📘Haloalkane and Haloarenes Subjective Important Questions Topic Wise
These questions are extremely important from the exam point of view and must be thoroughly prepared by every student. ✅
Q1: Write the structure of 2,4-dinitrochlorobenzene.
Q2: Out of 
Q3: Write the structure of 1-bromo-4-chlorobut-2-ene.
Q4: Write the structure of 3-bromo-2-methylprop-1-ene.
Q5: Draw the structure of 2-bromopentane.
Q6: Write the IUPAC name of (CH₃)₂CH–CH(Cl)–CH₃.
Q7: Write the IUPAC name of
Q8: Write the IUPAC name of
Q9: Write the IUPAC name of
Q10: Write the IUPAC name of the compound
Q11. Write the IUPAC name of the following compound: CH₃–CH(Br)–CH₂–CH(CH₃)–CH₃
Q12. Write the IUPAC name of the following compound:
Q13. Write the reaction when bromine is added to CH₂=CH–CH₂–C≡CH
Q14. Write the IUPAC name of the compound: (CH₃)₃CCH₂Br
Q15. Write the IUPAC name of the compound: CH₂=CHCH₂Br
Q16. Write the structure of 1,4-dibromobut-2-ene.
Q17. Write the structure of 2-(2-bromophenyl)butane.
Q18. Write the structure of 2-(2-chlorophenyl)-1-iodooctane.
Q19. Write the structure of 1-bromo-4-sec-butyl-2-methylbenzene.
Q20. Write the IUPAC name of the compound: 2-Chloro-3-methylpentane
Q21. Write the structure of 3-(4-chlorophenyl)-2-methylpropane.
Q22. Write the IUPAC name of
Q23. Write the IUPAC name of
Q24. Write the structure of 4-tert-butyl-3-iodoheptane.
Q25. Write the IUPAC name of the compound
Q26. Write the structure of 1-chloro-4-ethylcyclohexane.
Q27. Write the structure of 4-bromo-3-methylpent-2-ene.
Q28. Draw the structure of
(i) p-Bromochlorobenzene. (ii) 1-chloto-4-ethylcyclohexane
Q29. Give the IUPAC names of the following compounds:
Q30. Write the structural Formula and Common name of the following :
i) 2-chlorobutane
ii) 2-chloro-2-methyl propane
iii) Dichloromethane
iv) 1,2-dichloroethane
v) 1,5-diiodopentane
vi) TriIodomethane
vii) Tetrachloromethane
viii) 1-chloro-4-methyl Benzene
ix) 1,4-dibromobenzene
x) 1,3,5-tribromobenzene
Q31: What is the nature of the C–X bond? Why is it considered polar?
Answer : The C–X bond is a covalent bond but it is considered polar in nature.
This is because halogen atoms (X) are more electronegative than carbon, so they pull the shared pair of electrons towards themselves.
As a result, carbon gets a partial positive charge (δ+) and halogen gets a partial negative charge (δ−), making the C–X bond polar.
This polarity affects the physical and chemical properties of haloalkanes and haloarenes.
Q32: Why is the C–X bond in haloarenes stronger and shorter than in haloalkanes? Give reasoning based on structure and bonding.
Answer : The C–X bond in haloarenes is stronger and shorter than in haloalkanes.
This is because in haloarenes, the halogen atom is attached to an sp2 hybridized carbon of the aromatic ring, which is more electronegative and holds the bond pair more tightly.
Also, there is possible resonance between the halogen lone pair and the aromatic ring, giving partial double bond character to the C–X bond.
In contrast, in haloalkanes, the halogen is attached to an sp3 carbon, so the C–X bond is purely single and longer.
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Q33:Why is t-butyl bromide more reactive towards SN1 reaction as compared to n-butyl bromide?
Answer : t-Butyl bromide is more reactive towards SN1 reaction than n-butyl bromide.
This is because t-butyl bromide forms a stable tertiary (3°) carbocation after the departure of Br−, which is highly stabilized due to inductive and hyperconjugation effects.
In contrast, n-butyl bromide forms a primary (1°) carbocation, which is highly unstable and not favored in SN1 reactions.
Thus, the stability of the intermediate carbocation makes t-butyl bromide much more reactive in the SN1 mechanism.
Q34: Predict the major product formed when sodium ethoxide reacts with tert.butyl chloride.
Answer : The major product is 2-methylpropene (isobutene).
When tert-butyl chloride reacts with sodium ethoxide, an E2 elimination reaction occurs.
