JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 52

Multiple Choice Questions (MCQs) for JEE Chemistry: Organic Chemistry Some Basic Principles and Techniques

Access targeted multiple-choice questions for Organic Chemistry Some Basic Principles and Techniques designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Organic Chemistry Some Basic Principles and Techniques MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: In the Lassaigne test with FeSO4, FeCl\(_3\) and HCl, a red colour is due to
(a) \( \ce{Fe4[Fe(CN)6]3} \)
(b) \( \ce{[Fe(SCN)]^{2+}} \)
(c) \( \ce{Fe(CN)3} \)
(d) \( \ce{FeS} \)
Show Answer & Explanation

Answer: (b) \( \ce{[Fe(SCN)]^{2+}} \)

Explanation:
1. Red colour appears when both N and S are present: the thiocyanate complex.

Teacher's Note:
1. Prussian blue is Fe4[Fe(CN)6]3.
2. Sulphur interferes with the nitrogen test.

Question 2: In Kjeldahl's method nitrogen is estimated as
(a) ammonia
(b) N2 gas
(c) NO2
(d) nitric acid
Show Answer & Explanation

Answer: (a) ammonia

Explanation:
1. The compound is digested with conc. H2SO4, converting N to (NH4)2SO4; NaOH liberates NH3, absorbed in standard acid.

Teacher's Note:
1. Dumas gives N2, Kjeldahl gives NH3.
2. Titration finds the NH3.

Question 3: The catalyst used in Kjeldahl's method is
(a) sodium
(b) magnesium
(c) mercury (as Hg or HgO)
(d) zinc dust
Show Answer & Explanation

Answer: (c) mercury (as Hg or HgO)

Explanation:
1. Mercury (as HgO or Hg) with potassium sulphate speeds up digestion.

Teacher's Note:
1. K2SO4 raises the boiling point.
2. CuSO4 is also used.

Question 4: The equivalent weight of an acid is
(a) molecular weight multiplied by basicity
(b) molecular weight multiplied by acidity
(c) molecular weight divided by basicity
(d) molecular weight divided by acidity
Show Answer & Explanation

Answer: (c) molecular weight divided by basicity

Explanation:
1. Equivalent weight = M / n, where n is the number of replaceable H+.

Teacher's Note:
1. Basicity counts replaceable hydrogens.
2. For H2SO4 it is 98/2 = 49.

Question 5: The Liebig method is used to estimate
(a) carbon and hydrogen
(b) carbon only
(c) hydrogen only
(d) nitrogen
Show Answer & Explanation

Answer: (a) carbon and hydrogen

Explanation:
1. The compound is burnt in oxygen; water is absorbed in a desiccant and CO2 in KOH solution.

Teacher's Note:
1. Both are weighed from their absorbers.
2. KOH absorbs CO2.

Question 6: In the Carius method, 0.15 g of an organic compound gave 0.12 g of AgBr. The percentage of bromine is (Ag = 108, Br = 80)
(a) 18.05
(b) 53.19
(c) 63.10
(d) 34.04
Show Answer & Explanation

Answer: (d) 34.04

Explanation:
1. Br in AgBr = 0.12 x 80/188 = 0.05106 g. Percentage = 0.05106/0.15 x 100 = 34.04.

Teacher's Note:
1. Use the mass ratio Br/AgBr = 80/188.
2. Convert the precipitate mass to element mass.

Question 7: An organic compound of molecular mass 60 contains C = 20%, H = 6.67% and N = 46.67%. The compound is
(a) CH3NCO
(b) urea, (NH\(_2\))\(_2\)CO
(c) CH3CONH2
(d) CH3CH2CONH2
Show Answer & Explanation

Answer: (b) urea, (NH\(_2\))\(_2\)CO

Explanation:
1. O = 100 - (20 + 6.67 + 46.67) = 26.67%. Atoms in 60: C 1, H 4, N 2, O 1 giving CH4N2O.

Teacher's Note:
1. The empirical formula has mass 60.
2. Check mass percentages from the formula.

