Questions & explanations
1. Arrange the following in decreasing order of SN1 reactivity: chlorobenzene, benzyl chloride, allyl chloride, 1-chloropropane, 2-chloropropane, vinyl chloride.
- benzyl chloride > allyl chloride > 1-chloropropane > 2-chloropropane > chlorobenzene = vinyl chloride
- allyl chloride > benzyl chloride > 1-chloropropane > 2-chloropropane > chlorobenzene = vinyl chloride
- benzyl chloride > allyl chloride > 2-chloropropane > 1-chloropropane > chlorobenzene = vinyl chloride
- allyl chloride > benzyl chloride > 2-chloropropane > 1-chloropropane > chlorobenzene = vinyl chloride
Answer: benzyl chloride > allyl chloride > 2-chloropropane > 1-chloropropane > chlorobenzene = vinyl chloride
SN1 reactivity depends on carbocation stability. Benzyl and allyl carbocations are resonance-stabilised, so they are most reactive. Tertiary (2-chloropropane) forms a more stable carbocation than primary (1-chloropropane). Vinyl and aryl halides do not undergo SN1 because their carbocations are highly unstable. Thus the correct order is benzyl ≈ allyl > 3° > 2° > 1° >> vinyl ≈ aryl.
2. Which of the following statements about physical properties of haloarenes is correct?
- Chlorobenzene has a higher boiling point than n-propyl chloride.
- p-Dichlorobenzene has the highest melting point among dichlorobenzene isomers.
- Haloarenes are highly soluble in water.
- Haloarenes are less dense than water.
Answer: p-Dichlorobenzene has the highest melting point among dichlorobenzene isomers.
p-Dichlorobenzene has the highest melting point (53°C) among dichlorobenzene isomers due to its symmetrical structure, which allows better crystal packing and stronger intermolecular forces. In contrast, o-dichlorobenzene (m.p. -17°C) and m-dichlorobenzene (m.p. -24°C) have lower melting points due to less symmetrical packing.
3. Which of the following reactions proceeds via an SN2 mechanism?
- PhCl + NaOH(aq) 623 K, 300 atm → phenol
- (CH3)3CBr + EtOH/Δ → 2-methylpropene
- (CH3)3CBr + KOC(CH3)3 in tBuOH → 2-methylpropene
- CH3Br + KOH(aq) → CH3OH
Answer: CH3Br + KOH(aq) → CH3OH
CH3Br is a primary halide, and KOH(aq) provides a strong nucleophile (OH⁻) in a polar protic solvent. Primary halides favour SN2 because the backside attack is not sterically hindered. The product is methanol. Options b and c involve tertiary halides, which favour elimination (E1 or E2). Option d is an SNAr reaction, not SN2.
4. Which of the following correctly describes the stereochemical outcome of an SN1 reaction on a chiral alkyl halide?
- Complete inversion of configuration
- Complete retention of configuration
- Racemisation with a slight excess of retention
- Racemisation with a slight excess of inversion
Answer: Racemisation with a slight excess of inversion
In SN1, the planar carbocation intermediate allows attack from either face, leading to racemisation. However, due to ion-pair shielding, the nucleophile attacks preferentially from the side opposite the leaving group, giving a slight excess of inversion (typically 5-20%). NCERT discusses this in Chapter 6, Section 6.7.1.
5. Arrange the following in decreasing order of SN1 reactivity: (I) CH2=CH-CH2-Cl, (II) C6H5-CH2-Cl, (III) CH3-CH2-CH2-Cl, (IV) (CH3)3C-Cl.
- II > IV > I > III
- I > II > IV > III
- IV > II > I > III
- II > I > IV > III
Answer: II > I > IV > III
SN1 reactivity depends on carbocation stability. Benzylic carbocation (II) is most stable due to resonance. Allylic (I) is next, also resonance-stabilized but less than benzylic. Tertiary (IV) is less stable than allylic because allylic resonance is more effective. Primary (III) is least stable. Order: II > I > IV > III.
