(a) Write the mechanism of the following reaction: CH3CH2OH + HBr → CH3CH2Br + H2O
(b) Write the equation involved in Reimer-Tiemann reaction.

(a) Mechanism of the reaction between ethanol and HBr:

Step 1: Protonation of the hydroxyl group
The oxygen atom in the hydroxyl group of ethanol has lone pairs of electrons. These lone pairs can attack the proton (H+) from the HBr molecule. This step protonates the hydroxyl group, making it a better leaving group.

CH3CH2OH + HBr ⇌ CH3CH2OH2+ + Br-

Step 2: Formation of a carbocation
The protonated hydroxyl group (-OH2+) is a good leaving group and departs from the molecule, leaving behind a carbocation.

CH3CH2OH2+ → CH3CH2+ + H2O

Step 3: Nucleophilic attack
The bromide ion (Br-), which is a nucleophile, attacks the carbocation, forming a new carbon-bromine bond.

CH3CH2+ + Br- → CH3CH2Br

Overall reaction:
CH3CH2OH + HBr → CH3CH2Br + H2O

(b) Reimer-Tiemann Reaction:
The Reimer-Tiemann reaction is an example of electrophilic aromatic substitution. It involves the ortho-formylation of phenols using chloroform (CHCl3) in the presence of a strong base, usually aqueous NaOH. The reaction proceeds through the formation of a dichlorocarbene intermediate.

The overall equation is:

C6H5OH + CHCl3 + 3NaOH → C6H4(OH)CHO + 3NaCl + 2H2O

where C6H5OH represents phenol and C6H4(OH)CHO represents salicylaldehyde (ortho-hydroxybenzaldehyde).