Question 8: A gaseous hydrocarbon has 85% carbon and a vapour density of 28. A possible formula is
(a) \( \ce{C3H6} \)
(b) \( \ce{C4H8} \)
(c) \( \ce{C2H4} \)
(d) \( \ce{C2H2} \)
Show Answer & Explanation

Answer: (b) \( \ce{C4H8} \)

Explanation:
1. M = 2 x 28 = 56. Carbon mass = 0.85 x 56 = 47.6, about 4 atoms, so the formula is C4H8 (85.7% C).

Teacher's Note:
1. Molecular mass = 2 x vapour density.
2. Only C4H8 has M = 56.

Question 9: Kjeldahl's method of nitrogen estimation is NOT applicable to
(a) acetamide
(b) aliphatic amines
(c) amino acids
(d) diazo compounds
Show Answer & Explanation

Answer: (d) diazo compounds

Explanation:
1. Nitrogen in diazo, azo and nitro compounds is not converted completely to ammonium sulphate.

Teacher's Note:
1. The N must be in the form of amine or amide.
2. Dumas works for all.

Question 10: The Beilstein test cannot detect which organic halide?
(a) fluoride
(b) chloride
(c) bromide
(d) iodide
Show Answer & Explanation

Answer: (a) fluoride

Explanation:
1. The green flame comes from volatile copper halides (CuCl2, CuBr2, CuI2); copper fluoride is not volatile.

Teacher's Note:
1. Test with a copper wire in a flame.
2. Organic fluorides give no coloured flame.

Question 11: Nitric acid is added to the sodium extract before AgNO\(_3\) in testing for halogens because it
(a) oxidises halides to halogens
(b) reduces silver ions
(c) decomposes NaCN and Na2S, which would also give precipitates
(d) dissolves AgCl
Show Answer & Explanation

Answer: (c) decomposes NaCN and Na2S, which would also give precipitates

Explanation:
1. CN- and S2- would precipitate AgCN and Ag2S; HNO3 removes them as HCN and H2S (boiling).

Teacher's Note:
1. Test without nitric acid would give false positives.
2. Boiling drives off the gases.

Question 12: Lead acetate test for sulphur uses acetic acid, not sulphuric acid, because
(a) acetic acid is a stronger acid
(b) sulphuric acid oxidises S2-
(c) acetic acid reduces the sulphide
(d) sulphuric acid would precipitate white lead sulphate
Show Answer & Explanation

Answer: (d) sulphuric acid would precipitate white lead sulphate

Explanation:
1. PbSO4 is white and insoluble and would mask the black PbS.

Teacher's Note:
1. Acetic acid gives the soluble lead acetate.
2. Choose an acid that does not form an insoluble salt.

Question 13: Which of the following compounds will give a positive iodoform test?
(a) methanol
(b) ethanol
(c) 2,2-dimethylpropan-1-ol
(d) methanal
Show Answer & Explanation

Answer: (b) ethanol

Explanation:
1. Ethanol is oxidised by hypoiodite to acetaldehyde, which has the CH3CO- group and gives CHI3.

Teacher's Note:
1. Methanol, methanal and the neopentyl alcohol lack the CH3CH(OH)- or CH3CO- unit.
2. Iodoform needs CH3CH(OH)R or CH3COR.

Question 14: A mixture of products from ammonia and ethyl halide contains C\(_2\)H\(_5\)NH\(_2\), (C\(_2\)H\(_5\))\(_2\)NH and (C\(_2\)H\(_5\))\(_3\)N. The test used to distinguish or separate them is the
(a) Hinsberg test
(b) carbylamine test
(c) iodoform test
(d) Fehling's solution test
Show Answer & Explanation

Answer: (a) Hinsberg test

Explanation:
1. Carbylamine recognises only the primary amine; Hinsberg's reagent separates all three classes.

Teacher's Note:
1. Primary gives an alkali-soluble sulphonamide, secondary an insoluble one, tertiary none.
2. Use the reagent that responds differently to all three.

