Question:

Which compound in each of the following pairs will react faster in SN^2 reaction with OH^-? (i) CH₃Br or CH₃I (ii) (CH₃)3​CCl or CH₃Cl

Answer:

(i) CH₃Br will react faster in SN2 reaction with OH^-

(ii) (CH₃)3CCl will react faster in SN2 reaction with OH^-.

Question:

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

Answer:

(i) 1-Butanol + NaOH → 1-Butoxide + H₂O

(ii) 1-Butoxide + Cl₂ → 1-Chlorobutane + NaCl

(iii) 1-Chlorobutane + HI → 1-Iodobutane + HCl

Question:

Out of C₆H₅CH₂​Cl and C₆H₅CH(Cl)C₆H₅, which is more easily hydrolysed by aqueous KOH?

Answer:

1. Aqueous KOH is a strong base, so it will react with the acidic protons of the compound.

2. C₆H₅CH₂Cl has one acidic proton, while C₆H₅CH(Cl)C₆H₅ has two acidic protons.

3. Therefore, C₆H₅CH₂Cl is more easily hydrolysed by aqueous KOH.

Question:

Write the mechanism of the following reaction: nBuBr+KCNEtOH−H₂​O——–>nBuCN

Answer:

1. nBuBr + KCNEtOH → nBuNEt + KBr

2. nBuNEt + H₂O → nBuOH + NEtOH

3. nBuOH + HCN → nBuCN + H₂O

Question:

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.

Answer:

(i) 1-Bromo-1-methylcyclohexane Major alkene formed: 2-Methylcyclohexene

(ii) 2-Chloro-2-methylbutane Major alkene formed: 2-Methyl-2-butene

(iii) 2,2,3-Trimethyl-3-bromopentane Major alkene formed: 2,3-Dimethyl-2-pentene

Question:

Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: i) (CH₃)2​CHCH(Cl)CH₃ ii) CH₃CH₂​CH(CH₃)CH(C₂H₅)Cl iii) CH₃CH₂​C(CH₃)2​CH₂​I iv) (CH₃)3​CCH₂​CH(Br)C₆H₅ v) CH₃CH(CH₃)CH(Br)CH₃ vi) CH₃C(C₂H₅)2​CH₂​Br vii) CH₃C(Cl)(C₂H₅)CH₂​CH₃ viii) CH₃CH=C(Cl)CH₂​CH(CH₃)2​ ix) CH₃CH=CHC(Br)(CH₃)2​ x) p−ClC₆H₄CH₂​CH(CH₃)2​ xi) m−ClCH₂​C₆H₄CH₂​C(CH₃)3​ xii) o−Br−C₆H₄CH(CH₃)CH₂​CH₃

Answer:

i) (CH₃)2​CHCH(Cl)CH₃ - secondary alkyl halide

ii) CH₃CH₂​CH(CH₃)CH(C₂H₅)Cl - tertiary alkyl halide

iii) CH₃CH₂​C(CH₃)2​CH₂​I - tertiary alkyl halide

iv) (CH₃)3​CCH₂​CH(Br)C₆H₅ - benzyl halide (secondary)

v) CH₃CH(CH₃)CH(Br)CH₃ - secondary alkyl halide

vi) CH₃C(C₂H₅)2​CH₂​Br - tertiary alkyl halide

vii) CH₃C(Cl)(C₂H₅)CH₂​CH₃ - tertiary alkyl halide

viii) CH₃CH=C(Cl)CH₂​CH(CH₃)2​ - vinyl halide

ix) CH₃CH=CHC(Br)(CH₃)2​ - vinyl halide

x) p−ClC₆H₄CH₂​CH(CH₃)2​ - aryl halide (primary)

xi) m−ClCH₂​C₆H₄CH₂​C(CH₃)3​ - aryl halide (primary)

xii) o−Br−C₆H₄CH(CH₃)CH₂​CH₃ - aryl halide (primary)

Question:

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?

Answer:

(i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride because the resonance of the double bonds in the chlorobenzene molecule reduces the overall dipole moment of the molecule, whereas the single bond in the cyclohexyl chloride molecule does not.

(ii) Alkyl halides, though polar, are immiscible with water because the non-polar hydrocarbon chain of the alkyl halide molecule does not mix with the polar water molecules.

(iii) Grignard reagents should be prepared under anhydrous conditions because water reacts with Grignard reagents to form an alkoxide, which can reduce the reactivity of the Grignard reagent and thus reduce the efficiency of the reaction.

Question:

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

Answer:

1. Freon 12: Freon 12 is a type of chlorofluorocarbon (CFC) that was used as a refrigerant in air conditioning and refrigeration systems. It has since been phased out due to its ozone-depleting effects.

2. DDT: DDT (dichlorodiphenyltrichloroethane) is an insecticide used to control vector-borne diseases such as malaria, yellow fever, and typhus. It is also used to control agricultural pests, such as locusts and moths.

3. Carbon Tetrachloride: Carbon tetrachloride is a chemical used as a solvent and in fire extinguishers. It is also used to make pesticides and other products.

4. Iodoform: Iodoform is a chemical used as an antiseptic and disinfectant. It is also used in the production of pharmaceuticals and in the manufacture of dyes and pigments.

Question:

The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.

Answer:

1. Alkyl chlorides are organic compounds composed of one or more alkyl groups bonded to a chlorine atom.

2. When alkyl chlorides are treated with aqueous KOH (potassium hydroxide), the chlorine atom is replaced by a hydroxyl group, resulting in the formation of an alcohol.

3. However, when alkyl chlorides are treated with alcoholic KOH (potassium hydroxide dissolved in an alcohol solvent), the chlorine atom is replaced by an alkoxide ion, which is then protonated by the alcohol solvent. This results in the formation of an alkene.

4. Therefore, when alkyl chlorides are treated with aqueous KOH, alcohols are the major products, whereas when they are treated with alcoholic KOH, alkenes are the major products.

Question:

p-Dichlorobenzene has higher melting point than those of o- and m-isomers. Discuss.

Answer:

1. First, it is important to understand the structure of p-dichlorobenzene and the other two isomers, o- and m-dichlorobenzene. P-dichlorobenzene has two chlorine atoms on opposite sides of the benzene ring, while o- and m-dichlorobenzene have the two chlorine atoms on adjacent sides of the benzene ring.

2. The difference in the structure of these three isomers affects the strength of the intermolecular forces between them. P-dichlorobenzene has stronger intermolecular forces than o- and m-dichlorobenzene due to its structure, which increases the melting point of p-dichlorobenzene.

3. This explains why p-dichlorobenzene has a higher melting point than the other two isomers. The stronger intermolecular forces require more energy to break them apart, resulting in a higher melting point.

Question:

Write the number of structural isomers of the compound having formula C₄​H₉Br. A 4 B 5 C 6 D 7

Answer:

Answer: C 6

Question:

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

Answer:

(i)

CH₃CH₂CH(CH₃)CH₂CH₂Cl

(ii)

ClCH₂C₆H₄Br

(iii)

CH₃CH₂CH(CH₃)CH₂CH₂Cl

(iv)

C8H₉ICl

(v)

CH₃CH₂Br

(vi)

C₇H₁₅I

(vii)

CH₃C(CH₃)2C₆H₄Br

(viii)

C₄H₇Br₂

Question:

Write the structure of the major organic product in each of the following reactions: (i) CH₃CH₂​CH₂​Cl+NaI⟶acetone/heat​ (ii) (CH₃)3​CBr+KOH⟶ethanol/heat​ (iii) CH₃CH(Br)CH₂​CH₃+NaOH⟶water (iv) (CH₃)3​CBr+KOH⟶ethanol/heat​ (v) C₆H₅ONa+C₂H₅Cl⟶ (vi) CH₃CH₂​CH₂​OH+SOCl₂​⟶ (vii) CH₃CH₂​CH=CH₂​+HBr⟶peroxide (viii) CH₃CH=C(CH₃)2​+HBr⟶

Answer:

(i) CH₃CH₂CH₂Cl + NaI → CH₃CH₂CH₂I + NaCl (major product)

(ii) (CH₃)3CBr + KOH → (CH₃)3COH + KBr (major product)

(iii) CH₃CH(Br)CH₂CH₃ + NaOH → CH₃CH(OH)CH₂CH₃ + NaBr (major product)

(iv) (CH₃)3CBr + KOH → (CH₃)3COH + KBr (major product)

(v) C₆H₅ONa + C2H₅Cl → C₆H₅OCl + C2H₅ONa (major product)

(vi) CH₃CH₂CH₂OH + SOCl₂ → CH₃CH₂CH₂Cl + SO2 + HCl (major product)

(vii) CH₃CH₂CH=CH₂ + HBr → CH₃CH₂CH(Br)CH₃ + H₂O (major product)

(viii) CH₃CH=C(CH₃)2 + HBr → CH₃CH(Br)C(CH₃)2 + H₂O (major product)

Question:

Arrange the compounds of each set in order of reactivity towards SN​₂ displacement. (i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane (ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane (iii) 1-Bromobutane, 1-Bromo-2, 2-dimethylpropane, 1-Bromo-2-methylbutane,1-Bromo-3-methylbutane.

