Question

‘X’ \(\xrightarrow[(2)\ \mathrm{H_2O}]{(1)\ \mathrm{NaNO_2/HCl}}\) \(\xrightarrow[\ ]{\mathrm{NaOH/I_2}}\) ‘Y’ (gives positive iodoform test):
X has \(%C = 65.75%\), \(%H = 15.25%\), \(%N = 19%\).
Identify ‘X’ among the following:

• A. Option 1
• B. Option 2
• C. Option 3
• D. Option 4

Hint:

If diazotization of a primary amine followed by hydrolysis gives an alcohol that answers the iodoform test, the alcohol must be of the type \(\mathrm{CH_3{-}CH(OH){-}}\).

Answer 1

The Correct Option is D

Concept:


Primary aliphatic amines on treatment with \(\mathrm{NaNO_2/HCl}\) form alcohols.
A positive iodoform test requires the presence (or formation) of a \(\mathrm{CH_3CO{-}}\) group or a secondary alcohol of the type \(\mathrm{CH_3{-}CH(OH){-}}\).

Step 1: Determine Molecular Formula from Percent Composition
Assume \(100\,\text{g}\) of compound: \[ \text{C: } \frac{65.75}{12} \approx 5.48,\quad \text{H: } \frac{15.25}{1} = 15.25,\quad \text{N: } \frac{19}{14} \approx 1.36 \] Divide by the smallest value (\(\approx 1.36\)): \[ \text{C:H:N} \approx 4:11:1 \] So, molecular formula is: \[ \boxed{\mathrm{C_4H_{11}N}} \]
Step 2: Analyze Reaction Path


‘X’ is a primary amine
with formula \(\mathrm{C_4H_{11}N}\).
On diazotization and hydrolysis, it forms an alcohol.
The resulting alcohol gives a positive iodoform test
, hence must be a secondary alcohol
of type \(\mathrm{CH_3{-}CH(OH){-}R}\).

Step 3: Match with Given Options
Among the given structures:

Options (1), (2), and (3) lead to primary alcohols or alcohols that do not give iodoform test.
Option (4) is a primary amine which on diazotization forms a secondary alcohol capable of giving iodoform test.

Final Conclusion:
The compound ‘X’ is represented by option (4)
.