Organic Chemistry - Some Basic Principles And Techniques

C₆H₆

There are 6 C-C sigma bonds, 6 C-H sigma bonds and 3 C=C pi resonating bonds.

C₆H₁₂

There are 6 C-C sigma bonds and 12 C-H sigma bonds.

CH₂C₁₂

There are 2 C-H sigma bonds,and 2 C-Cl sigma bonds.

CH₂=C=CH₂

There are 1 C-N sigma bonds, 3 C-H sigma bonds, 1 N-O sigma bond and 2 C=C pi bonds.

CH₃NO₂

There are 6 C-C sigma bonds, 6 C-H sigma bonds and 1 N=O pi resonating bonds.

HCONHCH₃

There are 2 C-N sigma bonds, 4 C-H sigma bonds and 1 N-H bond AND 1 C=O pi bond.

Isopropyl alcohol

2,3-Dimethylbutanal,

Heptan-4- one.

Or

(a) Propyl benzene

(b) 3-methyl-pentanitrile

(c) 2,5-dimethyl heptane

(d) 3-bromo-3-chloroheptane

(e) 3-chloropropanal

(f) 2,2-dichloro-ethanol

(a) The prefix di in the IUPAC name indicates that two identical substituent groups are present in the parent chain. Since two methyl groups are present in the C-2 of the parent chain of the given compound, the correct IUPAC name is 2,2-dimethylpentane.

(b) Locant number 2,4,7 is lower than 2,5,7. Hence, the IUPAC name of the given compound is 2,4,7-trimethyloctane.

(c) If the substituents are present in the equivalent position of the parent chain, then the lower number is given to the one that comes first in the name according to the alphabetical order. Hence, the correct IUPAC name is 2-chloro-4-methylpentane.

(d) Two functional groups-alcoholic and alkyne are present in the given compound. The principal functional group is the alcoholic group. Hence, the parent chain will be suffixed with ol. The alkyne is present in the C-3 of the parent chain. Hence, the correct IUPAC is but-3-yn-1-ol.

(a) H–COOH: Methanoic acid

CH₂-COOH:ethanoic acid

CH₂ –CH₂-COOH: propanoic acid

CH₂-CH₂ –CH₂-COOH: butanoic acid

CH₂-CH₂ –CH₂-CH₂-COOH: pentanoic acid

(b) CH₃COCH₃: propanone

CH₃COCH₂CH₃ : butanone

CH₃COCH₂CH₂CH₃ : Pentan-2-one

CH₃COCH₂CH₂CH₂CH₃ : Hexan-2-one

CH₃COCH₂CH₂CH₂ CH₂CH₃ : Heptan-2-one

(c) H-CH=CH₂: Ethene

CH₂-CH=CH₂:propene

CH₃-CH₂-CH=CH₂:1-butene

CH₃- CH₂CH₂-CH=CH₂:1-pentene

CH₃-CH₂- CH₂CH₂-CH=CH₂:1-hexene

Condensed formula: (CH₃)₂CHCH₂C(CH₃)₃

Bond line formula:

(b) Condensed formula: (COOH)CH₂C(OH)(COOH)CH₂(COOH)

Bond line formula:

(c) Condensed formula: (CHO)(CH₂)₄(CHO)

Bond line formula:

The functional groups present in the given compound are:-

Aldehyde(-CHO)

Hydroxyl(-OH)

Methoxy (-OMe)

C=C double bond(-C=C-)

Amino(-NH₂)

Ketone(C=O)

Diethylamine(N(C₂H₅)₂)

Nitro(-NO₂)

C=C double bond(-C=C-)

NO₂ is the electron withdrawing group. Hence, it shows –I effect. By withdrawing the electrons toward it, the NO₂ group decreases the negative charge on the compound, thereby stabilising it, as as shown below:

On the other hand, ethyl group is an electron-releasing group. Hence, the ethyl group shows +I effect. This increases the negative charge on the compound, thereby destabilising it. Hence, O₂NCH₂CH₂O- is expected to be more stable than CH₂CH₂O^-.

