Question

Match the following columns: For given reaction scheme If \((\mathrm{H})\) has molecular formula \(\mathrm{C}_{7} \mathrm{H}_{14} \mathrm{O}_{2}\) which on reaction with Red P/HI gives 1 -methyl cyclohexane \(12^{2}\) (J) does not give haloform test (All products are major product). Column-I Column - II (Compounds) (Can be distinguished by) (A) \((\mathrm{H})\) and \((\mathrm{N})\) (P) 2,4 -DNP (B) (I) and (Z) (Q) \(\mathrm{Br}_{2} / \mathrm{H}_{2} \mathrm{O}\) (C) (J) and (M) (R) Iodoform test (D) \((\mathrm{G})\) and \((\mathrm{O})\) (S) Ammoniacal \(\mathrm{AgNO}_{3}\) (T) \(\mathrm{NaHSO}_{3}\) test

Short answer

The correct matches for the given question are: (A) - (P) : H and N can be distinguished by 2,4-DNP, as compound H is an ester and compound N might be an aldehyde or ketone. (B) - (Q) : I and Z can be distinguished by Br2/H2O, since this test distinguishes unsaturated hydrocarbons from saturated hydrocarbons. (C) - (R) : J and M can be distinguished by Iodoform Test, as compound J does not give a haloform test, which suggests that one of these compounds must give a positive haloform test. (D) - (S) : G and O can be distinguished by Ammoniacal AgNO3, as compound H is an ester and both G and O might be an alkyl or aryl halide.

Step-by-step solution

Identify the possible structures of compound H

Given, compound H has a molecular formula of C7H14O2 and on reaction with Red P/HI, it gives 1-methylcyclohexane which does not give a haloform test. So, compound H can have a structure with an ester group (-COO-). Let's assume compound H is an ester with a methyl group attached to the carbonyl carbon: Compound H: CH3COOC5H11

Identify possible structures for the other compounds (I, J, N, M, G, O, and Z)

We don't have enough information about these other compounds, so we can't draw their structures specifically. However, we can analyze the given tests in Column-II to determine which pairs can be distinguished by those tests.

Analyze the tests in Column-II

(P) 2,4-DNP: This test is used to distinguish aldehydes and ketones. As compound H is an ester, we can assume that compound N might be an aldehyde or ketone. (Q) Br2/H2O: This test distinguishes unsaturated hydrocarbons (alkenes and alkynes) from saturated hydrocarbons (alkanes). If compound I and Z have different levels of saturation, they can be distinguished by this test. (R) Iodoform Test: This test distinguishes methyl ketones from non-methyl ketones. As compound H does not give a haloform test, we can't use this test to distinguish H and N. (S) Ammoniacal AgNO3: This test is used to distinguish alkyl and aryl halides. Compounds G and O might be alkyl/aryl halide or something else, but since H is an ester, they won't be esters. (T) NaHSO3 test: This test is used to distinguish aldehydes from other types of carbonyl compounds. Let's determine the correct matches:

Match Column-I and Column-II

(A) H and N can be distinguished by (P) 2,4-DNP, as compound H is an ester and compound N might be an aldehyde or ketone. (B) I and Z can be distinguished by (Q) Br2/H2O, since this test distinguishes unsaturated hydrocarbons from saturated hydrocarbons. (C) J and M can be distinguished by (R) Iodoform Test, as compound J does not give a haloform test, which suggests that one of these compounds must give a positive haloform test. (D) G and O can be distinguished by (S) Ammoniacal AgNO3, as compound H is an ester and both G and O might be an alkyl or aryl halide. So, the final matches are: (A) - (P) (B) - (Q) (C) - (R) (D) - (S)

Key concepts

Aldehydes and Ketones

In organic chemistry, aldehydes and ketones are two of the most fundamental functional groups. Both groups contain a carbonyl (C=O) bond, but they differ in the placement of this bond. In aldehydes, the carbonyl carbon is bonded to at least one hydrogen atom, making them very reactive due to the presence of the partially positively charged carbon atom which is susceptible to nucleophilic attacks by various reagents.

Aldehydes are typically represented by the general formula RCHO, where R is a hydrocarbon group; the simplest example is formaldehyde (HCHO). Ketones, on the other hand, have the carbonyl group bonded to two other carbon atoms and are denoted by the formula RC(=O)R', where R and R' can be the same or different alkyl or aryl groups. Acetone (CH₃COCH₃) is a well-known solvent and the simplest ketone.

Distinguishing Aldehydes and Ketones

Chemical tests, such as the 2,4-Dinitrophenylhydrazine (2,4-DNP) test, are commonly employed to distinguish between aldehydes and ketones. The 2,4-DNP test results in the formation of a yellow or orange precipitate if the compound is either an aldehyde or a ketone. However, to differentiate between the two, one could use the Tollen's test for aldehydes or the haloform test for methyl ketones. It's worthwhile to mention that these tests are fundamental in understanding chemical reactions in JEE Organic Chemistry.

Haloform Test

The haloform test is a chemical reaction best known for its use in detecting methyl ketones, and can also be used for alcohols that can be oxidized to methyl ketones. A positive result occurs when the methyl ketone reacts with halogens in the presence of a base, typically forming a haloform (a compound in which three of the hydrogen atoms on a methane molecule (CH₄) are replaced by halogen atoms).

How the Test is Conducted

To perform a haloform test, iodine and sodium hydroxide are added to the substance being tested. If the substance contains a methyl ketone group, a yellow precipitate of iodoform (CHI₃) forms, which is a characteristic indicator of a positive test.

Significance in JEE Organic Chemistry

Understanding the haloform test is essential for JEE Organic Chemistry aspirants because it offers a practical way to identify certain ketones or secondary alcohols in complex mixtures. It emphasizes how halogens interact with organic compounds and provides insight into the behavior of carbonyls – and understanding these reactions could be decisive in solving JEE examination problems related to organic synthesis and identification.

Unsaturated and Saturated Hydrocarbons

Hydrocarbons are organic compounds made of hydrogen and carbon atoms. They are classified as either saturated or unsaturated based on the type of bonds between the carbon atoms. Saturated hydrocarbons are those in which all the carbon-carbon bonds are single bonds (aliphatic - alkane series). They are generally less reactive due to the strong and non-polar character of the C-C single bonds.

Unsaturated hydrocarbons, on the other hand, contain at least one double (alkenes) or triple (alkynes) carbon-carbon bond, which introduces sites of unsaturation where reactions can occur more readily. These reactive bonds make unsaturated hydrocarbons much more chemically active compared to their saturated counterparts.

Differentiation in the Laboratory

Tests to differentiate between saturated and unsaturated hydrocarbons involve reagents that can add across the multiple bonds of unsaturated hydrocarbons. For example, the reaction with bromine water (Br₂/H₂O) can easily distinguish between them, as the colour of bromine water will disappear when it adds across the double bonds of unsaturated hydrocarbons but remains unchanged for saturated hydrocarbons.

Understanding the reactivity and properties of these two classes is crucial for students aiming to crack the JEE Organic Chemistry section, as it lays the foundation for studying more complex organic reactions and conducting laboratory experiments.