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NCERT Solutions · Class 12 Chemistry · NCERT Chemistry Part I and II Part II · Chapter 6

Chapter 6: Haloalkanes and Haloarenes (Organic Chemistry)

Q: Arrange each set of compounds in order of increasing boiling points. (i) Bromomethane, Bromoform, Chloromethane, Dibromomethane. (ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.

(i) CH₃Cl < CH₃Br < CH₂Br₂ < CHBr₃. (ii) isopropyl chloride < 1-chloropropane < 1-chlorobutane.

Answers to all 9 intext questions and 22 exercises of Chapter 6, Haloalkanes and Haloarenes (NCERT Class 12 Chemistry Part II, 2026-27): IUPAC names and classification, structures, preparation, boiling points, SN1 and SN2 reactivity, elimination (Saytzeff rule), Grignard, Wurtz and Fittig reactions, polyhalogen compounds, and all the named conversions with reagents. All 31 questions are answered, with the key answer highlighted.

NCERT textbook Chapter 6 (PDF)

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Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-12-chemistry/chapter-6-haloalkanes-and-haloarenes

SN2 (one step, back-side attack, inversion): reactivity 1° > 2° > 3° (less crowding is faster). SN1 (via carbocation): 3° > 2° > 1°; benzylic and allylic halides are also fast. Leaving group: I⁻ > Br⁻ > Cl⁻. Aqueous KOH gives substitution (alcohols); alcoholic KOH gives elimination (alkenes, the more substituted alkene is the major product). Markovnikov addition of HX, but anti-Markovnikov for HBr with peroxide.

Intext questions

6.1

Write structures of the following compounds: (i) 2-Chloro-3-methylpentane (ii) 1-Chloro-4-ethylcyclohexane (iii) 4-tert-Butyl-3-iodoheptane (iv) 1,4-Dibromobut-2-ene (v) 1-Bromo-4-sec-butyl-2-methylbenzene.

Solution

(i) $CH X_{3} - CHCl - CH (CH X_{3}) - CH X_{2} - CH X_{3}$
(ii) a cyclohexane ring with Cl on C-1 and $- CH X_{2} CH X_{3}$ on C-4
(iii) $CH X_{3} CH X_{2} - CHI - CH (C (CH X_{3}) X_{3}) - CH X_{2} CH X_{2} CH X_{3}$
(iv) $BrCH X_{2} - CH = CH - CH X_{2} Br$
(v) a benzene ring with Br on C-1, $CH X_{3}$ on C-2 and $- CH (CH X_{3}) CH X_{2} CH X_{3}$ (sec-butyl) on C-4

As drawn above.

6.2

Why is sulphuric acid not used during the reaction of alcohols with KI?

Solution

$H X_{2} SO X_{4}$ converts KI into HI, but it is also an oxidising agent and oxidises HI to iodine ( $2 HI + H X_{2} SO X_{4} I X_{2} + SO X_{2} + 2 H X_{2} O$ ). So no HI is left to convert the alcohol to an alkyl iodide. A non-oxidising acid such as phosphoric acid is used instead.

H₂SO₄ oxidises the HI formed to I₂, so the alkyl iodide is not obtained.

6.3

Write structures of different dihalogen derivatives of propane.

Solution

Taking X as the halogen, there are four:

$CH X_{3} CH X_{2} CHX X_{2}$ (1,1-dihalopropane)
$CH X_{3} CHXCH X_{2} X$ (1,2-dihalopropane)
$XCH X_{2} CH X_{2} CH X_{2} X$ (1,3-dihalopropane)
$CH X_{3} CX X_{2} CH X_{3}$ (2,2-dihalopropane)

1,1-, 1,2-, 1,3- and 2,2-dihalopropanes.

6.4

Among the isomeric alkanes of molecular formula

C X_{5} H X_{12}

, identify the one that on photochemical chlorination yields (i) a single monochloride, (ii) three isomeric monochlorides, (iii) four isomeric monochlorides.

Solution

The number of monochlorides equals the number of different kinds of hydrogen.