This is because tert-butyl chloride is a bulky tertiary halide, and sodium ethoxide is a strong base. Due to steric hindrance, elimination (E2) is favored over substitution (SN2), leading to the formation of an alkene.
The β-hydrogen is removed, and a double bond forms, resulting in the formation of 2-methylpropene.
Also this (But you prefer 1st one)
Q35: Out of chlorobenzene and benzyl chloride which one gets easily hydrolysed by aqueous NaOH and why?
Answer : Benzyl chloride gets easily hydrolysed by aqueous NaOH.
Benzyl chloride contains a benzylic carbon, and the intermediate benzyl carbocation formed during the reaction is highly stabilized by resonance with the aromatic ring.
Whereas, chlorobenzene has a C–Cl bond with partial double bond character due to resonance, making it less reactive and difficult to break during hydrolysis.
Q36: Write the structure of an isomer of compound C4H9Br which is most reactive towards SN1 reaction.
Answer: The isomer is tert-butyl bromide (2-Bromo-2-methylpropane).
Structure:
CH3 — C(Br) — CH3
|
CH3
Reason: Tert-butyl bromide forms a stable 3° carbocation after the loss of Br-, making it the most reactive isomer towards SN1 reaction among C4H9Br compounds.
Q37: Which is more reactive towards SN1 reaction and why?
Answer : 2-Chlorobutane reacts faster towards SN1 reaction.
2-Chlorobutane is a secondary halide and forms a relatively stable 2° carbocation after the departure of Cl-.
Whereas, 1-Chloro-2-methylpropane is a primary halide and forms an unstable 1° carbocation, making it less reactive towards SN1 reaction.
Q38:Which would undergo SN1 reaction faster in the following pair?
Answer :
reacts faster towards SN1 reaction.Because 2-Bromopropane is a secondary halide, forms stable 2° carbocation easily.
Compound (I) is a primary halide, unstable carbocation, slower SN1.
Q39: Which would undergo SN2 reaction faster in the following pair and why
Answer : 1-Bromoethane reacts faster towards SN2 reaction.
1-bromopropane is a primary halide, less steric hindrance, SN2 occurs easily.
although 2-bromo-2-methylpropane is tertiary, highly hindered, difficult for SN2 attack.
Q40:Which would undergo SN2 reaction faster in the following pair and why?
Answer :
reacts faster towards SN2 reaction.
Because Iodide is a better leaving group than bromide due to its larger size and weaker C–I bond.
Hence,
undergoes SN2 faster.
Q41: What happens when CH3 -Br is treated with KCN?
Answer :
CH3Br + KCN → CH3CN + KBr
Methyl cyanide (Acetonitrile) is obtained .
It is a Nucleophilic Substituion Reaction, Nucleophile CN- substitutes Br-
Q42: What happens when ethyl chloride is treated with aqueous KOH?
Answer : CH₃CH₂Cl + KOH (aq) → CH₃CH₂OH + KCl
Q43: Which compound in the following pair undergoes faster towards SN1 reaction?
Answer : Compound (I) reacts faster towards SN1 reaction.
Reason: Compound (I) is a tertiary halide, forms stable 3° carbocation easily.
Compound (II) is a primary halide, unstable carbocation, slower SN1.
Q44: How methyl bromide be preferentially converted to methyl isocyanide.
Answer : Methyl bromide can be converted to methyl isocyanide by treating it with silver cyanide (AgCN).
Reaction:
CH3Br + AgCN(alc.) → CH3NC + AgBr
Reason: AgCN gives isocyanide (CH3NC) because Ag–CN bond is covalent, so nucleophile attacks via nitrogen.
Q45: Predict the order of reactivity of four isomeric bromobutanes in SN1 reaction.
Answer : Order of reactivity in SN1 reaction:
3-Bromo-2-methylpropane > 2-Bromobutane > 1-Bromo-2-methylpropane > 1-Bromobutane
Q46: Which will react faster towards SN2 reaction, 1-bromopentane or 2-bromopentane and why?
Answer: 1-Bromopentane will react faster in SN2.
Reason: Primary halide, less hindered, easy attack.
2-Bromopentane is secondary, more hindered.
Q47: A solution of KOH hydrolyses CH3CHClCH2CH3 and CH3CH2CH2CH2Cl is which one of these is more easily hydrolysed?
CH3CHClCH2CH3 is more easily hydrolysed.
Reason: It is a secondary halide and forms a more stable carbocation during the reaction, making hydrolysis faster than CH3CH2CH2CH2Cl, which is a primary halide and forms a less stable carbocation.
Q48: Explain the following reactions with an example: Friedel-Craft's reaction.
