Question

An alcohol X($C_4H_{10}O$) is converted to corresponding chloride Y by shaking it with conc. HCl at room temperature. Reaction of Y with Mg in dry ether and then with water gave Z. What are Y and Z?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

{Lucas Test Rate:} $3^\circ>2^\circ>1^\circ$. Immediate turbidity $\rightarrow$ Tertiary Alcohol. Grignard + Active H (Water/Alcohol/Acid) $\rightarrow$ Alkane.

Answer 1

The Correct Option is B

Step 1: Identify Alcohol X:
Formula $C_4H_{10}O$. Condition: Reacts with conc. HCl at room temperature to form Chloride Y. This indicates the Lucas Test. Tertiary alcohols react immediately at room temperature. Secondary take minutes, Primary require heat/ZnCl2. Structure of X (Tertiary Butyl Alcohol): $(CH_3)_3C-OH$ (2-Methylpropan-2-ol).
Step 2: Identify Chloride Y:
Reaction: $(CH_3)_3C-OH + HCl \rightarrow (CH_3)_3C-Cl + H_2O$. Y is t-Butyl Chloride (2-Chloro-2-methylpropane).
Step 3: Identify Z:
Reaction of Y with Mg/Ether $\rightarrow$ Grignard Reagent ($t-BuMgCl$). Reaction with Water $\rightarrow$ Hydrolysis. $(CH_3)_3C-MgCl + H_2O \rightarrow (CH_3)_3C-H + Mg(OH)Cl$. Z is Isobutane (2-Methylpropane).
Step 4: Check Options:
Y = 2-Chloro-2-methylpropane. Z = 2-Methylpropane. Matches Option (B) (Diagram corresponds to t-butyl chloride and isobutane).