(i) Neopentane, $C (CH X_{3}) X_{4}$ : all 12 H are equivalent, so one monochloride.
(ii) n-Pentane, $CH X_{3} CH X_{2} CH X_{2} CH X_{2} CH X_{3}$ : three kinds of H (C-1, C-2, C-3), so three monochlorides.
(iii) Isopentane, $(CH X_{3}) X_{2} CHCH X_{2} CH X_{3}$ : four kinds of H, so four monochlorides.

(i) Neopentane. (ii) n-Pentane. (iii) Isopentane (2-methylbutane).

6.5

Draw the structures of major monohalo products in each of the following reactions: (i) cyclohexanol +

SOCl X_{2}

(ii) 1-ethyl-4-nitrobenzene +

Br X_{2}

, heat or UV light (iii) 4-(hydroxymethyl)phenol + HCl, heat (iv) 1-methylcyclohexene + HI (v)

CH X_{3} CH X_{2} Br

+ NaI (vi) cyclohexene +

Br X_{2}

, heat or UV light.

Solution

(i) Chlorocyclohexane (the –OH is replaced by Cl; $SO X_{2}$ and HCl escape).
(ii) $O X_{2} N - C X_{6} H X_{4} - CH (Br) CH X_{3}$ : bromination at the benzylic carbon (free radical, light).
(iii) $HO - C X_{6} H X_{4} - CH X_{2} Cl$ : only the alcoholic –OH is replaced; the phenolic –OH is not.
(iv) 1-Iodo-1-methylcyclohexane: HI adds by Markovnikov's rule, I goes to the carbon carrying the methyl group.
(v) $CH X_{3} CH X_{2} I$ (Finkelstein reaction).
(vi) 3-Bromocyclohexene: substitution at the allylic position (free radical).

(i) chlorocyclohexane; (ii) 1-(1-bromoethyl)-4-nitrobenzene; (iii) 4-(chloromethyl)phenol; (iv) 1-iodo-1-methylcyclohexane; (v) iodoethane; (vi) 3-bromocyclohexene.

6.6

Arrange each set of compounds in order of increasing boiling points. (i) Bromomethane, Bromoform, Chloromethane, Dibromomethane. (ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.

Solution

(i) Chloromethane < bromomethane < dibromomethane < bromoform. Boiling point rises with molecular mass (stronger van der Waals forces).

(ii) Isopropyl chloride < 1-chloropropane < 1-chlorobutane. Branching lowers the surface area and the boiling point; a longer chain raises it.

(i) CH₃Cl < CH₃Br < CH₂Br₂ < CHBr₃. (ii) isopropyl chloride < 1-chloropropane < 1-chlorobutane.

6.7

Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism? Explain your answer. (i)

CH X_{3} CH X_{2} CH X_{2} CH X_{2} Br

CH X_{3} CH X_{2} CHBrCH X_{3}

(ii)

CH X_{3} CH X_{2} CHBrCH X_{3}

(CH X_{3}) X_{3} CBr

(iii)

(CH X_{3}) X_{2} CHCH X_{2} CH X_{2} Br

CH X_{3} CH X_{2} CH (CH X_{3}) CH X_{2} Br

Solution

SN2 needs the nucleophile to reach the back of the carbon bearing the halogen, so less crowded halides react faster.

(i) $CH X_{3} CH X_{2} CH X_{2} CH X_{2} Br$ (primary, less hindered than the secondary one).
(ii) $CH X_{3} CH X_{2} CHBrCH X_{3}$ (secondary, less hindered than the tertiary one).
(iii) $(CH X_{3}) X_{2} CHCH X_{2} CH X_{2} Br$ : the branching is two carbons away from the reacting carbon. In $CH X_{3} CH X_{2} CH (CH X_{3}) CH X_{2} Br$ the branch is on the very next (β) carbon and blocks the attack more.

(i) 1-bromobutane; (ii) 2-bromobutane; (iii) 1-bromo-3-methylbutane, because they are less sterically hindered.

6.8

In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction? (i)

(CH X_{3}) X_{3} CCl

CH X_{3} CH X_{2} CHClCH X_{3}

(ii) a secondary alkyl chloride (Cl on C-2 of a straight chain) or the primary 1-chloroalkane.

Solution

SN1 goes through a carbocation; the more stable the carbocation, the faster the reaction (3° > 2° > 1°).

(i) $(CH X_{3}) X_{3} CCl$ , which gives a tertiary carbocation.
(ii) The secondary chloride, which gives a secondary carbocation (more stable than primary).

