//M0//QN1//CS//DL0//EQ

Alkyl halide nucleophilic substitution reactions primarily follow the SN1 and SN2 mechanisms because the C–X bond is polar. The reaction rate of the SN1 mechanism depends on the stability of the intermediate carbocation, while the reaction rate of the SN2 mechanism depends on steric hindrance. Chirality plays a crucial role in understanding the mechanisms of SN1 and SN2. A chiral alkyl halide undergoes the SN1 mechanism to give a racemic mixture, whereas the SN2 mechanism is characterized by inversion of configuration.(a) Which of the following compounds is chiral?

(A) 2-methyl-2-chloropropane
(B) 2-chloropropane
(C) 2,2-dimethyl-1-chloropropane
(D) 2-chlorobutane
(b) Which of the following compounds is more reactive for the S N 2 mechanism?
(A) CH 3–CH 2–Br (B)
(C) (D) CH 3 – CH 2 – Cl
(c) Arrange the four butyl bromides in the ascending order of their reactivity for the S N 1 mechanism.
(A) I < II < III < IV
(B) IV < II < I < III
(C) II < I < III < IV
(D) IV < I < II < III
(d) Which of the following compounds will give a racemic mixture according to the S N 1 mechanism?
(A) (CH 3)2 CH – CH 2 – Br
(B) (CH 3)3 CH – Br
(C) CH 3 – CH (Br) – CH 2 – CH 3
(D) (CH 3)2 CH – Br

//X

(a) (D) 2-Chlorobutane
2-Chloro butane: The second carbon is bonded to a hydrogen atom (H), a chlorine atom (Cl), a methyl group (CH 3), and an ethyl group (C 2 H 5). Since all four are different, it is chiral.
(b) (A) CH 3–CH 2–Br
S N 2 reactivity depends on steric hindrance. The less crowded the carbon atom attached to the leaving group, the faster the reaction.
The order of reactivity: Primary (1°) > Secondary (2°) > Tertiary (3°).
Bromine is a better leaving group than chlorine because the C–Br bond is weaker and the Br− ion is more stable. Therefore, Ethyl bromide is the most reactive.
(c) (B) IV < II < I < III
S N 1 reactivity depends on the stability of the carbocation intermediate.
The stability order for carbocations: Tertiary (3°) > Secondary (2°) > Primary (1°).
I: CH 3–CH 2–CH(Br)–CH 3: 2° Carbocation.
II: (CH 3)2 CH–CH 2–Br: 1° Alkyl halide, but it has some branching nearby.
III: (CH 3)3 C–Br: 3° Carbocation (Most stable).
IV: CH 3–CH 2–CH 2–CH 2–Br: 1° Alkyl halide (Least stable).
Order: IV < II < I < III
(d) (C) CH 3 – CH (Br) – CH 2 – CH 3
A chiral alkyl halide undergoes S N 1 to produce a racemic mixture.
2-bromobutane: Upon losing the bromine atom, it forms a secondary carbocation. The subsequent attack by a nucleophile results in a 50:50 mixture of the (R) and (S) enantiomers, known as a racemic mixture.

//M0//QN2//CS//DL0//EQ

When a haloalkane with a β-hydrogen atom is heated with alcoholic KOH, a dehydrohalogenation reaction occurs, resulting in the formation of an alkene. This process is called β-elimination because the hydrogen atom present at the β-position of the haloalkane is removed.

If a haloalkane contains more than one β-hydrogen, multiple products may be formed. However, the major product is the one in which the alkene has the greatest number of alkyl groups attached. This rule is known as the Zaitsev's rule.
(a) Identify A and B
(A) (B)
(A)
(B)
(C)
(D)
(b) Identify the product A.
(A) 2-Methyl propene
(B) But-2-ene
(C) But-1-ene
(D) 2-Methyl-but-2-ene
(c)
Identify the product C.
(A) Propene (B) Propyne
(C) Propan-1-ol (D) Propan-2-ol
(d) The compound (X) C 6 H 13 Cl undergoes dehydrohalogenation to give CH 3–= CH–C 2 H 5. Identify compound X.
(A)
(B)
(C)
(D) None of these
(e)
Identify X and Y.
(X) (Y)
(A) Dilute NaOH HBr / Acetic acid
(B) Alcoholic NaOH HBr / Acetic acid
(C) Aqueous NaOH Br2 / Acetic acid
(D) Alcoholic NaOH Br2 / CHCl3

