1. Molecular formula and degree of unsaturation

Given compound $A$ has formula $\mathrm{C_{10}H_{12}O_2}$.

The degree of unsaturation (DBE) is:

So, total unsaturation $=5$.

This includes:

• $\pi$ bonds
• rings

2. Information from neutral $\mathrm{FeCl_3}$ test

Positive neutral $\mathrm{FeCl_3}$ test indicates a phenolic $-OH$ group.

So one oxygen is likely present as phenolic hydroxyl.

3. Reaction with $\mathrm{NaOH}$ and $\mathrm{CH_3Br}$

This gives $\mathrm{C_{11}H_{14}O_2}$.

Increase from $\mathrm{C_{10}H_{12}O_2}$ to $\mathrm{C_{11}H_{14}O_2}$ is:

• $+\mathrm{CH_2}$ overall

This is consistent with methylation of phenoxide:

So $A$ contains a phenolic $-OH$.

4. Reaction with HI gives methyl iodide

On cleavage with HI, formation of $\mathrm{CH_3I}$ indicates the presence of a methoxy group $(-OCH_3)$ already present in $A$.

Thus the two oxygens in $A$ are:

• one phenolic $-OH$
• one methoxy $-OCH_3$

Therefore $A$ is likely a methoxy phenol derivative with one more substituent.

5. Hot concentrated NaOH gives compound $B$ with same formula

$A$ gives $B$ having the same molecular formula $\mathrm{C_{10}H_{12}O_2}$, so this suggests isomerisation/rearrangement, not addition or removal of atoms.

6. Alkaline $\mathrm{KMnO_4}$ is decolorised

This indicates the presence of an oxidisable C=C bond outside the aromatic ring, such as an alkenyl side chain.

So $A$ likely contains:

• a benzene ring
• one phenolic $-OH$
• one methoxy $-OCH_3$
• one alkenyl substituent

Since it is trisubstituted, the benzene ring has 3 substituents total:

1. $-OH$
2. $-OCH_3$
3. alkenyl group

This fits the formula.

For example, a structure like propenyl/methyl-substituted guaiacol type derivative is consistent.

7. Count the number of $\pi$ bonds

Now count $\pi$ bonds in $A$:

• Benzene ring contributes 3 $\pi$ bonds
• One extra alkene side chain contributes 1 $\pi$ bond

Total:

8. Check with DBE

We found total unsaturation $=5$.

If $\pi$ bonds $=4$, then remaining unsaturation is:

That corresponds to one ring, namely the benzene ring.

So this is fully consistent.

Final Answer

The number of $\pi$ bonds in compound $A$ is: