Q: Which of the following reaction(s) can be used for the preparation of alkyl halides?

I and IV can be used due to presence of anhydrous ZnCl2 (III) gives alkyl halide due to formation of more stable carbocation.

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

Dichloromethane has highest dipole moment among CH2Cl2, CHCl3 and CCl4. The decreasing order of dipole moments is CH2Cl2>CHCl3>CCl4. These molecules have tetrahedral geometry due to sp3 hybridization of carbon atom.In CCl4, the individual C−Cl bond dipoles cancel each other which results in zero dipole moment. Hence, CCl4 is non polar.

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

The molecular formula of hydrocarbon is C5H10. It satisfies the general formula CnH2n . It suggests that the hydrocarbon is either alkene or cyclloalkane. It does not react with chlorine in dark. Hence, it cannot be alkene. Hence, it is cycloalkane. It gives a single monochloro compound C5H9Cl in bright sunlight. Hence, the hydrocarbon.is cyclopentane.

Q: Write the number of structural isomers of the compound having formula C4H9Br.

The isomers of the compound having formula C4H9Br are shown above. Among the above compounds, only the second compound shown is chiral. Hence, it has two optical isomers.

Question 1

Which of the following reaction(s) can be used for the preparation of alkyl halides?

Accepted Answer

I and IV can be used due to presence of anhydrous ZnCl2​ (III) gives alkyl halide due to formation of more stable carbocation.

Question 2

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) CH3​CH2​CH(CH3​)CH(C2​H5​)Cl   iii) CH3​CH2​C(CH3​)2​CH2​I           iv) (CH3​)3​CCH2​CH(Br)C6​H5​v) CH3​CH(CH3​)CH(Br)CH3​       vi) CH3​C(C2​H5​)2​CH2​Brvii) CH3​C(Cl)(C2​H5​)CH2​CH3​       viii) CH3​CH=C(Cl)CH2​CH(CH3​)2​ix) CH3​CH=CHC(Br)(CH3​)2​       x) p−ClC6​H4​CH2​CH(CH3​)2​xi) m−ClCH2​C6​H4​CH2​C(CH3​)3​   xii) o−Br−C6​H4​CH(CH3​)CH2​CH3​

Accepted Answer

The IUPAC names of various compounds are given below in order:(1) 2-Chloro-3-methylbutane (secondary alkyl halide)(ii) 3-Chloro-4-methylhexane (secondary alkyl halide)(iii) 1-Iodo-2,2-dimethylbutane (primary alkyl halide)(iv) 1-Bromo-3,3-dimethyl-1-phenylbutane (secondary benzyl halide)(v) 2-Bromo-3-methylbutane (secondary alkyl halide)(vi) 1-Bromo-2-ethyl-2-methylbutane (primary alkyl halide)(vii) 3-Chloro-3-methylpentane (tertiary alkyl halide)(viii)  3-Chloro-5-methylhex-2-ene (vinyl halide)(ix) 4-Bromo-4-methylpent-2-ene (allyl halide)(x) 1-Chloro-4-(2-methylpropyl) benzene (aryl halide)(xi) 1-Chloromethyl-3-(2,2-dimethylpropyl) benzene (primary benzyl halide)(xii) 1-Bromo-2-(1-methylpropyl) benzene (aryl halide)

Question 3

Give the IUPAC names of the following compounds:CH3​CH(Cl)CH(Br)CH3​

Accepted Answer

The IUPAC names of various compounds are given below in order:(i) 2-Bromo-3-chlorobutane(ii) 1-Bromo-1-chloro-1,22-trifluoroethane(iii) 1-Bromo-4-chlorobut-2-yne(iv) 2-(Trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane(v) 2-Bromo-3,3-bis (4-chlororphenyl) butane(vi) 1-Chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene

Question 4

Write the structures of the following organic halogen compounds.(i) 2-Chloro-3-methylpentane                             (ii) p-Bromochlorobenzene(iii) 1-Chloro-4-ethylcyclohexane                       (iv) 2r-(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

The structures of various compounds are given above.

Question 5

Which one of the following has the highest dipole moment?(i)CH2​Cl2​(ii)CHCl3​(iii)CCl4​

Accepted Answer

Dichloromethane has highest dipole moment among CH2​Cl2​, CHCl3​ and CCl4​. The decreasing order of dipole moments is CH2​Cl2​>CHCl3​>CCl4​. These molecules have tetrahedral geometry due to sp3 hybridization of carbon atom.In CCl4​, the individual C−Cl bond dipoles cancel each other which results in zero dipole moment. Hence, CCl4​ is non polar.

Question 6

A hydrocarbon C5​H10​ does not react with chlorine in dark but gives a single monochloro compound C5​H9​Cl in bright sunlight. Identify the hydrocarbon.

