Animated Solution for Chemistry - Organic Chemistry: The major products of the following reaction are

Organic chemistry is often like a beautifully choreographed dance, where one reaction perfectly sets the stage for the next. In this problem, we are presented with a classic two-step sequence that tests our ability to recognize standard reagents and predict regiochemistry.

We begin our journey with 4-chlorophenol. The first set of reagents thrown into the mix is chloroform ($CHC{l}_3$) and aqueous sodium hydroxide ($NaOH$).

Whenever you see a phenol reacting with chloroform in a basic medium, your mind should immediately jump to one of the most famous named reactions in organic chemistry.

The combination of $CHC{l}_3$ and a strong base generates a highly reactive, electron-deficient species known as a dichlorocarbene ($:CC{l}_2$). This acts as our electrophile.

Because the phenol ring is strongly activated by the $-OH$ group (which exists as a phenoxide ion in the basic medium), it is highly susceptible to electrophilic aromatic substitution. The phenoxide oxygen directs the incoming electrophile to the ortho and para positions.

Since the para position is already blocked by the chlorine atom, the dichlorocarbene has no choice but to attack the ortho position. After a series of proton transfers and hydrolysis steps, a formyl group ($-CHO$) is successfully installed on the ring.

Our intermediate is 5-chloro-2-hydroxybenzaldehyde.

Now, the plot thickens. In the second step, we introduce formaldehyde ($HCHO$) and concentrated $NaOH$ to our newly formed aromatic aldehyde.

Take a close look at both of these molecules. What do they have in common? Neither of them possesses an $\alpha$-hydrogen!

When aldehydes lacking $\alpha$-hydrogens are subjected to concentrated base, they cannot undergo an aldol condensation. Instead, they undergo a disproportionation reaction known as the Cannizzaro reaction. Because we have two different aldehydes, this is specifically a Crossed Cannizzaro reaction.

Who Does What?

In a Crossed Cannizzaro reaction, one aldehyde must be oxidized to a carboxylic acid (or its salt), and the other must be reduced to an alcohol. How do we decide which is which?

Formaldehyde is the smallest aldehyde. It is sterically unhindered and its carbonyl carbon is highly electrophilic. Therefore, the hydroxide ion from the base preferentially attacks the formaldehyde molecule.

This attack forms a tetrahedral intermediate that readily expels a hydride ion ($H^{-}$). Because formaldehyde acts as the hydride donor, it gets oxidized to a formate ion ($HCO{O}^{-}$).

Our larger, more sterically hindered aromatic aldehyde is forced to act as the hydride acceptor. It takes the hydride ion and gets reduced to an alkoxide ion.

At the end of the Cannizzaro reaction, we are left with salts in a highly basic medium. To isolate our final neutral products, we must perform an acidification.

The third step introduces hydronium ions ($H_3{O}^{+}$). This acidifies the solution, protonating the alkoxide to yield a primary alcohol group ($-C{H}_{2}OH$) and protonating the formate ion to yield formic acid ($HCOOH$).

Our final aromatic product is 5-chloro-2-hydroxybenzyl alcohol.

Looking at the given choices, this perfectly matches option (d). The sequence of Reimer-Tiemann followed by a Crossed Cannizzaro is a brilliant test of fundamental organic mechanisms!