NCERT Solution Class 11th Chemistry Chapter – 12 Organic Chemistry – Some Basic Principles and Techniques
| Textbook | NCERT |
| class | Class – 11th |
| Subject | Chemistry |
| Chapter | Chapter – 12 |
| Chapter Name | Organic Chemistry – Some Basic Principles and Techniques |
| Category | Class 11th Chemistry Question & Answer |
| Medium | English |
| Source | last doubt |
NCERT Solution Class 11th Chemistry Chapter – 12 Organic Chemistry – Some Basic Principles and Techniques
?Chapter – 12?
✍Organic Chemistry – Some Basic Principles and Techniques✍
?Question & Answer?
Q 1. What are hybridisation states of each carbon atom in the following compounds?
CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6
Answer:
(i) 
C–1 is sp2 hybridised.
C–2 is sp hybridised.
(ii) 
C–1 is sp3 hybridised.
C–2 is sp2 hybridised.
C–3 is sp2 hybridised.
(iii) 
C–1 and C–3 are sp3 hybridised.
C–2 is sp2 hybridised.
(iv) 
C–1 is sp2 hybridised.
C–2 is sp2 hybridised.
C–3 is sp hybridised.
(v) C6H6
All the 6 carbon atoms in benzene are sp2 hybridised.
Q 2. Indicate the σ and π bonds in the following molecules:
C6H6, C6H12, CH2Cl2, CH2 = C = CH2, CH3NO2, HCONHCH3
Answer:
(i) C6H6
There are six C – C sigma bonds, six C–H sigma bonds, and three C=C pi resonating bonds in the given compound.
(ii) C6H12
There are six C – C sigma bonds and twelve C–H sigma bonds in the given compound.
(iii) CH2Cl2
There are two C – Cl sigma bonds and two C–H sigma bonds in the given compound.
(iv) CH2 = C = CH2
There are two C–C sigma bonds, four C–H sigma bonds, and two C=C pi bonds in the given compound.
(v) CH3NO2
There are three C–H sigma bonds, one C–N sigma bond, one N–O sigma bond, and one N=O pi bond in the given compound.
(vi) HCONHCH3
There are four C–H sigma bonds, two C–N sigma bond, one N–H sigma bond, and one C=O pi bond in the given compound.
Q 3. Write bond-line formulas for: Isopropyl alcohol, 2, 3–dimethyl butanal, Heptan–4–one.
Answer:
Isopropyl alcohol
2, 3–dimethyl butanal
Heptan–4–one
Q 4. Give the IUPAC names of the following compounds:
(a) 
(b) 
(c) 
(d) 
(e) 
(f) Cl2CHCH2OH
Answer:
(a)
Propylbenzene
(b)
3–methylpentanenitrite
(c)
2, 5–dimethyl heptane
(d)
3–bromo–3–chloroheptane
(e)
3–chloropropanal
(f) Cl2CHCH2OH
2, 2–dichloroethanol
Q 5. Which of the following represents the correct IUPAC name for the compounds concerned?
(a) 2,2-Dimethylpentane or 2-Dimethylpentane
(b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane
(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane
(d) But-3-yn-1-ol or But-4-ol-1-yne
Answer: (a) The prefix di shows that there are two methyl groups in the chain. Thus, the correct IUPAC name would be 2,2-Dimethylpentane.
(b) The locant number should start from the minimum. Here, 2,4,7 is lower than 2,5,7. Thus, the correct IUPAC name would be 2,4,7-Trimethyloctane.
(c) If the substituents in the chain are in equivalent positions, then the lower number is given to the substituent group in alphabetical order. Thus, the correct IUPAC name would be 2-Chloro-4-methylpentane.
(d) Out of the two functional groups present in the given compound, the alcoholic group is the principal functional group. Thus, the parent chain will have an –ol suffix. Since the alkyne group is in C–3, the IUPAC name would be But–3–yn–1–ol.