Question 15: Which of the following is an incorrect match of reagent and test?
(a) carbohydrates with alpha-naphthol (Molisch)
(b) nitroethane with Zn, NH4Cl and AgNO3 (Mulliken-Barker)
(c) phenol with anhydrous ZnCl\(_2\) + conc. HCl (Lucas reagent)
(d) benzoic acid with NaHCO3
Show Answer & Explanation

Answer: (c) phenol with anhydrous ZnCl\(_2\) + conc. HCl (Lucas reagent)

Explanation:
1. Lucas reagent distinguishes alcohols; phenol is tested with neutral FeCl3.

Teacher's Note:
1. Molisch, Mulliken-Barker and NaHCO3 matches are correct.
2. Learn one test per group.

Question 16: A compound P reacts with Q in KOH to give ethyl isocyanide, CH\(_3\)CH\(_2\)NC. Q is prepared from acetone and calcium hypochlorite. P and Q are
(a) CH3CH2NO2 and CH3Cl
(b) CH3CH2NHCH3 and COCl2
(c) CH\(_3\)CH\(_2\)NH\(_2\) and CHCl\(_3\)
(d) (CH3CH2)3N and Cl2
Show Answer & Explanation

Answer: (c) CH\(_3\)CH\(_2\)NH\(_2\) and CHCl\(_3\)

Explanation:
1. Acetone with bleaching powder (haloform reaction) gives chloroform. Primary amine + CHCl3 + KOH gives the isocyanide (carbylamine reaction).

Teacher's Note:
1. Ethylamine supplies the C2H5N unit.
2. Carbylamine reaction.

Question 17: Which of the following can give the Hinsberg test?
(a) \( \ce{CH3CH2NH2} \)
(b) \( \ce{CH3CH2OH} \)
(c) \( \ce{CH3CH2NO2} \)
(d) \( \ce{CH3CONH2} \)
Show Answer & Explanation

Answer: (a) \( \ce{CH3CH2NH2} \)

Explanation:
1. Only an amine with N-H (or a tertiary amine for comparison) reacts with benzenesulphonyl chloride; ethylamine does.

Teacher's Note:
1. The other compounds are not amines.
2. Amides do not react normally.

Question 18: Two elements X (atomic mass 75) and Y (atomic mass 16) form a compound containing 75.8% of X. The formula is
(a) \( \ce{XY} \)
(b) \( \ce{X2Y} \)
(c) \( \ce{X2Y2} \)
(d) \( \ce{X2Y3} \)
Show Answer & Explanation

Answer: (d) \( \ce{X2Y3} \)

Explanation:
1. For X2Y3: 150/(150 + 48) = 75.76%, which matches.

Teacher's Note:
1. Test each formula.
2. Percentage = mass of X/total mass x 100.

Question 19: In the Carius method halogens in organic compounds are estimated as
(a) barium halide
(b) silver halide
(c) hydrogen halide
(d) halogen gas
Show Answer & Explanation

Answer: (b) silver halide

Explanation:
1. The compound is heated with fuming HNO3 and AgNO3, giving AgX which is filtered, dried and weighed.

Teacher's Note:
1. Sulphur is estimated as BaSO4.
2. AgX is weighed.

Question 20: The number of structural alkenes that on catalytic hydrogenation give 3-methylhexane (ignoring stereoisomerism) is
(a) 4
(b) 6
(c) 5
(d) 7
Show Answer & Explanation

Answer: (b) 6

Explanation:
1. Placing a double bond at each non-terminal C-C position with hydrogens on both carbons gives 6 structural alkenes.

Teacher's Note:
1. Count all bonds between carbons that carry hydrogen.
2. Do not count geometrical isomers.

Question 21: How many structural isomeric ketones of formula C\(_5\)H\(_{10}\)O give the iodoform test?
(a) 1
(b) 3
(c) 4
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. Pentan-2-one and 3-methylbutan-2-one each contain CH3CO-. Pentan-3-one does not.

Teacher's Note:
1. Methyl ketones give iodoform.
2. Check the three ketones of C5H10O.

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