Question

An alcohol \( X \) (\( C_4H_{10}O \)) does not give turbidity with conc. HCl and ZnCl\(_2\) at room temperature. \( X \) on reaction with reagent \( Y \) gives \( Z \). What are \( X \), \( Y \), and \( Z \) respectively?

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Lucas test distinguishes alcohols:
- Primary alcohols show no reaction at room temperature.
- Secondary alcohols react slowly.
- Tertiary alcohols react immediately.

Answer 1

The Correct Option is A

Step 1: Identify \( X \) - The given molecular formula \( C_4H_{10}O \) suggests a butanol isomer.
- Since \( X \) does not give turbidity with Lucas reagent (conc. HCl + ZnCl\(_2\)), it must be a primary alcohol.
- The correct structure for \( X \) is 1-Butanol (\chemfig{CCCOH}). Step 2: Identify \( Y \) - PCC (Pyridinium chlorochromate) is a mild oxidizing agent.
- It selectively oxidizes primary alcohols to aldehydes without further oxidation to carboxylic acids.
- Therefore, \( Y \) is PCC. Step 3: Identify \( Z \) - The oxidation of 1-Butanol (\chemfig{CCCOH}) using PCC yields butanal (\chemfig{CCC=O}), an aldehyde.
- Thus, \( Z \) is Butanal. \bigskip

Answer 2

Given the reaction:
An alcohol \( X \) with molecular formula \( C_4H_{10}O \) does not give turbidity with conc. HCl and ZnCl\(_2\) at room temperature.
\( X \) reacts with reagent \( Y \) (PCC) to give compound \( Z \).

Step 1: Alcohols that give turbidity with conc. HCl and ZnCl\(_2\) are typically 1° or 2° alcohols, as they form alkyl chlorides.
If \( X \) does not give turbidity, it suggests that \( X \) is a 3° alcohol.

Step 2: \( X \) has formula \( C_4H_{10}O \). Possible 3° alcohol with 4 carbons is 2-methyl-2-propanol (tert-butanol), but that gives turbidity too.
Alternatively, consider 1-butanol (primary) and 2-butanol (secondary) alcohols.

Step 3: PCC oxidizes primary alcohols to aldehydes and secondary alcohols to ketones.
Given the product \( Z \) is an aldehyde with 4 carbons (butanal), the starting alcohol \( X \) is 1-butanol.

Step 4: But 1-butanol should give turbidity with conc. HCl and ZnCl\(_2\), which contradicts the condition.
Hence, \( X \) is 2-butanol (secondary alcohol) that does not give turbidity easily.

Step 5: Oxidation of 2-butanol with PCC gives 2-butanone (a ketone), not an aldehyde.
Therefore, the alcohol is 1-butanol which does give turbidity.

Step 6: The image shows \( X = \text{butanol} \), \( Y = \text{PCC} \), and \( Z = \text{butanal} \).
Hence,
\[ X = \text{1-butanol}, \quad Y = \text{PCC}, \quad Z = \text{butanal} \]