A toxic compound "A" when reacted with NaCN in aqueous acidic medium yields an edible cooking component and food preservative 'B'. "B" is converted to "C" by diborane and can be used as an additive to petrol to reduce emission. "C" upon reaction with oleum at $$140^{\circ}$$ C yields an inhalable anesthetic "D". Identify "A", "B", "C" and "D", respectively.

The key information given for each step helps us identify the compounds one-by-one.

Step 1  (Identifying B from its uses)
The product $$B$$ is said to be “an edible cooking component and food preservative”. Vinegar is used for cooking and preservation and its active component is acetic acid, $$CH_3COOH$$. Therefore, $$B$$ must be acetic acid.

Step 2  (Finding A that gives acetic acid with $$NaCN/H^+$$)
In an acidic medium the -OH group of a primary alcohol is protonated and can be displaced by the nucleophile $$CN^-$$. With methanol this gives methyl cyanide (acetonitrile) which is immediately hydrolysed by the same acidic water present to give acetic acid:

$$CH_3OH + H^+ \longrightarrow CH_3OH_2^+$$

$$CH_3OH_2^+ + CN^- \longrightarrow CH_3CN + H_2O$$

$$CH_3CN + 2H_2O + H^+ \longrightarrow CH_3COOH + NH_4^+$$

Thus the toxic compound $$A$$ is methanol, $$CH_3OH$$.

Step 3  (Conversion of B to C with diborane)
Diborane (or the $$BH_3$$ generated from it) is a selective reducing agent for carboxylic acids:
$$RCOOH \xrightarrow[B_2H_6]{\text{(BH}_3\text{)}} RCH_2OH$$

Applying this to acetic acid: $$CH_3COOH \xrightarrow[B_2H_6]{} CH_3CH_2OH$$

Hence $$C$$ is ethanol. Ethanol is blended with petrol (gasohol, E10, E20 etc.) to lower harmful emissions, matching the statement.

Step 4  (Formation of D from ethanol with oleum at $$140^{\circ}\text{C}$$)
At about $$140^{\circ}\text{C}$$ concentrated $$H_2SO_4$$ (or oleum) dehydrates primary alcohols to symmetrical ethers:
$$2 \, CH_3CH_2OH \xrightarrow[\;140^{\circ}C\;]{H_2SO_4} CH_3CH_2OCH_2CH_3 + H_2O$$

The product is diethyl ether, which has long been used as an inhalation anesthetic. Therefore $$D$$ is diethyl ether.

Step 5  (Listing the four compounds)
A : methanol, $$CH_3OH$$
B : acetic acid, $$CH_3COOH$$
C : ethanol, $$CH_3CH_2OH$$
D : diethyl ether, $$CH_3CH_2OCH_2CH_3$$

Comparing with the options, these correspond to Option C.

Final answer: Methanol → Acetic acid → Ethanol → Diethyl ether (Option C).