(i) tert-butyl chloride; (ii) the secondary chloride, because they form more stable carbocations.

6.9

Identify A, B, C, D, E, R and R¹ in the following: (bromocyclohexane + Mg, dry ether → A; A +

H X_{2} O

→ B); (R–Br + Mg, dry ether → C; C +

D X_{2} O

→

CH X_{3} CH (D) CH X_{3}

); (R¹–X + Na/ether → 2,2,3,3-tetramethylbutane; R¹–X + Mg → D; D +

H X_{2} O

→ E).

Solution

A = cyclohexylmagnesium bromide, $C X_{6} H X_{11} MgBr$ ; B = cyclohexane, $C X_{6} H X_{12}$ (the Grignard reagent takes H from water).
C with $D X_{2} O$ gives $CH X_{3} CHDCH X_{3}$ , so the Grignard carbon was the middle carbon of propane: R = isopropyl, $(CH X_{3}) X_{2} CH X^{-}$ , and C = $(CH X_{3}) X_{2} CHMgBr$ .
The Wurtz product $(CH X_{3}) X_{3} C - C (CH X_{3}) X_{3}$ is two tert-butyl groups joined, so R¹ = tert-butyl, $(CH X_{3}) X_{3} C X^{-}$ . Then D = $(CH X_{3}) X_{3} CMgX$ and E = $(CH X_{3}) X_{3} CH$ (2-methylpropane).

A = C₆H₁₁MgBr, B = C₆H₁₂, R = (CH₃)₂CH–, C = (CH₃)₂CHMgBr, R¹ = (CH₃)₃C–, D = (CH₃)₃CMgX, E = (CH₃)₃CH.

Exercises

6.1

Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: (i)

(CH X_{3}) X_{2} CHCH (Cl) CH X_{3}

(ii)

CH X_{3} CH X_{2} CH (CH X_{3}) CH (C X_{2} H X_{5}) Cl

(iii)

CH X_{3} CH X_{2} C (CH X_{3}) X_{2} CH X_{2} I

(iv)

(CH X_{3}) X_{3} CCH X_{2} CH (Br) C X_{6} H X_{5}

(v)

CH X_{3} CH (CH X_{3}) CH (Br) CH X_{3}

(vi)

CH X_{3} C (C X_{2} H X_{5}) X_{2} CH X_{2} Br

(vii)

CH X_{3} C (Cl) (C X_{2} H X_{5}) CH X_{2} CH X_{3}

(viii)

CH X_{3} CH = C (Cl) CH X_{2} CH (CH X_{3}) X_{2}

(ix)

CH X_{3} CH = CHC (Br) (CH X_{3}) X_{2}

(x) p-

ClC X_{6} H X_{4} CH X_{2} CH (CH X_{3}) X_{2}

(xi) m-

ClCH X_{2} C X_{6} H X_{4} CH X_{2} C (CH X_{3}) X_{3}

(xii) o-

Br - C X_{6} H X_{4} CH (CH X_{3}) CH X_{2} CH X_{3}

Solution

(i) 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 benzylic 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: vinylic halide
(ix) 4-Bromo-4-methylpent-2-ene: allylic halide (tertiary)
(x) 1-Chloro-4-(2-methylpropyl)benzene: aryl halide
(xi) 1-Chloromethyl-3-(2,2-dimethylpropyl)benzene: primary benzylic halide
(xii) 1-Bromo-2-(1-methylpropyl)benzene: aryl halide

As listed above.

6.2

Give the IUPAC names of the following compounds: (i)

CH X_{3} CH (Cl) CH (Br) CH X_{3}

(ii)

CHF X_{2} CBrClF

(iii)

ClCH X_{2} C \equiv CCH X_{2} Br

(iv)

(CCl X_{3}) X_{3} CCl

(v)

CH X_{3} C (p - ClC X_{6} H X_{4}) X_{2} CH (Br) CH X_{3}

(vi)

(CH X_{3}) X_{3} CCH = CClC X_{6} H X_{4} I

-p.

Solution

(i) 2-Bromo-3-chlorobutane
(ii) 1-Bromo-1-chloro-1,2,2-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-chlorophenyl)butane
(vi) 1-Chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene

As listed above.