Answer:

(i) 2-Bromopentane, 1-Bromopentane, 2-Bromo-2-methylbutane

(ii) 2-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 1-Bromo-3-methylbutane

(iii) 1-Bromobutane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane, 1-Bromo-2, 2-dimethylpropane

Question:

Give the IUPAC names of the following compounds: i) CH₃CH(Cl)CH(Br)CH₃ ii) CHF₂CBrClF iii) ClCH₂​C≡CCH₂​Br iv) (CCl₃)3​CCl v) CH₃C(p−ClC₆H₄)2​CH(Br)CH₃ vi) (CH₃)3​CCH=CClC₆H₄I−p

Answer:

i) 1,1-Dichloro-1-bromo-2-methylpropane

ii) 2,2-Dibromo-1-chloro-1,1-difluoroethane

iii) 1-Bromo-2-chloro-1-ethyl-1-ene

iv) Trichlorotrifluoromethane

v) 1,1-Dibromo-2-(4-chlorophenyl)-2-methylpropane

vi) 2,2-Dichloro-3-methyl-1-(4-iodophenyl)butane

Question:

What happens when: (i) n-butyl chloride is treated with alcoholic KOH. (ii) bromobenzene is treated with Mg in the presence of dry ether. (iii) chlorobenzene is subjected to hydrolysis. (iv) ethyl chloride is treated with aqueous KOH. (v) methyl bromide is treated with sodium in the presence of dry ether. (vi) methyl chloride is treated with KCN.

Answer:

(i) When n-butyl chloride is treated with alcoholic KOH, an alkoxide is formed through an SN2 reaction.

(ii) When bromobenzene is treated with Mg in the presence of dry ether, an organomagnesium halide is formed through an SN2 reaction.

(iii) When chlorobenzene is subjected to hydrolysis, an aryl alcohol is formed through an SN1 reaction.

(iv) When ethyl chloride is treated with aqueous KOH, an alkoxide is formed through an SN2 reaction.

(v) When methyl bromide is treated with sodium in the presence of dry ether, an organosodium halide is formed through an SN2 reaction.

(vi) When methyl chloride is treated with KCN, a cyanide is formed through an SN2 reaction.

Question:

How the following conversions can be carried out? (i) Propene to propan-1-ol (ii) Ethanol to but-1-yne (iii) 1-Bromopropane to 2-bromopropane (iv) Toluene to benzyl alcohol (v) Benzene to 4-bromonitrobenzene (vi) Benzyl alcohol to 2-phenylethanoic acid (vii) Ethanol to propanenitrile (viii) Aniline to chlorobenzene (ix) 2-Chlorobutane to 3, 4-dimethylhexane (x) 2-Methyl-1-propene to 2-chloro-2-methylpropane (xi) Ethyl chloride to propanoic acid (xii) But-1-ene to n-butyliodide (xiii) 2-Chloropropane to 1-propanol (xiv) Isopropyl alcohol to iodoform (xv) Chlorobenzene to p-nitrophenol (xvi) 2-Bromopropane to 1-bromopropane (xvii) Chloroethane to butane (xviii) Benzene to diphenyl (xix) tert-Butyl bromide to isobutyl bromide (xx) Aniline to phenylisocyanide

Answer:

(i) Propene to propan-1-ol: Propene can be oxidized to propan-1-ol using a strong oxidizing agent such as potassium permanganate.

(ii) Ethanol to but-1-yne: Ethanol can be oxidized to acetaldehyde, which can then be reacted with hydrogen cyanide to form but-1-yne.

(iii) 1-Bromopropane to 2-bromopropane: 1-Bromopropane can be reacted with a base such as sodium hydroxide to form 2-bromopropane.

(iv) Toluene to benzyl alcohol: Toluene can be oxidized to benzoic acid, which can then be reduced to benzyl alcohol using a reducing agent such as sodium borohydride.

(v) Benzene to 4-bromonitrobenzene: Benzene can be brominated using a brominating agent such as N-bromosuccinimide, followed by nitration using a nitrating agent such as nitric acid.

(vi) Benzyl alcohol to 2-phenylethanoic acid: Benzyl alcohol can be oxidized to benzaldehyde, which can then be reacted with acetic acid to form 2-phenylethanoic acid.

(vii) Ethanol to propanenitrile: Ethanol can be reacted with a cyanide source such as sodium cyanide to form propanenitrile.

(viii) Aniline to chlorobenzene: Aniline can be reacted with chlorine to form chlorobenzene.

(ix) 2-Chlorobutane to 3, 4-dimethylhexane: 2-Chlorobutane can be reacted with an alcohol such as methanol, followed by hydrogenation using a catalyst such as palladium to form 3, 4-dimethylhexane.

