JEE Chemistry Purification And Characterisation Of Organic Compounds MCQs Set 02

Answer: (a) \( \ce{C2H6O} \)

Explanation:
1. % O = 34.76; moles C = 4.35, H = 13.04, O = 2.17, ratio 2 : 6 : 1
2. Empirical mass 46 = 2 x 23, so the molecular formula is \( \ce{C2H6O} \)

Teacher's Note:
1. Use the standard factor method: the fraction of the element in the weighed product times its mass gives the mass of the element.
2. Divide by the mass of the sample and multiply by 100 for the percentage.

Answer: (c) \( \ce{C4H8O4} \)

Explanation:
1. n = 120/30 = 4
2. Molecular formula = \( \ce{C4H8O4} \)

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (c) 80

Explanation:
1. 56 mL of vapour at STP has mass 0.2 g
2. M = 0.2 x 22400/56 = 80

Teacher's Note:
1. Write the working as mass of element, then moles, then ratio. This avoids slips with atomic masses.
2. Round only when a ratio is within about 0.1 of a whole number.

Answer: (b) Cold dilute alkaline \( \ce{KMnO4} \)

Explanation:
1. Baeyer's reagent (cold dilute alkaline \( \ce{KMnO4} \)) loses its purple colour with a C=C bond
2. The alkene is converted to a diol and brown \( \ce{MnO2} \) forms

Teacher's Note:
1. Match the technique to the property that differs between the components: boiling point, solubility, volatility or adsorption.
2. Questions often ask why one method is chosen over another.

Answer: (b) Chromatography

Explanation:
1. Chromatography is the most modern of the listed methods
2. It separates even very similar compounds from tiny samples

Teacher's Note:
1. Learn each test with its observation and the compound responsible for the colour or precipitate.
2. Remember which method suits which kind of sample (volatile, non-volatile, heat-sensitive).

Answer: (d) Steam distillation

Explanation:
1. o-Nitrophenol has intramolecular H-bonding and is steam volatile
2. p-Nitrophenol has intermolecular H-bonding and is not, so steam distillation separates them

Teacher's Note:
1. Learn each test with its observation and the compound responsible for the colour or precipitate.
2. Remember which method suits which kind of sample (volatile, non-volatile, heat-sensitive).

Answer: (b) A mixture of conc. \( \ce{H2SO4} \) and conc. \( \ce{HNO3} \)

Explanation:
1. Conc. \( \ce{H2SO4} \) protonates \( \ce{HNO3} \) to give the nitronium ion \( \ce{NO2^+} \)
2. This electrophile attacks the benzene ring

Teacher's Note:
1. Match the technique to the property that differs between the components: boiling point, solubility, volatility or adsorption.
2. Questions often ask why one method is chosen over another.

Answer: (b) Homolytic fission

Explanation:
1. Light supplies energy that splits a bond evenly
2. Each atom keeps one electron, giving free radicals (homolytic fission)

Teacher's Note:
1. Match the technique to the property that differs between the components: boiling point, solubility, volatility or adsorption.
2. Questions often ask why one method is chosen over another.

Answer: (a) Urea

Explanation:
1. Acid taken = 100 x 0.1 x 2 = 20 meq; NaOH used = 20 x 0.5 = 10 meq
2. Acid used by ammonia = 10 meq, so N = 10 x 14 = 140 mg and % N = 0.14/0.30 x 100 = 46.7
3. Urea (46.7% N) matches; the others are lower

Teacher's Note:
1. Use the standard factor method: the fraction of the element in the weighed product times its mass gives the mass of the element.
2. Divide by the mass of the sample and multiply by 100 for the percentage.

Answer: (d) \( \ce{C6H12O3} \)

Explanation:
1. Empirical mass = 44, and 132/44 = 3
2. Molecular formula = \( \ce{C6H12O3} \)

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (d) \( \ce{C3H6O3} \)

Explanation:
1. Molecular mass = 90 and empirical mass = 30
2. n = 3, so \( \ce{C3H6O3} \)

Teacher's Note:
1. Write the working as mass of element, then moles, then ratio. This avoids slips with atomic masses.
2. Round only when a ratio is within about 0.1 of a whole number.

Answer: (d) \( \ce{CH3COOH} \) and \( \ce{C6H12O6} \)

Explanation:
1. Both \( \ce{CH3COOH} \) (24/60) and \( \ce{C6H12O6} \) (72/180) contain 40% carbon
2. In the other pairs the values differ: ethanol 52.2%, sucrose 42.1%

Teacher's Note:
1. Use the standard factor method: the fraction of the element in the weighed product times its mass gives the mass of the element.
2. Divide by the mass of the sample and multiply by 100 for the percentage.

Answer: (c) \( \ce{(NH2)2CO} \)

Explanation:
1. O = 26.67%; moles C = 1.67, H = 6.67, N = 3.33, O = 1.67 give \( \ce{CH4N2O} \) (60)
2. This is urea, which on heating gives biuret and \( \ce{NH3} \), and biuret gives the violet colour

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (a) Chemical method

Explanation:
1. Both solids sublime, so sublimation does not separate them cleanly
2. Adding aqueous NaOH converts benzoic acid to water-soluble sodium benzoate, leaving camphor behind

Teacher's Note:
1. Match the technique to the property that differs between the components: boiling point, solubility, volatility or adsorption.
2. Questions often ask why one method is chosen over another.

