//M1//QN1//SUB//DL0//EQ

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)CH3 (ii) CH3CH2CH(CH3)CH(C2H5)Cl (iii) CH3CH2C(CH3)2CH2I (iv) (CH3)3CCH2CH(Br)C6H5 (v) CH3CH(CH3)CH(Br)CH3 (vi) CH3C(C2H5)2CH2Br (vii) CH3C(Cl)(C2H5)CH2CH3 (viii) CH3CH = C(Cl)CH2CH(CH3)2 (ix) CH3CH = CHC(Br)(CH3)2 (x) p-ClC6H4CH2CH(CH3)2 (xi) m-ClCH2C6H4CH2C(CH3)3 (xii) o-Br-C6H4CH(CH3)CH2CH3

//X

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

//M0//QN2//SUB//DL0//EQ

Give the IUPAC names of the following compounds: (i) CH3CH(Cl)CH(Br)CH3 (ii) CHF2CBrClF (iii) ClCH2C ≡ CCH2Br (iv) (CCl3)3CCl (v) CH3C(p – ClC6H4)2CH(Br)CH3 (vi) (CH3)3CCH = CClC6H4I – p

//X

(i)

(ii)

(iii)

(iv)

(v)

(vi)

//M0//QN3//SUB//DL0//EQ

Write the structure 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

//X

(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

Br – 1CH2 – 2CH = 3CH – 4CH2 – Br

//M0//QN4//SUB//DL0

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

//X

//M3//QN5//SUB//DL0//EQ

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

//X

A hydrocarbon with the molecular formula, C5H10 belongs to the group with a general molecular formula CnH2n. Therefore, it may either be an alkene or a cycloalkane.

Since hydrocarbon does not react with chlorine in the dark, it cannot be an alkene. Thus, it should be a cycloalkane.

Further, the hydrocarbon gives a single monochloro compound, C5H9Cl by reacting with chlorine in bright sunlight. Since a single monochloro compound is formed, the hydrocarbon must contain H-atoms that are all equivalent. Also, as all H-atoms of a cycloalkane are equivalent, the hydrocarbon must be a cycloalkane. Hence, the said compound is cyclopentane.

The reactions involved in the question are:

//M0//QN6//SUB//DL0//EQ

Write the isomers of the compound having formula C4H9Br.

//X

There are four isomers of the compound having the formula C4H9Br. These isomers are given below.

(i)

(ii)

(iii)

(iv)

//M0//QN7//SUB//DL0//EQ

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

//X

(i) 1-Iodobutane from 1-butanol

(ii) 1-Iodobutane from 1-chlorobutune

(iii) 1-Iodobutane from but-1-ene

//M0//QN8//SUB//DL0

What are ambident nucleophiles? Explain with an example.

//X

//M0//QN9//SUB//DL0//EQ

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

//X

(i) In the SN2 mechanism, the reactivity of halides for the same alkyl group increases in the order. This happens because as the size increases, the halide ion becomes a better leaving group.

R–F << R–Cl < R–Br < R–l

Therefore, CH3I will react faster than CH3Br in SN2 reactions with OH–.

(ii)

The SN2 mechanism involves the attack of the nucleophile at the atom bearing the leaving group. But, in case of (CH3)3CCl, the attack of the nucleophile at the carbon atom is hindered because of the presence of bulky substituents on that carbon atom bearing the leaving group. On the other hand, there are no bulky substituents on the carbon atom bearing the leaving group in CH3CI. Hence, CH3CI reacts faster than (CH3)3CCI in SN2 reaction with OH–.

//M0//QN10//SUB//DL0//EQ

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-l-methylcyclohexane (ii) 2-Chloro-2-methylbutane (iii) 2,2,3-Trimethyl-3-bromopentane.

//X

(i) 1-Bromo-l-methylcyclohexane:

(ii) 2-Chloro-2-methylbutane:

(iii) 2,2,3-Trimethyl-3-bromopentane:

//M0//QN11//SUB//DL0//EQ

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

//X

(i) Ethanol to but-l-yne:

(ii) Ethane to bromoethene:

(iii) Propene to l-nitropropane:

(iv) Toluene to benzyl alcohol:

(v) Propene to propyne:

(vi) Ethanol to ethyl fluoride:

(vii) Bromomethane to propanone:

(viii) But-l-ene to but-2-ene:[June 2025]

(ix) l-Chlorobutane to n-octane:

(x) Benzene to biphenyl: [March 2020, March2022, April 2022]

//M0//QN12//SUB//DL0//EQ

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?

//X

(i) In chlorobenzene, the Cl-atom is linked to a sp2 hybridized carbon atom.

In cyclohexyl chloride, the Cl-atom is linked to a sp3 hybridized carbon atom. Now, sp2 hybridized carbon has more s-character than sp3 hybridized carbon atom. Therefore, the former is more electronegative than the latter. Therefore, the density of electrons of C-Cl bond near the Cl-atom is less in chlorobenzene than in cyclohexyl chloride.

Moreover, the -R effect of the benzene ring of chlorobenzene decreases the electron density of the C-Cl bond near the Cl-atom. As a result, the polarity of the C-Cl bond in chlorobenzene decreases. Hence, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

(ii) To be miscible with water, the solute-water force of attraction must be stronger than the
solute-solute and water-water forces of attraction. Alkyl halides are polar molecules and so held together by
dipole-dipole interactions. Similarly, strong H- bonds exist between the water molecules.