6.3

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.

Solution

(i) $CH X_{3} CHClCH (CH X_{3}) CH X_{2} CH X_{3}$
(ii) a benzene ring with Br and Cl at para (1,4) positions
(iii) a cyclohexane ring with Cl on C-1 and $C X_{2} H X_{5}$ on C-4
(iv) $ICH X_{2} - CH (C X_{6} H X_{4} Cl) - (CH X_{2}) X_{5} CH X_{3}$ , the Cl on the ortho position of the phenyl ring
(v) $CH X_{3} CHBrCH X_{2} CH X_{3}$
(vi) $CH X_{3} CH X_{2} CHI - CH (C (CH X_{3}) X_{3}) - CH X_{2} CH X_{2} CH X_{3}$
(vii) a benzene ring with Br on C-1, $CH X_{3}$ on C-2 and $CH (CH X_{3}) CH X_{2} CH X_{3}$ on C-4
(viii) $BrCH X_{2} CH = CHCH X_{2} Br$

As drawn above.

6.4

Which one of the following has the highest dipole moment? (i)

CH X_{2} Cl X_{2}

(ii)

CHCl X_{3}

(iii)

CCl X_{4}

Solution

$CCl X_{4}$ is symmetrical (tetrahedral) and its bond dipoles cancel: 0 D. In $CHCl X_{3}$ the C–Cl dipoles partly cancel (1.04 D). In $CH X_{2} Cl X_{2}$ the two C–Cl dipoles reinforce the most (1.60 D).

(i) CH₂Cl₂

6.5

A hydrocarbon

C X_{5} H X_{10}

does not react with chlorine in dark but gives a single monochloro compound

C X_{5} H X_{9} Cl

in bright sunlight. Identify the hydrocarbon.

Solution

No reaction with chlorine in the dark means no C=C double bond, so $C X_{5} H X_{10}$ must be a cycloalkane. A single monochloro product means all ten hydrogens are equivalent: cyclopentane.

Cyclopentane.

6.6

Write the isomers of the compound having formula

C X_{4} H X_{9} Br

Solution

$CH X_{3} CH X_{2} CH X_{2} CH X_{2} Br$ : 1-bromobutane
$CH X_{3} CH X_{2} CHBrCH X_{3}$ : 2-bromobutane (chiral; exists as two enantiomers)
$(CH X_{3}) X_{2} CHCH X_{2} Br$ : 1-bromo-2-methylpropane
$(CH X_{3}) X_{3} CBr$ : 2-bromo-2-methylpropane

Four structural isomers (2-bromobutane also has two optical isomers).

6.7

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

Solution

(i) $CH X_{3} (CH X_{2}) X_{3} OH + KI H X_{3} PO X_{4} CH X_{3} (CH X_{2}) X_{3} I$ (or with red phosphorus and iodine)
(ii) $CH X_{3} (CH X_{2}) X_{3} Cl + NaI dry acetone CH X_{3} (CH X_{2}) X_{3} I + NaCl$ (Finkelstein reaction; NaCl precipitates)
(iii) $CH X_{3} CH X_{2} CH = CH X_{2} + HBr peroxide CH X_{3} CH X_{2} CH X_{2} CH X_{2} Br$ , then $CH X_{3} (CH X_{2}) X_{3} Br + NaI acetone CH X_{3} (CH X_{2}) X_{3} I + NaBr$

As in the equations above.

6.8

What are ambident nucleophiles? Explain with an example.

Solution

Nucleophiles that can attack through two different atoms are ambident. The cyanide ion, $[: C \equiv N :] X^{-}$ , can bond through carbon or nitrogen: with KCN (ionic) an alkyl halide gives an alkyl cyanide (R–CN), while with AgCN (mainly covalent) it gives an alkyl isocyanide (R–NC). Similarly, the nitrite ion gives alkyl nitrites (R–O–N=O) through O, or nitroalkanes ( $R - NO X_{2}$ ) through N with $AgNO X_{2}$ .

Nucleophiles with two donor sites, e.g. CN⁻ (gives R–CN or R–NC) and NO₂⁻ (gives R–ONO or R–NO₂).

