Important Questions on Organic Chemistry

Organic Chemistry is a branch of chemistry that deals with the study of hydrocarbons and their derivatives. Organic compounds are thought to be fundamentally different from inorganic compounds because they are derived from living natural sources. 

Some basic principles and techniques of organic chemistry cover important topics such as shapes of organic compounds, the nomenclature of organic compounds, isomerism, fundamental organic reaction mechanisms, electron displacement effects and electromeric effect. Questions include the qualitative analysis of organic compounds.

Read More: Classification of Organic Compounds

---

Very Short Answer Questions (1 Mark Questions)

Ques. Give the bond-line formula of Isopropyl alcohol.

Ans. The bond-line formula is as follows:

Ques. Why do we fuse an organic compound with sodium metal for testing halogens, nitrogen and sulphur?

Ans. There is a conversion from the covalent form into the ionic form when the compound is fused with sodium metal for the elements present in the compound.

Ques. What will be the chemical composition of the compound containing both N and S when ferric chloride is added to it.

Ans. The chemical composition of the compound is

Ques. In Lassaigne's test, can potassium be used in place of sodium?

Ans. No, because potassium is more reactive as compared to sodium.

Ques. Answer the following questions from the figure given below:
(a) Identify the functional group isomers.
(b) Give compounds representing
(i) Chain Isomerism
(ii) Position Isomerism

Ans.

(a) I - V; I - VI; I - VII; II - V; II - VI; II - VII; III - V; III - VI; III - VII; IV - V; IV - VI; IV - VI are functional group isomers.

(b) (i) Chain Isomerism: I and I, I and IV, II and III and IV.

(ii) Position Isomerism: I and III, III and IV, VI and VII (they have different positions of -OH and -O- group.)

Ques. Why do diazonium salts not show Lassaigne's Test?

Ans. Usually, the diazonium salts leave N₂ on the heating way before they get a chance to react with the fused sodium metal. That is why the diazonium salts do not show positive Lasseigne's test for nitrogen.

Read More: Preparation of Diazonium Salts

Ques. Give an example of a Non- Benzoid Compound.

Ans. An example of Non- Benzoid Compound is: Aniline

---

Short Answer Questions (2 Marks Questions)

Ques. Expanded the given condensed formulas into their structural formula.
(i) CH₃CH₂COCH₂CH₃.
(ii) CH₃CH=CH(CH₂)₃CH₃

Ans. The Structural formulas are as follows:

(i)

(ii)

Ques. Write the number of σ and π bonds in the molecules given below. Also, explain why electrons are available easily for attacking reagents in π – bonds?
(i) CH₂=C=CHCH₃
(ii) HC≡CC≡CCH₃

Ans. (i) σ C = C : 3

σ C – H : 6 σ C – H : 6

π C = C : 3 π C = C : 2

(ii) σ C = C : 4

The electrons are easily available for attacking reagent because the electron charge cloud of the π – bond is situated above and below the bonding plane of atoms.

Read More: Atoms & Molecules: Definition, Properties, Key Differences

Ques. Structural Isomers are those compounds that have the same molecular formula but are different in structures. What type of structural isomerism is shown by the following.

Ans.

These are position isomers as the functional groups are attached to different carbon atoms m carbon chain. They are not metamers because the number of C atoms on either side of-S—are same: 

CH₃​−S−CH₂​−CH_2​−CH−3.

Methyl n-propyl thioether

Ques. Give the structures of various carbonations that can result from 2-methyl butane and arrange these in the increasing order of stability.

Ans. These 4 structures can be obtained from 2-methyl butane

Order of stability according to increasing stability: (III) > (II) > (IV) > (I).

Reason: Because (I) and (IV) are primary, (II) has secondary carbocation and (III) has tertiary carbocation.

Read More: Carbocation Stability

Ques. Define:
(i) Homologous Series
(ii) Hetrolytic Cleavage

Ans. (i) Homologous Series: This is the series in which a group or series of organic compounds each have a characteristic functional group. The members of this series are called homologous.

(ii) Hetrolytic Cleavage: When a bond breaks and the shared pair of electrons remain with one of the fragments it is known as heterolytic cleavage.

Ques. Name the following compounds.

Ans.

(i) 3-ethyl-4-methylhept-5-en-2-one

(ii) 3-nitrocyclohex-1-en

Ques. Explain with resonance which of the following Ions is more stable.

Ans.

Structure I is more stable due to resonance. (See resonance structure ‘A’ and ‘B’). No resonance is possible in structure II.

Read More: Ion: Definition, Anions, Cations and Examples

---

Long Answer Type Questions (3 Marks Questions)

Ques. Differentiate between Inductive Effect and Resonance Effect.

Ans. The difference between the Inductive Effect and the Resonance Effect is as follows:

Ques. Write a short note on the Electromeric effect (E Effect).

Ans. The electromeric effect or the E Effect can be described as the polarity which is produced in a multiple bonded compound when it gets attacked by a reagent. When a double or a triple bond gets exposed to an attack by any electrophile E+ (a reagent), the two π electrons that formed the π bond get completely transferred to either one atom or the other.

This effect is temporary which means that it remains as long as the attacking reagent is present and exposed to the organic compound. The molecule that was polarized goes back to its original state, once this attacking reagent is removed from the system.

