Q.Match each reaction example in Column I with the name of the reaction in Column II.
Column I (reaction examples):

Let’s unpack each reaction separately, starting with the concept behind the reagent.

1. Butan-2-one with Zn(Hg) and conc. HCl

Concept: This is the Clemmensen reduction. It’s a classic method to reduce a carbonyl group ($\ce{C=O}$) in a ketone or aldehyde all the way to a methylene group ($\ce{CH2}$). The reagent — zinc amalgam in concentrated hydrochloric acid — provides a strongly acidic, reducing environment. The zinc metal donates electrons, and the acid protonates intermediates, ultimately replacing the oxygen with two hydrogens.

Why does this work? The carbonyl oxygen is first protonated, making the carbon more electrophilic. Zinc then transfers electrons, breaking the $\ce{C=O}$ bond and forming a carbene-like intermediate that gets further reduced. The net result: the ketone becomes an alkane.

Step-by-step:

1. Identify the substrate: Butan-2-one is $\ce{CH3COCH2CH3}$. The carbonyl is at the second carbon.
2. Apply the reduction: The $\ce{C=O}$ group is replaced by $\ce{CH2}$. So the product has the same carbon skeleton, but the carbonyl carbon becomes a $\ce{CH2}$ group.
3. Write the product: The carbon chain is $\ce{CH3-CH2-CH2-CH3}$, which is butane.

Chemical equation:

2. Two molecules of benzaldehyde with conc. NaOH

Concept: This is the Cannizzaro reaction. It occurs with aldehydes that have no α-hydrogen atoms (i.e., the carbon next to the carbonyl has no hydrogen). Benzaldehyde ($\ce{C6H5CHO}$) is the classic example. In concentrated base, one molecule of aldehyde is oxidized to a carboxylic acid (as its salt), and the other is reduced to a primary alcohol. It’s a disproportionation reaction.

Why does this happen? Without an α-hydrogen, the aldehyde cannot form an enolate (which would lead to an aldol reaction). Instead, the hydroxide ion attacks the carbonyl carbon of one aldehyde molecule, forming a tetrahedral intermediate. This intermediate transfers a hydride ion ($\ce{H-}$) to the carbonyl carbon of a second aldehyde molecule. The result: one aldehyde becomes a carboxylate (after deprotonation), and the other becomes an alcohol. …

The Concept Behind Each Conversion

These two problems test your understanding of how to transform functional groups using reagents that either build up or break down carbon chains. The key is to see the target molecule and work backwards: what functional group change is needed, and what reagent accomplishes it in one or two steps?

For (a), acetonitrile ($CH_3CN$) has a nitrile group, while acetone ($CH_3COCH_3$) has a ketone. The nitrile carbon is electrophilic and can be attacked by a Grignard reagent, adding an alkyl group. Hydrolysis then converts the resulting imine intermediate into the ketone. This is a classic two-step chain extension.

For (b), benzoic acid ($C_6H_5COOH$) has a carboxyl group attached to a benzene ring, and benzene ($C_6H_6$) is just the bare ring. The carboxyl group must be removed entirely. Decarboxylation — loss of $CO_2$ — is the direct route, but it requires heating with a strong base like soda lime ($NaOH + CaO$). This is a one-step conversion.

Step-by-Step Solution

(a) $CH_3CN$ to $CH_3COCH_3$

Step 1: Grignard addition to the nitrile

Acetonitrile has a polar $C \equiv N$ triple bond. The carbon is electrophilic. When we add methylmagnesium iodide ($CH_3MgI$, a Grignard reagent), the carbanion ($CH_3^-$) attacks the nitrile carbon. This forms an imine intermediate (a magnesium salt of an imine).

The reaction is:

Step 2: Acidic hydrolysis

Treating the imine intermediate with dilute acid (e.g., $H_3O^+$) hydrolyses the $C=N$ bond to a $C=O$ bond, giving the ketone. Water adds across the imine, and ammonia (as $NH_3$) is eliminated.