Question 1

1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
(i) (CH3)2CHCH(Cl)CH2 (ii) CH3CH2CH(CH3)CH(C2H5)Cl
(iii) CH3CH2C(CH3)2CH I (iv) (CH3)2CCH2CH(Br)C6H5
(v) CH3CH(CH3)CH(Br)CH2 (vi) CH3C(C2H5)CH Br
(vii) CH3C(Cl)(C2H5)CH CH3 (viii) CH3CH=C(Cl)CH CH(CH3)2
(ix) CH3CH=CHC(Br)(CH3) (x) p-ClC6H4CH2CH(CH3)
(xi) m-ClCH2C6H4CH C(CH3)3 (xii) o-Br-C6H4CH(CH3)CH CH3

Accepted Answer

Solutions:
(i) (CH3)2CHCH(Cl)CH2: 3-Chloro-2-methylbutane; Alkyl halide (secondary)
(ii) CH3CH2CH(CH3)CH(C2H5)Cl: 4-Chloro-3-methylpentane; Alkyl halide (secondary)
(iii) CH3CH2C(CH3)2CH I: 2-Iodo-2-methylbutane; Alkyl halide (tertiary)
(iv) (CH3)2CCH2CH(Br)C6H5: 4-Bromo-2-methylbutylbenzene; Benzyl halide (primary)
(v) CH3CH(CH3)CH(Br)CH2: 3-Bromo-2-methylbutane; Alkyl halide (secondary)
(vi) CH3C(C2H5)CH Br: 2-Bromo-3-methylbutane; Alkyl halide (secondary)
(vii) CH3C(Cl)(C2H5)CH CH3: 2-Chloro-3-methylbutane; Alkyl halide (secondary)
(viii) CH3CH=C(Cl)CH CH(CH3)2: 3-Chloro-3-methylpent-1-ene; Vinyl halide
(ix) CH3CH=CHC(Br)(CH3): 4-Bromo-3-methylbut-1-ene; Vinyl halide
(x) p-ClC6H4CH2CH(CH3): 1-(p-Chlorophenyl)-2-methylethane; Benzyl halide (primary)
(xi) m-ClCH2C6H4CH C(CH3)3: 1-(m-Chlorobenzyl)-2,2-dimethylpropane; Benzyl halide (primary)
(xii) o-Br-C6H4CH(CH3)CH CH3: 2-(o-Bromophenyl)-3-methylpropane; Benzyl halide (secondary)

Classification is based on the position of halogen and nature of carbon attached. — Each compound is named according to IUPAC rules by identifying the longest carbon chain, numbering to give the halogen the lowest possible number, and naming substituents. Classification is done based on the carbon to which halogen is attached: alkyl (primary, secondary, tertiary), allyl (adjacent to double bond), benzyl (attached to benzyl carbon), vinyl (attached to alkene carbon), or aryl (attached directly to aromatic ring).

Question 2

2 Give the IUPAC names of the following compounds:
(i) CH3CH(Cl)CH(Br)CH3
(ii) CHF CBrClF
(iii) ClCH CºCCH Br
(iv) (CCl3) CCl
(v) CH3C(p-ClC6H4)CH(Br)CH3
(vi) (CH3)3CCH=CClC6H5

Accepted Answer

Solutions:
(i) 2-Chloro-3-bromobutane
(ii) 1-Bromo-1,1,2,2-tetrafluoro-2-chloroethane
(iii) 4-Bromo-1-chlorobut-2-yne
(iv) Carbon tetrachloride (CCl4)
(v) 2-(p-Chlorophenyl)-3-bromobutane
(vi) 1-(tert-Butyl)-2-chloro-1-phenylethene

Each name is derived by identifying the parent chain, numbering to give substituents lowest numbers, and naming substituents accordingly. — IUPAC naming involves identifying the longest chain, numbering to give substituents lowest possible numbers, and naming substituents in alphabetical order. For compounds with multiple halogens, prefixes like di-, tri- are used. For alkynes, the position of triple bond is indicated. For aromatic compounds, substituents on benzene ring are named with their positions.