Accepted Answer

The molecular formula of hydrocarbon is C5​H10​. It satisfies the general formula Cn​H2n​ . It suggests that the hydrocarbon is either alkene or cyclloalkane. It does not react with chlorine in dark. Hence, it cannot be alkene. Hence, it is cycloalkane. It gives a single monochloro compound C5​H9​Cl in bright sunlight. Hence, the hydrocarbon.is cyclopentane.

Question 7

Write the number  of structural isomers of the compound having formula C4​H9​Br.

Accepted Answer

The isomers of the compound having formula C4​H9​Br are shown above. Among the above compounds, only the second compound shown is chiral. Hence, it has two optical isomers.

Question 8

What are ambident nucleophiles? Explain with an example.

Accepted Answer

The nucleophiles that can attack through two different sites are known as ambident nucleophiles.For example, cyanide ion is an ambident nucleophile. It can attack through either C atom or N atom to form alkyl cyanide or alkyl isocyanide.

Question 9

Which compound in each of the following pairs will react faster in SN2 reaction with OH−?(i) CH3​Br or CH3​I (ii) (CH3​)3​CCl or CH3​Cl

Accepted Answer

(i) Since, iodide ion is better leaving group than bromide ion, methyl iodide will react faster than methyl bromide in SN2 reaction with hydroxide ion.(ii) Due to steric hinderance in tert butyl chloride, methyl chloride is more reactive than tert butyl chloride towards SN2 reaction.

Question 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

The reactions are given above:(i) 1-methylcyclohexene is the major product.Only one alkene is formed at beta H atoms on either side of Br atoms are equivalent.(ii) 2-methyl-2-butene is the major product.Two different alkenes are formed. More substituted alkene is the major product as it is more stable. This is in accordance with Saytzeff's rule.(iii) 3,4,4-Trimethylpent-2-ene is the major product.Two different alkenes are formed. More substituted alkene is the major product as it is more stable. This is in accordance with Saytzeff's rule.

Question 11

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.

Accepted Answer

i) Ethanol to but-1-yneethanolCH3​−CH2​−OH​+SOCl2​Pyridine​ethyl chlorideCH3​−CH2​−Cl​+SO2​↑+HCl↑acetyleneHC≡CH​+NaNH2​liq.NH3​​sodium acetylideHC≡C−Na+​CH3​−CH2​−Cl+HC≡C−Na+→but−1−yneCH3​−CH2​−C≡CH​+NaClii) Ethane to bromoetheneethaneCH3​−CH3​​+Br2​UV light​CH3​−CH2​Bralc.KOH−HBr​CH2​=CH2​Br2​​Br−CH2​−CH2​−Br Br−CH2​−CH2​−Bralc.KOH−HBr​bromoetheneCH2​=CHBr​iii) Propene to 1-nitropropanepropeneCH3​−CH=CH2​​+HBrPeroxide​CH3​−CH2​−CH2​BrAgNO2​​1−nitropropaneCH3​−CH2​−CH2​NO2​​+AgBriv) Toluene to benzyl alcoholtolueneC6​H5​−CH3​​+Cl2​UV light​C6​H5​−CH2​ClNaOH​benzyl alcoholC6​H5​−CH2​OH​+NaClv) Propene to PropynepropeneCH3​−CH=CH2​​+Br2​CCl4​​CH3​−CH(Br)−CH2​(Br)NaNH2​liqNH3​​propyneCH3​−C≡CH​vi) Ethanol to ethyl fluorideethyl alcoholCH3​−CH2​OH​+HClZnCl2​​ethyl chlorideCH3​−CH2​−Cl​+H2​OCH3​CH2​−Cl+AgF→ethyl fluorideCH3​−CH2​−F​ (Swarts reaction)vii) Bromomthane to propanonebromomethaneCH3​−Br​KCN​CH3​−CNCH3​−Mg−Br​CH3​−CCH3​=NMgBrhydrolysis​propanoneCH3​−CO−CH3​​viii) But-1-ene to but-2-eneBut−1−eneCH3​−CH2​−CH=CH2​​HBrMarkownikoff′s rule​CH3​−CH2​−CH(Br)−CH3​alc.KOH−HBr​but−2−eneCH3​−CH=CH−CH3​​ix) 1-chlorobutane to n-octaneWurtz synthesis1−chlorobutane2CH3​−CH2​−CH2​−CH2​−Cl​+2Nadry ether​n−octaneCH3​−(CH2​)6​−CH3​​+2NaClx) Benzene to biphenylbenzeneC6​H6​​+Br2​dark,Fe​bromobenzeneC6​H5​−Br​Wurtz-Fittig reaction2C6​H5​−Br+2Nadry ether​biphenylC6​H5​−C6​H5​​+2NaBr

