Question
An alcohol $$(A)$$ gives Lucas test within $$5\,\min .$$ $$7.4\,g$$ of alcohol when treated with sodium metal liberates $$1120\,ml$$ of $${H_2}$$ at $$STP.$$ What will be alcohol $$(A)?$$
A.
$$C{H_3}{\left( {C{H_2}} \right)_3}OH$$
B.
$$C{H_3}CH\left( {OH} \right)C{H_2}C{H_3}$$
C.
$${\left( {C{H_3}} \right)_3}COH$$
D.
$$C{H_3}CH\left( {OH} \right)C{H_2}C{H_2}C{H_3}$$
Answer :
$$C{H_3}CH\left( {OH} \right)C{H_2}C{H_3}$$
Solution :
$$ROH + Na \to RONa + \frac{1}{2}{H_2} \uparrow $$
We have to get molecular mass of alcohol corresponding to half mole of $${H_2}$$ only.
$$\frac{{1120}}{{11200}} = \frac{{7.4}}{M} \Rightarrow M = 74$$
$${C_n}{H_{2n + 1}}OH = 74 \to {C_n}{H_{2n + 1}}$$ $$ = 74 - 17 = 57$$
$$ \Rightarrow {C_n}{H_{2n}} = 57 - 1 = 56\,\,\,\,{\text{i}}{\text{.e}}{\text{.}}$$ $$12n + 2n = 14n = 56$$
$$ \Rightarrow n = \frac{{56}}{{14}} = 4$$
Thus, molecular formula of $$(A)$$ is $${C_4}{H_9}OH.$$ As $$(A)$$ gives Lucas test within $$5\,\min .,$$ thus $${2^ \circ }$$ alcohol corresponding to molecular formula $${C_4}{H_9}OH$$ is $$C{H_3}CH\left( {OH} \right)C{H_2}C{H_3}$$ ( butan-2-ol ).