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NOTES

ASSIGNMENT

Section A: Multiple Choice Questions (1 mark each)

1. Which of the following is optically active?
   a) 2-chlorobutane
   b) 1-chlorobutane
   c) 2-chloropropane
   d) Chloromethane
2. A compound is optically active if:
   a) It has a double bond
   b) It has all atoms different
   c) It has a chiral carbon
   d) It has polar bonds only
3. SN2 reactions of optically active alkyl halides occur with:
   a) Retention of configuration
   b) Racemisation
   c) Inversion of configuration
   d) No stereochemical change
4. Reaction of alkyl halide with KCN gives:
   a) Isocyanide
   b) Cyanide
   c) Nitroalkane
   d) Alcohol
5. AgNO₂ reacts with alkyl halide to form:
   a) Alkyl nitrite (R–O–NO)
   b) Nitroalkane (R–NO₂)
   c) Alcohol
   d) Nitrile

Section B: Assertion and Reason Questions (1 mark each)

1. Assertion (A): A carbon attached to four different groups is called a chiral center.
   Reason (R): Such a molecule is always superimposable on its mirror image.
2. Assertion (A): SN1 reaction leads to racemisation.
   Reason (R): Carbocation formed is planar and can be attacked from either side.
3. Assertion (A): Reaction of alkyl halide with aq. KOH gives alcohol.
   Reason (R): OH⁻ replaces the halide ion by nucleophilic substitution.

Section C: One Word/One Line Answers (1 mark each)

1. Define a chiral carbon.
2. What happens to the configuration during SN2 reaction of a chiral halide?
3. Name the product when ethyl bromide reacts with AgCN.

Section D: Short Answer Questions (2 marks each)

1. Identify the chiral center in 2-chlorobutane and explain why it is optically active.
2. Distinguish between the products formed when ethyl bromide reacts with KCN and AgCN.

Section E: Short Answer Question (3 marks)

1. Explain the stereochemical outcomes of SN1 and SN2 reactions using optically active 2-bromobutane as an example. Mention the type of product formed in each case.

KEY

Section A: Multiple Choice Questions

1. Answer: a) 2-chlorobutane
2. Answer: c) It has a chiral carbon
3. Answer: c) Inversion of configuration
4. Answer: b) Cyanide
5. Answer: a) Alkyl nitrite (R–O–NO)

Section B: Assertion and Reason

1. Answer: c) Assertion is true, but Reason is false
2. Answer: a) Both Assertion and Reason are true, and Reason is the correct explanation
3. Answer: a) Both Assertion and Reason are true, and Reason is the correct explanation

Section C: One Word/One Line Answers

1. Answer: A carbon atom bonded to four different atoms or groups.
2. Answer: Inversion of configuration (Walden inversion)
3. Answer: Ethyl isocyanide (C₂H₅NC)

Section D: Short Answer Questions

1. Answer:
   In 2-chlorobutane, the second carbon is attached to –H, –Cl, –CH₃, and –CH₂CH₃. All are different, making it a chiral center. Thus, the compound is optically active.
2. Answer:
   – With **KCN**, nucleophile is CN⁻ (via C-atom) → gives **ethyl cyanide (C₂H₅CN)**.
   – With **AgCN**, the CN is covalently bonded to Ag, so nucleophile is –NC → gives **ethyl isocyanide (C₂H₅NC)**.

Section E: Short Answer Question (3 marks)

1. Answer:
   – In an **SN2 reaction**, nucleophile attacks from the backside, leading to **inversion** of configuration (e.g., 2-bromobutane → 2-butanol with inverted configuration).
   
   – In an **SN1 reaction**, a planar carbocation intermediate forms, and nucleophile can attack from either side, resulting in **racemisation** (mixture of enantiomers).
   
   – Therefore, SN2 gives a single stereoisomer (inverted), while SN1 gives a racemic mixture.