Answer : Friedel-Craft's Reaction:
It is an electrophilic substitution reaction where an alkyl or acyl group is introduced into an aromatic ring using alkyl halides or acyl halides in the presence of anhydrous AlCl3 as a catalyst.
Example (Friedel-Craft's Alkylation):
Example (Friedel-Craft's Acylation):
Q49:Which one of the following compounds is more reactive towards SN2 reaction and why?
CH3CH(Cl)CH2CH3 or CH3CH2CH2Cl
Answer : CH3CH2CH2Cl is more reactive towards SN2 reaction.
Reason: CH3CH2CH2Cl is a primary halide with less steric hindrance, allowing the nucleophile to easily attack the carbon atom bearing the leaving group.
On the other hand, CH3CH(Cl)CH2CH3 is a secondary halide with more steric hindrance, making SN2 attack difficult.
Q.50. Out of SN1 and SN2 which reaction occurs with
(a) Inversion of configuration (b) Racemisation
Answer (a) Inversion of configuration occurs in SN2 reaction.
Reason: SN2 mechanism involves a backside attack by the nucleophile, resulting in complete inversion of configuration (Walden inversion).
(b) Racemisation occurs in SN1 reaction.
Reason: SN1 mechanism proceeds via a planar carbocation intermediate, allowing the nucleophile to attack from either side with equal probability, leading to the formation of both enantiomers (racemic mixture).
Q51: Write chemical equations when Chlorobenzene is treated with CH3COCl in the presence of anhydrous AlCl3.
Answer : When chlorobenzene is treated with CH3COCl (acetyl chloride) in the presence of anhydrous AlCl3 (Friedel-Crafts acylation), p-chloroacetophenone and o-chloroacetophenone are formed:
Q52: (i) Which alkyl halide from the following pair would you expect to reat more rapidaly by an SN2 mechanism and why ?
(ii) Racemisation occurs in SN1 reactions Why ?
Answer (i) Between the two given alkyl halides:
1. 2-Bromopentane (CH3-CH2-CHBr-CH2-CH3)
2. 1-Bromopentane (CH3-CH2-CH2-CH2-Br)
Answer: 1-Bromopentane reacts faster via SN2 mechanism.
Reason: 1-Bromopentane is a primary alkyl halide with less steric hindrance, allowing the nucleophile to attack easily from the backside, which is essential for the SN2 mechanism. 2-Bromopentane is a secondary halide and has more steric hindrance, making SN2 attack difficult.
(ii) Racemisation occurs in SN1 reactions because the carbocation intermediate formed during the reaction is planar (sp2 hybridised).
Reason: The nucleophile can attack the planar carbocation from either side with equal probability, resulting in the formation of both enantiomers (dextro and laevo), thus producing a racemic mixture (optically inactive).
Q53: Which compound in each of the following pairs will react faster towards SN2 reaction with -OH ? Why ?
(i) CH3Br or CH3I
(ii) (CH3)3CCl or CH3Cl
Answer (i) CH3I will react faster than CH3Br in SN2 reaction.
Reason: The C–I bond is weaker and longer than the C–Br bond, making it easier to break during the SN2 mechanism.
(ii) CH3Cl will react faster than (CH3)3CCl in SN2 reaction.
Reason: CH3Cl is a primary halide with less steric hindrance, allowing the nucleophile (-OH) to attack easily. In contrast, (CH3)3CCl is a tertiary halide with bulky groups, making SN2 attack difficult.
Q54: Write the chemical equations when,
(i) methylchloride is treated with AgN02
(ii) bromobenzene is treated with CH3Cl in the presence of anhydrous AlCl3.
Answer (i) When methyl chloride is treated with AgNO2, nitromethane is formed:
CH3Cl + AgNO2 → CH3NO2 + AgCl
(ii) When bromobenzene is treated with CH3Cl in the presence of anhydrous AlCl3 (Friedel-Crafts alkylation), methylbromobenzene is formed (mixture of ortho and para isomers):
Q55: What are ambident nucleophiles? Give example.
Answer : Ambident Nucleophiles: Ambident nucleophiles are nucleophiles that have two or more nucleophilic sites, meaning they can attack an electrophile from two different atoms, giving rise to different products.
Example: The cyanide ion (CN-) is an ambident nucleophile. It can attack through the carbon atom to form alkyl cyanides (R-CN) or through the nitrogen atom to form alkyl isocyanides (R-NC).
Q56: Chlorobenzene is extremely less reactive towards a nucleophilic substitution reaction. Give two reasons for the same.
