Download Ch 10 Haloalkanes n haloarenes

Chapter 10 -Halo alkane and halo arenes
1-Mark questions
Q.1. Give the IUPAC name of the following compound.
Br
CH3- C = C-CH2-OH
CH3
Q2 Write the structure of the following organic compound:
2- Chloro-3- methyl pentane
Q.3 Why is chloro benzene less reactive towards nucleophilic substitution reactions?
Q4. Which of the following will undergo SN2 reaction faster and why?
Br
and
I
Q5.Out of C6H5-CH2-Cl&C6H5-CH(Cl)-CH3 which is more easily hydrolyzed with aq. KOH & why?
Q6. Give reason for the following observation:
p-Dichlorobenzene has higher m.p. that its o-and m- isomers.
Q7. Why is (d/l) butane-2-ol is optically inactive?
Q8. C-Cl bond length in chlorobenzene is smaller than the C-Cl bond length in halo alkane. Why?
Q9. Identify the chiral molecule in the following pair.
l
Cl
Br
Q10. Complete the following equation:
Acetone
CH3-CH2-Br +NaI --------------------
Q11. Why Grignard reagent should be prepared under anhydrous conditions
Q12.What is difference between d and l? (1)
Q13.What is racemic mixtures? (1)
Q14.Give one use of each of the following: (1)
(a)Freon
(b)D.D.T
Q15.How can prepare Chlorobenzene from Benzene? (1)
Q16.Why are enantiomers called optical isomers? (1)
Q17.What happen when Sodium acetylide is treated with ethyl iodide? (1)
Answer of 1 mark questions
Ans 1: 2-Bromo-3-methyl but-2en-1-ol
Ans 2: CH3-CH-CH-CH2-CH3
Cl
CH3
Ans 3: Because of partial double bond character between halogen and the ring carbon.
Ans 4: Ethyl iodide (II structure)as iodide is better leaving group.
Ans 5: 1- Chloro-1-phenyl ethane (2nd structure) because the carbocation formed in the 2nd structure
will be more stable.
+
C6H5- CH- CH3
Ans 6: Since p- isomer is more symmetrical so offers effective packing in the crystal lattice.
Ans 7: It is racemic mixture.
Ans 8: Because of partial double bond character between Chlorine and the ring carbon. Also, the Cl is
Attached to SP2 hybridized carbon atom.
Ans 9: Structure 1st
Ans10:
Acetone
CH3-CH2-Br +NaI
CH3-CH2-I + NaBr
Ans 11: Since it reacts with water to form alkane.
Ans12 .d and l stands for dextro-rotatory and levo-rotatory.
Ans13 .A mixtures of 2 enantiomers.
Ans 14 .DDT is used as an insecticide and Freon is used as refrigerant in refrigerator and air conditioner.
Ans 15.Chlorination of Benzene in presence of AlCl3.
Ans 16.enantiomers rotate the plane of polarized light either towards right or towards left.
Ans17.1- butyne and sodium Iodide is obtained.
2-Mark questions:
Q1. When an alkyl halide is treated with ethanolic solution of KCN, the major product is alkyl cyanide
whereas if alkyl halide is treated with ethanolicAgCN,the major product is alkyl isocyanide. Why?
Q2. Although chlorine is an electron- withdrawing group, yet it is ortho and para directing in
electrophillic aromatic substitution reactions.Why
Q3. Draw the major monohalo product in each of the following products:
SOCl2
- OH -----------------
(a)
Peroxide
(b)
CH2-CH =CH2 + HBr -------------------
Q4. Complete the following equations:
OH
HCl
(a)
(b)
- CH2-OH -----------------
CH2-CH =CH2 + HBr -------------------
Q5. Predict the reactivity in SN1
(a) C6H5CH2Br, C6H5CH (C6H5)Br, C6H5CH(CH3)Br, C6H5C(CH3)( C6H5)Br
Q6. Give reason for the following:
(a) Chloroform is stored in dark coloured bottles completely filled so that air is kept out. Also
1% ethanolic solution is added to it.
(b)Chloro ethane is immiscible with water.
Q7. Write following name reactions:
(a)Sandmayer reaction
(b)Swart’sreaction
Q8. Give reasons:
(a)Halo alkanes easily dissolve in organic solvents.
(b) Halogen compounds used in industries as solvent are alkyl chlorides and not bromides
and iodides.
Q9. How does Br2 react with CH3- C
CH? How would you justify this reaction?
Q10. What are the enantiomers? Draw the possible enantiomers of 3-Methylpent-1-ene.
Answer of 2 mark questions
Ans 1: With KCN reaction is ionic but with AgCN it is molecular.
OR
KCN is ionic they can attach through C or N but C-C bond is strong than C-N bond. So alkyl cyanide is the major
product but AgCN is covalent so more electronegative N can attach to C and forms isocyanides.
Ans 2. Because Cl donates electron to the benzene ring by + R effect and electron density increases
at o- and p- positions.
Ans 3: (a)
Cl
(b)
CH2-CH2 -CH2-Br
Ans 4:
