Question

This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. A compound (A) reduces \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{OH}\) and \(\mathrm{HgCl}_{2}\). On warming (A) with (B) in presence of conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) gives a product with pleasant fruity smell. When (B), containing two carbons, is heated with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to \(180^{\circ} \mathrm{C},(\mathrm{C})\) is obtained. Identify correct statement regarding (C). (a) It decolourizes bromine water (b) It is oxidized to (B) by hot acidified \(\mathrm{KMnO}_{4}\) (c) It is oxidized to (A) by cold alkaline dil.KMnO (d) It is unaffected by \(\mathrm{H}_{2}\) and \(\mathrm{Pt}\)

Short answer

Answer: It decolorizes bromine water.

Step-by-step solution

Understanding Reactions involving A and B

We have the following reactions: 1. A reduces [Ag(NH3)2]OH and HgCl2. 2. A reacts with B in presence of conc. H2SO4 to produce a fruity smell. From the above information: - Reducing silver and mercury salts indicates that compound A most likely has an aldehyde or a reducing sugar group. - The fruity smell of the product of the reaction between A and B in the presence of H2SO4 strongly suggests that it is an ester formation reaction.

Analyzing Reaction involving B and Its Structure

We are given that compound B contains two carbon atoms and its reaction with conc. H2SO4 at 180°C forms compound C. A common reaction with two carbon compounds at increased temperatures with concentrated H2SO4 is dehydration to form alkenes. Therefore, B is likely an alcohol (C2H5OH, ethanol).

Deduce Compound A and the product of A and B Reaction

Given that compound B is ethanol (C2H5OH), and A reacts with B to form a fruity-smelling ester, we can assume that compound A is an organic acid (e.g., CH3COOH, acetic acid). The product of this esterification reaction will be ethyl acetate (CH3COOC2H5).

Analyzing Compound C and Its Reactions

We know that compound C is obtained when compound B (ethanol) is heated with conc. H2SO4 at 180°C, producing an alkene. Therefore, compound C should be C2H4 (ethylene). Now, we examine each option to determine the correct statement about compound C: (a) It decolorizes bromine water: True. Alkenes like ethylene react with bromine water to form dibromoalkanes and decolorize the solution. (b) It is oxidized to (B) by hot acidified KMnO4: False. Hot acidified KMnO4 would oxidize an alkene to a ketone or an aldehyde, but not to an alcohol as in compound B. (c) It is oxidized to (A) by cold dilute alkaline KMnO4: False. Cold alkaline KMnO4 would oxidize ethylene to a diol, not an organic acid as in compound A. (d) It is unaffected by H2 and Pt: False. Ethylene (an alkene) will undergo hydrogenation to form ethane in the presence of hydrogen gas and a Pt catalyst. Based on this analysis, the correct option is: (a) It decolorizes bromine water.

Key concepts

Aldehyde Reactions

Aldehydes are organic compounds with the functional group \(-CHO\) and they are known for their distinct chemical behaviors. One notable reaction is their ability to reduce certain metal ions. For example, aldehydes can reduce

• Silver ions in Tollens' reagent, where complexed silver ions (\[ [\text{Ag(NH}_3)_2]^+ \]) are reduced to metallic silver forming a silver mirror on the flask.
• Mercuric chloride (HgCl\(_2\)) in solution can also be reduced by aldehydes, which further supports the presence of an aldehyde group in a compound if a similar reaction is observed.

This reducing ability is central to how aldehydes participate in reactions and identifies them from other functional groups. In typical organic chemistry problems, especially in exams like IIT JEE, identifying these reactions can help in determining that an unknown compound might contain an aldehyde group.

Ester Formation

Ester formation, or esterification, is a critical reaction in organic synthesis, often producing compounds with fruity odors, which makes them easily recognizable. Esters are formed by the reaction between an acid and an alcohol. The general esterification process can be represented by:

• The acid (usually a carboxylic acid) and the alcohol compound react in the presence of a catalyst, often concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)).
• During the reaction, a molecule of water is eliminated, thus linking the two reactants together to form an ester.

For instance, when ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) and acetic acid (\(\text{CH}_3\text{COOH}\)) react together, ethyl acetate (\(\text{CH}_3\text{COOC}_2\text{H}_5\)) is produced with a pleasant fruity smell. This reaction exemplifies one of the ways organic chemists synthesize esters, making it an important concept for exams like IIT JEE.

Alkene Reactions

Alkenes are hydrocarbons characterized by at least one carbon-carbon double bond \(\text{C=C}\). They are unsaturated compounds, which means they can participate in addition reactions. One of their defining features is their ability to react with bromine water, a property often used in chemical tests to confirm the presence of an alkene.

• In the presence of bromine water, alkenes decolorize the solution as they form dibromo compounds.

This ability makes alkenes easily identifiable. Alkenes can also undergo dehydration reactions, where, for example, ethanol is heated with sulfuric acid at 180°C to yield ethylene (\(\text{C}_2\text{H}_4\)), which is a simple alkene. These reactions indicate how vital understanding of alkenes is in organic chemistry.

Oxidation Reactions

Oxidation reactions are transformations where a molecule loses electrons, often reacting with oxygen or an oxidizing agent. In organic chemistry, various oxidizing agents are used to convert different functional groups.

• Cold, dilute alkaline \(\text{KMnO}_4\) is known to oxidize alkenes to diols. This reagent adds hydroxyl groups (\[ -OH \]) across the double bond of alkenes.
• Hot, acidified \(\text{KMnO}_4\) is a more vigorous oxidizing agent, converting alkenes to carboxylic acids or ketones, depending on the structure of the alkene.

These reactions highlight the importance of understanding oxidation in organic chemistry, as they allow for the prediction and control over the transformation of molecules. Such knowledge is crucial for students tackling competitive exams like IIT JEE.