The new force of attraction between the alkyl halides and water molecules is weaker than the alkyl halide-alkyl halide and water-water forces of attraction. Hence, alkyl halides (though polar) are immiscible with water.

• (iii) Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes.

Therefore, Grignard reagents should be prepared under anhydrous conditions.

//M0//QN13//SUB//DL0

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

//X

//M0//QN14//SUB//DL0//EQ

Write the structure of the major organic product in each of the following reactions:

(i) CH3– CH2 – CH2 – Cl + NaI (ii) (CH3)3CBr + KOH (iii) CH3CH(Br)CH2CH3 + NaOH (iv) CH3CH2Br + KCN (v) C6H5ONa + C2H5Cl (vi) CH3CH2CH2OH + SOCl2 (vii) CH3CH2CH = CH2 + HBr (viii) CH3CH = C(CH3)2 + HBr

//X

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

//M0//QN15//SUB//DL0//EQ

Write the mechanism of the following reaction: n BuBr + KCN n BuCN.

//X

The given reaction is: n BuBr + KCN n BuCH

The given reaction is an SN2 reaction. In this reaction, CN– acts as the nucleophile and attacks the carbon atom to which Br is attached. CN– ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.

//M0//QN16//SUB//DL0//EQ

Arrange the compounds of each set in order of reactivity towards SN2 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, l-Bromo-2,2-dimethylpropane, l-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.

//X

(i)

SN2 reactivity order: 1o > 2o > 3o.

(ii)

(iii) 1-Bromobutane > 1-Bromo-3-Methylbutane >

1-Bromo-2-Methylbutane > 1-Bromo-2,2-dymethylbutane

The steric hindrance to the nucleophile in the SN2 mechanism increases with decrease in the distance of the substituents from the atom containing the leaving group. Further, the steric hindrance increases with an increase in the number of substituents.

Hence, the increasing order of reactivity of the given compounds towards SN2 displacement is:

1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane < 1- Bromobutane

//M0//QN17//SUB//DL0//EQ

Out of C6H5CH2CI and C6H5CI(Cl)–C6H5, which is more easily hydrolysed by aqueous KOH.

//X

2° – carbocation is more stable than 1°– carbocation, so C6H5CH(CI)C6H5 gets easily hydrolysed.

//M0//QN18//SUB//DL0//EQ

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

//X

Three isomers of p-Dichlorobenzene

p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, it fits more closely than o- and m-isomers in the crystal lattice.

Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene.
As a result, p-dichlorobenzene has a higher melting point and lower solubility than o- and m-isomers.

//M0//QN19//SUB//DL0//EQ

How the following conversions can be carried out? (i) Propene to propan-l-ol (ii) Ethanol to but-l-yne (iii) l-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-l-propene to 2-chloro-2-methylpropane (xi) Ethyl chloride to propanoic acid (xii) But-l-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

//X

(i) Propene to propan-l-ol

(ii) Ethanol to but-l-yne

(iii) l-Bromopropane to 2-bromopropane

(iv) Toluene to benzyl alcohol

(v) Benzene to 4-bromonitrobenzene[July 2023]

(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-l-propene to 2-chloro-2-methylpropane

(xi) Ethyl chloride to propanoic acid

(xii) But-l-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[March 2024, March/April-2022, March 2020]

(xix) tert-Butyl bromide to isobutyl bromide

(xx) Aniline to phenylisocyanide

//M0//QN20//SUB//DL0//EQ

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.

//X

In an aqueous solution, KOH almost completely ionizes to give OH– ions. OH– ion is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol.

On the other hand, an alcoholic solution of KOH contains alkoxide (RO–) ion, which is a strong base. Thus, it can abstract a hydrogen from the b-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCI.

OH– ion is a much weaker base than RO– ion. Also, OH– ion is highly solvated in an aqueous solution and as a result, the basic character of OH– ion decreases. Therefore, it cannot abstract a hydrogen from the β-carbon.

//M0//QN21//SUB//DL0//EQ

Primary alkyl halide C4H9Br (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), C8H18 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.

//X

There are two primary alkyl halides having the formula, C4H9Br. They are n - butyl bromide and isobutyl bromide.

Therefore, compound (a) is either n-butyl bromide or isobutyl bromide.

Now, compound (a) reacts with Na metal to give compound (b) of molecular formula, C8H18 which is different from the compound formed when n-butyl bromide reacts with Na metal. Hence, compound (a) must be isobutyl bromide.

Thus, compound (d) is 2, 5-dimethylhexane.

It is given that compound (a) reacts with alcoholic KOH to give compound (b). Hence, compound (b) is 2-Methylpropane.

Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a).

Hence, compound (c) is 2-bromo-2-methylpropane.

//M0//QN22//SUB//DL0//EQ

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?

//X

(i) When n-butyl chloride is treated with alcoholic KOH, the formation of but-l-ene takes place. This reaction is a dehydrohalogenation reaction.

(ii) When bromobenzene is treated with Mg in the presence of dry ether, phenylmagnesium bromide is formed.

(iii) Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm to form phenol.

(iv) When ethyl chloride is treated with aqueous KOH, it undergoes hydrolysis to form ethanol.

(v) When methyl bromide is treated with sodium in the presence of dry ether, ethane is formed. This reaction is known as the Wurtz reaction.

(vi) When methyl chloride is treated with KCN, it undergoes a substitution reaction to give methyl cyanide.