When an alkyl group is attached to a ∏ system, it acts as an electron-donor group by the process of hyper conjugation. To understand this concept better, let us take the example of propene.

In hyper conjugation, the sigma electrons of the C-H bond of an alkyl group are delocalised. This group is directly attached to an atom of an unsaturated system. The delocalisation occurs because of the partial overlap of a sp³-s sigma bond orbital with an empty p orbital of the ∏ bond of an adjacent carbon atom.

The process of hyper conjugation in propene is shown as :

This type of overlap leads to delocalisation (also known as no-bond resonance) of the ∏ electrons, making the molecule more stable.

1. An electrophile is a reagent that takes away an electron pair. In other words, an electron-seeking reagent is called electrophile (E⁺).

2. Electrophiles are electron deficient and can receive an electron pair.

3. Carbocations and neutral molecules having functional groups such as carbonyl group are examples of electrophiles.

4. A nucleophile is a reagent that brings an electron-pair. In other words, a nucleus seeking reagent is called nucleophile(Nu^-)

5. For example: OH^-, NC^-, carbanions etc.

6. Neutral molecules such as H₂O and ammonia also act as nucleophiles because of the presence of a lone pair.

(i) Here, HO^- acts as nucleophile as it is an electron rich species i.e., it is a nucleus seeking species.

(ii) Here, CN^- acts as nucleophile as it is an electron rich species i.e., it is a nucleus seeking species.

(iii) Here, CH₃C⁺Oacts as electrophile as it is an electron deficient species.

(a) It is an example of substitution reaction as in this reaction the bromine group in bromo ethane is substituted by the –SH group.

(b) It is an example of addition reaction as in this reaction two reactant molecules combine to form a single product.

(c) It is an example of elimination reaction as in this reaction hydrogen and bromine are removed from bromo ethane to give ethane.

(d) In this reaction, substitution takes place, followed by a rearrangement of atoms and groups of atoms.

Compounds having the same molecular formula but different structures are called structural isomers. The given compounds have the same molecular formula but they differ in the positions of the functional group(ketone group).

In structure 1, the ketone group is at the C-3 of the parent chain(hexane chain) and in structure 2, the ketone group is at the C-2 of the parent chain(hexane chain). Hence, the given pairs represents structural isomers.

Compounds having the same molecular formula, the same constituition , and the same sequence of covalent bonds,but with different relative position of their atoms in space are called geometrical isomers.

In structures 1 and 2, the relative position of Deuterium (D) and hydrogen(H) in space are different. Hence, the given pairs represent geometrical isomers.

The given structures are canonical structures or contributing structures. They are hypothetical and individually do not represent any real molecule. Hence, the given pairs represents resonance structures called resonance isomers.

(a) The bond cleaves using curved arrows to show the electron flow of the given reaction can be represented as

It is an example of homolytic cleavage as one of the shared pair in a covalent bond goes with the bonded atom. The reaction intermediate formed is free radical.

(b) The bond cleaves using curved arrows to show the electron flow of the given reaction can be represented as

It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the carbon of propanone. The reaction intermediate is carbanion.

(c) The bond cleaves using curved arrows to show the electron flow of the given reaction can be represented as

It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the bromine ion. The reaction intermediate is carbocation.

(d) The bond cleaves using curved arrows to show the electron flow of the given reaction can be represented as

It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with one of the fragments. The reaction intermediate is carbcation.

Inductive effect

The permanent displacement of sigma electrons along a saturated chain, whenever an electron withdrawing or donating group is present, is called inductive effect.

Inductive effect could be +I effect or –I effect. When an atom or group electrons towards itself more strongly than hydrogen, it is said to possess –I effect. For example.

When an atom or group attracts electrons towards itself less strongly than hydrogen, it is said to possess +I effect. For example.

Electrometric effect

It involves the complete transfer of the shared pair of ∏ electrons to the two atoms linked by multiple bonds in the presence of an attacking agent. For example

Electronic effect could be +E effect or –E effect.

+E effect: When the electrons are transferred towards the attacking reagent.

