Preparation of Bakelite proceeds via reactions:

We start by recalling what Bakelite is. Bakelite is the very first synthetic polymer, technically called phenol-formaldehyde resin. It is obtained when phenol $$\left(C_{6}H_{5}OH\right)$$ reacts with formaldehyde $$\left(HCHO\right)$$ under either acidic or basic conditions and then undergoes further cross-linking.

The first stage of the process is the introduction of $$-CH_{2}OH$$ (hydroxymethyl) groups into the aromatic ring of phenol. In organic chemistry, whenever an aromatic ring such as benzene or phenol is attacked by an electron-deficient species, the mechanism is called electrophilic substitution. Here the electrophile is the carbon of formaldehyde, which carries a partial positive charge $$\left(\delta ^+\right)$$ because oxygen is more electronegative than carbon.

So, step one is an electrophilic substitution:

$$C_{6}H_{5}OH + HCHO \;\xrightarrow{acid / base\; C6H4(OH)(CH2OH)}$$

Because more than one $$-CH_{2}OH$$ group can enter the ring, we often get mixtures such as $$p$$- and $$o$$-hydroxymethyl phenol.

Now, once two phenol molecules each carry $$-CH_{2}OH$$ groups, they can link together. Two $$-CH_{2}OH$$ groups condense by eliminating a molecule of water $$\left(H_{2}O\right)$$ and form a methylene bridge $$\left(-CH_{2}-\right)$$ between two aromatic rings. This removal of water is literally a dehydration.

The dehydration step can be shown schematically as:

$$\bigl[C_{6}H_{4}(OH)(CH_{2}OH)\bigr]_2 \;\xrightarrow[-H_{2}O]{\; C6H4(OH)-CH2-C6H4(OH)}$$

So we clearly have two fundamental types of reactions taking place:

1. Electrophilic substitution (introduction of $$-CH_{2}OH$$ on the aromatic ring).

2. Dehydration (elimination of water to form methylene bridges, leading to polymer growth).

Looking at the options, Option C exactly states “Electrophilic substitution and dehydration,” which matches our mechanistic description.

Hence, the correct answer is Option C.