6.9

Which compound in each of the following pairs will react faster in SN2 reaction with

OH X^{-}

? (i)

CH X_{3} Br

CH X_{3} I

(ii)

(CH X_{3}) X_{3} CCl

CH X_{3} Cl

Solution

(i) $CH X_{3} I$ : iodide is a better leaving group (weaker C–I bond).
(ii) $CH X_{3} Cl$ : it is far less crowded than the tertiary halide.

(i) CH₃I. (ii) CH₃Cl.

6.10

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.

Solution

By Saytzeff's rule, the more substituted alkene is the major product.

(i) 1-Methylcyclohexene (major) and methylenecyclohexane.
(ii) 2-Methylbut-2-ene, $(CH X_{3}) X_{2} C = CHCH X_{3}$ (major) and 2-methylbut-1-ene, $CH X_{2} = C (CH X_{3}) CH X_{2} CH X_{3}$ .
(iii) The bromide is $(CH X_{3}) X_{3} C - CBr (CH X_{3}) - CH X_{2} CH X_{3}$ . The C-2 carbon has no hydrogen, so H can come from the C-3 methyl or from C-4: 3,4,4-trimethylpent-2-ene, $(CH X_{3}) X_{3} C - C (CH X_{3}) = CHCH X_{3}$ (major), and 2-ethyl-3,3-dimethylbut-1-ene, $(CH X_{3}) X_{3} C - C (= CH X_{2}) CH X_{2} CH X_{3}$ .

(i) 1-methylcyclohexene (major) + methylenecyclohexane. (ii) 2-methylbut-2-ene (major) + 2-methylbut-1-ene. (iii) 3,4,4-trimethylpent-2-ene (major) + 2-ethyl-3,3-dimethylbut-1-ene.

6.11

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.

Solution

(i) $C X_{2} H X_{5} OH SOCl X_{2} C X_{2} H X_{5} Cl$ ; $HC \equiv CH NaNH X_{2} HC \equiv C X^{-} Na X^{+}$ ; $HC \equiv CNa + C X_{2} H X_{5} Cl HC \equiv C - CH X_{2} CH X_{3}$ (but-1-yne)
(ii) $C X_{2} H X_{6} Br X_{2} h ν C X_{2} H X_{5} Br alc \cdot KOH CH X_{2} = CH X_{2} Br X_{2} BrCH X_{2} CH X_{2} Br alc \cdot KOH (1 mol) CH X_{2} = CHBr$
(iii) $CH X_{3} CH = CH X_{2} HBr peroxide CH X_{3} CH X_{2} CH X_{2} Br AgNO X_{2} CH X_{3} CH X_{2} CH X_{2} NO X_{2}$
(iv) $C X_{6} H X_{5} CH X_{3} Cl X_{2} h ν C X_{6} H X_{5} CH X_{2} Cl aq \cdot KOH C X_{6} H X_{5} CH X_{2} OH$
(v) $CH X_{3} CH = CH X_{2} Br X_{2} CH X_{3} CHBrCH X_{2} Br alc \cdot KOH, then NaNH X_{2} CH X_{3} C \equiv CH$
(vi) $C X_{2} H X_{5} OH SOCl X_{2} C X_{2} H X_{5} Cl AgF C X_{2} H X_{5} F$ (Swarts reaction)
(vii) $CH X_{3} Br KCN CH X_{3} CN$ ; $CH X_{3} Br Mg dry ether CH X_{3} MgBr$ ; $CH X_{3} MgBr + CH X_{3} CN (CH X_{3}) X_{2} C = NMgBr H X_{3} O X^{+} CH X_{3} COCH X_{3}$
(viii) $CH X_{3} CH X_{2} CH = CH X_{2} HBr CH X_{3} CH X_{2} CHBrCH X_{3} alc \cdot KOH CH X_{3} CH = CHCH X_{3}$ (major, Saytzeff)
(ix) $2 CH X_{3} (CH X_{2}) X_{3} Cl + 2 Na dry ether CH X_{3} (CH X_{2}) X_{6} CH X_{3} + 2 NaCl$ (Wurtz reaction)
(x) $C X_{6} H X_{6} Br X_{2} FeBr X_{3} C X_{6} H X_{5} Br$ ; $2 C X_{6} H X_{5} Br + 2 Na dry ether C X_{6} H X_{5} - C X_{6} H X_{5} + 2 NaBr$ (Fittig reaction)

As in the reaction sequences above.

