Practice free Halkoarenes & Haloarenes (Chemistry) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
The given compound $\text{C}_6\text{H}_5\!-\!\text{CH}\!=\!\text{CH}\!-\!\text{CH}(X)\!-\!\text{CH}_2\text{CH}_3$ is an example of _____.
X is on the sp³ carbon adjacent to a C=C → allylic halide.
Identify the final product [D] obtained in the following sequence:
$$\text{CH}_3\text{CHO} \xrightarrow[(ii)\,\text{H}_3\text{O}^+]{(i)\,\text{LiAlH}_4} [A] \xrightarrow[\Delta]{\text{H}_2\text{SO}_4} [B] \xrightarrow{\text{HBr}} [C] \xrightarrow{\text{C}_6\text{H}_5\text{Br},\ \text{Na/dry ether}} [D]$$
CH₃CHO → CH₃CH₂OH (A) → CH₂=CH₂ (B) → CH₃CH₂Br (C) → Wurtz with PhBr/Na → PhCH₂CH₃ = ethylbenzene.
The compound that will undergo $S_N^1$ reaction with the fastest rate is
(A) forms a stable 2° benzylic carbocation (resonance-stabilised).
Major products A and B formed in the following reaction sequence are:
2-methyl-1-cyclohexanol (i.e. 1-methylcyclohexanol — OH and CH₃ on the same ring carbon)
$\xrightarrow{PBr_3}$ A (major) $\xrightarrow[\Delta]{alc.\ KOH}$ B (major)
PBr₃ replaces OH with Br at the same (tertiary) carbon → 1-bromo-1-methylcyclohexane. Alcoholic KOH gives E2 to the Saytzeff alkene → 1-methylcyclohexene.
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): 1-iodobutane undergoes $S_N2$ reaction faster than 1-chlorobutane.
Reason (R): Iodine is a better leaving group because of its large size.
Choose the correct answer:
I⁻ is a weaker base and a better leaving group than Cl⁻ (larger, more polarizable), so the C–I bond breaks more readily in $S_N2$. R correctly explains A.
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