NEET Haloalkanes and Haloarenes Previous Year Questions with Solutions

2-Methylbutane has the structure (CH₃)₂CH-CH₂-CH₃. There are four distinct types of hydrogen atoms available for substitution:

• • This yields 4 constitutional (structural) isomers.

Examine each constitutional isomer for chiral stereocenters (asymmetric carbons):

• • (2) CH₃-C(Cl)(CH₃)-CH₂-CH₃: C2 is bonded to two identical methyl groups, hence achiral ⟹ 1 isomer.

Examine the remaining two constitutional isomers:

• • (4) (CH₃)₂CH-CH₂-CH₂Cl: Contains no asymmetric carbon atom, hence achiral ⟹ 1 isomer.
• • Total number of products including stereoisomers:
• • \text{Total} = 2 + 1 + 2 + 1 = 6.
• • Matches Option (4) perfectly.

In a bimolecular nucleophilic substitution (S_N2), the rate equation is Rate = k [Substrate][Nu⁻]. The rate constant k depends directly on the leaving group ability of the departing halide ion:

• • Because 1-iodobutane and 1-chlorobutane are both primary (1°) alkyl halides with identical steric profiles, the difference in rate is determined purely by the leaving group.
• • 1-Iodobutane reacts substantially faster than 1-chlorobutane.
• • Assertion (A) is completely TRUE.

A good leaving group is one that can stably accommodate the electron pair acquired upon cleavage:

• • The large diffuse electron cloud of iodide stabilizes the negative charge effectively, making I⁻ a very weak conjugate base and hence an exceptional leaving group.
• • Reason (R) is completely TRUE.

Connect cause and effect:

• • The large size and weak C-I bond are the precise mechanistic reason why 1-iodobutane reacts faster.
• • Therefore, Both A and R are true and R is the correct explanation of A, matching Option (1).

The unimolecular nucleophilic substitution (SN1) proceeds through a two-step mechanism. The slow, rate-determining step is the heterolytic cleavage of the C-Br bond to form a planar carbocation:

Analyzing the stability of the intermediate generated from each given substrate: 1. Bromobenzene: Phenyl cation C6H5⁺ has the positive charge in an sp² orbital orthogonal to the π-system; it cannot be stabilized by resonance and is energetically prohibitive. 2. (1-Bromoethyl)benzene: The resulting carbocation Ph-C⁺H-CH3 is a secondary benzylic carbocation. The positive charge is extensively delocalized into the benzene ring across four contributing resonance structures, enhanced by hyperconjugation from the methyl group. 3. (Bromomethyl)cyclohexane: Generates an unstable 1° carbocation. 4. Bromocyclohexane: Generates a 2° aliphatic carbocation without any resonance stabilization.

By the Hammond postulate, the transition state leading to the 2° benzylic carbocation has the lowest activation energy (ΔG‡). Consequently, (1-bromoethyl)benzene undergoes solvolysis/SN1 with the fastest rate.

Reaction of secondary alcohols with phosphorus tribromide (PBr3) cleanly converts the -OH into a bromide leaving group without carbocation rearrangement, yielding 1-bromo-2-methylcyclohexane:

Alcoholic potassium hydroxide (alc. KOH) is a strong dehydrohalogenating reagent. Upon heating, the ethoxide/hydroxide base abstracts a proton from a β-carbon concurrent with the expulsion of the bromide ion in a concerted anti-coplanar E2 pathway:

The base can abstract a proton from either: - The tertiary β-carbon (C-2): Abstraction of the single tertiary hydrogen yields 1-methylcyclohex-1-ene (a trisubstituted alkene with endocyclic double bond). - The secondary β-carbon (C-6): Abstraction yields 3-methylcyclohex-1-ene (a disubstituted alkene). By Zaitsev’s rule, the more substituted alkene possesses more hyperconjugative α-hydrogens and is thermodynamically more stable, predominating as the major product B:

Phosphorus trichloride possesses phosphorus in the +3 oxidation state. Nucleophilic attack by three molecules of alcohol displaces all three chloride atoms in sequence, producing three moles of alkyl chloride and one mole of orthophosphorous acid (H3PO3):

Phosphorus pentachloride possesses phosphorus in the +5 oxidation state. The alcohol oxygen coordinates to phosphorus, releasing HCl and collapsing through a four-membered transition state to form one mole of alkyl chloride and one mole of phosphoryl chloride / phosphorus oxychloride (POCl3):

Product A is H3PO3 (phosphorous acid), and Product B is POCl3 (phosphoryl chloride). This pairing uniquely matches Option (2).

