Organic Chemistry

Which one of the following products is formed by the oxidation of phenol with chromic acid?

[amp_mcq option1=”1,4-Benzoquinone” option2=”1,2-Benzoquinone” option3=”Benzoic acid” option4=”Diphenyl ether” correct=”option1″]

The correct answer is 1,4-Benzoquinone.

Phenol can be oxidized to quinones. Chromic acid (H₂CrO₄), typically generated in situ from K₂Cr₂O₇ or CrO₃ in acidic solution, is a strong oxidizing agent. Oxidation of phenol with strong oxidizing agents like chromic acid generally results in the formation of p-benzoquinone (1,4-benzoquinone). The hydroxyl group is oxidized, and the aromatic ring is modified to form a conjugated cyclic diketone structure.

1,2-Benzoquinone (o-benzoquinone) is usually formed by milder oxidation methods, such as using Fremy’s salt. Benzoic acid involves cleavage of the ring, which is not the primary product of phenol oxidation with chromic acid under typical conditions. Diphenyl ether (C₆H₅OC₆H₅) is formed through a different type of reaction involving phenol, not simple oxidation.

The reaction of 1,2-dibromoethane with alcoholic KOH yields

[amp_mcq option1=”ethene” option2=”ethyne” option3=”1-bromo-2-hydroxyethane” option4=”1-bromoethene” correct=”option2″]

1,2-dibromoethane is a vicinal dibromide (bromine atoms on adjacent carbons). Alcoholic KOH is a strong base that promotes elimination reactions (dehydrohalogenation). When 1,2-dibromoethane is treated with alcoholic KOH, a double dehydrohalogenation occurs. First, one molecule of HBr is eliminated to form 1-bromoethene (vinyl bromide). Then, a second molecule of HBr is eliminated from 1-bromoethene to form ethyne (acetylene).

Reaction of a vicinal dihalide with alcoholic KOH results in double dehydrohalogenation, yielding an alkyne.

Step 1: CH₂Br-CH₂Br + KOH (alc.) → CH₂=CHBr + KBr + H₂O
Step 2: CH₂=CHBr + KOH (alc.) → HC≡CH + KBr + H₂O
The final product is ethyne. Ethene would result from single dehydrohalogenation of a mono-halogenated alkane or dehalogenation of a vicinal dihalide using a different reagent (e.g., Zn dust). 1-bromo-2-hydroxyethane would be a substitution product, which is less favored with alcoholic KOH, which promotes elimination. 1-bromoethene is an intermediate in the reaction, not the final product under excess alcoholic KOH.

The electrophile generated in sulphonation of benzene from fuming sulphuric acid is

[amp_mcq option1=”SO₃⁺” option2=”SO₃H” option3=”SO₃” option4=”SO₂H” correct=”option3″]

Sulphonation of benzene is an electrophilic aromatic substitution reaction. When fuming sulfuric acid (H₂SO₄ containing dissolved SO₃) is used, the active electrophile is sulfur trioxide (SO₃). SO₃ is a highly electron-deficient species due to the electron-withdrawing nature of the oxygen atoms and the formal positive charge on sulfur (though it can be considered neutral overall with resonance structures). It is sufficiently electrophilic to attack the pi electron system of the benzene ring.

The electrophile in the sulphonation of benzene by fuming sulfuric acid is SO₃.

In concentrated sulfuric acid without fuming (less SO₃), the electrophile is still primarily SO₃, generated from the equilibrium 2H₂SO₄ ⇌ H₃O⁺ + HSO₄⁻ + SO₃. However, fuming sulfuric acid has a much higher concentration of SO₃, making the reaction faster. SO₃H and SO₂H are not the active electrophilic species attacking the benzene ring.

Which one of the following reagents is used to carry out the transformation given below?
CH₃(CH₂)₉COOC₂H₅ → CH₃(CH₂)₉CHO

[amp_mcq option1=”DIBAL-H/H₂O” option2=”H₂/Pd-BaSO₄” option3=”SnCl₂/HCl” option4=”LiAlH₄” correct=”option1″]

The transformation involves reducing an ester (CH₃(CH₂)₉COOC₂H₅) to an aldehyde (CH₃(CH₂)₉CHO). Reducing an ester to an aldehyde requires a specific reducing agent that can stop the reduction at the aldehyde stage, as complete reduction of an ester yields a primary alcohol. Diisobutylaluminium hydride (DIBAL-H) is a reducing agent that, when used at low temperatures and in controlled stoichiometric amounts, can selectively reduce esters to aldehydes.

DIBAL-H is a useful reagent for the partial reduction of esters and nitriles to aldehydes.

LiAlH₄ (Lithium aluminium hydride) is a strong reducing agent that would reduce the ester completely to a primary alcohol (CH₃(CH₂)₉CH₂OH). H₂/Pd-BaSO₄ (Rosenmund catalyst) is used for the reduction of acid chlorides to aldehydes, not typically esters. SnCl₂/HCl (Stephen reduction) is a method for reducing nitriles to aldehydes via an imine intermediate. Thus, DIBAL-H is the appropriate reagent for this transformation.

