Class 11 - Chemistry - Organic Chemistry
What are hybridization states of each carbon atom in the following compounds?
Indicate the σ and π bonds in the following molecules:
Write bond line formulas for Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.
Give the IUPAC names of the following compounds:
Which of the following represent the correct IUPAC name of the compounds concerned?
Draw formulas for the first five members of each homologous seris beginning with the following compounds.
(a) H-COOH (b) CH3COCH3 (c) HCH=CH2
Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for:
Identify the functional groups in the functional compounds.
Which of the two: O2NCH2CH2O- or CH3CH2O- is expected to be more stable and why?
O2NCH2CH2O- is more stable than CH3CH2O- because NO2 has I-effect and hence, it tends to disperse the negative charge on oxygen atom.
CH3CH2 exerts a +Ieffect and it tends to intensify the negative charge and destabilize it.
Explain why alkyl groups act as electron donors when attached to a π system.
Due to hyperconjugation, alkyl groups act as electron donor which when attached to pi-system.
The hydrogen which is shown by dotted line is firmly attached.
Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
What are electrophiles and nucleophiles? Explain with examples.
A nucleophile is a reagent that has an electron pair and is willing to donate it.
It is also known as a nucleus-loving reagent. Ex: NC–, OH–, R3C– (carbanions) etc.
An electrophile is a reagent which is in need of an electron pair and is also known as an electron-loving pair.
Ex: Carbonyl groups, CH3CH2+(Carbocations), Neutral molecules( due to the presence of a lone pair).
Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
Nucleophiles (OH–) and (CN– ) are (a) and (b)
Electrophiles(CH3C+O) are (c)
Classify the following reactions in one of the reaction type studied in this unit.
What is the relationship between the members of following pairs of structures?
Are they structural or geometrical isomers or resonance contributions?
Structural isomers. They have the same molecular formula but have different structures.
For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis.
Identify reactive intermediate produced as free radical, carbocation and carbanion.
Explain the terms Inductive and Electromeric effects.
Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?
Inductive effect is polarization of sigma bond due to electron donating or withdrawing nature of group attached.
For example: Attachment of chlorine (which is electronegative atom) to ethane molecule causes carbons to carry a partial positive charge.
This is induced polarity due to presence of Cl atoms.
Electromeric effect is the polarity produced in a multiple bonded compound as it is approached by a reagent.
It is temporary effect and original condition is restored if the reagent is removed.
As the number of halogen atoms decreases, overall negative inductive effective decreases. Hence, the acid strength decreases.
As the number of alkyl groups increases, positive inductive effect increases and acid strength decreases.
Give a brief description of the principles of the following techniques taking an example in each case.
Crystallization is used to purify solid organic compounds.
Principle: The principle on which it works is the difference in the solubility of the compound and impurities in a given solvent.
The impure compound is made to dissolve in the solvent at a higher temperature since it is sparingly soluble at lower temperatures.
This is continued till we get an almost saturated solution.
On cooling and filtering it, we get its’ crystals. Ex: By crystallizing 2-4g of crude aspirin in 20mL of ethyl alcohol, we get pure aspirin.
It is heated if needed and left undisturbed until it crystallizes. The crystals are then separated and dried.
This method is used to separate non-volatile liquids from volatile impurities.
It is also used when the components have a considerable difference in their boiling points.
Principle: The principle on which it works is that liquids having different boiling points vaporise at different temperatures.
They are then cooled and the formed liquids are separated.
Ex: A mixture of aniline (boiling point = 457 K) and chloroform (boiling point = 334 K) is taken in a round bottom flask having a condenser.
When they are heated, Chloroform, vaporizes first due to its high volatility and made to pass through a condenser where it cools down.
The aniline is left behind in the round bottom flask.
It is widely used for the separation and purification of organic compounds.
Principle: The principle on which it works is that individual components of a mixture
move at different paces through the stationary phase under the influence of mobile phase.