Question 3

3 Write the structures of the following organic halogen compounds.
(i) 2-Chloro-3-methylpentane
(ii) p-Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-1-iodooctane
(v) 2-Bromobutane
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2-ene

Accepted Answer

Solutions:
(i) Structure of 2-Chloro-3-methylpentane: Pentane chain with Cl at C2 and methyl at C3
(ii) p-Bromochlorobenzene: Benzene ring with Br and Cl at para positions
(iii) 1-Chloro-4-ethylcyclohexane: Cyclohexane ring with Cl at C1 and ethyl at C4
(iv) 2-(2-Chlorophenyl)-1-iodooctane: Octane chain with I at C1 and 2-chlorophenyl at C2
(v) 2-Bromobutane: Butane chain with Br at C2
(vi) 4-tert-Butyl-3-iodoheptane: Heptane chain with tert-butyl at C4 and I at C3
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene: Benzene ring with Br at C1, sec-butyl at C4, methyl at C2
(viii) 1,4-Dibromobut-2-ene: But-2-ene chain with Br at C1 and C4

Structures involve drawing the carbon skeleton with substituents at specified positions. — Structures are drawn by identifying the parent hydrocarbon chain or ring, numbering carbons, and placing halogen and other substituents at the correct positions as per IUPAC names.

Question 4

4 Which one of the following has the highest dipole moment?
(i) CH2Cl2
(ii) CHCl3
(iii) CCl4

Accepted Answer

Answer: (ii) CHCl3 has the highest dipole moment.

Explanation:
- CH2Cl2 is polar but has two Cl atoms and two H atoms, dipoles partially cancel.
- CHCl3 has three Cl atoms and one H atom, dipole moments add up more effectively.
- CCl4 is symmetrical tetrahedral with four identical Cl atoms, dipoles cancel out, net dipole moment is zero.

Hence, CHCl3 has the highest dipole moment among the three. — Dipole moment depends on the molecular geometry and electronegativity differences. Symmetrical molecules like CCl4 have zero dipole moment. Molecules with asymmetrical distribution of polar bonds have higher dipole moments.

Question 5

5 A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.

Accepted Answer

Answer: The hydrocarbon is cyclopentane.

Explanation:
- C5H10 corresponds to either cyclopentane or an alkene.
- It does not react with chlorine in dark (no substitution), so no allylic or benzylic hydrogens.
- Gives a single monochloro compound in sunlight indicating substitution at one type of hydrogen.
- Cyclopentane has all hydrogens equivalent; substitution gives one monochloro derivative.
- Hence, the hydrocarbon is cyclopentane. — Alkenes react with chlorine by addition, giving multiple products. Cycloalkanes undergo substitution giving single monochloro derivative if all hydrogens are equivalent.

Question 6

6 Write the isomers of the compound having formula C4H9Br.

Accepted Answer

Answer: The isomers of C4H9Br are:

1. 1-Bromobutane (n-butyl bromide)
2. 2-Bromobutane (sec-butyl bromide)
3. 1-Bromo-2-methylpropane (isobutyl bromide)
4. 2-Bromo-2-methylpropane (tert-butyl bromide)

These are four isomers differing in the position of Br and branching of carbon chain. — Isomers are structural isomers differing in the position of bromine and carbon skeleton branching. The formula C4H9Br corresponds to four possible isomers as above.

Question 7

7 Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-1-ene.

Accepted Answer

Solutions:
(i) From 1-butanol:
C4H9OH + PI3 → C4H9I + H3PO3
Usually prepared in situ by reaction of red phosphorus and iodine.

(ii) From 1-chlorobutane:
C4H9Cl + 2NaI → C4H9I + 2NaCl (Finkelstein reaction)

(iii) From but-1-ene:
C4H8 + HI → C4H9I
Addition of HI to the double bond (Markovnikov addition).