(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane: 2-Methyl-1-propene can be chlorinated using a chlorinating agent such as phosphorus pentachloride to form 2-chloro-2-methylpropane.

(xi) Ethyl chloride to propanoic acid: Ethyl chloride can be hydrolyzed using a base such as sodium hydroxide to form propanoic acid.

(xii) But-1-ene to n-butyliodide: But-1-ene can be reacted with an iodine source such as iodine monochloride to form n-butyliodide.

(xiii) 2-Chloropropane to 1-propanol: 2-Chloropropane can be reacted with a base such as sodium hydroxide to form 1-propanol.

(xiv) Isopropyl alcohol to iodoform: Isopropyl alcohol can be oxidized to acetone, which can then be reacted with an iodine source such as iodine monochloride to form iodoform.

(xv) Chlorobenzene to p-nitrophenol: Chlorobenzene can be nitrated using a nitrating agent such as nitric acid to form p-nitrophenol.

(xvi) 2-Bromopropane to 1-bromopropane: 2-Bromopropane can be reacted with a base such as sodium hydroxide to form 1-bromopropane.

(xvii) Chloroethane to butane: Chloroethane can be hydrolyzed using a base such as sodium hydroxide to form butane.

(xviii) Benzene to diphenyl: Benzene can be reacted with phenyl magnesium bromide to form diphenyl.

(xix) tert-Butyl bromide to isobutyl bromide: tert-Butyl bromide can be reacted with an alcohol such as methanol, followed by hydrogenation using a catalyst such as palladium to form isobutyl bromide.

(xx) Aniline to phenylisocyanide: Aniline can be reacted with phosgene to form phenylisocyanide.

Question:

Primary alkyl halide C₄​H₉Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C8​H18​ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.

Answer:

(a) Structural formula of C₄H₉Br:

CH₃CH₂CH₂CH₂Br

(b) Reaction of C₄H₉Br with alcoholic KOH:

C₄H₉Br + KOH (alcoholic) → CH₃CH₂CH₂CH₂OH + KBr

(c) Reaction of (b) with HBr:

CH₃CH₂CH₂CH₂OH + HBr → CH₃CH₂CH₂CH₂Br + H₂O

(d) Reaction of (a) with sodium metal:

C₄H₉Br + 2Na → CH₃CH₂CH₂CH₂CH₂CH₃ + 2NaBr

Question:

How will you bring about the following conversions? (i) Ethanol to but-1-yne (ii) Ethane to bromoethene (iii) Propene to1-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.

Answer:

(i) Ethanol to but-1-yne: Oxidation of ethanol using an oxidizing agent such as potassium permanganate to form acetaldehyde, followed by reaction of acetaldehyde with hydrogen in the presence of a catalyst to form but-1-yne.

(ii) Ethane to bromoethene: Reaction of ethane with bromine to form bromoethene.

(iii) Propene to1-nitropropane: Reaction of propene with nitric acid followed by reduction with hydrogen in the presence of a catalyst to form 1-nitropropane.

(iv) Toluene to benzyl alcohol: Reaction of toluene with hydrobromic acid followed by reduction with hydrogen in the presence of a catalyst to form benzyl alcohol.

(v) Propene to propyne: Reaction of propene with hydrogen in the presence of a catalyst to form propyne.

(vi) Ethanol to ethyl fluoride: Reaction of ethanol with hydrofluoric acid to form ethyl fluoride.

(vii) Bromomethane to propanone: Reaction of bromomethane with sodium hydroxide followed by oxidation with chromic acid to form propanone.

(viii) But-1-ene to but-2-ene: Reaction of but-1-ene with hydrogen in the presence of a catalyst to form but-2-ene.

(ix) 1-Chlorobutane to n-octane: Reaction of 1-chlorobutane with sodium hydroxide followed by reduction with hydrogen in the presence of a catalyst to form n-octane.

(x) Benzene to biphenyl: Reaction of benzene with chlorine to form dichlorobenzene, followed by reaction of dichlorobenzene with sodium hydroxide to form biphenyl.

Question:

What are ambident nucleophiles? Explain with an example

Answer:

Step 1: An ambident nucleophile is a molecule or ion that can act as both a nucleophile and an electrophile.

Step 2: A nucleophile is a species that donates a pair of electrons to form a new bond, while an electrophile is a species that accepts a pair of electrons to form a new bond.

Step 3: An example of an ambident nucleophile is the thiocyanate ion, SCN^-. It has a lone pair of electrons on the sulfur atom that can be used to form new bonds with electrophilic species, and it also has a negative charge that can be used to attract positively charged electrophiles.

Question:

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

Answer:

Answer: The hydrocarbon is an alkene, specifically pent-1-ene.