Answer : Reasons why chlorobenzene is extremely less reactive towards nucleophilic substitution reaction:
1. Resonance Stabilisation: In chlorobenzene, the lone pair of electrons on the chlorine atom delocalises into the benzene ring, creating partial double bond character between carbon and chlorine. This strengthens the C–Cl bond, making its cleavage difficult during the substitution reaction.
2. sp2 Hybridisation of Carbon: The carbon atom bonded to chlorine in chlorobenzene is sp2 hybridised (part of the aromatic ring), which holds the chlorine atom more tightly compared to the sp3 carbon in alkyl halides. This also makes the bond stronger and harder to break in a nucleophilic substitution reaction.
Q57: Explain. Why
(i) alkyl halides, though polar, are immiscible with water?
(ⅱ) Grignard's reagent should be prepared under anhydrous conditions?
Answer (i) Alkyl halides, though polar, are immiscible with water because they cannot form hydrogen bonds with water molecules. Water molecules are strongly hydrogen-bonded to each other, and alkyl halides cannot break these bonds due to their inability to form such interactions, resulting in their separation as a different layer.
(ii) Grignard's reagents should be prepared under anhydrous (completely dry) conditions because they are highly reactive towards water. Even a small amount of moisture reacts with the Grignard reagent to form alkanes, thereby destroying the reagent and making it ineffective for further reactions.
Q58: (i) Write equation for preparation of 1-iodobutane from 1-chlorobutane
(ii) Out of 2-bromopentane, 2-bromo-2-methylbutane and 1-bromopentane which compound is most reactive towards elimination reaction and why?
Answer (i) Preparation of 1-iodobutane from 1-chlorobutane:
(via Finkelstein Reaction)
(ii) Most reactive towards elimination reaction: 2-bromo-2-methylbutane
Reason: It is a tertiary halide, which forms a stable tertiary carbocation intermediate that favours the E1 mechanism or undergoes E2 easily due to β-hydrogen availability and steric hindrance that disfavors substitution, promoting elimination instead.
Q59: Give reasons for the following:
(a) The presence of - NO2 group at ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution reactions.
(b) p-dichlorobenzene has higher melting point than that of ortho or meta isomer.
(c) Thionyl chloride method is preferred for preparing alkyl chloride from alcohols.
Answer (a) The presence of the -NO2 group at ortho or para positions increases the reactivity of haloarenes towards nucleophilic substitution reactions because the -NO2 group is an electron-withdrawing group. It stabilizes the negative charge developed on the intermediate (Meisenheimer complex) during the reaction, thereby facilitating the substitution process.
(b) p-Dichlorobenzene has a higher melting point than its ortho or meta isomers because its molecules are more symmetrical. This allows them to pack more efficiently in the crystal lattice, resulting in stronger intermolecular forces and, hence, a higher melting point.
(c) The thionyl chloride (SOCl2) method is preferred for preparing alkyl chlorides from alcohols because it produces gaseous by-products (SO2 and HCl), which escape from the reaction mixture, driving the reaction to completion and making the purification of the product easier.
Q60: (i) Out of (CH3)3C-Br and (CH3)3C- I which one is more reactive SN1 and why?
(ii) Write the product formed when p-nitrochlorobenzene is heated with aqueous NaOH at 443 K followed by acidification.
(ii) Why dextro and laevo-rotatory isomers of butan-2-ol are difficult to separate by fractional distillation
(i) (CH3)3C-I is more reactive towards SN1 reaction than (CH3)3C-Br because the C–I bond is weaker than the C–Br bond due to the larger size of iodine, making it easier to break and form the carbocation intermediate required for the SN1 mechanism.
(ii) When p-nitrochlorobenzene is heated with aqueous NaOH at 443 K followed by acidification, p-nitrophenol is formed.
(iii) Dextro and laevo-rotatory isomers of butan-2-ol are difficult to separate by fractional distillation because they have **identical boiling points and physical properties** (except for their ability to rotate plane-polarised light in opposite directions). Fractional distillation relies on differences in boiling points, which do not exist between these enantiomers.
Q1: (6) Identify the chiral molecule in the following pair:
(ⅱ) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.
(ⅲ) Write the structure of the alkene formed by dehydrohalogenation of 1-bromo- 1-methylcyclohexane with alcoholic KOH.
Answer (ii) When chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether (Wurtz-Fittig reaction), **toluene (methylbenzene)** is formed.
Structure:
C6H5Cl + CH3Cl + 2Na → C6H5CH3 + 2NaCl
(iii) When 1-bromo-1-methylcyclohexane undergoes dehydrohalogenation with alcoholic KOH, the product is **1-methylcyclohexene**.
