Hydrogen bonding makes ethanol boil highest.

Answer 1

Step 1: Acetophenone and benzophenone (a-i):
Use the iodoform test. Warm each compound with iodine and sodium hydroxide solution.
Acetophenone, \( \mathrm{C_6H_5COCH_3} \), has a \( \mathrm{CH_3CO-} \) group. It gives a yellow precipitate of iodoform, \( \mathrm{CHI_3} \).
Benzophenone, \( \mathrm{C_6H_5COC_6H_5} \), has no \( \mathrm{CH_3CO-} \) group, so it gives no yellow precipitate.

Step 2: Propanal and propanone (a-ii):
Use Tollens' test. Warm each with ammoniacal silver nitrate.
Propanal is an aldehyde, so it reduces \( \mathrm{Ag^+} \) to silver and gives a silver mirror. Propanone, a ketone, gives no mirror.

Step 3: Pentan-2-one and pentan-3-one (a-iii):
Use the iodoform test again.
Pentan-2-one, \( \mathrm{CH_3COCH_2CH_2CH_3} \), has a methyl ketone group and gives a yellow precipitate of \( \mathrm{CHI_3} \).
Pentan-3-one, \( \mathrm{CH_3CH_2COCH_2CH_3} \), has no methyl ketone group, so it gives no precipitate.

Step 4: Which acid is stronger (b-i):
\( \mathrm{CH_2FCOOH} \) is the stronger acid.
Fluorine is highly electronegative and shows a strong electron-withdrawing (-I) effect. It spreads the negative charge of the carboxylate ion \( \mathrm{FCH_2COO^-} \) and makes it more stable.
In acetic acid, the methyl group pushes electrons (+I effect) towards the carboxylate, which makes it less stable. So \( \mathrm{CH_3COOH} \) is the weaker acid.

Step 5: Boiling points (b-ii):
Propane, \( \mathrm{CH_3CH_2CH_3} \), is non-polar and has only weak forces. It boils at about 231 K.
Dimethyl ether, \( \mathrm{CH_3OCH_3} \), is slightly polar and boils at about 249 K.
Acetaldehyde, \( \mathrm{CH_3CHO} \), is more polar and has strong dipole forces. It boils at about 294 K.
Ethanol, \( \mathrm{CH_3CH_2OH} \), forms intermolecular hydrogen bonds. It boils at about 351 K.
\[ \mathrm{CH_3CH_2CH_3 < CH_3OCH_3 < CH_3CHO < CH_3CH_2OH} \]

Final Answer:
(a) Iodoform test for (i) and (iii), Tollens' test for (ii). (b) (i) \( \mathrm{CH_2FCOOH} \) is stronger because of the -I effect of F. (ii) Propane < dimethyl ether < acetaldehyde < ethanol. \[ \boxed{\mathrm{CH_3CH_2CH_3 < CH_3OCH_3 < CH_3CHO < CH_3CH_2OH}} \]