-E effect: When the electrons are transferred away from the attacking reagent.

(a) Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH

The order of acidity can be explained on the basis of inductive effect(-I) effect. As the number of chlorine atoms increases, the –I effect increases. With the increase in-I effect, the acid strength also increases accordingly.

(b) CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃C.COOH
The order of acidity can be explained on the basis of inductive effect(+I) effect. As the number of alkyl group increases, the +I effect increases. With the increase in +I effect, the acid strength also increases accordingly.

It is one of the most commonly used techniques for the purification of solid organic compounds.

Principle:- It is based in the difference in the solubilities of the compound and the impurities in a given solvent. The impure compound gets dissolved in the solvent in which it is sparingly soluble at room temperature, but appreciably soluble at higher temperature. The solution is concentrated to obtain a nearly saturated solution. On cooling the solution, the pure compound crystallises out and is removed by filteration.

For example, pure asprin is obtained by recrystallising crude aspirin. Approximately 2-4 g of crude aspirin is dissolved in about 20ml of ethyl alcohol. The solution is heated (if necessary) to ensure complete dissolution. The solution is then left undisturbed until some crystals start to separate out. The crystals are then filtered and dried.

1. This method is used to separate volatile liquids from non-volatile impurities or a mixture of those liquids that have a sufficient difference in their boiling points.

2. Principle:- It is based on the fact that liquids having different boiling points vaporises at different temperatures. The vapours are then cooled and the liquids so formed are collected separately.

3. For example:- a mixture of chloroform(b.p.=334K) and aniline(b.p.=457K) can be separated by the method of distillation. The mixture is taken in a round bottom flask fitted with a condenser. It is then heated. Chloroform, being more volatile vaporises first and passes into the condenser. In the condenser, the vapours condense and chloroform trickles down. In the round bottom flask, aniline is left behind.

4. The distillation set up is shown here:

1. It is one of the most useful methods for the separation and purification of organic compounds.

2. Principle:- It is based on the difference in movement of individual components of a mixture through the stationary phase under the influence of mobile phase.

3. For example- A mixture of red and blue ink can be separated by chromatography. A drop of the mixture is placed on the chromatogram. The component of the ink, which is less adsorbed on the chromatogram moves with the mobile phase while the less adsorbed component remains almost stationary.

4. The diagram is shown below:

1. Fractional Crystallisation is the method used separating two compounds with different solubilities in a solvent S. The process of fractional crystallisation is carried out in 4 steps.

(a) Preparation of the solution:- the powdered mixture is taken in the flask and the solvent is added to it slowly and stirred simultaneously. The solvent is added till the solute is just dissolved in the solvent. The saturated solution is then heated.

(b) Filtration of the solution:- the solution is then filtered through a filter paper in a china dish.

(c) Fractional crystallisation:- the solution in the china dish is now allowed to cool. The less soluble compound crystallises first, while the more soluble compound remains in the solution. After separating these crystals from the mother liquor, the latter is concentrated once again. The hot solution is allowed to cool and consequently the crystals of the more soluble compounds are obtained.

(d) Isolation and drying:- these crystals are separated from the mother liquor by filtration. Finally, the crystals are dried.

The set up is shown below:

Lassaigne’s test

This test is employed to detect the presence of nitrogen,sulphur,halogens and phosphorus in an organic compound. These elements are present in the covalent form in an organic compound. These are converted into ionic form by fusing the compound with sodium metal.

Na+ C +N→ NaCN

Na+ S+C+N→ NaSCN

2Na+S→ Na₂S

Na+ X→ NaX

(X= Cl, Br, I)

The cyanide,sulphide and halide of sodium formed are extracted from the fused mass by boiling it in distilled water. The extract so obtained is called lassaigne’s extract. This lassaigne’s extract is then tested for the presence of nitrogen, sulphur, halogens and phosphorus.

Estimation of halogens: It involves oxidising the organic substance with fuming nitric acid in the presence of silver nitrate. The halogen of the substance is thus converted to silver halide which is separated and weighed.