Structure:
Reaction:
Q61: Write the product(s) formed when
(i) 2-bromopropane undergoes dehydrohalogenation reaction.
(ⅱ) Chlorobenzene undergoes nitration reaction.
(iii) Methylbromide is treated with KCN.
Answer (i) When 2-bromopropane undergoes dehydrohalogenation (with alcoholic KOH), **propene** is formed.
CH3CHBrCH3 + alc. KOH → CH2=CHCH3 + HBr
(ii) When chlorobenzene undergoes nitration (with conc. HNO3 and conc. H2SO4), a mixture of **ortho- and para-nitrochlorobenzene** is formed.
C6H5Cl + HNO3 → o-C6H4ClNO2 + p-C6H4ClNO2
(iii) When methylbromide is treated with KCN, **methyl cyanide (acetonitrile)** is formed.
CH3Br + KCN → CH3CN + KBr
Q62: Following compounds are given to you:
2-bromopentane, 2-bromo-2-methylbutane, 1-bromopentane
(i) Write the compound which is most reactive towards SN2 reaction.
(ii) Write the compound which is optically active.
(iii) Write the compound which is most reactive towards ẞ-elimination reaction.
Answer (i) The compound most reactive towards SN2 reaction is 1-bromopentane because it is a primary halide with the least steric hindrance, which favors the SN2 mechanism.
(ii) The compound that is optically active is 2-bromopentane because the carbon atom bearing the bromine is attached to four different groups, making it a chiral centre.
(iii) The compound most reactive towards β-elimination reaction is 2-bromo-2-methylbutane because it is a tertiary halide, and tertiary halides readily undergo β-elimination to form alkenes.
Q63: How do you convert the following?
(i) Chlorobenzene to biphenyl.
(ii) Propene to 1-iodopropane.
(iii) 2-bromobutane to but-2-ene.
Answer (i) Chlorobenzene to Biphenyl:
Chlorobenzene reacts with sodium metal in dry ether (Wurtz-Fittig reaction) to form biphenyl.
2C6H5Cl + 2Na ⟶ C6H5-C6H5 + 2NaCl
(ii) Propene to 1-iodopropane:
Propene is first reacted with HBr in presence of peroxide (anti-Markovnikov addition) to form 1-bromopropane, which is then treated with NaI in acetone (Finkelstein reaction) to give 1-iodopropane.
CH2=CHCH3 + HBr (peroxide) ⟶ CH3CH2CH2Br
CH3CH2CH2Br + NaI ⟶ CH3CH2CH2I + NaBr
(iii) 2-Bromobutane to But-2-ene:
2-Bromobutane reacts with alcoholic KOH to undergo dehydrohalogenation, forming but-2-ene.
CH3CHBrCH2CH3 + alc. KOH ⟶ CH3CH=CHCH3 + HBr
Q64: Write the major product(s) in the following:
Q65: How can the following conversions be carried out?
(i) Aniline to bromobenzene
(ii) Chlorobenzene to 2-chloroacetophenone
(iii) Chloroethane to butane
(i) Aniline to Bromobenzene:
Aniline is first converted to diazonium salt using NaNO2/HCl at 0–5°C, followed by reaction with CuBr (Sandmeyer reaction) to form bromobenzene.
C6H5NH2 + NaNO2 + HCl ⟶ C6H5N2Cl + NaCl + H2O
C6H5N2Cl + CuBr ⟶ C6H5Br + N2
(ii) Chlorobenzene to 2-chloroacetophenone:
Chlorobenzene undergoes Friedel-Crafts acylation with CH3COCl (acetyl chloride) in the presence of AlCl3 to form 2-chloroacetophenone.
C6H5Cl + CH3COCl ⟶ Cl-C6H4-COCH3 (2-chloroacetophenone)
(iii) Chloroethane to Butane:
Chloroethane reacts with sodium metal in dry ether (Wurtz reaction) to form butane.
2C2H5Cl + 2Na ⟶ C4H10 + 2NaCl
Q66: What happens when
(i) chlorobenzene is treated with Cl2 / FeCl3 ?
(ii) ethyl chloride is treated with AgNO2 ?
(iii) 2-bromopentane is treated with alcoholic KOH?
Write the chemical equations in support of your answer.
Answer (i) When chlorobenzene is treated with Cl2 in the presence of FeCl3, it undergoes electrophilic aromatic substitution to give a mixture of ortho- and para-chlorobenzene.