Answer: (b) \( \ce{N2} \)

Explanation:
1. The compound is heated with \( \ce{CuO} \) and nitrogen oxides are reduced over hot copper
2. Free \( \ce{N2} \) is collected over KOH solution and its volume is measured

Teacher's Note:
1. Learn each test with its observation and the compound responsible for the colour or precipitate.
2. Remember which method suits which kind of sample (volatile, non-volatile, heat-sensitive).

Answer: (d) 30

Explanation:
1. S in 1.158 g \( \ce{BaSO4} \) = 32/233 x 1.158 = 0.159 g
2. % S = 0.159/0.53 x 100 = 30

Teacher's Note:
1. Use the standard factor method: the fraction of the element in the weighed product times its mass gives the mass of the element.
2. Divide by the mass of the sample and multiply by 100 for the percentage.

Answer: (c) \( \ce{Fe4[Fe(CN)6]3} \)

Explanation:
1. NaCN in the extract gives \( \ce{Na4[Fe(CN)6]} \) with \( \ce{FeSO4} \)
2. Adding \( \ce{Fe^{3+} \)} then forms ferric ferrocyanide \( \ce{Fe4[Fe(CN)6]3} \), Prussian blue

Teacher's Note:
1. Learn each test with its observation and the compound responsible for the colour or precipitate.
2. Remember which method suits which kind of sample (volatile, non-volatile, heat-sensitive).

Answer: (c) 58

Explanation:
1. 44.8 mL at STP corresponds to 44.8/22400 = 0.002 mol
2. M = 0.116 g / 0.002 mol = 58

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (c) 90

Explanation:
1. Ag = 0.162/108 = 0.0015 mol; a dibasic acid gives \( \ce{Ag2A} \), so the salt is 0.00075 mol
2. M(salt) = 0.228/0.00075 = 304; acid = 304 - 2 x 108 + 2 = 90

Teacher's Note:
1. Use the standard factor method: the fraction of the element in the weighed product times its mass gives the mass of the element.
2. Divide by the mass of the sample and multiply by 100 for the percentage.

Answer: (d) \( \ce{C6H12O6} \)

Explanation:
1. 1 g of H per 0.0833 mol means 12 g (12 atoms) of H per mole
2. \( \ce{(CH2O)n} \) with 2n = 12 gives n = 6, so \( \ce{C6H12O6} \)

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (b) 80.0

Explanation:
1. H = 0.9 x 2/18 = 0.1 g
2. C = 0.5 - 0.1 = 0.4 g, so % C = 0.4/0.5 x 100 = 80

Teacher's Note:
1. Use the standard factor method: the fraction of the element in the weighed product times its mass gives the mass of the element.
2. Divide by the mass of the sample and multiply by 100 for the percentage.

Answer: (b) \( \ce{NH2NH2.HCl} \)

Explanation:
1. In Lassaigne's test sodium fusion must form NaCN, which needs carbon in the compound
2. Hydrazine hydrochloride has no carbon, so no cyanide forms and the test fails

Teacher's Note:
1. Match the technique to the property that differs between the components: boiling point, solubility, volatility or adsorption.
2. Questions often ask why one method is chosen over another.

Answer: (d) \( \ce{3.6 x 10^{21} \)}

Explanation:
1. Moles of \( \ce{C2H5OH} \) = 0.046/46 = 0.001
2. H atoms = 0.001 x 6 x 6.022 x 10^23 = 3.6 x 10^21

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Plant pigments are coloured substances that move at different rates over the adsorbent
2. So the reason states the very principle that makes the separation possible

Teacher's Note:
1. Match the technique to the property that differs between the components: boiling point, solubility, volatility or adsorption.
2. Questions often ask why one method is chosen over another.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are true
2. The liberation of ammonia is the next step of the estimation and does not explain why nitrogen becomes ammonium sulphate

Teacher's Note:
1. Match the technique to the property that differs between the components: boiling point, solubility, volatility or adsorption.
2. Questions often ask why one method is chosen over another.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Thiophene is sulphonated more easily than benzene and dissolves in the acid layer
2. Both statements are true, but being heterocyclic and aromatic does not itself explain this greater reactivity

Teacher's Note:
1. Learn each test with its observation and the compound responsible for the colour or precipitate.
2. Remember which method suits which kind of sample (volatile, non-volatile, heat-sensitive).

Answer: (b) Mass spectrometer

Explanation:
1. A mass spectrometer measures the mass-to-charge ratio of ions very precisely
2. This gives accurate isotopic and atomic masses

Teacher's Note:
1. Learn each test with its observation and the compound responsible for the colour or precipitate.
2. Remember which method suits which kind of sample (volatile, non-volatile, heat-sensitive).

Answer: (c) \( \ce{C6H8N2} \)

Explanation:
1. Moles: C = 0.75, H = 1, N = 0.25, ratio 3 : 4 : 1
2. Empirical formula \( \ce{C3H4N} \) (54), and 108/54 = 2 gives \( \ce{C6H8N2} \)

Teacher's Note:
1. Write the working as mass of element, then moles, then ratio. This avoids slips with atomic masses.
2. Round only when a ratio is within about 0.1 of a whole number.

Answer: (c) 16.2

Explanation:
1. % N = 1.4 x normality x volume / mass = 1.4 x 0.2 x 58 / 1
2. = 16.2

Teacher's Note:
1. Convert every percentage or mass to moles first and divide by the smallest value to get the simplest ratio.
2. Check the answer by recomputing the formula mass and comparing it with the given molecular mass.