Which of the following compounds reacts with ethyl magnesium bromide and also decolourises bromine w

Q: Why doesn't option (a) decolourise bromine water even though it has a phenol group?

While phenols typically decolourise bromine water via electrophilic aromatic substitution, the presence of the strongly electron-withdrawing $-CN$ group in option (a) severely deactivates the benzene ring. This makes the reaction with bromine water extremely slow or non-existent under standard test conditions.

Q: Does option (b) react with the Grignard reagent?

No. Option (b) is an ether (anisole derivative) and an alkene. It lacks any acidic hydrogen (like $-OH$, $-NH_2$, or terminal alkyne), so it cannot undergo an acid-base reaction with ethyl magnesium bromide.

Q: What exactly is the 'active hydrogen' in option (d)?

In option (d), the hydrogen atom on the carbon between the benzene ring and the lactone carbonyl group is highly acidic. This is because the resulting carbanion is resonance-stabilised by both the carbonyl group and the $-CN$ substituted benzene ring. The Grignard reagent acts as a base and abstracts this proton.

Q: Why does option (c) decolourise bromine water?

Option (c) contains a vinyl group ($-CH=CH_2$), which is an aliphatic double bond. Bromine water readily undergoes an electrophilic addition reaction across this double bond, breaking the pi bond and adding two bromine atoms, which consumes the bromine and decolourises the solution.

Q: Can Grignard reagents act as nucleophiles instead of bases here?

Grignard reagents are both strong bases and good nucleophiles. However, acid-base reactions are generally much faster than nucleophilic addition reactions. If a molecule has an acidic hydrogen, the Grignard reagent will almost always act as a base first, destroying the reagent before it can act as a nucleophile.

Question

Which of the following compounds reacts with ethyl magnesium bromide and also decolourises bromine water solution

Sigma Percentile · Accepted Answer

### The Dual Challenge

Organic chemistry often presents us with beautiful puzzles where a single molecule must satisfy multiple, sometimes conflicting, chemical conditions. In this problem, we are on a hunt for a specific compound that can successfully pass two distinct chemical tests: reacting with ethyl magnesium bromide and decolourising bromine water.

To solve this, we must first deeply understand the nature of our two reagents.

### The First Challenge: Ethyl Magnesium Bromide

Ethyl magnesium bromide ($CH_3CH_2MgBr$) is a classic Grignard reagent. Grignard reagents are fascinating because the carbon-magnesium bond is highly polarized, making the carbon atom both a powerful nucleophile and an exceptionally strong base.

When a Grignard reagent encounters a molecule, it faces a choice: should it attack an electrophilic center (like a carbonyl carbon) as a nucleophile, or should it steal a proton as a base? Acid-base reactions are kinetically much faster than nucleophilic additions. Therefore, if the target molecule possesses any **acidic hydrogen** (such as the hydrogen in $-OH$, $-NH_2$, $-COOH$, terminal alkynes, or active methylene groups), the Grignard reagent will instantly act as a base. It will abstract the proton to form an alkane gas—in this case, ethane ($C_2H_6$).

So, our first condition simply translates to: **The molecule must contain an acidic hydrogen.**

### The Second Challenge: Bromine Water Test

Bromine water ($Br_2/H_2O$) is a reddish-brown solution used primarily as a test for unsaturation. When it encounters an **aliphatic double or triple bond** (like an alkene or alkyne), it undergoes a rapid electrophilic addition reaction. The pi bond breaks, bromine atoms are added across the carbons, and the reddish-brown colour vanishes, leaving a colourless solution.

Interestingly, bromine water can also be decolourised by highly activated aromatic rings, such as phenol or aniline. The electron-donating groups push so much electron density into the ring that it undergoes rapid electrophilic aromatic substitution, forming a white precipitate (like 2,4,6-tribromophenol) and consuming the bromine.

However, if the aromatic ring is attached to strong electron-withdrawing groups (like $-CN$ or $-NO_2$), the ring becomes deactivated, and this substitution reaction is severely hindered.

So, our second condition translates to: **The molecule must contain an aliphatic $>C=C<$ bond OR a highly activated aromatic ring without strong deactivating groups.**

### Analyzing the Contenders

Now, let's put our four options to the test.

**Option (a): 3-(methoxycarbonylmethyl)-4-hydroxybenzonitrile**
This molecule has a phenolic $-OH$ group. This provides the acidic hydrogen needed to react with the Grignard reagent. Condition 1 is met! However, look at the aromatic ring. It is flanked by a strongly electron-withdrawing cyano ($-CN$) group. This deactivates the ring, preventing it from rapidly reacting with bromine water. Since there are no aliphatic double bonds either, it fails Condition 2.

**Option (b): 3-vinylanisole**
This molecule features a vinyl group ($-CH=CH_2$), which is an aliphatic double bond. It will eagerly undergo addition with bromine water, decolourising it instantly. Condition 2 is met! But wait, where is the acidic hydrogen? The oxygen is part of a methoxy group ($-OCH_3$), an ether. Ethers do not have acidic protons. Therefore, it will completely ignore the Grignard reagent. It fails Condition 1.

**Option (c): 3-vinylphenol**
Let's examine this beautiful molecule. It has a phenolic $-OH$ group, providing the perfect acidic proton to react with ethyl magnesium bromide and release ethane gas. Condition 1 is met! Furthermore, it possesses a vinyl group ($-CH=CH_2$). This aliphatic double bond will rapidly react with bromine water, decolourising the solution. Condition 2 is also met! This molecule is our winner.

**Option (d): 3-(2-oxotetrahydrofuran-3-yl)benzonitrile**
This molecule looks tricky. It doesn't have an $-OH$ group. Does it have an acidic hydrogen? Yes! The hydrogen atom on the carbon situated between the benzene ring and the lactone carbonyl group is an **active methine hydrogen**. The carbanion formed by removing this proton is highly resonance-stabilised. Thus, it will react with the Grignard reagent. Condition 1 is met! However, just like option (a), it lacks an aliphatic double bond, and its aromatic ring is deactivated by the $-CN$ group. It will not decolourise bromine water. It fails Condition 2.

### Conclusion

By systematically applying our knowledge of functional group reactivity, we can confidently conclude that **3-vinylphenol (Option c)** is the only compound that satisfies both the Grignard and the bromine water tests. It is a perfect example of how different functional groups within the same molecule operate independently to give a unique chemical fingerprint.

Animated Solution for Chemistry - Alcohols, Phenols, Ethers: Which of the following compounds reacts with ethyl magnesium bromide and also decolourises bromine water solution

Select Answer:

Visualized Solution

$R e a g e n t C o n d i t i o n s$

$A na l y s i s o f O pt i o n (a)$

$A na l y s i s o f O pt i o n (b)$

$A na l y s i s o f O pt i o n (c)$

$A na l y s i s o f O pt i o n (d)$

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

The Sigma Insight: Alcohols, Phenols, Ethers

The Dual Challenge

The First Challenge

The Second Challenge

Analyzing the Contenders

Conclusion

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