Cl
(a)
- CH2-OH
(b)
CH2-CH(Br)-CH3
Ans 5. 30 > 20> 10 (SN1)
C6H5C(CH3)(C6H5)Br > C6H5(C6H5)Br > C6H5CH(CH3)Br > C6H5CH2Br
(30)
(20)
(20)
(10)
Ans 6: (a)- Chloroform slowly oxidises to form phosgene gas(COCl2) in the presence of air and UV light.
(b)Chloro ethane does not break intermolecular H- bonding of the water molecule thus it is
Immiscible with water
Ans 7: (a)
(b) CH3-CH2-Br + AgF -------------- CH3-CH2-F + AgBr
Ans 8: (a) Because of Vanderwalls’ forces between haloalkanes and solvents.
(c) (b)Alkyl chlorides are more stable.
Ans 9:Br
CH3- C-CH-Br
Br Br
1,1,2,2-Tetrabromo propane
Justification: It is electrophilic addition reaction in which Br2 gets ionized and attacks on the
triple bond as an electrophile.
Ans 10: The non-superimposable mirror images of an organic compound.
CH3-CH2
CH2-CH3
C
CH3
c
CH=CH2
CH2=CH
H
CH3
H
3-Mark questions:
Q1. How would you distinguish between following pairs of organic compounds:
(a) Cyclohexyl chloride and chlorobenzene
(b) Ethyl bromide and ethyl iodide
(c) Bromobenzene and bromo ethane
Q2. Arrange in order of boiling points.
(a) Bromobenzene, Bromoform, chloromethane,Dibromo-methane
(b) 1-chloropropane, Isopropyle chloride, 1-Chlorobutane.
Q3) How the following conversions can be carried out?
(i) Chloro ethane to n-butane
(ii)Benzene to diphenyl (iii) But-1-ene to n-butyliodide
Q4. Write following name reactions:
(1) Wurtz’sraction (2) Gatterman reaction (3) Fittig reaction
Q5 Explain following reactions:
(1) Friedel craft’s alkylation (2) Wurtz- fittig reaction (3) Dow’s process
Answers of 3 mark questions
Ans 1: (a) Addition of aq KOH followed by AgNO3 will give white ppt with Cyclohexyl chloride and not
withchlorobenzene
(b)Addition of aq KOH followed by AgNO3 will give dull yellow ppt with ethyl bromide and
Bright yellow with ethyl iodide
(c)Addition of aq KOH followed by AgNO3 will give yellow ppt with brome ethane and not
withbromobenzene
Ans2. (a) chloromethane<Bromobenzene<Dibromo methane <Bromoform
(b) Isopropyl chloride <1-chloropropane <1-Chlorobutane
(As Branching increases , boiling point decreases)
Ans3: (i) CH3-CH2-Cl ------------------ CH3-CH2-CH2-CH3
Anh. FeCl3
Na/ether
(ii)C6H6 + Cl2 --------------- C6H5-Cl ----------------------- C6H5-C6H5
(iii)
Ans 4.See name reaction in haloalkane and haloarenes.
Ans.5. See name reaction in haloalkane and haloarenes .
5-Mark questions:
Q. 1.(1)
p-Dichlorbenzene has higher melting point and lower solubility than those of
o- and m- isomers. Discuss.
(2)
The treatment of alkyl chlorides with aq KOH leads to the formation of alcohols but in
presence of alcoholic KOH, alkenes are the major products. Explain.
( 3).
Explain why is Chlorobenzene difficult to hydrolyse than ethyl chloride ?
( 4).
Tert-butyl chloride reacts with aq. NaOH by SN¹ mechanism while n-butyl chloride
reacts by SN² mechanism. Why ?
(5.)
Allyl Chloride is more reactive than n-propyl Chloride towards nucleophilic
substitution reactions. Explain.
Q. 2. (1) R—Cl is hydrolysed to R—OH slowly but the reaction is rapid if a catalytic amount of
KI is added to the reaction mixture.
(2).
Why alkyl halides are generally not prepared in laboratory by free radical halogenation
of alkanes ?
(3).
Why preparation of aryl iodide by electrophilic substitution requires presence of an
oxidisingagent ? Why can aryl flouride not be prepared by this method ?
(4).
Why aryl halides are extremely less reactive towards nucleophilic substitution ?
(5). Why does 2 bromopentane gives pent-2-ene as major product in elimination reaction ?
Q3.A sweet smelling organic compound A is slowly oxidized by air in the presence of light to a
high poisonous gas.
On warming with silver powder, it form a gaseous substance B, Which is also produce by
action of calcium carbide on
water. Identify A and B and write the chemical equation of the reaction involved. (5)
Q4.What happens when: (5)
(i)Methyl chloride treated with KCN.
(ii)Methyl bromide treated with Sodium in the presence of dry ether.
(iii)Ethyl chloride treated with aqueous KOH.
(iv)n-Butyl chloride treated with alcoholic KOH.
Q5.How the following conversion carried out: (5)
(i)Propane to Propan-1-ol
(ii)Ethane to but-1-yne
(iii)1-bromopropane to 2-bromopropane.
(iv)Chloroethane to Butane
(v)Benzene to di-phenyl
Answers of 5 Marks Questions
Ans.1.(1)

