An organic compound (A) (molecular formula $$C_6H_{12}O_2$$) was hydrolysed with dil. $$H_2SO_4$$ to give a carboxylic acid (B) and an alcohol (C). 'C' gives white turbidity immediately when treated with anhydrous $$ZnCl_2$$ and conc. HCl. The organic compound (A) is:

We have to identify an organic compound (A) whose molecular formula is $$C_6H_{12}O_2$$. On hydrolysis with dilute $$H_2SO_4$$ it gives a carboxylic acid (B) and an alcohol (C). This description exactly fits the behaviour of an ester, because an ester undergoes acid-catalysed hydrolysis according to the general equation

$$\text{RCOOR'}\;+\;H_2O\;\xrightarrow[\;]{\;H_2SO_4\;}\;\text{RCOOH}\;+\;\text{R'OH}$$

So, compound (A) must be an ester made up of an acid part $$\text{RCO-}$$ and an alcohol part $$\text{-OR'}$$.

Now, alcohol (C) gives white turbidity immediately with anhydrous $$ZnCl_2$$ and conc. HCl. This is the well-known Lucas test for classifying alcohols:

• Tertiary alcohol → turbidity within seconds (immediate).
• Secondary alcohol → turbidity in 5-10 min.
• Primary alcohol → no turbidity at room temperature.

Because turbidity is obtained at once, alcohol (C) must be a tertiary alcohol.

Let us search for the smallest tertiary alcohol that can arise from an ester whose total formula is $$C_6H_{12}O_2$$. The simplest tertiary alcohol is tert-butyl alcohol (2-methyl-2-propanol):

$$\text{(CH}_3)_3\text{C-OH}$$

Its molecular formula is $$C_4H_{10}O$$ (four carbons, one oxygen, ten hydrogens).

If alcohol (C) contributes four carbon atoms, the remaining two carbon atoms (because the ester has six carbons in all) must belong to the acid (B). The only common two-carbon carboxylic acid is acetic acid (ethanoic acid):

$$\text{CH}_3\text{COOH}$$

Acetic acid has the formula $$C_2H_4O_2$$.

Now we must check that these two fragments really fit the molecular formula of (A) when they combine to form an ester. Combining an acid and an alcohol produces an ester with the loss of one molecule of water ($$H_2O$$). Let us add their formulas algebraically and then subtract the elements of water.

• Acid part: $$C_2H_4O_2$$
• Alcohol part: $$C_4H_{10}O$$
• Sum: $$C_{2+4}H_{4+10}O_{2+1}=C_6H_{14}O_3$$
• Subtract water $$\left(H_2O\right)$$: $$C_6H_{14-2}O_{3-1}=C_6H_{12}O_2$$

Exactly the required molecular formula. Therefore our choice of fragments is correct.

Hence compound (A) must be the ester formed from acetic acid and tert-butyl alcohol, i.e. tert-butyl acetate:

$$\text{CH}_3\text{COO-C(CH}_3)_3$$

Among the given structural options, this structure corresponds to Option (1).

Hence, the correct answer is Option (1).