Which of the following compounds will show retention in configuration on nucleophilic substitution b

Q: Why doesn't option (a) show retention if it undergoes SN1?

In an $S_N1$ reaction, the leaving group departs to form a planar carbocation. When the nucleophile attacks, it can do so from either side, leading to a mixture of both configurations (racemization), not pure retention.

Q: What exactly is a chiral center?

A chiral center is typically a carbon atom bonded to four different groups. If any two groups are identical, it is achiral, and the concept of optical configuration doesn't apply.

Q: Why is option (c) achiral?

The central carbon in option (c) is bonded to a hydrogen, a bromine, and *two* methyl groups. Because two of the groups are identical, it lacks chirality.

Q: Does the nucleophile matter for retention in this case?

Not really! Whether it's $OH^-$ or any other nucleophile, the key is *where* it attacks. Since it attacks a primary carbon away from the chiral center in option (b), retention is guaranteed regardless of the nucleophile's strength.

Q: Can SN2 ever lead to retention?

If the $S_N2$ reaction happens *at* the chiral center, it always leads to inversion. It only leads to retention if the reaction happens at a different atom, leaving the chiral center's bonds intact.

Question

Which of the following compounds will show retention in configuration on nucleophilic substitution by $OH^-$ ion?

Options:

(A) $CH_3-CH(C_6H_5)-Br$

(B) $CH_3-CH(C_2H_5)-CH_2Br$

(C) $CH_3-CH(CH_3)-Br$

(D) $CH_3-CH(C_6H_{13})-Br$

Sigma Percentile · Accepted Answer

### The Mystery of Retention

Stereochemistry is one of the most fascinating aspects of organic chemistry. It forces us to think in three dimensions. When a molecule undergoes a reaction, we don't just care about *what* is formed; we care about *how* it looks in space.

In this problem, we are asked to find a compound that shows **retention of configuration** during a nucleophilic substitution reaction with a hydroxide ion ($OH^-$).

To solve this, we need to understand what retention actually means. Imagine a chiral center as a highly secure, locked safe. The spatial arrangement of the groups around it is the combination to the lock. If a reaction happens directly *at* the safe (breaking a bond to the chiral center), the lock is tampered with. Depending on the mechanism, the combination might be completely reversed (inversion in $S_N2$) or scrambled (racemization in $S_N1$).

However, if the reaction happens somewhere else—say, you are just painting the door of the room where the safe is kept—the safe itself is completely untouched. The combination remains exactly the same. This is **retention of configuration**. It occurs when no bonds to the chiral center are broken during the reaction.

### Analyzing the Suspects

Let's look at our options and identify where the reaction is taking place.

In option (a), the molecule is $CH_3-CH(C_6H_5)-Br$. The bromine atom is directly attached to the chiral carbon. If $OH^-$ attacks, it must break the $C-Br$ bond. Because the reaction is happening directly at the chiral center, the spatial arrangement will be altered. It will undergo inversion or racemization, but definitely not retention.

Similarly, in option (d), $CH_3-CH(C_6H_{13})-Br$, the leaving group is again directly bonded to the chiral center. Just like in option (a), the bond to the chiral center is broken, meaning retention is impossible.

### The Achiral Imposter

Now, let's examine option (c): $CH_3-CH(CH_3)-Br$.

At first glance, it looks like the others. But look closely at the central carbon. It is bonded to a hydrogen, a bromine, and **two identical methyl groups** ($CH_3$).

For a carbon to be chiral, it must be bonded to four *different* groups. Because this carbon has two identical groups, it is **achiral**. The concepts of optical activity, inversion, and retention simply do not apply to achiral molecules. It's a trick option!

### The Perfect Alibi

Finally, we arrive at option (b): $CH_3-CH(C_2H_5)-CH_2Br$.

Let's map out the molecule. The bromine is attached to a primary carbon ($-CH_2-$). Is this carbon chiral? No, it has two identical hydrogen atoms.

Where is the chiral center? It's the adjacent $\alpha$-carbon, which is bonded to a methyl group, an ethyl group, a hydrogen atom, and the $-CH_2Br$ group.

When the $OH^-$ nucleophile attacks, it targets the primary carbon to kick out the bromine. The reaction happens entirely at the $-CH_2-$ group. The chiral center is completely untouched! Not a single bond connected to the chiral carbon is broken.

Because the chiral center is merely a bystander to the reaction happening next door, its spatial arrangement remains perfectly intact. Thus, it exhibits **retention of configuration**.

### Conclusion

This problem beautifully illustrates that stereochemistry isn't just about memorizing $S_N1$ and $S_N2$ outcomes. It's about looking at the molecule as a whole and identifying the exact site of the reaction. By realizing that the reaction in option (b) happens away from the chiral center, we can confidently conclude that the configuration is retained.

Animated Solution for Chemistry - Organic Chemistry: Which of the following compounds will show retention in configuration on nucleophilic substitution by $O H^{-}$ ion?

Select Answer:

Visualized Solution

$R e t e n t i o n o f C o n f i g u r a t i o n$

$A na l y z in g O pt i o n s (a) an d (d)$

$A na l y z in g O pt i o n (c)$

$A na l y z in g O pt i o n (b)$

$C o n c l u s i o n$

The Sigma Insight: Haloalkanes & Haloarenes

The Mystery of Retention

Analyzing the Suspects

The Achiral Imposter

The Perfect Alibi

Conclusion

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Animated Solution for Chemistry - Organic Chemistry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following compounds will show retention in configuration on nucleophilic substitution by OH− ion?