Halogen derivatives of hydrocarbons are classified into distinct structural categories based on the hybridization and environment of the carbon atom bearing the halogen:

In the provided structure C₆H₅–CH=CH–CH(X)–CH₂CH₃, locate the halogen X:

The halogen X is bonded to carbon C-γ, which is an sp³-hybridized carbon atom. C-γ is bonded directly to carbon C-β, which is an sp²-hybridized carbon participating in an aliphatic carbon-carbon double bond (C=C). This matches the exact definition of an allylic halide.

The benzene ring (Ar) is attached to C-α, not to C-γ. Because C-γ is not directly attached to the aromatic ring, the compound CANNOT be classified as a benzylic halide.

Lithium aluminium hydride reduces acetaldehyde to primary alcohol ethanol:

Heating ethanol with concentrated sulphuric acid at 443 K undergoes intramolecular dehydration to form ethene:

Addition of hydrogen bromide across the symmetrical alkene double bond produces bromoethane:

Reaction of bromoethane and bromobenzene with sodium metal in dry ether produces alkylarene (ethylbenzene) via the Wurtz-Fittig reaction:

When 2-bromopentane is heated with alcoholic potassium hydroxide ($alc.\; KOH$), base-induced $\beta$-elimination of $HBr$ occurs across the $C_\alpha-C_\beta$ bond:

- Pathway A (Loss of $\beta_1$-H): Yields pent-1-ene ($CH_3CH_2CH_2CH=CH_2$), which has only 1 alkyl substituent and 2 hyperconjugative $\alpha$-hydrogens (19% yield, minor). - Pathway B (Loss of $\beta_2$-H): Yields pent-2-ene ($CH_3CH_2CH=CHCH_3$), which has 2 alkyl substituents and 5 hyperconjugative $\alpha$-hydrogens (81% yield, major).

Alexander Zaitsev (Saytzeff) formulated in 1875 that: *"In dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms."* This is Saytzeff's Rule, corresponding to Option (1).

Halogen atoms in Group 17 increase in size as new quantum electron shells are added down the column: $F < Cl < Br < I$. Carbon uses its $2sp^3$ hybrid orbital to bond with valence $p$ orbitals of increasing principal quantum number ($2p$ in F, $3p$ in Cl, $4p$ in Br, $5p$ in I).

As orbital size increases, orbital overlap becomes more diffuse and less effective. Experimental bond lengths from NCERT Table 10.2: - $C-F$: $139\text{ pm}$ - $C-Cl$: $178\text{ pm}$ - $C-Br$: $193\text{ pm}$ - $C-I$: $214\text{ pm}$

Because shorter bonds possess stronger electrostatic attraction between nuclei and shared electrons, bond dissociation enthalpy decreases monotonically down the group:

When 2-bromopentane is heated with alcoholic potassium hydroxide ($$\text{alc. KOH}$$), the base abstracts a proton from a $$\beta$$-carbon while the bromide leaving group departs from the $$\alpha$$-carbon: $$\text{CH}_3-\text{CH(Br)}-\text{CH}_2-\text{CH}_2-\text{CH}_3 \xrightarrow{\text{alc. KOH, } \Delta} \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_3 + \text{KBr} + \text{H}_2\text{O}$$ Because hydrogen is abstracted from the $$\beta$$-position, this is universally termed a $$\beta$$-elimination reaction (Statement a).

2-Bromopentane has two different $$\beta$$-carbons: $$\beta_1$$ ($$\text{C}_1$$, methyl group) and $$\beta_2$$ ($$\text{C}_3$$, methylene group). According to Zaitsev's rule, in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms. Hence, pent-2-ene (disubstituted alkene, 81%) is formed as the major product over pent-1-ene (monosubstituted alkene, 19%). Thus, it follows Zaitsev's rule (Statement b).

- Dehydrohalogenation: Elimination of a hydrogen halide ($$\text{HX}$$, here $$\text{HBr}$$) from an alkyl halide. Statement (c) is TRUE. - Dehydration: Elimination of a water molecule ($$\text{H}_2\text{O}$$) from an alcohol to form an alkene. Here, $$\text{HBr}$$ is lost, not $$\text{H}_2\text{O}$$. Thus, Statement (d) is FALSE.

Statements (a), (b), and (c) are all correct descriptions, corresponding to Option 1.