Ex: Chromatography can be used to separate a mixture of blue and red ink.
This mixture is placed on chromatogram where the component which is less absorbed by the chromatogram
moves faster up the paper than the other component which is almost stationary.
Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
Fractional crystallization is used to separate two compounds with different solubilities in solvent.
A hot saturated solution of two compounds is allowed to cool.
Less soluble crystallizes out while the more soluble remains in the solution.
Crystals are separated from mother liquor and mother liquor is again concentrated and hot solution again
allowed to cool down when crystals of second compound are obtained. These are filtered and dried.
What is the difference between distillation, distillation under reduced pressure and steam distillation?
Distillation is used for those liquids which are sufficiently stable at their boiling points and contain non-volatile impurities.
Distillation under reduced pressure is used to purify liquid which have very
high boiling point and decompose at or below their boiling points.
Steam distillation is used to purify volatile liquids associated with water immiscible impurities.
Discuss the chemistry of Lassaigne’s test.
Lassaigne’s test tests for nitrogen, sulphur, halogens and phosphorous present in an organic compound.
Initially, all the compounds are converted to ionic form by fusing compound with sodium metal.
Here, X = Cl, Br, I
Cyanide, sulphide or halide of sodium are extracted from fused mass by boiling with distilled water.
This is known as sodium fusion extract.
Difference between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl’s method
Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
Estimation of halogens:
On heating an organic compound with fuming HNO3 in the presence of AgNO3,
the halogen present in the compound forms silver halide which is collected, filtered, washed, dried and weighed.
If, weight of organic compound = W grams
Weight of silver halide = x gram
Estimation of sulphur:
The organic compound is first heated with fuming nitric acid.
It oxidizes sulphur to sulphuric acid which is precipitated as BaSO4 upon reaction with Ba(OH)2.
Estimation of Phosphorous:
A known mass of organic compound is heated with fuming nitric acid.
Phosphorous present in the compound is oxidised to phosphoric acid.
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.
Explain the principle of paper chromatography.
Paper chromatography is the simplest form of chromatography. A strip of paper acts as an adsorbent.
It is based on the principle of partial adsorption. The paper is made of cellulose fibres with molecules of water adsorbed on them.
This acts as stationary phase. Mobile phase is the mixture of the components to be identified prepared in s suitable solvent.
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Nitric acid is added to sodium extract.
NaCN + HNO3 ---> NaNO3 + HCN
Na2S + 2HNO3 ---> 2NaNO3 + H2S
Explain the reason for the fusion f an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Organic compound is fused with sodium metal so as to convert organic compounds into NaCN, Na2S, Na3PO4.
These ionic compounds become more reactive and can be easily tested by suitable reagents.
Name a suitable technique for separation of the components from a mixture of calcium sulphate and camphor.
This is the technique of sublimation because camphor can sublime whereas CaSO4 does not.
Explain, why an organic liquid vaporizes at a temperature below its boiling point in its steam distillation?
Organic liquid vaporizes at a temperature below its boiling point in its steam distillation
because in steam distillation the sum of vapour pressure of organic compound and steam should be equal to atmospheric pressure.
Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
CCl4 is completely non-polar covalent compounds whereas AgNO3 is ionic in nature.
They do not react and thus white precipitates of silver chloride will not be formed. As CCl4 is a covalent
compound, therefore, it does not ionize to give Cl- ions needed for the formation of ppt. of AgCl.
Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during
the estimation of carbon present in an organic compound?
Carbon dioxide is acidic in nature and it reacts with the strong base KOH to form K2CO3.
2KOH + CO2 --->K2CO3 + H2O
The increase in the mass of U-tube containing KOH then gives the mass of CO2 produced and from its mass, the percentage
of carbon in the organic compound can be calculated as:-
%C = (12/44) x (Mass of CO2 formed )/(Mass of substance taken) x 100
Why is it necessary to use acetic acid and not sulphuric acid for acidification of
sodium extract for testing sulphur by lead acetate test?