Q 6. Draw formulas for the first five members of each homologous series beginning with the following compounds
(a) H–COOH
(b) CH3COCH3
(c) H–CH=CH2
Answer:
The first five members of each homologous series beginning with the given compounds are
(a)
H–COOH: Methanoic acid
CH3–COOH: Ethanoic acid
CH3–CH2–COOH: Propanoic acid
CH3–CH2–CH2–COOH: Butanoic acid
CH3–CH2–CH2–CH2–COOH: Pentanoic acid
(b)
CH3COCH3: Propanone
CH3COCH2CH3: Butanone
CH3COCH2CH2CH3 : Pentan-2-one
CH3COCH2CH2CH2CH3: Hexan-2-one
CH3COCH2CH2CH2CH2CH3 : Heptan-2-one
(c)
H–CH=CH2: Ethene
CH3–CH=CH2: Propene
CH3–CH2–CH=CH2: 1-Butene
CH3–CH2–CH2–CH=CH2: 1-Pentene
CH3–CH2–CH2–CH2–CH=CH2: 1-Hexene
Q 7. Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for :
(a)2,2,4-Trimethylpentane
(b)2-Hydroxy-1,2,3-propanetricarboxylic acid
(c) Hexanedial
Answer: (a) 2, 2, 4–trimethylpentane
Condensed formula
(CH3)2CHCH2C (CH3)3
Bond line formula :
(b) 2–hydroxy–1, 2, 3–propanetricarboxylic acid
Condensed Formula
(COOH)CH2C(OH) (COOH)CH2(COOH)
Bond line formula:
Functional groups:
Carboxylic acid (-COOH) and Hydroxyl (-OH) groups
(c) Hexanedial
Condensed Formula
(CHO)(CH2)4(CHO)
Bond line formula:
Functional groups:
Aldehyde (-CHO)
Q 8. Identify the functional groups in the following compounds.
(a) 
(b) 
(c) 
Answer: (a) Hydroxyl (–OH), Aldehyde (–CHO), Methoxy (–OMe),
C=C double bond
(b) Ester -(O = C – O)-, 1oAmino (–NH2) (aromatic), Diethylamine CH2N(C2H5)2 – 30 Amine group
(c) Nitro (–NO2),
C=C Ethylenic double bond
Q 9. Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and why?
Answer: Since NO2 belongs to the electron-withdrawing group, it shows –I effect. NO2 tries to decrease the negative charge on the compound by withdrawing the electrons toward it. This stabilizes the compound whereas the ethyl group belongs to the electron-releasing group and shows +I effect. This results in an increase in the negative charge on the compound thus destabilizing the compound. Hence, I would expect O2NCH2CH2O–to be more stable than CH3CH2O–.
Q 10. Explain why alkyl groups act as electron donors when attached to a π system.
Answer:
Due to hyperconjugation, an alkyl group behaves as an electron-donor group when attached to a π system. For example, look at propene.
The sigma electrons of the C-H bond get delocalized due to hyperconjugation. The alkyl is attached directly to an unsaturated system. The delocalization happens due to the partial overlap of an sp3-s sigma bond orbital with an empty p orbital of the n bond of an adjacent carbon atom.
This process can be shown as:
The overlap as shown above results in delocalization making the compounds more stable.
Q 11. Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
(a) C6H5OH
(b) C6H5NO2
(c) CH3CH = CH – CHO
(d) C6H5CHO
(e) C6H5-CH2
(f)CH3CH=CHCH2
Answer:
(a) The structure of C6H5OH is: 
Resonating structures: 
(b) The structure of C6H5NO2 is: 
Resonating structures: 
(c) CH3CH = CH – CHO
Resonating structures: 
(d) The structure of C6H5CHO is: 
Resonating structures: 
(e) C6H5-CH2
Resonating structures: 
(f) CH3CH=CHCH2
Resonating structures:
Q 12. What are electrophiles and nucleophiles? Explain with examples.
Answer: 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 electron deficiency atom).
Q 13. Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
(a) CH3COOH + HO–[latex]\rightarrow[/latex] CH3COO– + H2O
(b) CH3COCH3+ C–N [latex]\rightarrow[/latex] (CH3)2C(CN) + (OH)–
(c) C6H5 + CH3C+O [latex]\rightarrow[/latex] C6H5COCH3
Answer: 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.
An electrophile is a reagent which is in need of an electron pair and is also known as an electron-loving pair.
(a) CH3COOH + HO–[latex]\rightarrow[/latex] CH3COO– + H2O
It is a nucleophile since HO- is electron rich in nature.
