Questions & explanations
1. What is the major product and the mechanism of the reaction when 2-bromo-2-methylbutane is treated with alcoholic potassium hydroxide?
- A. 2-methylbut-1-ene via E1 mechanism
- B. 2-methylbut-2-ene via E2 mechanism
- C. 2-methylbutan-2-ol via SN1 mechanism
- D. 2-methylbut-2-ene via E1 mechanism
Answer: B. 2-methylbut-2-ene via E2 mechanism
2-Bromo-2-methylbutane is a tertiary alkyl halide. Alcoholic potassium hydroxide provides a strong base (ethoxide ion) and is a polar protic solvent. With a strong base, elimination proceeds via an E2 mechanism rather than E1. The major product is the more substituted alkene according to Saytzeff's rule, which is 2-methylbut-2-ene. Option A is incorrect because the product is the less substituted alkene. Option C is incorrect because substitution does not occur under these strongly basic conditions; an alcohol would not form. Option D is incorrect because the mechanism is E2, not E1, due to the strong base favoring a concerted bimolecular elimination.
2. Which condition favors an SN1 reaction over an SN2 reaction?
- Use of a strong nucleophile
- Use of a polar protic solvent
- Use of a primary alkyl halide
- Use of a high concentration of nucleophile
Answer: Use of a polar protic solvent
Polar protic solvents stabilize the carbocation intermediate and favor SN1 by solvating the leaving group and the carbocation. Strong nucleophile, primary alkyl halide, and high nucleophile concentration all favor SN2. SN1 is unimolecular, so its rate is independent of nucleophile concentration.
3. Which of the following best describes the SN2 mechanism?
- It is a two-step mechanism involving a carbocation intermediate.
- It is a single-step mechanism with simultaneous bond formation and bond breaking, and it follows first-order kinetics.
- It is a single-step mechanism with simultaneous bond formation and bond breaking, and it follows second-order kinetics.
- It is a two-step mechanism involving a free radical intermediate.
Answer: It is a single-step mechanism with simultaneous bond formation and bond breaking, and it follows second-order kinetics.
The SN2 reaction is a concerted, single-step process where the nucleophile attacks as the leaving group departs, resulting in second-order kinetics (rate ∝ [substrate][nucleophile]). Option A describes SN1, option B has incorrect kinetics (first-order), and option D describes radical reactions.
4. Which statement about chiral molecules is correct?
- Chiral molecules are always optically inactive.
- Enantiomers have identical physical properties except for their interaction with plane-polarized light.
- A molecule with a plane of symmetry is chiral.
- Racemic mixture rotates plane-polarized light.
Answer: Enantiomers have identical physical properties except for their interaction with plane-polarized light.
Enantiomers rotate plane-polarized light in opposite directions, but other physical properties (melting point, boiling point, etc.) are identical in an achiral environment. Chiral molecules are optically active; a plane of symmetry implies achirality; a racemic mixture is optically inactive.
5. The rate-determining step in an SN1 reaction is:
- Formation of the carbocation intermediate
- Attack of the nucleophile on the carbocation
- Departure of the leaving group from the carbocation
- Rearrangement of the carbocation
Answer: Formation of the carbocation intermediate
In SN1, the first step is slow ionization to form a carbocation, which is the rate-determining step. The subsequent nucleophilic attack is fast. Departure of the leaving group is part of the first step. Rearrangement may occur but is not the rate-determining step in the basic mechanism.
6. Which stereochemical outcome is typically observed in an SN1 reaction involving a chiral substrate?
- Complete retention of configuration
- Complete inversion of configuration
- Racemization (partial or complete)
- Formation of a single enantiomer
Answer: Racemization (partial or complete)
In SN1, the planar carbocation intermediate allows the nucleophile to attack from either face with equal probability, leading to racemization (a mixture of both enantiomers). Retention and inversion are not typical; single enantiomer is not formed unless a chiral environment is present.
7. Walden inversion is associated with which of the following reaction mechanisms?
- SN1 reaction
- SN2 reaction
- E1 reaction
- E2 reaction
Answer: SN2 reaction
Walden inversion refers to the inversion of configuration at a chiral center that occurs in SN2 reactions due to backside attack. SN1 gives racemization, E1 and E2 give mixtures or specific stereochemistry but not typically termed Walden inversion.
8. The assignment of R/S configuration to a chiral center is based on:
- The optical rotation of the compound (dextrorotatory or levorotatory)
- The priority of substituents according to the Cahn-Ingold-Prelog rules
- The absolute configuration relative to D-glyceraldehyde
- The number of carbon atoms in the molecule
Answer: The priority of substituents according to the Cahn-Ingold-Prelog rules
R/S system uses priority rules based on atomic number (Cahn-Ingold-Prelog). d/l is based on relative configuration to glyceraldehyde. Optical rotation (d or l) is unrelated to R/S; the number of carbons does not determine configuration.
9. The nucleophilic substitution of 2-bromobutane with aqueous KOH predominantly gives which product?
- Butan-2-ol (racemic mixture)
- But-1-ene
- Butan-1-ol
- 2-Methoxybutane
Answer: Butan-2-ol (racemic mixture)
2-Bromobutane is a secondary alkyl halide. With aqueous KOH (a strong nucleophile in a protic solvent), substituted predominantly follows an SN1 mechanism via a planar carbocation, leading to a racemic mixture of butan-2-ol.
10. Dehydrohalogenation of 2-bromopentane with alcoholic KOH gives the major alkene according to Saytzeff's rule. That alkene is:
- Pent-1-ene
- Pent-2-ene
- 2-Methylbut-2-ene
- Pent-1-yne
Answer: Pent-2-ene
Saytzeff rule states that the most substituted alkene is the major product. Elimination of HBr from 2-bromopentane can give pent-1-ene (less substituted) or pent-2-ene (more substituted). Pent-2-ene is the major product.
11. The conversion of ethanol to bromoethane can be carried out using which of the following reagents?
- Br₂ in CCl₄
- HBr gas in the presence of ZnCl₂
- NaBr and conc. H₂SO₄
- Both (2) and (3)
Answer: Both (2) and (3)
Ethanol reacts with HBr (in presence of ZnCl₂ as catalyst) to give bromoethane. Also, NaBr with conc. H₂SO₄ generates HBr in situ, which then reacts with ethanol. Option (1) gives addition to alkenes, not substitution.
12. Which of the following alkyl halides will undergo an SN2 reaction most readily?
- tert-Butyl bromide
- sec-Butyl bromide
- n-Butyl bromide (1°)
- Methyl bromide
Answer: Methyl bromide
SN2 reactivity decreases with increasing steric hindrance: methyl > primary > secondary > tertiary. Methyl bromide has the least steric hindrance, so it reacts fastest. Tertiary is slowest due to steric hindrance.