6. DDT is prepared by reacting chlorobenzene with which compound in the presence of concentrated H2SO4?
- Tetrachloromethane (carbon tetrachloride)
- Trichloromethane (chloroform)
- Trichloroethanal (chloral)
- Dichloromethane (methylene chloride)
Answer: Trichloroethanal (chloral)
DDT (p,p'-dichlorodiphenyltrichloroethane) is synthesised by the condensation of two molecules of chlorobenzene with one molecule of trichloroethanal (chloral, CCl3CHO) in the presence of concentrated H2SO4. This reaction is described in NCERT Class 12 Chemistry, Chapter 6, Section 6.9.6.
7. In the catalytic ozone depletion cycle, one chlorine atom can destroy up to how many ozone molecules before being removed?
- 100
- 100,000
- 1000
- 10,000
Answer: 100,000
In the stratosphere, a chlorine radical from CFC photolysis reacts with ozone: Cl• + O3 → ClO• + O2, then ClO• + O → Cl• + O2. The chlorine radical is regenerated, allowing it to destroy many ozone molecules. NCERT states that one chlorine atom can destroy up to 100,000 ozone molecules.
8. Which of the following undergoes nucleophilic aromatic substitution most readily?
- Chlorobenzene
- 2,4,6-Trinitrochlorobenzene
- p-Nitrochlorobenzene
- 2,4-Dinitrochlorobenzene
Answer: 2,4,6-Trinitrochlorobenzene
Electron-withdrawing nitro groups at ortho and para positions stabilise the Meisenheimer intermediate via -M effect. More nitro groups increase stabilisation, making substitution easier. 2,4,6-Trinitrochlorobenzene reacts with warm water, while chlorobenzene requires harsh conditions.
9. Which of the following is the best leaving group in an SN2 reaction in acetone (a polar aprotic solvent)?
- I-
- Cl-
- Br-
- F-
Answer: I-
Leaving group ability depends on the stability of the anion: weaker bases are better leaving groups. I- is the conjugate base of HI (strongest acid), so it is the weakest base and thus the best leaving group. In polar aprotic solvents, leaving group order remains I- > Br- > Cl- > F-.
10. Which of the following statements is correct about the reactivity of haloalkanes and haloarenes?
- Haloalkanes undergo nucleophilic substitution readily, while haloarenes undergo electrophilic substitution readily.
- Haloalkanes undergo electrophilic substitution readily, while haloarenes undergo nucleophilic substitution readily.
- Both haloalkanes and haloarenes undergo nucleophilic substitution with equal ease.
- Haloarenes are more reactive than haloalkanes in nucleophilic substitution.
Answer: Haloalkanes undergo nucleophilic substitution readily, while haloarenes undergo electrophilic substitution readily.
Haloalkanes have a polar C-X bond and can form carbocations, so they undergo SN1/SN2 readily. Haloarenes have a deactivated ring due to the -I effect of halogen, but they undergo electrophilic substitution (EAS) as the halogen is ortho/para-directing. Thus statement a is correct.
11. Chlorobenzene reacts with NaOH(aq) at 623 K and 300 atm to give which major product?
- o- and p-dichlorobenzene via EAS
- Aniline via benzyne
- Phenol via SNAr
- Sodium phenoxide via SN2
Answer: Phenol via SNAr
Under drastic conditions (high temperature and pressure), chlorobenzene undergoes nucleophilic aromatic substitution (SNAr) with NaOH to give phenol. The mechanism is addition-elimination, not benzyne, because the conditions are not strongly basic enough for benzyne.
12. Which condition favours E1 over SN1 for a tertiary alkyl halide?
- Low temperature and strong nucleophile
- High temperature and weak base
- Polar aprotic solvent and strong base
- Primary substrate and weak base
Answer: High temperature and weak base
E1 elimination is favoured at high temperature (due to positive entropy change) and with a weak base (which cannot effectively act as a nucleophile). Both E1 and SN1 share the carbocation intermediate, but high temperature shifts the equilibrium toward elimination.