These reactions convert the respective starting materials to 1-iodobutane. — 1-Iodobutane can be prepared by substitution of hydroxyl or chloro groups by iodine using PI3 or NaI respectively, or by addition of HI to but-1-ene.

Question 8

8 What are ambident nucleophiles? Explain with an example.

Accepted Answer

Answer: Ambident nucleophiles are nucleophiles that can attack through two different atoms, both having lone pairs.

Example: Cyanide ion (CN–) can attack through carbon or nitrogen.

Explanation:
- CN– can react via carbon to give alkyl cyanides (R–CN)
- Or via nitrogen to give alkyl isocyanides (R–NC)

Thus, ambident nucleophiles have two nucleophilic centers and can give different products depending on the site of attack. — Ambident nucleophiles possess two nucleophilic sites; their reactivity depends on reaction conditions and electrophile, leading to different products.

Question 9

9 Which compound in each of the following pairs will react faster in SN2 reaction with –OH?
(i) CH3Br or CH3I
(ii) (CH3)3CCl or CH3Cl

Accepted Answer

Answer:
(i) CH3I reacts faster than CH3Br in SN2 reaction because I– is a better leaving group than Br–.

(ii) CH3Cl reacts faster than (CH3)3CCl because SN2 reactions occur faster with primary alkyl halides than tertiary ones due to steric hindrance.

Explanation:
- SN2 reaction rate depends on the leaving group ability and steric hindrance.
- Iodide is a better leaving group than bromide.
- Primary alkyl halides are less hindered and react faster than tertiary ones. — Better leaving groups and less steric hindrance increase SN2 reaction rates.

Question 10

10 Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:
(i) 1-Bromo-1-methylcyclohexane
(ii) 2-Chloro-2-methylbutane
(iii) 2,2,3-Trimethyl-3-bromopentane.

Accepted Answer

Solutions:
(i) 1-Bromo-1-methylcyclohexane:
Possible alkenes: 1-methylcyclohexene (double bond between C1-C2 or C1-C6)
Major product: More substituted alkene (Saytzeff product) - 1-methylcyclohexene

(ii) 2-Chloro-2-methylbutane:
Possible alkenes: 2-methyl-1-butene and 2-methyl-2-butene
Major product: 2-methyl-2-butene (more substituted alkene)

(iii) 2,2,3-Trimethyl-3-bromopentane:
Possible alkenes: Multiple alkenes depending on elimination site
Major product: The most substituted alkene formed by elimination of H from adjacent carbons

Explanation:
Dehydrohalogenation proceeds via E2 mechanism; major alkene is the more substituted (Saytzeff) alkene. — Dehydrohalogenation with strong base removes HX to form alkenes. The major product is the more substituted alkene due to greater stability.

Question 11

11 How will you bring about the following conversions?
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethene
(iii) Propene to 1-nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl fluoride
(vii) Bromomethane to propanone
(viii) But-1-ene to but-2-ene
(ix) 1-Chlorobutane to n-octane
(x) Benzene to biphenyl.

Accepted Answer

Solutions:
(i) Ethanol to but-1-yne:
Ethanol → Ethyl bromide → Coupling with acetylide ion → But-1-yne

(ii) Ethane to bromoethene:
Ethane → Bromination to bromoethane → Dehydrohalogenation → Bromoethene

(iii) Propene to 1-nitropropane:
Propene + HNO2 (nitrous acid) under suitable conditions → 1-nitropropane

(iv) Toluene to benzyl alcohol:
Toluene → Chloromethylbenzene (with Cl2/hv) → Hydrolysis → Benzyl alcohol

(v) Propene to propyne:
Propene → 1,2-Dibromopropane → Treatment with alcoholic KOH → Propyne

(vi) Ethanol to ethyl fluoride:
Ethanol + HF or treatment with SF4 → Ethyl fluoride

(vii) Bromomethane to propanone:
Bromomethane + Mg → Methylmagnesium bromide + Ethyl formate → Propanone

(viii) But-1-ene to but-2-ene:
But-1-ene + HBr (peroxide) → 2-Bromobutane → Dehydrohalogenation → But-2-ene

(ix) 1-Chlorobutane to n-octane:
1-Chlorobutane + 4Na → Coupling → n-Octane

(x) Benzene to biphenyl:
Benzene + Chlorobenzene + AlCl3 (Friedel-Crafts) → Biphenyl — Each conversion involves standard organic reactions such as substitution, elimination, coupling, Grignard formation, and Friedel-Crafts reactions.

