Solution
HALOALKANES AND HALOARENES
Class 12 - Chemistry
1. Read the text carefully and answer the questions:
When haloalkanes with β -hydrogen atom are boiled with alcoholic solution of KOH, they undergo elimination of hydrogen halide
resulting in the formation of alkenes. These reactions are called β -elimination reactions or dehydrohalogenation reactions. These
reactions follow Saytzeff's rule. Substitution and elimination reactions often compete with each other. Mostly bases behave as
nudeophiles and therefore can engage in substitution or elimination reactions depending upon the alkyl halide and the reaction
conditions.
(i) (d) t - Bul > t-BuBr > t-BuCl > t-BuF
Explanation: The order of reactivity of alkyl halides:
iodide > bromide > chloride (nature of the halogen atom) tertiary > secondary > primary (type of halogen atom).
(ii) (d) iso-butylene
Explanation: so-butylene is obtained.
Br
na
H3 C C H3 + C H3 ON a ⟶ C H3
| |
CH3 C H3 C = C H2 +C H3 OH+N aBr
(iii) (d) R-I > R-Br > R-Cl > R-F
C - I bond breaks more easily than C - F bond.
So reactivity order of halides
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Explanation: The order of bond dissociation energy: R - F > R - Cl > R - Br > R - I. During dehydrohalogenation
R - I > R - Br > R - Cl > R - F
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(iv) (d) 3o > 2o > 1o
Explanation: The ease of dehydrohalogenation of alkyl halide with alcoholic KOH is 3o > 2o > 1o. 'This order of
alkyl halides can be explained on the basis of the stability of the alkene formed after dehydrohalogenation of
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haloalkanes. 3o alkylhalides on dehydrohalogenation forms more substituted alkenes, which being more stable and
formed at faster rate, while primary alkyl halides yield least substituted alkenes, which being less stable and formed at
slower rate.
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(v) (a) CH3CH2CH2Br
Explanation: In alkyl halides, polarity of C - Br bond increases with increase in chain length.
Ga
2.
(d)
Explanation:
The products obtained can be explained by Markonikov’s rule. According to the rule, the hydrogen from HCl will get added to
the carbon directly bonded to the most hydrogen atoms(addition reaction), and the –Cl will get bonded to the carbon directly
bonded to the least hydrogens.
3. (a) 3, 4, 4 - trimethylpent - 2 - ene
Explanation: The longest chain contains a double bond and five carbon i.e pent-2-ene and 2 methyl is attached to the 4th
1/3
carbon and one is attached to 3rd carbon. Therefore IUPAC name is 3, 4, 4 - trimethylpent - 2 - ene.
4.
(c) 1 – Bromobutane, 2 – Bromobutane, 1 – bromo- 2 – methylpropane, and 2 – bromo-2 – methylpropane
Explanation: Isomers are compounds having the same molecular formula but different structures. In the given straight-chain
4-C compounds, Br is placed at different positions but the molecular formula is same.
C H3 C H2 C H(Br)C H3 , C H3 C H2 C H2 C H2 Br
2−Bromobu tan e 1−Bromobu tan e
In the given branched 4-C compounds, there is a branching of -CH3 at carbon 2 but Br is placed at positions 1 and 2
respectively.
C H3 C H(C H3 )C H2 Br , C H3 C (Br)(C H3 )C H3
1−Bromo−2−methylpropane 2−Bromo−2−methylpropane
So, all of these are isomers since they have the same molecular formula but different structures.
5.
(c) Sandmeyer's reaction
Explanation: C H N C l
6 5
+
2
−
+ C u2 C l2 /HC l → C6 H5 C l
Mixing the solution of the freshly prepared diazonium salt with cuprous chloride or cuprous bromide results in the replacement
of the diazonium group by –Cl or –Br. This is called Sandmeyer's reaction.
6.
na
(b) CH3CH2CH2CH2Cl
Explanation: The forces of attraction between the molecules of a compound get stronger as they get bigger in size and have
more electrons. Also, for a straight-chain compound, the points of interaction between the molecules are more than for a
7.
longest chain compound among the given options.
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branched compound having the same molecular formula. Thus CH3CH2CH2CH2Cl has the highest melting point since it is the
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(c) A is true but R is false.
Explanation: For the same alkyl group, the boiling point of haloalkanes decreases in the order RI > RBr > RCl > RF. This is
due to the increase in van der Waals forces when the size and mass of the halogen atom increases. For isomeric haloalkanes, the
boiling point decrease with an increase in branching.
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8.
(d) A is false but R is true.
Explanation:
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Halobenzenes become reactive to nucleophilic substitution reactions when electron-withdrawing groups (nitro, cyano) are
present at ortho/para position. This is evident from the milder conditions required for hydrolysis in 2,4-dinitrochlorobenzene
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than chlorobenzene.
9.
(c) A is true but R is false.
Explanation: CH2=CH−Cl has some partial double bond character between carbon and a chlorine atom. So, nucleophilic
substitution is difficult to carry as it is difficult to break the partial double bond in vinyl chloride than ethyl chloride CH3CH2
−Cl where there is no double bond character.
The vinyl group is not electron-donating. The carbon halogen bond in vinyl halides has some double-bond character and thus a
little difficult to break.
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10.
C H3 C H3 C H3
| KOH(alc),Δ | HBr,P eroxide |
11. i. C H 3 − C − C H3 −−−−−−−−−−−−−→ C H3 − C = C H2 −−−−−−−−−−−−→ C H3 − C H − C H2 Br
| Dehydroha log enation Isobutyl bromide
Anti Mark. Addition
Br
tert−Butylbromide
ii.
12. The compound ‘A’, C4H9Br when treated with aq. KOH proceeds with the rate of the reaction depend on the concentration of ‘A’
only. This means that the reaction follows first-order kinetics, which is the characteristic of SN1 reactions. This means C4H9Br is
a tertiary halide because tert-halides undergo SN1 reaction mechanism. On the other hand, the optically active isomer ‘B’, when
subjected to treatment with aq. KOH undergoes SN2 substitution reaction due to the rate of reaction depending on both the
reactants. This means that the isomer is a secondary halide.
i. The structure of ‘A’ is
CH3
na
C H3 − C − C H3
|
Br
The structure of ‘B’ is C H 3 − CH − CH − C H3
|
Br
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ii. Compound ‘B’ will undergo inversion of configuration and give an inverted product because it undergoes SN2 substitution
reaction.
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13. i.
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ii.
u
iii.
Ga
iv.
v. CH 3 CH2 Br + NaI → CH3 CH2 I + NaBr
Bromoethane Iodoethane
vi.
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