The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is:
ARetention of configuration occurs in SN1 and inversion in SN2
BRacemisation occurs in SN1 and inversion of configuration occurs in SN2
CRacemisation occurs in both SN1 and SN2
DInversion of configuration occurs in SN1 only
✅ Correct: B
SN1: forms planar carbocation intermediate → nucleophile attacks from both sides → racemisation. SN2: concerted backside attack → inversion of configuration (Walden inversion). This is one of the most fundamental distinctions in organic chemistry.
Q2
2-Chlorobutane reacts with Cl₂ in the presence of UV light to give C₄H₈Cl₂ isomers. The total number of optically active isomers among the products is:
A4
B5
C6
D8
✅ Correct: C
Chlorination at different positions of 2-chlorobutane: 1-position, 2-position, 3-position, 4-position. Products with chiral centers and non-superimposable mirror images are optically active. Systematic counting gives 6 optically active isomers among the C₄H₈Cl₂ products.
Q3
151 g of 2-bromopentane undergoes elimination with alcoholic KOH giving alkene P in 80% yield. Alkene P then reacts with Br₂ in CCl₄ (100% yield). The mass of the final dibromo product obtained is:
A168 g
B184 g
C200 g
D160 g
✅ Correct: B
Moles of 2-bromopentane = 151/151 = 1 mol. Moles of alkene P (pent-2-ene) = 1×0.80 = 0.8 mol. Dibromo product (2,3-dibromopentane, M = 230 g/mol) at 100% = 0.8×230 = 184 g. \(\mathbf{184\text{ g}}\).
Q4
The isomeric deuterated bromide (C₄H₈DBr) having two chiral carbon atoms is:
A2-Bromo-1-deuterobutane
B2-Bromo-1-deutero-2-methylpropane
C2-Bromo-3-deuterobutane
D2-Bromo-2-deuterobutane
✅ Correct: C
2-Bromo-3-deuterobutane: CH₃-CHBr-CHD-CH₃. C2 (bonded to Br, CH₃, CH₃, CHD — actually C2: CHBr with 4 different groups) and C3 (CHD with 4 different groups: D, H, CH₃, CHBrCH₃). Both C2 and C3 are chiral centers. Answer: C.
Q5
Among the following, the compound having the maximum number of chlorine atoms is:
AGammaxene (BHC, benzene hexachloride)
BChloropicrin
CDDT
DCarbon tetrachloride
✅ Correct: A
Gammaxene (BHC/lindane): C₆H₆Cl₆ — 6 Cl atoms. Chloropicrin: CCl₃NO₂ — 3 Cl atoms. DDT: (ClC₆H₄)₂CHCCl₃ — has 5 Cl atoms. CCl₄: 4 Cl atoms. Gammaxene has maximum 6 Cl atoms.
Q6
The reactivity order of halides in SN2 reactions is:
ARF > RCl > RBr > RI
BRI > RBr > RCl > RF
CRCl > RBr > RI > RF
DAll haloalkanes show equal reactivity
✅ Correct: B
In SN2: leaving group ability (ease of C−X bond breaking) determines reactivity. Weaker C−X bond = better leaving group. Bond strength: C−F > C−Cl > C−Br > C−I. So SN2 reactivity: RI > RBr > RCl > RF. I⁻ is best leaving group.
Q7
Finkelstein reaction involves the conversion of:
AAlkyl chloride to alkyl iodide using NaI in dry acetone
BAlkyl iodide to alkyl chloride using NaCl
CAlkyl bromide to alkyl fluoride using KF
DAryl halide to phenol using NaOH
✅ Correct: A
Finkelstein reaction: RCl + NaI (in dry acetone) → RI + NaCl↓. NaCl precipitates in acetone (insoluble), driving the equilibrium forward. This reaction is used to prepare alkyl iodides. Similarly: RBr → RI with NaI/acetone.
Q8
Which of the following undergoes SN1 reaction most readily?
ACH₃Cl (methyl chloride)
BC₂H₅Cl (ethyl chloride)
C(CH₃)₂CHCl (isopropyl chloride)
D(CH₃)₃CCl (tert-butyl chloride)
✅ Correct: D
SN1 reactivity depends on carbocation stability. Tertiary carbocations are most stable (3 electron-donating alkyl groups). (CH₃)₃CCl (tert-butyl chloride) undergoes SN1 most readily → forms stable (CH₃)₃C⁺ carbocation.
Q9
The Grignard reagent (RMgX) reacts with formaldehyde (HCHO) followed by hydrolysis to give:
AA secondary alcohol
BA tertiary alcohol
CA primary alcohol (with one more carbon than R)
DA carboxylic acid
✅ Correct: C
Grignard reagent (RMgX) + HCHO (formaldehyde) → RCH₂OMgX → hydrolysis → RCH₂OH (primary alcohol with one extra carbon). With aldehydes (RCHO): secondary alcohol. With ketones: tertiary alcohol. With CO₂: carboxylic acid.
Q10
Which of the following halocompounds readily undergoes hydrolysis?
AChlorobenzene (C₆H₅Cl)
BVinyl chloride (CH₂=CHCl)
CAllyl chloride (CH₂=CHCH₂Cl)
DAryl chloride in general
✅ Correct: C
Allyl chloride (CH₂=CHCH₂Cl) readily undergoes hydrolysis because the allylic carbocation (CH₂=CH−CH₂⁺) formed is resonance stabilized. Chlorobenzene and vinyl chloride are resistant to hydrolysis due to resonance delocalization of lone pairs into the ring/π bond.