C6H5Cl + Cl2 ⟶ o-C6H4Cl2 + p-C6H4Cl2 (in presence of FeCl3)
(ii) When ethyl chloride is treated with AgNO2, nitroethane is formed.
C2H5Cl + AgNO2 ⟶ C2H5NO2 + AgCl
(iii) When 2-bromopentane is treated with alcoholic KOH, dehydrohalogenation occurs to form pent-2-ene and pent-1-ene (major and minor products respectively).
CH3CHBrCH2CH2CH3 + alc.KOH ⟶ CH3CH=CHCH2CH3 + HBr (Pent-2-ene, major)
CH3CHBrCH2CH2CH3 + alc.KOH ⟶ CH2=CHCH2CH2CH3 + HBr (Pent-1-ene, minor)
Q67: Give reasons:
(i) n-butyl bromide has higher boiling point than t-butyl bromide.
(ii) Racemic mixture is optically inactive.
(iii) The presence of nitro group (-NO2) at o/p-positions increases the reactivity of haloarenes towards nucleophilic substitution reactions.
Answer (i) n-Butyl bromide has a higher boiling point than t-butyl bromide because n-butyl bromide has a more extended and linear structure, which allows stronger van der Waals forces (greater surface area) compared to the compact, branched structure of t-butyl bromide.
(ii) Racemic mixture is optically inactive because it contains equal amounts of two enantiomers (d- and l- forms), whose optical rotations cancel each other, resulting in no net rotation of plane-polarized light.
(iii) The presence of a nitro group (-NO2) at ortho or para positions increases the reactivity of haloarenes towards nucleophilic substitution because the nitro group is electron-withdrawing, which stabilizes the negative charge developed in the transition state or intermediate (Meisenheimer complex), making the substitution easier.
Q68: (i) Why are alkyl halides insoluble in water?
(ii) Why is butan-1-ol optically inactive but butan-2-ol is optically active?
(iii) Although chlorine is an electron withdrawing group, yet it is ortho, para- directing in electrophilic aromatic substitution reactions. Why?
Answer (i) Alkyl halides are insoluble in water because they cannot form hydrogen bonds with water molecules. Additionally, their large hydrophobic alkyl groups make them incompatible with the polar nature of water.
(ii) Butan-1-ol is optically inactive because the carbon atom carrying the –OH group is not chiral; it is attached to two hydrogen atoms. Butan-2-ol is optically active because the carbon atom bearing the –OH group is attached to four different groups, making it a chiral centre.
(iii) Although chlorine is an electron-withdrawing group due to its –I (inductive) effect, it has lone pairs that can participate in resonance with the aromatic ring. This resonance effect increases the electron density at the ortho and para positions, making these positions more reactive towards electrophilic substitution.
Q69: Draw the structures of major monohalo products in each of the following reactions:
(ii) Which halogen compound in each of the following pairs will react faster towards SN2 reaction?
(a) CH3Br or CH3I
(b) (CH3)3C-Cl or CH3-Cl
Answer(ii)(a) CH3I will react faster towards SN2 reaction because the C-I bond is weaker and iodine is a better leaving group than bromine.
(b) CH3-Cl will react faster towards SN2 reaction because it is a primary halide with minimal steric hindrance, while (CH3)3C-Cl is a tertiary halide, which is highly sterically hindered and thus less reactive in SN2 reactions.
Q70: Which compound in each of the following pairs will react faster towards SN2 reaction with -OH group?
(a) CH3Br or CH3I
(b) (CH3)3C-Cl or CH3-Cl
(ii) Write the product(s) of the following reactions.
Answer(a) CH3I will react faster towards SN2 reaction because iodine is a better leaving group than bromine due to its larger size and weaker C-I bond.
(b) CH3-Cl will react faster towards SN2 reaction because it is a primary halide with less steric hindrance, whereas (CH3)3C-Cl is a tertiary halide, which is highly hindered and not favorable for SN2 mechanism.
Q71: Give reasons for the following:
(i) Ethyl iodide undergoes SN2 reaction faster than ethyl bromide.
(ii) (+_) 2 - butanol is optically inactive.
(iii) C-X bond length in halobenzene is smaller than C - X bond length in CH3 - X
Answer(i) Ethyl iodide undergoes SN2 reaction faster than ethyl bromide because the C-I bond is weaker and longer than the C-Br bond. As a result, iodine leaves more easily as a leaving group, facilitating the SN2 reaction.
(ii) (+–) 2-butanol is optically inactive because it is a racemic mixture containing equal amounts of both d- (dexter) and l- (laevorotatory) forms. The optical rotations of these two forms cancel each other, making the mixture optically inactive.
