Question

Lucas test is given by an alcohol within 5 minutes. 0.22 g of which liberates 56 mL of $C{H}_4$ at STP on treating with $C{H}_{3}MgI$.The structure of alcohol is:

• A. $C{H}_3-\underset{OH}{\underset{|}{C}}H-C{H}_3$
• B. $(C{H}_3{)}_{2}CH-\underset{OH}{\underset{|}{C}}H-C{H}_3$
• C. $C{H}_3-C{H}_2-\underset{OH}{\underset{|}{C}}H-C{H}_3$
• D. $(C{H}_3{)}_{3}C-\underset{OH}{\underset{|}{C}}H-C{H}_3$

Answer 1

The correct option is B $(C{H}_3{)}_{2}CH-\underset{OH}{\underset{|}{C}}H-C{H}_3$

Lucas test will be given by all given alcohols within 5 minutes as all are $2^0$ alcohols .Exact structure can be known by molecular mass or formula of alcohol which will be obtained as follows:
Let molecular mass of alcohol be M .

$\frac{56}{22400}=\frac{0.22}{M}$ or , $\frac{22400\times 0.22}{56}=88$

The general formula of alcohols is $C_n{H}_{2n}{+}_{1}OH$.The molecular mass 88 corresponds to the value of n=5.

Thus, the $2^0$ alcohol is $\left(C{H}_3{)}_{2}CHCH\right(OH)C{H}_3$