Chemistry - STRUCTURAL ISOMERISM Question with Solution | TestHub
The compound has 2 isomers and they can be distinguished by the reaction with
Options:
Answer:
Solution:
1. Reaction of 1,2-dichloroethane (vicinal dihalide) with aq. KOH:
In 1,2-dichloroethane, the two chlorine atoms are on adjacent carbon atoms. When it reacts with aq. KOH, both chlorine atoms are substituted by hydroxyl groups, leading to the formation of 1,2-ethanediol (ethylene glycol).
1,2-ethanediol is a viscous, syrupy liquid with a relatively high boiling point (197 °C) due to extensive hydrogen bonding.
2. Reaction of 1,1-dichloroethane (geminal dihalide) with aq. KOH:
In 1,1-dichloroethane, both chlorine atoms are attached to the same carbon atom. When it reacts with aq. KOH, both chlorine atoms are substituted by hydroxyl groups on the same carbon atom, initially forming an unstable geminal diol.
Geminal diols, where two hydroxyl groups are on the same carbon, are generally unstable and readily lose a molecule of water to form a carbonyl compound (an aldehyde or a ketone). In this case, the unstable 1,1-ethanediol immediately dehydrates to form ethanal (acetaldehyde).
Ethanal is a volatile liquid with a much lower boiling point (20 °C) compared to 1,2-ethanediol. It also has a characteristic pungent odor.
Therefore, the distinct products (1,2-ethanediol vs. ethanal) formed upon reaction with aq. KOH allow for the differentiation of the two isomers.
Why other options are incorrect:
B. alc. KOH (alcoholic KOH): Alcoholic KOH is a strong base used for dehydrohalogenation (elimination reactions), leading to the formation of alkenes or alkynes.
- 1,2-dichloroethane with alc. KOH would undergo double dehydrohalogenation to form ethyne (acetylene).
- 1,1-dichloroethane with alc. KOH would also undergo double dehydrohalogenation to form ethyne (acetylene).
Since both isomers yield the same product (ethyne), alc. KOH cannot distinguish between them.
C. Bromine water: Bromine water is used to test for unsaturation (presence of C=C or C≡C bonds). Both 1,2-dichloroethane and 1,1-dichloroethane are saturated compounds and would not react with bromine water, so it cannot distinguish them.
D. NaNH₂ (sodamide): Sodamide is a very strong base primarily used for dehydrohalogenation reactions, especially to form alkynes from dihalides. Similar to alc. KOH, both isomers would react with NaNH₂ to form ethyne, thus not distinguishing them.
The final answer is A.