Animated Solution for Chemistry - Organic Compounds Containing Halogens: Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to decolourise the colour of bromine ?

Substitution

Imagine you are a bulky, powerful base like sodium tert-butoxide ($tert-BuONa$). You are massive, sterically hindered, and you have one primary mission: to rip protons off molecules and create double bonds through $E2$ elimination. You generally despise acting as a nucleophile in $S_{N}2$ reactions because you are simply too big to squeeze into tight spaces.

In this problem, we are looking for a compound that, after facing the wrath of $tert-BuONa$, fails to decolourise bromine water.

What does this mean? Bromine water ($B{r}_2/H_{2}O$) is the classic chemical detective for unsaturation. If a molecule has a double or triple bond (a $\pi$-bond), it will react with the reddish-brown bromine, turning the solution colorless. Therefore, if a compound fails to decolourise it, the final product must be completely saturated.

Let's look at the molecules that are practically begging to be eliminated:

Option (b): (Bromo(phenyl)methyl)cyclohexane Here, the $\beta$-hydrogen is benzylic. It is sitting right next to a phenyl ring, making it highly acidic. When $tert-BuONa$ approaches, it effortlessly removes this proton, creating a highly stable, conjugated alkene. This product is unsaturated and will rapidly decolourise bromine water.

Options (c) and (d): The Carbonyl Siblings Both 2-bromocyclohexanone and 2-(bromomethyl)cyclohexanone possess a carbonyl group ($C=O$). The $\beta$-hydrogens available for elimination are located at the $\alpha$-position relative to this carbonyl. These $\alpha$-hydrogens are notoriously acidic. The bulky base will strip them away, yielding highly stable $\alpha ,\beta$-unsaturated ketones. Again, these are unsaturated products that will easily pass the bromine water test.

Bromocyclohexane

Now we arrive at Option (a): Bromocyclohexane.

Unlike its competitors, bromocyclohexane lacks any activating groups. There is no phenyl ring or carbonyl group to increase the acidity of its $\beta$-hydrogens. They are just standard, unactivated aliphatic hydrogens.

According to the specific pedagogical logic of this problem, this stark lack of activation means that the bulky base struggles to perform its preferred $E2$ elimination. Instead, it is forced into a substitution pathway, replacing the bromine atom to form tert-butyl cyclohexyl ether.

Ethers are completely saturated molecules. They possess no $\pi$-bonds. Therefore, when you add bromine water to this ether, absolutely nothing happens. The reddish-brown color remains. It has successfully failed to decolourise the bromine water.

It is important to acknowledge a slight divergence between textbook logic and laboratory reality. In a real flask, a secondary alkyl halide like bromocyclohexane would undergo significant $E2$ elimination when treated with a strong, bulky base like $tert-Bu{O}^{-}$, forming cyclohexene.

However, competitive exams often test relative concepts. The examiner designed this question to highlight the extreme reactivity of activated $\beta$-hydrogens (benzylic and $\alpha$-carbonyl) compared to unactivated ones. By recognizing this stark contrast, we can confidently identify option (a) as the intended outlier.