(iii) The C-X bond length in halobenzene is smaller than the C-X bond length in CH3-X because in halobenzene, the C-X bond has partial double bond character due to resonance with the aromatic ring. This partial double bond character shortens the bond length compared to the purely single C-X bond in CH3-X.
Q72: Explain the following:
(i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(ii) Alkyl halides, though polar, are immiscible with water.
(iii) Grignard's reagents should be prepared under anhydrous conditions.
Answer (i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride because, in chlorobenzene, the lone pair of electrons on chlorine gets delocalized into the benzene ring through resonance, reducing the polarity of the C-Cl bond. In cyclohexyl chloride, such delocalization does not occur, and the C-Cl bond remains purely polar, resulting in a higher dipole moment.
(ii) Alkyl halides, though polar, are immiscible with water because they cannot form hydrogen bonds with water molecules. Also, their molecules are larger and have a non-polar hydrocarbon part, which prevents them from dissolving in the polar solvent, water.
(iii) Grignard reagents should be prepared under anhydrous (completely dry) conditions because they react violently with water to form alkanes and lose their reactivity. Presence of water would destroy the Grignard reagent by converting it into the corresponding hydrocarbon, making it useless for further reactions.
Q73: Answer the following questions:
(i) What is meant by chirality of a compound? Give an example.
(ii) Which one of the following compounds is more easily hydrolysed by KOH and why?
CH3CHClCH2CH3 or CH3CH2CH2CH2Cl
(iii) Which one undergoes SN2 substitution reaction faster and why ?
Answer(i) Chirality refers to the property of a molecule having a non-superimposable mirror image, usually due to the presence of an asymmetric (chiral) carbon atom bonded to four different groups.
Example: 2-Butanol (CH3CH(OH)CH2CH3) is a chiral compound.
(ii) CH3CHClCH2CH3 is hydrolysed more easily by KOH because it is a secondary alkyl halide and forms a more stable carbocation intermediate compared to CH3CH2CH2CH2Cl which is a primary halide and less reactive towards hydrolysis (SN1).
(iii) The compound containing iodine (1-iodobutane) undergoes SN2 substitution faster than the compound containing chlorine (1-chlorobutane) because the C-I bond is weaker and easier to break than the C-Cl bond, making iodine a better leaving group than chlorine.
Q74: Answer the following:
(i) Haloalkanes easily dissolve in organic solvents. Why?
(ii) What is the racemic mixture? Give an example.
(iii) Of the two bromo derivatives, C6H5CH(CH3)Br and (C6H5)CH(C6H5)Br, which one is towards SN1 substitution reaction why?
Answer (i) Haloalkanes easily dissolve in organic solvents because they are covalent compounds and can form van der Waals interactions with organic solvents. "Like dissolves like" principle favors their solubility in non-polar or weakly polar organic solvents.
(ii) A racemic mixture contains equal amounts of two enantiomers (d- and l- forms) of a chiral compound and shows no optical activity because the rotations due to both isomers cancel each other.
Example: A mixture of (+) and (–) lactic acid.
(iii) (C6H5)CH(C6H5)Br reacts faster in SN1 reaction because the carbocation formed after the removal of Br– is stabilized by resonance through two phenyl groups. In C6H5CH(CH3)Br, the carbocation is less stabilized as only one phenyl group provides resonance stabilization.
Q75: Rearrange the compounds of each of the following sets 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,
3-bromo-2-methylbutane.
(iii) 1-bromobutane,
1-bromo-2, 2-dimethylpropane,
1-bromo-2-methylbutane,
1-bromo-3-methylbutane.
Answer (i) 2-bromo-2-methylbutane < 2-bromopentane < 1-bromopentane
(ii) 2-bromo-2-methylbutane < 3-bromo-2-methylbutane < 1-bromo-3-methylbutane
(iii) 1-bromo-2,2-dimethylpropane < 1-bromo-3-methylbutane < 1-bromo-2-methylbutane < 1-bromobutane
Q76: Write the mechanism of the following reaction:
Q77: (i) Write a chemical test to distinguish between
(a) chlorobenzene and benzyl chloride
(b) chloroform and carbon tetrachloride
(ii) Why is methyl chloride hydrolysed more easily than chlorobenzene?
(i) (a) Benzyl chloride gives a white precipitate with alcoholic AgNO3 due to the formation of AgCl, while chlorobenzene does not react under similar conditions.