6.12

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?

Solution

(i) In chlorobenzene the C–Cl carbon is $s p^{2}$ hybridised, which is more electronegative than the $s p^{3}$ carbon of cyclohexyl chloride, so the C–Cl bond is less polar. Also, resonance gives the C–Cl bond partial double-bond character and reduces the charge separation. Hence a lower dipole moment.

(ii) Alkyl halides cannot form hydrogen bonds with water. The energy needed to break the hydrogen bonds between water molecules is not regained by the weak attractions between the halide and water, so they do not mix.

(iii) Grignard reagents react instantly with water (and any source of $H X^{+}$ ) to give alkanes: $RMgX + H X_{2} O RH + Mg (OH) X$ . So any moisture would destroy them.

(i) sp² C–Cl is less polar and has partial double-bond character. (ii) They cannot form H-bonds with water. (iii) Water decomposes RMgX to RH.

6.13

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

Solution

Freon-12 ( $CCl X_{2} F X_{2}$ ): refrigerant in fridges and air conditioners; propellant in aerosols.
DDT: insecticide, used especially against malaria-carrying mosquitoes (now restricted because it persists in the environment).
Carbon tetrachloride: solvent for oils and fats; used in making refrigerants and propellants; formerly in fire extinguishers.
Iodoform: antiseptic (because of the iodine it releases).

Freon-12: refrigerant and propellant. DDT: insecticide. CCl₄: solvent and raw material for refrigerants. Iodoform: antiseptic.

6.14

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

CH X_{3} CH X_{2} CH X_{2} Cl + NaI

(ii)

(CH X_{3}) X_{3} CBr + KOH

(iii)

CH X_{3} CH (Br) CH X_{2} CH X_{3} + NaOH

(iv)

CH X_{3} CH X_{2} Br + KCN

(v)

C X_{6} H X_{5} ONa + C X_{2} H X_{5} Cl

(vi)

CH X_{3} CH X_{2} CH X_{2} OH + SOCl X_{2}

(vii)

CH X_{3} CH X_{2} CH = CH X_{2} + HBr

(viii)

CH X_{3} CH = C (CH X_{3}) X_{2} + HBr

Solution

(i) $CH X_{3} CH X_{2} CH X_{2} I$ (in acetone, Finkelstein)
(ii) $(CH X_{3}) X_{2} C = CH X_{2}$ (ethanolic KOH: elimination from the tertiary halide)
(iii) $CH X_{3} CH (OH) CH X_{2} CH X_{3}$ (aqueous NaOH, substitution)
(iv) $CH X_{3} CH X_{2} CN$
(v) $C X_{6} H X_{5} OC X_{2} H X_{5}$ (ethoxybenzene, Williamson synthesis)
(vi) $CH X_{3} CH X_{2} CH X_{2} Cl$
(vii) $CH X_{3} CH X_{2} CHBrCH X_{3}$ (Markovnikov)
(viii) $CH X_{3} CH X_{2} CBr (CH X_{3}) X_{2}$ (Markovnikov)

As listed above.

6.15

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

Solution

It is an SN2 reaction (primary halide, good nucleophile). The cyanide ion, through its carbon atom, attacks the carbon bearing Br from the side opposite to the Br. In one step a transition state forms in which C–CN is forming and C–Br is breaking:

$NC X^{-} + CH X_{3} CH X_{2} CH X_{2} CH X_{2} - Br [NC \cdot \cdot \cdot C \cdot \cdot \cdot Br] X^{-} CH X_{3} CH X_{2} CH X_{2} CH X_{2} - CN + Br X^{-}$

The bromide ion leaves and the configuration at carbon is inverted. The rate depends on both the halide and the cyanide concentrations.

A one-step SN2 reaction: CN⁻ attacks from the back, a five-coordinate transition state forms and Br⁻ leaves (inversion).

6.16

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, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.

Solution

SN2 is fastest for the least crowded carbon:

(i) 1-Bromopentane > 2-bromopentane > 2-bromo-2-methylbutane (1° > 2° > 3°)
(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 (branching nearer the reacting carbon slows the attack)

As ordered above (least hindered fastest).

