A trisubstituted compound 'A', C$$_{10}$$H$$_{12}$$O$$_2$$ gives neutral FeCl$$_3$$ test positive. Treatment of compound 'A' with NaOH and CH$$_3$$Br gives C$$_{11}$$H$$_{14}$$O$$_2$$, with hydroiodic acid gives methyl iodide and with hot conc. NaOH gives a compound B, C$$_{10}$$H$$_{12}$$O$$_2$$. Compound 'A' also decolourises alkaline KMnO$$_4$$. The number of $$\pi$$ bond/s present in the compound 'A' is ______.

A trisubstituted compound A with formula $$C_{10}H_{12}O_{2}$$ gives a positive $$FeCl_{3}$$ test and decolourises $$KMnO_{4}$$. We wish to determine the number of pi bonds in this molecule.

First, the degree of unsaturation is calculated by the formula

$$DBE = \frac{2(10) + 2 - 12}{2} = 5$$

From the positive $$FeCl_{3}$$ test, a phenolic -OH group is indicated. Treatment with $$NaOH/CH_{3}Br$$ produces $$C_{11}H_{14}O_{2}$$, showing that methylation of the phenol occurs. Further reaction with HI yields $$CH_{3}I$$, confirming the presence of a methoxy (-OCH₃) group. Finally, decolourisation of alkaline $$KMnO_{4}$$ reveals an unsaturation (C=C double bond) outside the benzene ring.

These observations point to a benzene ring (accounting for four degrees of unsaturation: three pi bonds plus one ring) substituted by a phenolic -OH, an -OCH₃ group, and an allyl side chain containing a C=C double bond (one additional degree of unsaturation). The resulting structure corresponds to eugenol, namely 4-allyl-2-methoxyphenol, which can be written as $$\text{HO-}C_{6}H_{3}(OCH_{3})(CH_{2}CH=CH_{2})$$.

In this structure, the benzene ring contributes three pi bonds and the allylic C=C double bond contributes one pi bond, giving a total of 4 pi bonds. Therefore, the answer is 4.