(b) CH3COCH3+ C–N [latex]\rightarrow[/latex] (CH3)2C(CN) + (OH)–
It is a nucleophile since C–N is electron rich in nature.
(c) C6H5 + CH3C+O [latex]\rightarrow[/latex] C6H5COCH3
It is an electrophile since CH3C+O is electron-deficient in nature.
Q 14. Classify the following reactions in one of the reaction type studied in this unit.
(a) CH3CH2Br + HS–[latex]\rightarrow[/latex]CH3CH2SH + Br–
(b) (CH3)2 C=CH2 + HCl[latex]\rightarrow[/latex](CH3)2ClC–CH3
(c) CH3CH2Br + HO–[latex]\rightarrow[/latex]CH2=CH2 + H2O + Br–
(d) (CH3)3C–CH2 OH + HBr[latex]\rightarrow[/latex](CH3)2 CBrCH2CH3 + H2O
Answer: (a) Substitution (nucleophilic) reaction since bromine group gets substituted by –SH group.
(b) Addition (electrophilic) reaction since two reactant molecules combines to form a single product.
(c) Elimination (bimolecular) reaction since reaction hydrogen and bromine are removed to form ethene.
(d) Substitution (nucleophilic) reaction since rearrangement of atoms takes place.
Q 15. What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?
(a) 
(b) 
(c) 
Answer: (a) The given compounds are a pair of structural isomers since they have the same molecular formula but have different structures. These compounds differ in the position of the ketone group. For the first structure, it is in C-3 whereas, for the 2nd one, it is in C-2.
(b) The given compounds are a pair of geometrical isomers since they have the same molecular formula, sequence of covalent bonds and the same constitution but differ in the relative positioning of the atoms in space. These compounds differ in the positioning of the Deutrium and Hydrogen.
(c) The given compounds are a pair of contributing structures or canonical structures. They do not represent any real molecule and are purely hypothetical. They are also called resonance isomers.
Q 16. 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.
(a) 
(b)
(c) 
(d) 
Answer: (a) The bond cleavage can be shown as:
It comes under homolytic cleavage since one of the shared pair in a covalent bond goes with the bonded atom. A free radical is formed as the reaction intermediate.
(b) The bond cleavage can be shown as:
It comes under heterolytic cleavage since the shared remains with the carbon atom of propanone. A carbanion is formed as the reaction intermediate.
(c) The bond cleavage can be shown as:
It comes under heterolytic cleavage since the shared remains with the bromine ion. A carbocation is formed as the reaction intermediate.
(d) The bond cleavage can be shown as:
It comes under heterolytic cleavage since the shared remains with one of the fragments. A carbocation is formed as the reaction intermediate.
Q 17. Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?
(a) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH
(b) CH3CH2COOH > (CH3)2CHCOOH > (CH3)3C.COOH
Answer: Electrometric effect
The complete transfer of the shared pair of π electrons to either of the two atoms linked by multiple bonds in the presence of an attacking agent is called the electrometric effect. It can either be – E effect or +E effect.
– E effect: Occurs when electrons are moved away from the attacking agent
+ E effect: Occurs when electrons are moved towards the attacking agent
Inductive effect
Inductive effect involves the permanent displacement of sigma (σ) electrons along a saturated chain, whenever an electron withdrawing or electron donating group is present.
It can either be + I effect or – I effect. When an atom or group attracts electrons towards itself more strongly than hydrogen, it is said to possess – I effect.
When the force with which an atom attracts electrons towards itself is greater than that of hydrogen, it is said to exhibit +I effect.
(a) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH
The acidity increases with the increase in –I effect which is directly proportional to the number of chlorine atoms.
(b) CH3CH2COOH > (CH3)2 CHCOOH > (CH3)3C.COOH
The acidity increases with the increase in +I effect which is directly proportional to the number of alkyl groups.
Q 18. Give a brief description of the principles of the following techniques taking an example in each case.
(a) Crystallisation
(b) Distillation
(c) Chromatography
Answer: (a) Crystallisation
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.
(b) Distillation
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 (b.p = 457 K) and chloroform (b.p = 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.
(c) Chromatography
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.