Practice MCQs for JEE Chemistry Organic Chemistry Some Basic Principles and Techniques
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Multiple Choice Questions
(a) Fe(CN)2
(b) K3[Fe(CN)5NS]
(c) Na4[Fe(CN)5NOS]
(d) Na4[Fe(CN)6]
Show Answer & Explanation
Answer: (c) Na4[Fe(CN)5NOS]
Explanation:
1. Sulphur is converted to Na2S in the extract.
2. Na2S combines with nitroprusside to give the violet sodium thionitroprusside, Na4[Fe(CN)5NOS].
3. Answer: (c)
Teacher's Note:
1. Recall what each reagent or step does in the detection or estimation of elements in organic compounds.
(a) An oxidising agent
(b) A reducing agent
(c) A hydrolysing agent
(d) A catalyst
Show Answer & Explanation
Answer: (d) A catalyst
Explanation:
1. Nitrogen is converted to (NH4)2SO4 on digestion with conc. H2SO4.
2. CuSO4 speeds up this digestion as a catalyst, and K2SO4 raises the boiling point.
3. Answer: (d)
Teacher's Note:
1. Recall what each reagent or step does in the detection or estimation of elements in organic compounds.
(a) Increase the ionisation of the compound
(b) Decrease the melting point of the compound
(c) Increase the reactivity of the compound
(d) Convert the covalent compound into a mixture of ionic compounds
Show Answer & Explanation
Answer: (d) Convert the covalent compound into a mixture of ionic compounds
Explanation:
1. Fusion with sodium converts N, S and halogens into NaCN, Na2S and NaX.
2. These are ionic and dissolve in water, so they can be tested by ordinary ionic tests.
3. Answer: (d)
Teacher's Note:
1. Recall what each reagent or step does in the detection or estimation of elements in organic compounds.
(a) Sodium hydroxide
(b) Sodium sulphate
(c) Calcium chloride
(d) Sodium bicarbonate
Show Answer & Explanation
Answer: (d) Sodium bicarbonate
Explanation:
1. The carboxylic acid is stronger than carbonic acid, so it dissolves in NaHCO3 as its sodium salt with evolution of CO2.
2. A phenol is weaker than carbonic acid and stays undissolved.
3. NaOH would dissolve both.
4. Answer: (d)
Teacher's Note:
1. Recall what each reagent or step does in the detection or estimation of elements in organic compounds.
(a) Dumas test
(b) Beilstein's test
(c) Liebig's test
(d) Lassaigne's test
Show Answer & Explanation
Answer: (b) Beilstein's test
Explanation:
1. In Beilstein's test an organic compound on a copper wire loop is heated in a flame.
2. Volatile copper halides give a green or blue-green flame.
3. Answer: (b)
Teacher's Note:
1. Recall what each reagent or step does in the detection or estimation of elements in organic compounds.
(a) 73.29%
(b) 78.45%
(c) 83.23%
(d) 89.50%
Show Answer & Explanation
Answer: (a) 73.29%
Explanation:
1. % C = (12/44) × (0.147/0.2) × 100 = 20.05.
2. % H = (2/18) × (0.12/0.2) × 100 = 6.67.
3. % O = 100 − 20.05 − 6.67 = 73.29.
4. Answer: (a)
Teacher's Note:
1. Recall what each reagent or step does in the detection or estimation of elements in organic compounds.
(a) Silver halides are soluble in HNO3
(b) Na2S and NaCN are decomposed by HNO3
(c) Ag2S is soluble in HNO3
(d) AgCN is soluble in HNO3
Show Answer & Explanation
Answer: (b) Na2S and NaCN are decomposed by HNO3
Explanation:
1. Na2S and NaCN would otherwise give black Ag2S and white AgCN with AgNO3.
2. Boiling with HNO3 expels H2S and HCN, so only the silver halides precipitate.
3. Answer: (b)
Teacher's Note:
1. Recall what each reagent or step does in the detection or estimation of elements in organic compounds.
(a) Chromatography
(b) Steam distillation
(c) Addition of oxalic acid
(d) Fractional crystallisation
Show Answer & Explanation
Answer: (b) Steam distillation
Explanation:
1. Aniline is volatile in steam and immiscible with water, and it decomposes if distilled alone at its boiling point.
2. Steam distillation therefore separates it from non-volatile impurities.
3. Answer: (b)
Teacher's Note:
1. Recall what each reagent or step does in the detection or estimation of elements in organic compounds.
(a) 26.32%
(b) 40%
(c) 53.6%
(d) 63.6%
Show Answer & Explanation
Answer: (a) 26.32%
Explanation:
1. Milliequivalents of NH3 = 20 × 0.5 = 10, i.e. 10 millimoles of N.
2. Mass of N = 10 × 14 mg = 0.14 g.
3. Percentage = 0.14 / 0.532 × 100 = 26.32.
4. Answer: (a)
Teacher's Note:
1. Recall what each reagent or step does in the detection or estimation of elements in organic compounds.
(a) 75.0
(b) 93.5
(c) 100
(d) 80.0
Show Answer & Explanation
Answer: (b) 93.5
Explanation:
1. The salt is (BH)2PtCl6, so its molar mass = 0.75 × 195 / 0.245 = 597.
2. The PtCl6 part weighs 195 + 213 = 408.
3. 2(B + 1) = 597 − 408 = 189, so B = 93.5.
4. Answer: (b)
Teacher's Note:
1. Recall what each reagent or step does in the detection or estimation of elements in organic compounds.
(a) A mixture of Na2S and CS2
(b) FeCl3
(c) FeSO4
(d) Na2SO3
Show Answer & Explanation
Answer: (b) FeCl3
Explanation:
1. Thiocyanate forms the blood-red [Fe(SCN)]2+ complex with Fe3+ ions.
2. Fe2+ (FeSO4) does not give this colour.
3. Answer: (b)
Teacher's Note:
1. Recall what each reagent or step does in the detection or estimation of elements in organic compounds.
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Multiple Choice Questions (MCQs) for JEE Chemistry Organic Chemistry Some Basic Principles and Techniques
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FAQs
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