//X

(a) (B)
(b) (A) 2-Methyl propene
(c) (D) Propan-2-ol
(d) (C)
(e) (B) Alcoholic NaOH HBr / Acetic acid

//M0//QN3//CS//DL0//EQ

Alkyl magnesium halides are Grignard reagents. In the Grignard reaction, the carbon-magnesium bond is covalent but it is more polar because the carbon pulls electrons from the electron-rich magnesium. The magnesium-halogen bond is essentially ionic. The hydrocarbon part of the Grignard reagent acts as a source of carbon anions. Therefore, the Grignard reaction rapidly undergoes nucleophilic addition with aldehydes and ketones, resulting in the formation of nucleophilic products, which upon hydrolysis yield alcohols.(a) Which of the following Grignard reagents is required to convert propanone to 2-methylpropan-2-ol?

(A) CH 3–CH 2–Mg Br
(B) CH 3–Mg Br
(C)
(D) Above all
(b) CH 3 – CH 2 – OH Identify the product B.
(A) CH 3–CH 2–CH 2 –OH
(B) CH 3–CH 2–CH 2–CH 2–OH
(C) CH 3–CH 2–OH
(D)
(c) Identify the product E.
(A)
(B) (CH 3)3 – CH
(C) CH 3 – CH 2 – CH 3
(D) CH 3 – CH 2 – CH 2 – CH 3
(d) Identify the product B.
(A) (B)
(C) (D) None of these
(e) Which of the following Grignard reactions with methanal can produce ?
(A)
(B)
(C)
(D)

//X

(a) (B) CH 3–Mg Br
(b) (D)
(c) (B) (CH 3)3 – CH
It’s a Wortz reaction. The product is formed by dimensation of two alkyl groups.
So, R' = (CH 3)3 Ic – X (X = Cl/Br)
(d) (A)
(e) (D)

//M0//QN4//CS//DL0

A pharmaceutical company is synthesizing an important intermediate using chloroethane (C 2 H 5 Cl) and bromoethane (C 2 H 5 Br) as alkylating agents. The reaction conditions (solvent, base strength, temperature) determine whether the reaction proceeds by S N 1 or S N 2 mechanism. The following experimental observations were made:

Substrate | Solvent used | Reaction rate
C 2 H 5 Br | Stong base in polar aprotic solvent (DMSO) | Very fast
C 2 H 5 Cl | Weak base in protic solvent (Ethanol) | Slow
tert-Butyl chloride (+ - Bucl) | Ethanol | Moderate
tert-Butyl chloride (+ - BuCl) | Water | Very fast

Additional data:
C2H5Br bond is weaker than C2H5Cl
Polar protic solvents stabilize carbocations.
Polar aprotic solvents enhance nucleophilicity.
The company wants to understand which reactions follow S N 1 or S N 2, and which substrate - solvent combination gives maximum yeild.(a) Identify which reactions above proceeds via S N 2 mechanism. Give reason.(b) Explain why C 2 H 5 Br reacts much faster than C 2 H 5 Cl in S N 2 reaction.(c) Why does tert-butyl chloride react faster in water than in ethanol?(d) Predict whether aryl halides (like chlorobenzene) undergo S N 1 or S N 2. Explain.

//X

(a) Only C 2 H 5 Br in polar aprotic solvent (DMSO) undergoes S N 2.
Reason:
Primary halide favors backside attack.
polar aprotic solvent → nucleophile becomes stronger.
(b) Because:
C––Br bond is weak (lower bond energy).
Br– is better leaving group than Cl–.
Transition state stabilizes more effectively.
(c) S N 1 reaction → Rate depends on carbocation stability.
Water is most polar solvent, stabilizes carbocation and increases ionization rate.
(d) Aryl halides do not undergo S N 1 or S N 2.
S N 1 not possible because, carbocation (aryl+ is highly unstable and positive charge cannot form in benzene ring.
S N 2 not possible because, backside attack is blocked by ring structure.