(b) Chloroform gives a positive carbylamine test (formation of foul-smelling isocyanide), whereas carbon tetrachloride does not respond to this test.
(ii) Methyl chloride is hydrolysed more easily than chlorobenzene because the C-Cl bond in methyl chloride is a simple polar covalent bond, whereas in chlorobenzene, the C-Cl bond has partial double bond character due to resonance, making it stronger and less reactive towards nucleophilic attack.
Q78: (i) State one use of DDT and iodoform.
(ⅱ) Which compound in the following couples will react faster towards SN2 displacement and why?
(a) 1-bromopentane or 2-bromopentane
(b) 1-bromo-2-methyl butane or 2-bromo-2-methyl butane
Answer: (i)
- DDT (Dichlorodiphenyltrichloroethane): It is used as an insecticide to control mosquitoes and other insects.
- Iodoform (CHI3): It is used as an antiseptic for dressing wounds.
(ii) (a) 1-bromopentane or 2-bromopentane
Answer: 1-bromopentane will react faster in SN2 reaction because it is a primary alkyl halide and has less steric hindrance, allowing the nucleophile to attack easily. On the other hand, 2-bromopentane is a secondary alkyl halide, which has more steric hindrance.
(iii) (b) 1-bromo-2-methylbutane or 2-bromo-2-methylbutane
Answer: 1-bromo-2-methylbutane will react faster in SN2 reaction because it is a primary halide with less steric hindrance. 2-bromo-2-methylbutane is a tertiary halide, which is highly hindered and therefore not favorable for SN2 mechanism.
Q79: How would you differentiate between SN1 and SN2 mechanism of substitution reactions ? Give one example of each.
Answer : Difference Between SN1 and SN2 Mechanisms
| Factor | SN1 Reaction | SN2 Reaction |
|---|---|---|
| Full Form | Substitution Nucleophilic Unimolecular | Substitution Nucleophilic Bimolecular |
| Molecularity | Unimolecular (rate depends only on substrate concentration) | Bimolecular (rate depends on both substrate & nucleophile concentration) |
| Rate Equation | Rate = k [Alkyl halide] | Rate = k [Alkyl halide] [Nucleophile] |
| Order of Reaction | First-order | Second-order |
| Mechanism Steps | Two-step (Carbocation intermediate formed) | One-step (single transition state) |
| Intermediate Formed | Carbocation | No intermediate (only transition state) |
| Carbocation Stability Requirement | Tertiary > Secondary > Primary | Methyl > Primary > Secondary (Tertiary is unfavorable) |
| Stereochemistry | Racemization (mixture of products) | Inversion of configuration (Walden inversion) |
| Favored Solvent | Polar protic (e.g., water, alcohol) | Polar aprotic (e.g., acetone, DMSO) |
| Example | (CH3)3CBr + H2O → (CH3)3COH + HBr (Tert-butyl bromide with water) |
CH3Br + OH- → CH3OH + Br- (Methyl bromide with hydroxide ion) |
Q80: 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) in the pair, (CH3)3C-Cl and CH3Cl, CH3Cl will react SN2 with -OH?
Answer: (i) In chlorobenzene, the chlorine atom is attached to an sp²-hybridized carbon of the benzene ring. Due to resonance, the lone pair of electrons on chlorine delocalizes into the aromatic ring, reducing the polarity of the C–Cl bond.
In cyclohexyl chloride, the chlorine atom is bonded to an sp³-hybridized carbon without such resonance delocalization. Hence, the C–Cl bond remains more polar.
The effective dipole in chlorobenzene decreases due to resonance, while cyclohexyl chloride retains full bond polarity. Therefore, chlorobenzene has a lower dipole moment than cyclohexyl chloride.
(ii) Alkyl halides possess a polar C–X bond, but the alkyl (R) group is non-polar and hydrophobic in nature.
Water forms strong hydrogen bonds, but alkyl halides cannot form such bonds with water molecules. Thus, the interaction between water and alkyl halide molecules is weak.
Due to these weak interactions and the hydrophobic alkyl group, alkyl halides do not dissolve in water despite their polarity.
(iii) SN2 reactions require a backside attack by the nucleophile (–OH⁻), and the approach is hindered by bulky groups.
In (CH₃)₃C–Cl (tert-butyl chloride), the carbon is tertiary and heavily hindered, preventing the nucleophile from approaching easily.
In CH₃Cl (methyl chloride), the carbon is least hindered, making it very reactive towards SN2 attack.
CH₃Cl undergoes SN2 reaction easily with –OH⁻, whereas (CH₃)₃C–Cl does not, because of steric hindrance.