Question 12

12 Explain why
(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
(ii) alkyl halides, though polar, are immiscible with water?
(iii) Grignard reagents should be prepared under anhydrous conditions?

Accepted Answer

Answers:
(i) Dipole moment of chlorobenzene is lower than cyclohexyl chloride because in chlorobenzene, the lone pair on Cl is delocalized into the aromatic ring reducing the polarity of C–Cl bond. In cyclohexyl chloride, no such resonance exists, so dipole moment is higher.

(ii) Alkyl halides are polar but immiscible with water because they are largely nonpolar hydrocarbons with only a small polar C–X bond, so they do not form hydrogen bonds with water and are insoluble.

(iii) Grignard reagents are highly reactive towards water and moisture; they react with water to give alkanes and destroy the reagent. Hence, they must be prepared under anhydrous conditions to prevent decomposition. — Resonance reduces dipole moment; polarity alone does not ensure solubility; moisture destroys Grignard reagents.

Question 13

13 Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

Accepted Answer

Answers:
- Freon 12 (CCl2F2): Used as refrigerant and propellant in aerosol sprays.
- DDT (Dichlorodiphenyltrichloroethane): Used as an insecticide.
- Carbon tetrachloride (CCl4): Used as a solvent, fire extinguisher, and in refrigeration.
- Iodoform (CHI3): Used as an antiseptic and disinfectant. — These compounds have specific industrial and medicinal applications due to their chemical properties.

Question 14

14 Write the structure of the major organic product in each of the following reactions:
(i) CH3CH2CH2Cl + NaI
(ii) (CH3)3CBr + KOH
(iii) CH3CH(Br)CH2CH3 + NaOH
(iv) CH3CH2Br + KCN
(v) C2H5ONa + C2H5Cl
(vi) CH3CH2CH2OH + SOCl2
(vii) CH3CH2CH=CH2 + HBr
(viii) CH3CH= C(CH3)2 + HBr

Accepted Answer

Solutions:
(i) CH3CH2CH2I (Finkelstein reaction)
(ii) (CH3)3COH (tert-butanol) (Elimination to alcohol)
(iii) CH3CH(OH)CH2CH3 (Substitution to alcohol)
(iv) CH3CH2CN (Nucleophilic substitution)
(v) C2H5OC2H5 (Diethyl ether) (Williamson ether synthesis)
(vi) CH3CH2CH2Cl (Substitution of OH by Cl)
(vii) CH3CH2CH2Br (Markovnikov addition of HBr)
(viii) CH3CHBrC(CH3)3 (Markovnikov addition of HBr)

Structures correspond to the products formed by substitution, elimination, or addition reactions as per reagents. — Each reaction involves typical organic transformations: nucleophilic substitution, elimination, addition, or ether formation.

Question 15

15 Write the mechanism of the following reaction:
nBuBr + KCN → nBuCN

Accepted Answer

Answer:
The reaction proceeds via SN2 mechanism:
1. The nucleophile CN– attacks the electrophilic carbon attached to Br in nBuBr from the backside.
2. The C–Br bond breaks simultaneously as C–CN bond forms.
3. The product n-butyl cyanide (nBuCN) is formed with inversion of configuration at carbon.

Mechanism steps:
- Nucleophilic attack by CN–
- Transition state with partial bonds
- Departure of Br– leaving group

This is a classic bimolecular nucleophilic substitution reaction. — SN2 reactions involve backside attack, simultaneous bond making and breaking, and inversion of stereochemistry.

Objectives

Objectives — Study Notes

Objectives

Practice Questions — Objectives

All 5 Chapters in Chemistry-II