Animated Solution for Chemistry - Organic Chemistry: Consider all possible isomeric ketones including stereoisomers of MW = 100, All these isomers are independently reacted with ${\text{NaBH}}_4$ (NOTE : stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are

Every great organic chemistry problem starts with a bit of detective work. We are given a ketone with a molecular weight of exactly $100$. But what does this molecule look like?

Let's recall the general formula for an acyclic, saturated ketone: $C_n{H}_{2n}O$. By plugging in the atomic weights of carbon ($12$), hydrogen ($1$), and oxygen ($16$), we can set up a simple algebraic equation:

Solving for $n$, we subtract $16$ from $100$ to get $84$, and dividing by $14$ gives us $n=6$. The mystery is solved! Our ketone has the molecular formula $C_6{H}_{12}O$.

The Carbon Skeleton

Now comes the fun part—drawing the structures. A ketone features a carbonyl group ($C=O$) sandwiched between two alkyl groups ($R$ and $R^{\prime }$). Since the carbonyl carbon takes up one carbon atom, we have exactly $5$ carbon atoms left to distribute between $R$ and $R^{\prime }$.

There are two main ways to split $5$ carbons: a $1+4$ split (Methyl and Butyl) or a $2+3$ split (Ethyl and Propyl).

Let's explore the Methyl + Butyl combination first. The butyl group is highly versatile and exists in four isomeric forms: normal butyl, isobutyl, secondary butyl, and tertiary butyl. This gives us four distinct structural isomers. However, the secondary butyl group contains a chiral center! Because the question explicitly asks us to consider stereoisomers, the sec-butyl methyl ketone actually exists as two distinct enantiomers: the $(R)$ and $(S)$ forms. This brings our count from this group to $5$ unique ketones.

Next, we look at the Ethyl + Propyl combination. The propyl group can be either normal propyl or isopropyl. This gives us $2$ more structural isomers, neither of which has a chiral center.

In total, we have successfully mapped out $7$ isomeric ketones.

With our $7$ ketones lined up, it's time to introduce them to our reagent: Sodium Borohydride ($NaB{H}_4$). This is a classic, mild reducing agent that transforms ketones into secondary alcohols.

The mechanism is elegant. The carbonyl carbon is $s{p}^2$ hybridized, meaning it sits perfectly flat in a plane. The nucleophilic hydride ion ($H^{-}$) from $NaB{H}_4$ can attack this flat surface from either the top face or the bottom face.

If the starting ketone is achiral (meaning it has a plane of symmetry or lacks a chiral center), the top and bottom faces are chemically equivalent. The hydride attacks both faces with equal $50\%$ probability. The result? A perfect $1:1$ mixture of $(R)$ and $(S)$ enantiomers. This is what we call a racemic mixture.

Racemic vs Diastereomeric

Let's evaluate our $7$ candidates.

Ketones (1) through (5)—which include hexan-2-one, 4-methylpentan-2-one, 3,3-dimethylbutan-2-one, hexan-3-one, and 2-methylpentan-3-one—are all completely achiral. When $NaB{H}_4$ attacks them, a brand new chiral center is born, and because there is no pre-existing asymmetry to bias the attack, they all yield beautiful racemic mixtures.

But what about ketones (6) and (7)? These are the $(R)$ and $(S)$ enantiomers of 3-methylpentan-2-one. They already possess a chiral center. When the hydride ion approaches the carbonyl group, it encounters a bulky, asymmetric environment. Due to steric hindrance (often predicted by Cram's Rule or the Felkin-Anh model), one face of the carbonyl is blocked more than the other.

The attack is no longer equal! The reaction will heavily favor one face, producing a pair of diastereomers in unequal amounts. Because the products are diastereomers and not enantiomers, this is not a racemic mixture.

We were asked to find the total number of ketones that give a racemic product upon reduction. Out of our $7$ total isomers, the $5$ achiral ketones perfectly fit the bill, while the $2$ chiral ketones fail the test.

Therefore, the final answer is $5$.