//M0//QN5//CS//DL0

A chemical industry manufactures chlorobenzene, bromobenzene and phenyl chloride derivatives. They study the effect of substituents on:

Reactivity of haloarenes
Nucleophilic substitution
Electrophilic substitution
Bond stength and stability
Some observations:

| Compound | Reactivity toward nucleophiles
(A) | Chlorobenzene | Very low
(B) | p-Nitrochlorobenzene | High
(C) | o-Nitrochlorobenzene | Even higher
(D) | p-Methoxy chlorobenzene | Lowest

Important points:
NO2 group = Strong electron withdrawing → activates nucleophilic substitution.
OCH3 group = Electron donating → deactivates nucleophilic substitution.
Bromine is better leaving group than chlorine.
(a) Arrange the compounds in increasing order of nucleophilic substitution reactivity.(b) Why does p-nitrochlorobenzene react faster than chlorobenzene?(c) Why is o-nitrobromobenzene more reactive than p-nitrochlorobenzene?(d) Explain why p-methoxychlorobenzene shows the lowest reactivity.

//X

(a) D < A < B < C (Lowest → Highest)
(b) Because of NO 2 group strongly electron with drawing, stabilizes Meisen Heimer complex and facilitates nucleophilic attack.
(c) Because, NO 2 at ortho position withdraws electrons strongly. Bromine is better leaving group then chlorine and ortho effect increases stabilization.
(d) Because, OCH 3 group (i) Electron donating (+M) (ii) Increasing electron density on the ring (iii) Repels nucleophiles, (iv) Decreases Meisen\Heimer complex formation.
Haloalkane: An alkane in which one or more H atoms are replaced by halogen atoms (R–X).
Haloarene (Aryl Halide): An aromatic compound where a halogen is bonded directly to the aromatic ring.
Alkyl Halide: A haloalkane where the halogen is attached to an sp 3 carbon (same as haloalkane in practice).
Aryl Halide: Halogen bonded to an sp 2 carbon of an aromatic ring (C–X is stronger, less reactive).
Vinylic Halide: A halogen attached to an sp 2 carbon of a C=C (halogen on the double bond carbon).
Allylic Halide: A halogen bonded to a carbon next to a double bond (stabilized by resonance).
Benzylic Halide: A halogen attached to the carbon next to an aromatic ring (resonance-stabilized intermediate).
Nucleophile: An electron-rich species that donates an electron pair to form a new bond (Nu–,:OH,:CN, etc.).
Electrophile: An electron-deficient species that accepts an electron pair (E+, H+, Br+, carbocations).
S N 1 Reaction: Unimolecular nucleophilic substitution; rate ∝ [R–X]; proceeds via carbocation intermediate; favours 3° centres.
S N 2 Reaction: Bimolecular nucleophilic substitution; rate ∝ [R–X][Nu]; backside attack with inversion of configuration; favours 1° centres.
Carbocation: Positively charged carbon intermediate (R–C+), stabilized by alkyl groups via hyperconjugation.
Radical Halogenation: Free-radical substitution (initiation, propagation, termination) commonly used to halogenate alkanes (UV light required).
Grignard Reagent: Organomagnesium halide (R–MgX) formed in dry ether; strong nucleophile/base used for C–C bond formation.
Sandmeyer Reaction: Conversion of aryl diazonium salts to aryl halides (CuX mediated) — key route to introduce Cl/Br/CN/I onto benzene.
Saytzeff (Zaitsev) Rule: In elimination, the more substituted (more stable) alkene is the major product.
Wurtz/Fittig Reactions: Coupling of alkyl/aryl halides with sodium in dry ether to form C–C bonds (Wurtz: alkyl–alkyl; Fittig: aryl–aryl).
Polyhalogen Compounds: Molecules with multiple halogens (e.g., CHCl3, CCl4, freons, DDT) — note uses and environmental/health hazards.
C–X Bond Trends: Bond length ↑ and bond enthalpy ↓ down the halogen group (C–I longest/weakest, C–F shortest/strongest); influences reactivity and ease of substitution.
Reactivity Order (Alkyl Halides): For nucleophilic substitution: SN1 (3° > 2° > 1°); SN2 (methyl > 1° > 2° > 3°).