Which one of the following statements about Fehling’s test is not correct?

[amp_mcq option1=”Fehling’s A solution is aqueous copper sulphate.” option2=”Fructose gives a positive Fehling’s test.” option3=”Red-brown ppt of CuO is obtained in the reaction.” option4=”Aromatic aldehydes do not respond to this test.” correct=”option3″]

Fehling’s test is a qualitative test used to distinguish between reducing and non-reducing sugars, and also to detect aldehydes. It involves Fehling’s solution, which is prepared fresh by mixing Fehling’s A (aqueous copper(II) sulfate) and Fehling’s B (aqueous sodium potassium tartrate and a strong alkali like NaOH or KOH). Reducing sugars (like glucose, fructose – due to isomerization) and aldehydes reduce the blue copper(II) ions (Cu²⁺) in Fehling’s solution to red-brown copper(I) oxide (Cu₂O) precipitate. Statement C claims the precipitate is red-brown CuO (copper(II) oxide), which is black. The precipitate formed is red-brown Cu₂O (copper(I) oxide). Therefore, statement C is incorrect.

Fehling’s test involves the reduction of Cu²⁺ ions to Cu⁺ ions, forming a red-brown precipitate of Cu₂O.

Fehling’s test is positive for all reducing sugars and most aliphatic aldehydes. Ketones generally do not give a positive Fehling’s test, except for alpha-hydroxy ketones like fructose, which isomerize under alkaline conditions to aldoses. Aromatic aldehydes, with a few exceptions, do not give a positive Fehling’s test because they are not easily oxidized and may undergo competing reactions like the Cannizzaro reaction in the presence of strong alkali. Statement A and B are correct. Statement D is generally correct for typical aromatic aldehydes.

Lassaigne’s test is used for the detection of which of the following elements?

[amp_mcq option1=”N, S, P, Cl” option2=”C, N, P, Br” option3=”C, N, S, I” option4=”C, N, S, Br” correct=”option1″]

Lassaigne’s test is a qualitative test used in organic chemistry to detect the presence of elements like Nitrogen (N), Sulfur (S), Halogens (Chlorine, Bromine, Iodine – Cl, Br, I), and Phosphorus (P) in an organic compound. The test involves fusing the organic compound with metallic sodium, which converts these elements into ionic compounds (NaCN, Na₂S, NaX, Na₃P) that can be detected by specific tests in the aqueous extract (Lassaigne’s extract). Option A correctly lists Nitrogen, Sulfur, Phosphorus, and Chlorine, all of which are detectable by Lassaigne’s test.

Lassaigne’s test detects elements *other than* carbon and hydrogen present in organic compounds.

After fusion with sodium, the compound is plunged into distilled water and boiled to create the Lassaigne’s extract. This extract is then tested for specific ions: CN⁻ (for N), S²⁻ (for S), X⁻ (for halogens), and PO₄³⁻ (for P). Carbon and hydrogen are the primary constituents of organic compounds and are not typically detected by this method.

Animal fats generally contain:

[amp_mcq option1=”unsaturated fatty acids.” option2=”saturated fatty acids.” option3=”unsaturated long chain alcohols.” option4=”lipids.” correct=”option2″]

Animal fats, such as butter and lard, are typically solid or semi-solid at room temperature. This physical state is primarily due to a higher proportion of saturated fatty acids in their composition. Saturated fatty acids have no double bonds between carbon atoms in the hydrocarbon chain, allowing them to pack tightly together, resulting in a higher melting point.

Animal fats generally contain a higher proportion of saturated fatty acids, which contributes to their solid or semi-solid state at room temperature.

Plant oils, on the other hand, contain a higher proportion of unsaturated fatty acids (with one or more double bonds), which introduces kinks in the hydrocarbon chains, preventing tight packing and resulting in lower melting points (liquid at room temperature). Fats and oils are both types of triglycerides, which are lipids. While animal fats are lipids, option D is too broad; option B specifically addresses the type of fatty acids characteristic of animal fats.

Which one of the following is the functional group of Methyl butanoate?

[amp_mcq option1=”Ester” option2=”Ether” option3=”Carboxylic acid” option4=”Ketone” correct=”option1″]

Methyl butanoate is an organic compound that belongs to the class of esters. The name “butanoate” indicates that it is derived from butanoic acid, and “methyl” indicates the alkyl group attached to the oxygen atom (R’ in R-COO-R’). The structure of methyl butanoate is CH₃CH₂CH₂COO-CH₃. The characteristic functional group present in esters is the ester group (-COO-).

Methyl butanoate is an ester, identified by the presence of the -COO- functional group derived from a carboxylic acid and an alcohol.

Ethers have the functional group R-O-R’. Carboxylic acids have the functional group -COOH. Ketones have the functional group R-CO-R’, where the carbonyl group is located within the carbon chain.