Why is ethanol the major product Z in the NEET 2026 sequence from ethane via monochlorination and ammonolysis?
The major product Z is ethanol. Ethane undergoes free-radical monochlorination under UV light to give X. X is then treated with ammonia to produce Y. Y is subjected to nitrous acid followed by water to give major product Z. The options are nitroethane, C2H5-N=N-OH, ethanol, and ethylamine.
The official answer is ethanol because the aliphatic diazonium salt cannot survive.
What is the official step-by-step solution for this NEET 2026 sequence?
The official solution identifies Z as C₂H₅OH.
Aliphatic 1° amines form unstable diazonium salts that lose N2 to give alcohols. The alkyl carbocation reacts immediately with water. This matches the verified official worked solution exactly.
What exact method mistake leads to selecting the wrong option in this question?
Treating the aliphatic diazonium salt as if it were aromatic and stable enough to isolate as C2H5-N=N-OH selects option B. Students write the diazohydroxide because they recall the aromatic pathway and stop there.
Stopping the sequence at Y and writing the primary amine as the final product because the nitrous-acid step is forgotten selects ethylamine. Failing to apply the rule that aliphatic diazonium ions decompose at the temperature of formation while aromatic ones require heating produces the remaining errors.
Why do aliphatic diazonium salts decompose while aromatic ones last longer?
Aliphatic diazonium ion lacks resonance stabilisation possible in aryl diazonium ions. It decomposes instantly at 0–5 °C to alkyl carbocation which is attacked by water to give alcohol.
Aromatic diazonium salts are stable below 5 °C and only on heating give phenol. The benzene ring supplies resonance that spreads the charge across ortho and para positions.
What related questions from the same chapter test the instability rule?
What is the major organic product when n-propylamine is treated with NaNO2/HCl followed by water?
n-Propanol. The unstable diazonium salt loses N₂ at once and the primary carbocation is captured by water. No rearrangement occurs under these dilute conditions.
How does benzenediazonium chloride behave when warmed with water versus when an aliphatic analogue is used?
Benzenediazonium chloride must be heated to 50–60 °C to evolve N₂ and form phenol. The aliphatic version decomposes at the moment of formation even in ice-cold solution and directly yields the alcohol.
Predict Z in the sequence CH3Br → (NH3) → Y → (NaNO2/HCl, H2O) → Z and explain the gas evolved.
Y is methylamine. Z is methanol. The gas evolved is nitrogen. The same unstable aliphatic diazonium ion releases N₂ quantitatively and leaves the carbocation for water to attack.
Work these three before you close the page. Write the equations from memory each time.
How does this NEET 2026 question improve accuracy across the linked chapters?
Once the instability rule is internalised, you can expect to solve the full chain in 90 seconds because you will spot the aliphatic primary amine and directly write alcohol plus N2. The question links haloalkane, amine, and nitrogen-compound chapters.
Solve every past NEET aliphatic-amine nitrous-acid question from the archive. Search by chapter name and year, read the official solution each time, then close the book and reproduce it. That repetition removes the confusion between aliphatic and aromatic pathways for good.
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Frequently asked questions
Why is ethanol the major product when aliphatic primary amine is treated with NaNO2/HCl and water?
Aliphatic primary amines form unstable diazonium salts that cannot be isolated. These salts decompose immediately at 0-5 °C releasing N2 gas and forming a carbocation that is attacked by water to yield the alcohol. In the given sequence ethane yields chloroethane (X), then ethylamine (Y), and finally ethanol (Z).
Why do aliphatic diazonium salts decompose while aromatic diazonium salts are stable?
Aliphatic diazonium ions lack resonance stabilisation from a benzene ring. They lose N2 instantly even in ice-cold solution to give alcohols. Aromatic diazonium salts receive resonance from the ring, remain stable below 5 °C, and require heating to form phenols.
What is the product when ethylamine reacts with nitrous acid followed by water?
The major product is ethanol. The unstable aliphatic diazonium salt decomposes at the temperature of its formation, evolving nitrogen gas. The resulting ethyl carbocation is immediately captured by water to give C2H5OH. This is the exact official NEET 2026 answer.
Why do many students choose C2H5-N=N-OH instead of ethanol in this NEET question?
Students wrongly apply the aromatic diazonium pathway and stop at the diazohydroxide stage. Aliphatic diazonium salts cannot survive long enough to form such intermediates. The correct reaction is immediate decomposition to alcohol plus N2.
What is the major product when n-propylamine is treated with NaNO2/HCl followed by water?
The major product is n-propanol. The unstable diazonium salt loses N2 at once and the primary carbocation is captured by water without rearrangement under dilute conditions. This follows the same rule as the ethanol example in the NEET 2026 question.