Weight of organic compound= W gm

Weight of silver halide=x gm

Estimation of sulphur: the organic substance is heated with fuming nitric acid but no silver nitrate is added. The sulphur of the substance is oxidised to sulphuric acid which is then precipitated as barium sulphate by adding excess of barium chloride solution. From the weight of BaSO₄ so obtained the percentage of sulphur can be calculated.

% Sulphur =

⇒ % Sulphur =

Estimation of phosphorous: The organic substance is heated with fuming nitric acid whereupon phosphorous is oxidised to phosphoric acid. The phosphoric acid is precipitated as ammonium phosphomolybdate. (NH₄)₃PO₄.12MoO₃ by the addition of ammonia and ammonium molybdate solution which is then separated, dried and weighed.

Where Molar mass of (NH₄)₃PO₄.12MoO₃ =1877g

If phosphorus is estimated as Mg₂P₂O₇

% of P=

1. In paper chromatography, chromatography paper is used. This paper contains water trapped in it, which acts as the stationary phase.

2. On the base of this chromatography paper, the solution of the mixture is spotted.

3. The paper strip is then suspended in a suitable solvent, which acts as the mobile phase. This solvent rises up the chromatography paper by capillary action and in the procedure, it flows over the spot.

4. The components are selectively retained on the paper (according to their differing position in these two phases). The spots of different components travel with the mobile phase to different heights.

5. The paper so obtained is known as a chromatogram.

While testing the lassaigne’s extract for the presence of halogens, it is first boiled with dilute nitric acid. This is done to decompose NaCN to HCN and Na₂S to H₂S and to expel these gases. That is, if any nitrogen and sulphur are present in the form of NaCN and Na₂S, then they are removed. The chemical equations involved in the reaction are represented as

NaCN + HNO₃ → NaNO₃ + HCN

Na₂S + 2HNO₃ → 2NaNO₃ + H₂S

Nitrogen, sulphur and halogens are covalently bonded in organic compounds. For their detection, they have to be first converted to ionic form. This is done by fusing the or. The chemical equations involved in the test are

Na+ C +N→ NaCN

Na+ S+C+N→ NaSCN

2Na+S→ Na₂S

Na+ X→ NaX

(X= Cl, Br, I)

Carbon, nitrogen, sulphur and halogen come from organic compounds.

The process of sublimation is used to separate a mixture of camphor and calcium sulphate. In this process, the sublimation compound changes from solid to vapour state without passing through the liquid state. Camphor is a sublimable compound and calcium sulphate is a non-sublimable solid. Hence, on heating, camphor will sublime while calcium sulphate will be left behind.

In steam distillation, the organic liquid starts to boil when the sum of vapour pressure due to the organic liquid(p₂) and the vapour pressure due to water(p₁) becomes equal to the atmospheric pressure(p), that is, p=p₁+p₂

Since,p₁>p₂,organic liquid will vapourise at a lower temperature than its boiling point.

CCl₄ will not give the white precipitate of AgCl on heating it with silver nitrate. This is because the chlorine atoms are covalently bonded to carbon in CCl_4. To obtain the precipitate, it should be present in ionic form and for this, it is necessary to prepare Lassaigne’s extract of CCl₄.

Carbon dioxide is acidic in nature and potassium hydroxide is a strong base. Hence,carbon dioxide reacts with potassium hydroxide to form potassium carbonate and water as

2KOH+CO₂→K₂CO₃+H₂O

Thus, the mass of the U-tube containing KOH increases. This increase in the mass of U tube gives the mass of CO₂ produced. From its mass,the percentage of carbon in the organic compound can be estimated.

Although the addition of sulphuric acid will precipitate lead sulphate, the addition of acetic acid will ensure a complete precipitation of sulphur in the form of lead sulphate due to common ion effect. Hence, it is necessary to use acetic acid for acidification of sodium extract for testing sulphur by lead acetate test.

Percentage of carbon in organic compound=69%

That is,100g of organic compound contains 69g of carbon

So, 0.2g of organic compound will contain==0.138g of C

Molecular mass of carbon dioxide, CO₂=44g

i.e., 12g of C will be contained in =0.506g of CO₂.