Question 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

(i) In chlorobenzene, the C atom of the C-Cl bond is sp2 hybridized and has more s character and electronegativity. In cyclohexyl chloride, the C atom of the C-Cl bond is sp3 hybridized, has less s character and less electronegative. Hence, the C-Cl bond in chlorobenzene is less polar than the C-Cl bond in cyclohexyl chloride.Also, in chlorobenzene, the lone pair of electron on the Cl atom is in resonance with a benzene ring. This gives partial double bond character to C-Cl bond and there is a decrease in the magnitude of partial negative charge on Cl atom. However, in cyclohexyl chloride, there is a C-Cl single bond. The dipole moment is a product of charge and distance. Also C-Cl bond with partial double bond character has lower bond length than with C-Cl single bond. Hence, chlorobenzene has lower dipole moment than cyclohexyl chloride.(ii) Inter-molecular attractive forces present in alkyl halides are dipole dipole interactions and in case of water molecules, intermolecular hydrogen bonds are present. When alkyl halides are added to water, the intermolecular attractive force present between an alkyl halide and water molecules are weaker than the forces that are already present between two alkyl halide molecules and two water molecules. Hence, alkyl halides are water-immiscible.(iii) Grignard reagents are highly reactive with moisture that is present in apparatus or starting materials. Hence, we should prepare Grignard reagents in the anhydrous condition.RMgX+HOH→RH+MgOHX

Question 13

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

Accepted Answer

The uses of various compounds ar given below.Freon 121) It is used as a refrigerant in refrigerators and air conditioners.2) It is used in aerosol spray propellants such as body sprays, hair sprays etc.DDT :1) It is used as an insecticide.2) It is effective against mosquitoes and lice.Carbon tetrachloride:1)  It is used in manufacturing refrigerants and propellants for aerosol cans.2) It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals.3) It is used as a solvent in the manufacture of pharmaceutical products.4) Till 1960's, it was used as a cleaning fluid, a degreasing agent and a fire extinguisher.Iodoform.:It was used as an antiseptic as it liberates free iodine when it comes in contact with the skin.

Question 14

Write the structure of the major organic product in each of the following reactions:CH3​CH2​CH2​Cl+NaIacetoneheat​(ii) (CH3​)3​CBr+KOHethanol heat​(CH3​)3​CBr+KOHethanolheat​(iv) (CH3​)3​CBr+KOHethanol heat​CH3​CH(Br)CH2​CH3​+NaOHwater​(vi) CH3​CH2​CH2​OH+SOCl2​ ⟶(CH3​)3​CBr+KOHethanolheat​(viii) CH3​CH=C(CH3​)2​+HBr ⟶

Accepted Answer

The major organic products are shown below:(i) CH3​CH2​CH2​Cl+NaI acetoneheat ​CH3​CH2​CH2​I(1−iodopropane)+NaCl(ii) (CH3​)3​CBr+KOHethanol heat​CH3​C(CH3​)=CH2​(2−methylpropene)+KBr+H2​O(iii) CH3​CH(Br)CH2​CH3​+NaOH water​CH3​−CHOHCH2​CH3​(2−butanol)(iv) (CH3​)3​CBr+KOHethanol heat​CH3​CH2​CN(propanenitrlie)+KBr(v)  C6​H5​ONa+C2​H5​Cl ⟶C6​H5​OCH2​H3​+NaCl(vi) CH3​CH2​CH2​OH+SOCl2​ ⟶CH3​CH2​CH2​−Cl(1−chloropropane)+HCl+SO2​(vii) CH3​CH2​CH=CH2​+HBr peroxide​CH3​CH2​CH2​CH2​−Br(1−bromobutane)(viii) CH3​CH=C(CH3​)2​+HBr ⟶CH3​CH2​C(Br)(CH3​)CH3​(2−bromo−2−methylbutane)

Question 15

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

Accepted Answer

This is an example of bimolecular nucleophilic substitution reaction SN​2. Cyanide ion acts as a nucleophile and bromide ion acts as leaving group. It is one step reaction and the mechanism is as shown above.

Question 16

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.

Accepted Answer

In SN2 reaction, steric factors determine the reactivity. more reactive alkyl halides have less steric hinderance. Hence, the decreasing order of the reactivity of alkyl halides is 1o>2o>3o.(i)  1-Bromopentane >  2-Bromopentane > 2-Bromo-2-methylbutane(ii) 1-Bromo-3-methylbutane >  2-Bromo-3-methylbutane >  2-Bromo-2-methylbutane(iii) 1-Bromobutane >  1-Bromo-3-methylbutane >  1-Bromo-2-methylbutane > 1-Bromo-2, 2-dimethylpropane

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Solution for Class 12, Chemistry, Haloalkanes and Haloarenes

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