Q11
The reaction of chloroform (CHCl₃) with aqueous NaOH followed by addition of aniline gives:
ACarbylamine (isocyanide)
BSodium formate
CDichlorocarbene
DChloroaniline
✅ Correct: A
Carbylamine reaction: CHCl₃ + 3KOH → :CCl₂ (dichlorocarbene) → reacts with amine. With primary amine (RNH₂): CHCl₃ + 3KOH + RNH₂ → RNC (isocyanide/carbylamine) + 3KCl + 3H₂O. Carbylamine formed has characteristic foul smell — test for primary amines.
Q12
In which reaction is an aryl halide most reactive?
ASN1 reaction
BSN2 reaction
CNucleophilic aromatic substitution (SNAr) with electron-withdrawing groups
DElectrophilic substitution at the halogen position
✅ Correct: C
Aryl halides are generally unreactive in SN1 and SN2. They undergo nucleophilic aromatic substitution (SNAr, addition-elimination) when electron-withdrawing groups (−NO₂, −CN) are present at ortho/para positions. These groups stabilize the Meisenheimer complex intermediate.
Q13
Swarts reaction is used to prepare:
AAlkyl chlorides from alkyl fluorides
BAlkyl fluorides from alkyl chlorides or bromides using AgF or SbF₃
CAryl fluorides from aryl diazonium salts
DAlkyl iodides from alkyl chlorides
✅ Correct: B
Swarts reaction: RCl/RBr + AgF or SbF₃ → RF (alkyl fluoride). Since alkyl fluorides cannot be made easily by direct methods (F₂ is too reactive), this exchange reaction using metal fluorides is used. Example: CH₃Cl + AgF → CH₃F + AgCl.
Q14
Dehalogenation of vicinal dihalides (1,2-dihalides) with Zn dust gives:
AAlkane
BAlkyne
CAlkene
DAlcohol
✅ Correct: C
Dehalogenation of 1,2-dihalides (vicinal, geminal) using Zn dust in alcoholic solution: CH₂BrCH₂Br + Zn → CH₂=CH₂ + ZnBr₂. Removal of two halogen atoms from adjacent carbons → alkene. This is an elimination reaction.
Q15
The boiling points of alkyl halides follow which order for the same alkyl group?
ARF > RCl > RBr > RI
BRI > RBr > RCl > RF
CAll have same boiling points
DRF > RI > RBr > RCl
✅ Correct: B
Boiling points of alkyl halides increase with molecular mass and polarisability of halogen: RI > RBr > RCl > RF. Iodo compounds have highest BP due to largest molecular mass and highest polarisability (strongest London dispersion forces).
Q16
The reaction of benzene diazonium chloride with CuCl in HCl gives chlorobenzene. This reaction is known as:
ABalz-Schiemann reaction
BSandmeyer reaction
CGattermann reaction
DWurtz reaction
✅ Correct: B
Sandmeyer reaction: ArN₂⁺Cl⁻ + CuX → ArX + N₂. Replacing the diazonium group by halide using copper(I) halide (CuCl, CuBr, CuCN). Gattermann reaction: uses Cu powder instead of Cu salts. Balz-Schiemann: gives ArF using BF₄⁻.
Q17
The compound that reacts with sodium metal to produce hydrogen gas and also gives a positive Baeyer's test is:
AEthyl alcohol
BAcetic acid
CPropyne
D1-propanol with propyne
✅ Correct: C
Propyne (CH₃C≡CH): terminal alkyne reacts with Na → NaC≡CCH₃ + ½H₂ (acidic terminal H). Also decolorises Baeyer's reagent (cold alkaline KMnO₄) — alkenes and alkynes give positive Baeyer's test. Actually propyne reacts with Na (acidic H) and decolorises KMnO₄.
Q18
In the nucleophilic substitution reaction of haloalkane R−X with NaOH, which factor increases the rate of SN2 reaction?
AUsing a tertiary alkyl halide instead of primary
BUsing a polar aprotic solvent (like DMSO) instead of polar protic solvent
CUsing a weaker nucleophile
DIncreasing the concentration of substrate only
✅ Correct: B
SN2 is favored by: (1) primary substrates, (2) strong nucleophiles, (3) polar aprotic solvents (DMSO, acetone, DMF) — these solvate cations but not anions, leaving nucleophile free and reactive. Polar protic solvents (water, alcohol) solvate and reduce nucleophile reactivity.
Q19
Which of the following haloalkanes undergoes predominantly elimination (E2) rather than substitution (SN2) with KOH/alcohol?
ACH₃Cl
BC₂H₅Cl
C(CH₃)₃CBr
DCH₃CH₂CH₂Cl
✅ Correct: C
(CH₃)₃CBr (tert-butyl bromide) undergoes predominantly elimination (E2/E1) rather than SN2 due to steric hindrance. Primary substrates undergo SN2. Tertiary substrates: bulky nucleophile → E2. With KOH/alcohol (strong base, heat) → elimination is favored. Product: isobutylene ((CH₃)₂C=CH₂).
Q20
The major product of the reaction of 1-bromopropane with alcoholic KCN is:
APropan-1-amine
BButanenitrile (propionitrile with CN)
CPropan-1-ol
D1-cyanopropane (butyronitrile)
✅ Correct: D
CH₃CH₂CH₂Br + KCN → CH₃CH₂CH₂CN + KBr. CN⁻ is a nucleophile that attacks through carbon → gives alkyl cyanide (nitrile). Product: CH₃CH₂CH₂CN = butanenitrile (1-cyanopropane/n-propyl cyanide). Chain length increases by one C. This is a useful method to extend carbon chain.
R
Roshan
Expert · 5 Years Experience
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