6.17

Out of

C X_{6} H X_{5} CH X_{2} Cl

and

C X_{6} H X_{5} CHClC X_{6} H X_{5}

, which is more easily hydrolysed by aqueous KOH?

Solution

$C X_{6} H X_{5} CHClC X_{6} H X_{5}$ . Hydrolysis goes by SN1, and the carbocation $(C X_{6} H X_{5}) X_{2} CH X^{+}$ is stabilised by resonance with two phenyl rings, so it forms more easily than $C X_{6} H X_{5} CH X_{2} X^{+}$ .

C₆H₅CHClC₆H₅ (its carbocation is stabilised by two phenyl rings).

6.18

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

Solution

The para isomer is symmetrical and packs more closely and efficiently in the crystal lattice. More energy is needed to break this lattice, so its melting point is higher.

Its symmetry lets it pack more tightly in the crystal.

6.19

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.

Solution

(i) $CH X_{3} CH = CH X_{2} HBr peroxide CH X_{3} CH X_{2} CH X_{2} Br aq \cdot KOH CH X_{3} CH X_{2} CH X_{2} OH$
(ii) $C X_{2} H X_{5} OH SOCl X_{2} C X_{2} H X_{5} Cl$ ; $HC \equiv CNa + C X_{2} H X_{5} Cl HC \equiv CCH X_{2} CH X_{3}$
(iii) $CH X_{3} CH X_{2} CH X_{2} Br alc \cdot KOH CH X_{3} CH = CH X_{2} HBr CH X_{3} CHBrCH X_{3}$
(iv) $C X_{6} H X_{5} CH X_{3} Cl X_{2} h ν C X_{6} H X_{5} CH X_{2} Cl aq \cdot KOH C X_{6} H X_{5} CH X_{2} OH$
(v) $C X_{6} H X_{6} Br X_{2} FeBr X_{3} C X_{6} H X_{5} Br conc \cdot HNO X_{3} / H X_{2} SO X_{4} p - BrC X_{6} H X_{4} NO X_{2}$ (Br is ortho, para directing; para is major)
(vi) $C X_{6} H X_{5} CH X_{2} OH SOCl X_{2} C X_{6} H X_{5} CH X_{2} Cl KCN C X_{6} H X_{5} CH X_{2} CN H X_{3} O X^{+} C X_{6} H X_{5} CH X_{2} COOH$
(vii) $C X_{2} H X_{5} OH SOCl X_{2} C X_{2} H X_{5} Cl KCN C X_{2} H X_{5} CN$
(viii) $C X_{6} H X_{5} NH X_{2} NaNO X_{2} / HCl 273 - 278 K C X_{6} H X_{5} N X_{2} X^{+} Cl X^{-} CuCl / HCl C X_{6} H X_{5} Cl$ (Sandmeyer)
(ix) $2 CH X_{3} CH X_{2} CHClCH X_{3} + 2 Na dry ether CH X_{3} CH X_{2} CH (CH X_{3}) CH (CH X_{3}) CH X_{2} CH X_{3}$ (Wurtz)
(x) $(CH X_{3}) X_{2} C = CH X_{2} + HCl (CH X_{3}) X_{3} CCl$ (Markovnikov)
(xi) $C X_{2} H X_{5} Cl KCN C X_{2} H X_{5} CN H X_{3} O X^{+} C X_{2} H X_{5} COOH$
(xii) $CH X_{3} CH X_{2} CH = CH X_{2} HBr peroxide CH X_{3} (CH X_{2}) X_{3} Br NaI acetone CH X_{3} (CH X_{2}) X_{3} I$
(xiii) $CH X_{3} CHClCH X_{3} alc \cdot KOH CH X_{3} CH = CH X_{2} HBr peroxide CH X_{3} CH X_{2} CH X_{2} Br aq \cdot KOH CH X_{3} CH X_{2} CH X_{2} OH$
(xiv) $CH X_{3} CH (OH) CH X_{3} I X_{2}, NaOH CHI X_{3}$ (iodoform reaction)
(xv) $C X_{6} H X_{5} Cl conc \cdot HNO X_{3} / H X_{2} SO X_{4} p - ClC X_{6} H X_{4} NO X_{2} NaOH, 443 K then H X^{+} p - HOC X_{6} H X_{4} NO X_{2}$ (the nitro group activates the Cl for substitution)
(xvi) $CH X_{3} CHBrCH X_{3} alc \cdot KOH CH X_{3} CH = CH X_{2} HBr peroxide CH X_{3} CH X_{2} CH X_{2} Br$
(xvii) $2 C X_{2} H X_{5} Cl + 2 Na dry ether CH X_{3} CH X_{2} CH X_{2} CH X_{3} + 2 NaCl$ (Wurtz)
(xviii) $C X_{6} H X_{6} Br X_{2} FeBr X_{3} C X_{6} H X_{5} Br$ ; $2 C X_{6} H X_{5} Br + 2 Na dry ether C X_{6} H X_{5} - C X_{6} H X_{5}$ (Fittig)
(xix) $(CH X_{3}) X_{3} CBr alc \cdot KOH (CH X_{3}) X_{2} C = CH X_{2} HBr peroxide (CH X_{3}) X_{2} CHCH X_{2} Br$
(xx) $C X_{6} H X_{5} NH X_{2} + CHCl X_{3} + 3 KOH (alc \cdot) C X_{6} H X_{5} NC + 3 KCl + 3 H X_{2} O$ (carbylamine reaction)