Ques. What will be the shape of the following molecules?
(i) H₂ C=O
(ii) CH3F
(iii) HC≡N
Ans. (i) sp² hybridized carbon, trigonal planar
(ii) sp³ hybridized carbon, tetrahedral
(iii) sp hybridized carbon, linear.

Read More: Hybridized Orbitals

Ques. Two liquids (P) and (Q) can be easily separated by using the method of fractional distillation. The boiling point of a liquid (P) is less than the boiling point of a liquid (Q). Then which of the following liquids is expected to come out first in the distillate? Explain with a diagram.

Ans. We use fractional distillation for separating the components of any mixture if their boiling points differ by 20^oC or less. In the fractional distillation method, a fractionating column is used between a flask and a condenser. This fractionating column is supposed to provide the hurdles for the ascending vapours and provide the ample surface area required for condensing the higher boiling liquid. Then, the low boiling liquid (P)'s vapours will start moving up whereas those of the higher boiling liquid will condense and fall back into the flask. Therefore, as a result of this fractionation, the liquid P with a lower boiling point will distil first, and liquid Q with a higher boiling point afterwards.

Fractional Distillation

Ques. The amount of ammonia produced when 0.75g of any substance was kjeldahlized, neutralized 30cm³ of 0.25 N H₂SO4. Then, calculate the percentage of nitrogen in this compound.

Ans. Mass of organic compound = 0.75g

Vol. of H₂SO4 to be used = 30cm3

Normality of H₂SO4 = 0.25N

30cm³ of H₂SO4 of normality 0.25N ≡ 30ml of NH₃ solution of normality 0.25N

But 1000cm² of NH₃ of normality 1 contains 14g of nitrogen

∴ 30cm³ of 0.25N NH₃ contains nitrogen 

 

 

---

Very Long Answer Type Questions (5 Marks)

Ques. When is the bubble plate type fractioning columns required for separating 2 liquids? Explain the principle involved in the separation of the components of any mix of liquids using fractioning columns. Give its industrial applications and draw the diagram of a bubble plate type fractionating column.

Ans. Fractional distillation is a process that is used in the separation and purification of organic liquids from their non-volatile impurities or it is used for separating two or more two volatile liquids from any liquid mixture which have boiling points close to each other. As, in the process, the distillate gets collected infractions under different temperatures. This process is defined as fractional distillation.

When these liquids are present in the mixture and have their boiling points close to each other, the process of separation of pure liquids is difficult. The fractions which are collected are always contaminated. The repeated distillations are responsible for further purifying the fractions. Then, to decrease the number of distillations, this separation is done by fitting the distillation flask with a fractionating column which, in turn, is then connected to a condenser.

There are various types of fractionating columns as shown in the figure given below. The liquids forming a constant boiling mixture known as the azeotropic mixture cannot be separated using this method. Fractional distillation is used these days in the industry, specifically in the distillation of petroleum, coal tar and crude alcohol. A mixture using methanol (b.pt. 338K) and propanone (b.pt. 330K) or a mixture using benzene and toluene can be separated using fractional distillation.

Fractionating Columns

Read More: Steam Distillation: Principle, Applications, Examples

Ques. 2. 0.12g of an organic compound that contained phosphorus gave 0.22g of Mg₂P₂O7 by the usual analysis. Calculate the percentage of phosphorus in the compound.

Ans. Given is the mass of the compound = 0.12g

Also, the Mass of Mg₂P₂O7 formed = 0.22g of atoms of P

Then, 1 mole of Mg₂P₂O7 = (2x24+2x31+1687)

= 222g of Mg₂P₂O7

= 62%

i.e; 222g of Mg₂P₂O7 contain phosphorus = 62g.

∴ 0.22g of Mg₂P₂O7 will contain phosphorus.

 

This is equivalent to the amount of phosphorus present in 0.12g of organic compound

Hence, the percentage of phosphorus

 

Ques. Draw the resonance structures of
(i) CH₃COO-
(ii) C₆H₅NH2
(iii) C₆H₅OH
(iv) C₆ H₅ - C+H2

Ans. (i) CH₃COO-

The resonance structure of CH₃COO- is as follows:

(ii) C₆H₅NH₂

The resonance structure of C₆H₅NH₂ is as follows:

(iii) C₆H₅OH

The resonating structure of phenol is represented as follows:

(iv) C₆ H₅ - C + H₂

The resonance structure of C₆ H₅ - C + H₂ is as follows:

Ques. C₅H₁₁Br (A) is a given alkyl halide that reacts with ethanolic KOH to give an alkene ‘B’, this then reacts with Br2 and gives a compound ‘C’, which on dehydrobromination gives an alkyne ‘D’. When this is treated with sodium metal in liquid ammonia, 1 mole of ‘D’ give us 1 mole of the sodium salt of ‘D’ and 1/2 mole of the hydrogen gas. Complete the hydrogenation of ‘D’ which yields a straight-chain alkane. Identify A, B, C and D. Write the reactions involved.

Ans. 

The reaction suggests that (D) is a terminal alkyne. This implies that there is a triple bond at the end of the chain. It could be either one of (I) or (II).

As alkyne D on hydrogenation gives a straight-chain alkane, that is why structure I is the structure of alkyne D.

Structures of A, B and C are given below:

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates

---

Read Also: