Chemistry : Organic Chemistry – Some Basic Principles and Techniques

Aryl halides doesn't undergo nucleophilic substitution reactions under ordinary conditions because of

1. approach of nucleophile is retarded
2. carbon carrying halogen atoms is sp³-hybridised
3. the substrate molecule is destabilised due to resonance
4. partial double bond character between carbon and halogen

• 2 and 4 only

• 1 and 4 only

• 2 and 3 only

• 2, 3 and 4 only

A solution containing 1.8 g of a compound (empirical formula CH₂O) in 40 g of water is observed to freeze at -0.465°C. The molecular formula of the compound is (k_f of water = 1.86 kg K mol^-1)

• C₂H₄O₂

• C₃H₆O

• C₆H₁₂O₆

• C₅H₁₀O₅

The IUPAC name of the compound CH₃-CH(CH₃) CO- CH₃ is

• 3-methyl 2-butanone

• 2-methyl 3-butanone

• isopropyl methyl ketone

• methyl isopropyl ketone

Among the following pairs, the pair that illustrates stereoisomerism is

• 1-butanol and 2-butanol

• cis - 2-butene and trans - 2-butene

• dimethyl ether and ethanol

• acetone and propanal

The compound CHCl=CHCHOHCOOH with molecular formula C₄H₅O₃Cl can exhibit

• geometric, optical, position and functional isomerism

• geometric, optical and functional isomerism

• position and functional isomerism only

• geometric and optical isomerism only

In Lassaigne's test for the detection of halogens, the sodium fusion extract is first boiled with concentrated nitric acid. This is

• to remove silver halides

• to decompose Na₂S and NaCN, if present

• to dissolve Ag₂S

• to dissolve AgCN, if formed

In Lassaigne's test for the detection of halogens, the sodium fusion extract is first boiled with concentrated nitric acid. This is

• to remove silver halides

• to decompose Na₂S and NaCN, if present

• to dissolve Ag₂S

• to dissolve AgCN, if formed

0.25 g of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralize 10 cm³ of 0.5 M H₂SO₄. The percentage of nitrogen in the compound is

• 28

• 56

• 14

• 112

Lassaigne's test for the detection of nitrogen fails in

• H₂N-CO-NHNH₂.HCl

• NH₂-NH₂.HCl

• NH₂-CO-NH₂

• C₆H₅-NH-NH₂.HCl

When tetrahydrafuran is treated with excess HI, the product formed is

• 1, 4-diiodobutane

• 1,4-butanediol

• 2-iodotetrahydrofuran

• 4-iodo-1-butanol