Thus,0.506g of CO₂ will be produced on complete combustion of 0.2g of organic compound.

Percentage of hydrogen in organic compound is 4.8

i.e., 100 g of organic compound contains 4.8g of hydrogen.

Therefore,0.2 g of organic compound will contain=(

=0.0096g of H

It is known that molecular mass of water=18g

Thus,2 g of hydrogen will be contained in=

=0.0864g of water.

Thus,0.0096g of water will be produced on complete combustion of 0.2g of the organic compound.

Given that, total mass of organic compound=0.50g

60ml of 0.5M solution of NaOH was required by residual acid for neutralisation.

60ml of 0.5M NaOH solution =ml of 0.5M H₂SO₄

=30ml of 0.5M H₂SO₄

Therefore, acid consumed in absorption of evolved ammonia is (50-30)ml=20ml

Again,20ml of 0.5M H₂SO₄=40ml of 0.5M NH₃

Therefore, 40ml of 0.5M NH₃ will contain=

=0.28g of N

Therefore, the % of nitrogen in 0.50g of organic compound==56%.

Given:-

Mass of organic compound=0.3780g

Mass of AgCl formed=0.5740g

1 mol of AgCl contains 1 mol of Cl.

Molecular wt. Of AgCl=143.32

(Mol. Wt. Of Ag=107.8 and Cl=35.5)

Thus, mass of chlorine in 0.5740g of AgCl

=

=0.1421g

Therefore, percentage of chlorine==37.59%

Hence, the percentage of chlorine present in the given organic chloro compound is 37.59%

Given:

Total mass of organic compound=0.468g

Mass of barium sulphate formed=0.688g

Since, 1 mol of BaSO₄=233g of BaSO₄=32g of sulphur

{molecular wt. Of Ba=137g, S=32g, O=16g}

{So, Molecular wt. Of BaSO₄ =137+32+4(16)=233g}

Thus, 0.688g of BaSO₄ contains g of sulphur=0.0917g of sulphur.

Therefore, percentage of sulphur==19.59%

Hence, the percentage of sulphur in the given compound is 19.59%

Correct option is (b)

C⁶H₂ = C⁵H – C⁴H₂ – C³H₂– C²≡ C¹H

In the given organic compound, the carbon atoms are numbered as 1,2,3,4,5,6 are sp,sp³,sp³,sp²,sp² hybridised respectively. Thus the pair of hybridised orbitals in the formaytion of C₂ – C₃ bond is: sp – sp_3.

(for simplicity for calculating hybridisation: double bond is count as 1, single bond is count as 1 and triple bond is count as 1)

Correct option is (b)

In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extract is boiled with iron(II) sulphate and then acidified with sulphuric acids. In the process, sodium cyanide first reacts with iron(II) sulphate and forms sodium hexacyanoferrate(II). Then, on heating with sulphuric acid, some iron(II) gets oxidised to form iron(III) hexacyanoferrate(III), which is Prussian blue in color. The chemical equations involved in the reaction is:

6CN^- + Fe⁺²→[Fe(CN)₆]^4-

Hence, the Prussian blue color is due to the formation of Fe₄[Fe(CN)₆]_3.

Correct option is (b)

is a tertiary carbocation(the carbon with a positive charge is attached with 3 methyl groups). A tertiary carbocation is the most stable carbocation due to the electron releasing effect of three methyl groups (+I effect). An increased +I effect by three methyl groups stabilizes the positive charge of the carbocation.

Order of stability of carbocation:- primary<secondary<tertiary.

Correct option is (d)

Chromatography is the most useful technique of separation and purification of organic compounds. It was first used to separate a mixture of coloured substances.

Correct option is (a)

CH₃CH₂I + KOH (aq) →CH₃CH₂OH + KI

The reaction is an example of nucleopihlic substitution reaction. The hydroxyl group of KOH (OH^-) with a lone pair of itself acts as a nucleophile and substitutes iodide ion in CH₃CH₂I to form ethanol.