As in the reaction sequences above.

6.20

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.

Solution

In water, $OH X^{-}$ is heavily hydrated; it acts mainly as a nucleophile and replaces the halogen: substitution gives an alcohol. In alcohol, KOH gives the alkoxide ion ( $RO X^{-}$ ), a strong base that is less solvated. It removes a β-hydrogen and the halide leaves: elimination gives an alkene.

Aqueous OH⁻ acts as a nucleophile (substitution → alcohol); alcoholic KOH gives RO⁻, a strong base (elimination → alkene).

6.21

Primary alkyl halide

C X_{4} H X_{9} Br

(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),

C X_{8} H X_{18}

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.

Solution

(a) is primary but not n-butyl bromide, so (a) = 1-bromo-2-methylpropane, $(CH X_{3}) X_{2} CHCH X_{2} Br$ .

$(CH X_{3}) X_{2} CHCH X_{2} Br alc \cdot KOH (CH X_{3}) X_{2} C = CH X_{2}$ : (b) = 2-methylpropene
$(CH X_{3}) X_{2} C = CH X_{2} + HBr (CH X_{3}) X_{3} CBr$ : (c) = 2-bromo-2-methylpropane (an isomer of a)
$2 (CH X_{3}) X_{2} CHCH X_{2} Br + 2 Na dry ether (CH X_{3}) X_{2} CHCH X_{2} CH X_{2} CH (CH X_{3}) X_{2} + 2 NaBr$ : (d) = 2,5-dimethylhexane

(n-Butyl bromide would give n-octane instead.)

(a) (CH₃)₂CHCH₂Br; (b) (CH₃)₂C=CH₂; (c) (CH₃)₃CBr; (d) 2,5-dimethylhexane.

6.22

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?

Solution

(i) Elimination gives but-1-ene: $CH X_{3} CH X_{2} CH X_{2} CH X_{2} Cl alc \cdot KOH CH X_{3} CH X_{2} CH = CH X_{2}$
(ii) A Grignard reagent forms: $C X_{6} H X_{5} Br + Mg dry ether C X_{6} H X_{5} MgBr$ (phenylmagnesium bromide)
(iii) Under ordinary conditions nothing happens (the C–Cl bond is strong). Under drastic conditions (NaOH, 623 K, 300 atm) phenol is formed: $C X_{6} H X_{5} Cl NaOH 623 K, 300 atm C X_{6} H X_{5} ONa H X^{+} C X_{6} H X_{5} OH$
(iv) Substitution gives ethanol: $C X_{2} H X_{5} Cl + KOH (aq) C X_{2} H X_{5} OH + KCl$
(v) Wurtz reaction gives ethane: $2 CH X_{3} Br + 2 Na dry ether CH X_{3} CH X_{3} + 2 NaBr$
(vi) Methyl cyanide (ethanenitrile) forms: $CH X_{3} Cl + KCN CH X_{3} CN + KCl$

(i) but-1-ene; (ii) C₆H₅MgBr; (iii) phenol only under drastic conditions; (iv) ethanol; (v) ethane; (vi) CH₃CN.

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