Explore topic-wise InterviewSolutions in .

(c) 9

There are 9 cyclic and acyclic isomers are possible for the molecular formula C₃H₆O.

(a) None, carbon-carbon bonds in benzene are delocalized around the ring

(C)  benzene

(D)   Votamers

(b) 2 – Hydroxy Propanoic acid

Simple distillation can be used because the two compounds have a difference of more than 20° in their boiling points and both the liquids can be distilled without any decomposition.

It can be separated by steam distillation o-nitrophenol being volatile distills over along with water while p-nitrophenol being non-volatile remains in the flask.

Steam of distillation

Column chromatography

Steam distillation.

O-nitrophenol being volatile distils over along with water while p-nitrophenol being non-volatile remains in the flask.

The process Distillation under reduced pressure is used to purify a liquid which decomposes at its boiling point. 

1-methoxypropane, 

2-methoxypropane

Distillation under reduced pressure, i.e. vacuum distillation

The sodium fusion extract is boiled with iron(II) sulphate and then acidified with concentrated sulphuric acid. The formation of Prussian blue colour confirms the presence of nitrogen. Sodium cyanide first reacts with iron(II) sulphate and forms sodium hexacyanoferrate(II). On heating with concentrated sulphuric acid some iron(II) ions are oxidised to iron(III) ions which react with sodium hexacyanoferrate(II) to produce iron(III) hexacyanoferrate(II) (ferriferrocyanide) which is Prussian blue in colour.

6CN^– + Fe²⁺ → [Fe(CN)6]^4–

3[Fe(CN)6]₄– + 4Fe3⁺       Fe4[Fe(CN)₆]3.xH₂O

Prussian blue

O₂NCH₂CH₂O– is more stable than CH₃CH₂O– because NO₂ group has –I effect and hence it tends to disperse the negative charge on the oxygen atom. In contrast, CH₃CH₂ has +I effect. It, therefore, tends to intensify the negative charge and hence destabilizes it.

By Fractional Crystallisation.

Two compounds with different solubilities in a solvent S can be separated by fractional crysatllisation. When a hot saturated solution of these two compounds is alloed to cool, the less soluble compound crystallises out first while the more soluble remains in the solution. The crystals are separated from the mother liquor and the mother liquor is again concentrated and the hot solution is allowed to cool when the crystals of the second compound are obtained. These are again filtered and dried.

The greater the s character of the hybrid orbitals, the greater is the electronegativity. Thus, a carbon atom having an sp hybrid orbital with 50% s character is more electronegative than that possessing sp² or sp³ hybridised orbitals. 

Hybridisation influences the bond length and bond enthalpy (strength) in organic compounds. The sp hybrid orbital contains more s character and hence it is closer to its nucleus and forms shorter and stronger bonds than the sp³ hybrid orbital. The sp² hybrid orbital is intermediate in s character between sp and sp³ and, hence, the length and enthalpy of the bonds it forms, are also intermediate between them.

(a) 2-Butene

2-Butene will exhibit cis-trans isomerism.

(d) NH₄OH

NH₄OH reagent is used to distinguish between halogens (Cl, Br, I) in an organic compound.

(b) Thiourea

Thiourea sodium fusion extract of organic compound gives brilliant violet colour with sodium nitroprusside solution.

Weight of organic compound = 0.32 g

Weight of BaSO₄ formed = 0.466 g

233 g of BaSO₄ contains = 32 g of sulphate

0.466 g of BaSO₄ contain = \(\frac{32}{233}\) x \(\frac{0.466}{2.32}\) x 100

= \(\frac{32}{233}\) x \(\frac{46.6}{0.32}\) = 19.999 g of sulphur

% of sulphur = 20 %

(c) [Fe(CN)₅NOS]^4-

Weight of organic compound = 0.24 g

Weight of silver chloride = 0.287 g

143.5 g of AgCl contains = 35.5 g of Cl

0.287 g of AgCl contains = \(\frac{35.5}{143.5}\) x 0.287 g of Cl

% of chlorine = \(\frac{35.5}{143.5}\) x \(\frac{0.287}{0.24}\) x 100 = 29.58%

(b) 34%

An organic compound weighing 0.15 g gave on carius estimation, 0.12 g of silver bromide. The percentage of bromine in the compound will be close to 34%.

Weight of organic substance (w) = 0.40 g

Weight of silver iodide (x) = 0.235 g

Percentage of iodine = \(\frac{127}{235}\) x \(\frac{x}{w}\) x 100

= \(\frac{x}{w}\) x \(\frac{0.235}{0.40}\) x 100 = 31.75%

Alkanes are insoluble in water because alkanes are called hydrophobic hydrocarbons. They have phobia for water. These are insoluble because these cannot make hydrogen bonds with water molecules.

Because methane is formed by the decomposition of plant and animal matter lying under water in marshy areas. 

1M of 10 mL H₂SO₄=1M of 20 mL NH₃

1000 mL of 1M ammonia contains 14 g nitrogen

20 mL of 1M ammonia contains14*20/1000 g nitrogen

Percentage of nitrogen = 14*20*100/ 1000 *0.5 =56.0%

When both double bond and triple bonds are present, double bond is given preference while numbering the chain.

Thus option (c) is correct.

The answer is (c) Carboxy

The answer is (b) R-OH

The answer is (d) (CH₃)₃C⁺

(d) CH₃ - CH₃

The answer is (a) Br

Two examples of -I effect are: -NO₂, -F.

Neutral nucleophiles ⇒ Ammonia\((\overset{..}{N}H_3)\)

Water \((H_2\overset{..}{\underset{..}{O}})H\)

Negatively Charged nucleophiles ⇒ Cyanide ion(CN^–) Chloride ion(Cl^–)

The carbon and hydrogen present in the organic compound is detected by heating the compound with copper (II) oxide in a hard glass tube when carbon present in the compound is oxidized to CO₂ which can be tested with lime water and hydrogen is converted to water which can be tested with anhydrous copper sulphate which turns blue.

2C + 2CuO → 2Cu +CO₂

2H+CuO → Cu + H₂O

CO₂ +Ca (OH₎₂ → CaCO₃ + H₂O

5H₂O + CuSO₄ → CuSO₄.5H₂O

Carius method is used for estimation of Halogens.

(i) 2-Bromo-4-methylentane-3-one

(ii) 4-Methyl-2-nitropentane-3-one

(iii) 2-Ethoxy-4-methylhexane-3-one

(iv) 2-Methoxy,4-methylhexane-3-one

(v) 2-Methylpropanoic acid

Liebig method is used for estimation of carbon and hydrogen.

The answer is (c) NaCN

sp³ is the hybridisation of carbon atom in carbanion.

4 – ethyl – 5 – methyl – hex – 2 – ene

C₂H₂ (120 pm) < C₂H₄ (134 pm) < C₂H₆(154pm)

It is sp² hybridized.

CH₃C⁺H₂ < CH₃CH₂C⁺H₂ < CH₃CH₂H₂C⁺H₂ < CH₃CH₂C⁺HCH₃ < (CH₃)₃C⁺.

I < −F < −COOH < −CN < −NO₂.

C₂H₆ < C₂H₄ < C₂H₂.