  
The p-isomer
lattice breaks, therefore a larger amount of energy is needed to melt or dissolve the pisomer than the corresponding o- and metaisomers.

(2)





In aq. solution, KOH is almost completely ionised to give OH– ions which being a
strong nucleophile brings about a substitution reaction to form alcohols. Further in aq.
solution, OH– ions are highly solvated (hydrated).
This solution reduces the basic character of OH– ions which fail to abstract a hydrogen
from the -carbon of the alkyl halide to form an alkene.
However an alcoholic solution of KOH contains alkoxide (RO –) ions which being a much
stronger base than OH– ions preferentially abstracts a hydrogen from the -carbon of
the alkyl halide to form alkene.

(3)
The lone pair of electrons of Chlorine is Chlorbenzene participates into
resonance with the benzene ring.
As a result C — Cl bond acquires a partial double bond character. Therefore, this C —
Cl bond is stronger than C — Cl bond in ethyl chloride which is a pure single bond. As
such the Chlorobenzene is difficult to hydrolyse than ethyl chloride.
 












(4)
Tert-butyl chloride reacts via SN¹ mechanism because the heterolytic cleavage of C —
Cl bond in tert-butyl chloride gives 3 carbocation which is highly stable and favourable for SN¹
mechanism. Moreover, tert-butyl chloride (3°) bring a bulky molecule has steric hindrance
which will not allow SN² mechanism to take place. Hence only SN1 mechanism can occur in
tert-butyl chloride. However n-butyl chloride (1°) reacts via SN² because ‘C’ of C — Cl bond is
less crowded and favourable for nucleophile to attack from back side results in the formation of
transition state. It has less steric hindrance which is a favourable factor for SN² mechanism
(5)
Allyl Chloride readily undergoes ionization to produce resonance stabilized allyl
carbocation. Since carbocations are reactive species they readily combine with
OH– ions to form allyl alcohol.
Ionization
+
CH2 = CH — CH2 — Cl ———— CH2 = CH — CH2
Slow



OH–
CH2 = CH — CH2OH ———— +CH2 — CH = CH2
Fast
In contrast, n-propyl chloride does not undergo ionization to produce n-propyl
carbocation and hence allyl chloride is hydrolysed more readily than n-propyl chloride.
Ans.2(1)
Iodide ion is a powerful nucleophile and hence reacts rapidly with RCl to form RI.
KI ——— K+ + I–; R — Cl + I– ——— R — I + Cl–
Further I– is a better leaving group than CI– ion, therefore, RI is more rapidly hydrolysed
than RCl to form ROH.
HO– + R — I ——— R — OH + I–
(2)
It is because :
(i)
It gives a mixture of isomeric monohalogenated products whose boiling points
are so close that they cannot be separated easily.
(ii)
Polyhalogenation may also take place, thereby making the mixture more complex
and hence difficult to separate.
(3)
Reactions with I2 are reversible in nature and require presence of oxidising agent
(HNO3, etc.) to oxidise HI formed during iodination and promote forward reaction.
Fluoro compounds cannot be prepared due to high reactivity of flourine.
(4)
(i)
Resonance effect :
Due to resonance C — Cl bond acquires partial double bond character.
(ii)
Difference in hybridisation of Carbon in C —X bond :
inhaloarene C-atom attached to halogen in sp² hybrid while sp³ in haloalkane.
C — X bond length in sp² hybrid is shorter and hence stronger and difficult to
break.
(iii)
Instability of phenyl cation
(iv)
Possible repulsion of nucleophile to approach e– rich arenes.
(v)-This is because of Saytzeff’s rule — Indehydrohalogen reactions, the preferred
product is that alkene which has the greater number of alkyl groups attached to the
doubly bonded carbon atoms.
Ans3 -CaC2+2H2O------
C2H2+Ca(OH)2
2CHCl3+6Ag------------ C2H2+6AgCl
2CHCl3+O2-------------
2COCl2+2HCl
Ans 4. (i)Methyl cyanide and KCl.
(ii)Ethane and sodium Bromide.
(iii)Ethyl alcohol and KCl.
(iv)But-1-ene + KCl + H2O
Ans 5- . Answer is given in NCERT book.