For testing sulphur, sodium extract is acidified with acetic acid.
This is because lead acetate is soluble and does not interfere with the test.
Pb(OCOCH3)2 + H2SO4 ---> PbSO4(ppt) + 2CH3COOH
If H2SO4 were used, lead acetate itself react with H2SO4 to form white ppt. of lead sulphate which will interfere
with the test.
An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen.
Calculate the masses of carbon dioxide and water produced
when 0.20 g of this substance is subjected to complete combustion.
Percentage of C = 69
Percentage of H = 4.8
Percentage of O = 100 – (69 + 4.8) = 26.2
Mass of compound = 0.2 g
100 g of compound contains 69 g of C.
0.2 g of compound contains
=(69 x 0.2)/(100)
= 0.138 g of C
12g of C produces 44 g of CO2.
0.138 g of C will give
=(44 x 0.138)/(12)
= 0.506 g of CO2
100 g of compound contains 4.8 g of H
Therefore, 0.2 g of compounds contains =(48 x 0.2)/(100)
= 0.0096 g of H
2 g of H produces 18 g of H2O
0.0096 g of H produces
=(4.8 x 0.2)/(100)
= 0.0864 g of H2O
Mass of CO2 produced = 0.506 g
Mass of H2O produced = 0.0864 g
A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method.
The ammonia evolved gas absorbed in 50 mL of 0.5 M H2SO4.
The residual acid required 60 Ml of 0.5 M solution of NaOH for neutralization.
Find the percentage composition of nitrogen in the compound.
Mass of compound taken = 0.50 g
Volume of H2SO4 = 50 mL
Molarity of H2SO4 = 0.5 M
Volume of NaOH required = 60 mL
Required, Molarity of NaOH = 0.5 M
Using Kjeldahl’s method
% of N = [1.4 x Mx 2[V-(V1 /2)]/(m)
Put, M = 0.5 M
V = 50 mL
V1 = 60 mL
m = 0.5 g
% of N = [1.4 x 0.5 x 2(50-(60/2)]/(0.5)
Percentage of N in given compound = 56.
0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation.
Calculate the percentage of chlorine present in the compound.
Mass of organic compound = 0.3780 g
Mass of silver chloride = 0.5740 g
1 mol of AgCl = 143.5 g
143.5 g of AgCl have 35.5 g of Cl
0.5740 g of AgCl have
= (35.5 x 0.574)/(143.5)= 0.142 g
0.142 g of Cl present in 0.3780 g of compound =[(0.142)/(0.378)] x 100 = 37.56%
In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate.
Find out the per centage of sulphur in the given compound.
Mass of organic sulphur compound = 0.468 g
Mass of BaSO4 = 0.668 g
1 mole of BaSO4 = 233 g
233 g of BaSO4 conatisn 32 g of S.
0.668 g of BaSO4 contains
=(0.668 x 32)/(233)
= 0.09174 g
Percentage of S in compound
In the organic compound CH2=CH-CH2-CH2-C=CH, the pair of hybridized orbitals involved in the formation of C2-C3 bond is:
(a) sp-sp2 (b) sp-sp3 (c) sp2-sp3 (d) sp3-sp3
Correct option is (c)
In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue color is obtained due to the formation of
Correct option is (b)
Prussian blue color in the Lassaigne’s test is due to the formation of Fe4 [Fe(CN)6]3
Which of the following carbocation is most stable?
Correct option is (b)
Most stable is tertiary carbocation.
The best and latest techniques for isolation, purification and separation of organic compounds is
Correct option is (d)
CH3CH2I + KOH (aq) --> CH3CH2OH + KI
Is classified as
(a)electrophilic substitution (b)nucleophilic substitution (c) elimination (d) addition
Correct option is (b). As the nucleophile I- is replaced by the nucleophile OH- ion.
KOH provides OH- ion for nucleophile attack.