(a) Give chemical tests to distinguish between :
(i) Propanol and propanone
(ii) Benzaldehyde and acetophenone
(b) Arrange the following compounds in an increasing order of their property as indicated:
(i) Acetaldehyde, acetone, methyl tertbutyl ketone (reactivity towards $H C N$ )
(ii) Benzoic acid, 3,4-dinitrobenzoicacid, 4-methoxybenzoic acid (acid strength)
(iii) $C H_{3} C H_{3} C H (B r) C O O H, C H_{3} C H (B r) C H_{2} C O O H, (C H_{3})_{2} C H C O O H$
(acid strength)

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Solution

(a)
(i) Propanone gives iodoform test while propanol does not give this test.
$C H_{3} C O C H P r o p a n o n e + 3 I_{2} + 2 N a_{2} C O_{3} \to C H I_{3} Y e l l o w p p t + C H_{3} C O O N a + 2 N a I + 2 C O_{2} + H_{2} O$
(ii) Benzaldehyde $(C_{6} H_{5} C H O)$ and acetophenone $(C_{6} H_{5} C O C H_{3})$ can be distinguished by iodoform test.
Acetophenone, being a methyl ketone on treatment with $I_{2} / N a O H$ undergoes iodoform reaction to give a yellow ppt. of iodoform. on the other hand, benzaldehyde does not give this test.
$C_{6} H_{5} C O C H_{3} + 3 N a O I$
Acetophenone
$↓$
$C_{6} H_{5} C O O N a + C H I_{3} ↓ I o d o f o r m + 2 N a O H$
$C_{6} H_{5} C H O N a O I - -- \to$ No yellow ppt of iodoform
Benzaldehyde
(b)
(i) Methyl ter-butyl ketone < Acetone < Acetaldehyde
When HCN reacts with a compound, the attacking species is a nucleophile, $C N^{-}$ .
Therefore, as the negative charge on the compound increases, its reactivity with $H C N$ decreases. In the given compounds, the $+ I$ effect increases as shown below. It can be observed the steric hindrance also increase in the same.
(ii) 4-Methoxybenzoic acid < Benzoic acid < 3, 4-Dinitrobenzoic acid
Electron-donating group decrease the strengths of acids, while electron-withdrawing groups increases the strengths of acids. As methoxy group is an electron-donating group, 4-ethoxybenzoic acid is a weaker acid than benzoic acid. Nitro group is an electron-withdrawing group and will increase the strengths of acids.
(iii) $(C H_{3})_{2} C H C O O H < C H_{3} C H (B r) C H_{2} C O O H < C H_{3} C H_{2} C H (B r) C O O H$
After losing a proton, carboxylic acids gain a negative charge as shown:
$R - C O O H \to R - C O O^{-} + H^{+}$
Now, any group that will help stabilise the negative charge will increase the stability of the carboxyl ion and as a result, will increase the strength of the acid. Thus, groups having $+ I$ effect will decrease the strength of the acids and groups having -I effect will increase the strength of the acids. In the given compounds, $- C H_{3}$ group has $+ I$ effect and $B r^{-}$ group has $- I$ effect. Thus, acid containing $B r^{-}$ are stronger.
The $- I$ effect grows weaker as distance increase. Hence, $C H_{3} C H (B r) C H_{2} C O O H$ is weaker acid than $C H_{3} C H_{2} C H (B r) C O O H .$

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Arrange the following compounds in increasing order of their property as indicated :

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(iii)

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