Question

An alcohol (A) gives Lucas test within 5 minutes. 7.4 g of alcohol when treated with sodium metal liberates 1120 mL of ${ H }_{ 2 }$ at STP. What will be the alcohol (A)?
(A) ${ CH }_{ 3 }{ (CH }_{ 2 }{ ) }_{ 3 }OH$
(B) ${ CH }_{ 3 }{ CH }(OH){ CH }_{ 2 }{ CH }_{ 3 }$
(C) ${ (CH }_{ 3 }{ ) }_{ 3 }COH$
(D) ${ CH }_{ 3 }CH(OH){ CH }_{ 2 }{ CH }_{ 2 }{ CH }_{ 3 }$

Answer 1

The Lucas test is used to distinguish between primary, secondary and tertiary alcohols. Lucas reagent consists of an equimolar mixture of concentrated hydrochloric acid and anhydrous Zinc chloride. The reaction is done at room temperature and the time it takes for the turbidity to appear is noted. Alcohols react with the sodium metal and liberate hydrogen gas.

Complete step by step solution:
Lucas test is used to distinguish between primary, secondary and tertiary alcohols.
The Oxygen atom and the carbon atom attached to it in an alcohol are both ${ sp }^{ 3 }$ hybridised. There are a total of four ${ sp }^{ 3 }$ hybrid orbitals present on the oxygen atom two of which contain lone pairs. One ${ sp }^{ 3 }$ hybrid orbital overlaps with the ${ sp }^{ 3 }$ orbital of the carbon while the remaining ${ sp }^{ 3 }$ orbital overlaps with the 1s orbital of the H atom. Let us look at the different types of alcohols:
Primary alcohols: In such alcohols the carbon atom which is directly attached to the oxygen atom of the –OH group is bonded with two hydrogen atoms and one alkyl/aryl group.
Secondary alcohols: In such alcohols the carbon atom which is directly attached to the oxygen atom of the –OH group is bonded with one hydrogen atom and two alkyl/aryl groups.
Tertiary alcohols: In such alcohols the carbon atom which is directly attached to the oxygen atom of the –OH group is bonded to three alkyl/aryl groups.
For doing the Lucas test, the alcohol is treated with the Lucas reagent which consists of an equimolar mixture of concentrated hydrochloric acid and anhydrous Zinc chloride. The reaction is done at room temperature and the time it takes for the turbidity (which is due to the formation of the insoluble alkyl chloride) to appear is noted.
When tertiary alcohols are reacted with the Lucas reagent, the turbidity appears immediately. The reaction is given below:
$\begin{matrix} { R }_{ 3 }COH \\\ Tertiary\quad alcohol \end{matrix}\xrightarrow [ Lucas\quad reagent ]{ RT }\begin{matrix} { R }_{ 3 }CCl \\\ Insoluble \end{matrix}+\begin{matrix} { H }_{ 2 }O \\\ Water \end{matrix}$
When secondary alcohols are reacted with the Lucas reagent, the turbidity appears within 5 minutes. The reaction is given below:
$\begin{matrix} { R }_{ 2 }CHOH \\\ Secondary\quad alcohol \end{matrix}\xrightarrow [ Lucas\quad reagent ]{ RT } \begin{matrix} { R }_{ 2 }CHCl \\\ Insoluble \end{matrix}+\begin{matrix} { H }_{ 2 }O \\\ Water \end{matrix}$
But when primary alcohols are reacted with the Lucas reagent no turbidity appears which implies that the reaction does not happen.
$\begin{matrix} { R }CH_{ 2 }OH \\\ Primary\quad alcohol \end{matrix}\xrightarrow [ Lucas\quad reagent ]{ RT } No\quad reaction$
Alcohol reacts with sodium metal and liberates hydrogen. The reaction is given below:
$\begin{matrix} { R }OH \\\ Alcohol \end{matrix}+\begin{matrix} Na(s) \\\ Sodium\quad metal \end{matrix}\rightarrow \begin{matrix} RONa \\\ Sodium\quad alkoxide \end{matrix}+\begin{matrix} 1/2{ H }_{ 2 }(g)\uparrow \\\ Hydrogen\quad gas \end{matrix}$
From the above reaction it is clear that 1 mole of the alcohol reacts with 1 mole (23 g) of the sodium metal to liberate half mole (1 g) of the hydrogen gas.
Now at STP conditions, 1 mole of an ideal gas has a volume of 22.4 L. Considering the Hydrogen gas to be behaving like an ideal gas, we will calculate its number of moles:
22.4 L of Hydrogen gas=1 mole of Hydrogen gas.
1120 mL of Hydrogen gas=$\cfrac { 1\quad mol\quad of\quad { H }_{ 2 }\quad gas }{ 22.4\quad L } \times 1120\times { 10 }^{ -3 }L=0.05\quad mole$
Now, 0.5 moles of hydrogen gas is released from=1 mole of the alcohol
0.05 moles of Hydrogen gas will be released from=$\cfrac { 1\quad mol\quad of\quad alcohol }{ 0.5\quad mol } \times 0.05\quad mol=0.1\quad mol$
Now we can calculate the molar mass of the alcohol:
Since, $Molar\quad mass=\cfrac { Mass\quad in\quad g }{ No.\quad of\quad moles }$
$\Rightarrow Molar\quad mass=\cfrac { 7.4\quad g }{ 0.1\quad mol } =74\quad g/mol$
Now, the general molecular formula for an alcohol is ${ C }_{ n }{ H }_{ 2n+1 }OH$, therefore:
$12n+1\times (2n+1)+16+1=74g/mol$
$12n+2n=74-18$
$2n(6+1)=56$
$\Rightarrow n=4$
Hence the molecular formula for the alcohol is ${ C }_{ 4 }{ H }_{ 9 }OH$
So the answer should be a secondary alcohol with the molecular formula ${ C }_{ 4 }{ H }_{ 9 }OH$.
Hence the correct answer is (B) ${ CH }_{ 3 }{ CH }(OH){ CH }_{ 2 }{ CH }_{ 3 }$.

Note: The primary alcohols do not react with the Lucas reagent at room temperature, hence the solution remains colourless. But when the solution is subjected to high temperature, an oily layer is formed. Hence primary alcohols react with the Lucas reagent at high temperatures.