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Among the given compounds I–III, the correct order of bond dissociation energy of the C–H bond marked with * is:
C–H bond strength rises with s-character: $sp$ (II, alkyne) > $sp^2$ (I, benzene) > $sp^3$ (III, cyclopropane). Hence II > I > III.
Which one of the following compounds does NOT decolourize bromine water?
Cyclohexane is a saturated hydrocarbon and does not react with bromine water. Phenol and aniline brominate the ring; styrene adds across the C=C.
The major product of the following reaction is: $\text{C}_6\text{H}_5\text{COCH}_2\text{CH}_2\text{CN} \xrightarrow{\text{(i) CH}_3\text{MgBr (excess)}}\xrightarrow{\text{(ii) H}_3\text{O}^+}$
Excess CH₃MgBr adds to the ketone (→ tertiary alcohol) and to the nitrile (→ imine, then ketone on hydrolysis). Product: a tertiary alcohol at the original carbonyl and a methyl ketone where the –CN was.
Which of the following aqueous solution will exhibit highest boiling point?
Boiling-point elevation $\propto i\times m$. Effective concentrations: urea 0.01, KNO₃ 0.02, Na₂SO₄ 0.03, glucose 0.015. Na₂SO₄ ($i=3$) is highest.
Match List-I with List-II:
List-I — A. Haber process, B. Wacker oxidation, C. Wilkinson catalyst, D. Ziegler catalyst
List-II — I. Fe catalyst, II. PdCl₂, III. [(PPh₃)₃RhCl], IV. TiCl₄ with Al(CH₃)₃
Choose the correct answer:
Haber → Fe (I); Wacker → PdCl₂ (II); Wilkinson → [(PPh₃)₃RhCl] (III); Ziegler → TiCl₄ with Al(CH₃)₃ (IV).
5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of 70 torr. The vapour pressures of pure X and Y are 63 torr and 78 torr respectively. Which of the following is true regarding the described solution?
Ideal (Raoult) pressure $= \tfrac13(63)+\tfrac23(78) = 73$ torr. Observed 70 torr < 73 torr → negative deviation.
Sugar 'X': A. is found in honey. B. is a keto sugar. C. exists in α and β anomeric forms. D. is laevorotatory. 'X' is:
A keto sugar found in honey, existing in α/β anomers and laevorotatory is D-Fructose.
Identify the suitable reagent for the following conversion: methyl benzoate (C₆H₅COOCH₃) → benzaldehyde (C₆H₅CHO).
DIBAL-H [AlH(iBu)₂] partially reduces an ester to the aldehyde at low temperature; aqueous work-up gives benzaldehyde. LiAlH₄ would over-reduce to the alcohol.
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.
The standard heat of formation, in kcal/mol of $\text{Ba}^{2+}$ is: [Given: standard heat of formation of $\text{SO}_4^{2-}$ ion (aq) $= -216$ kcal/mol, standard heat of crystallisation of $\text{BaSO}_4(s) = -4.5$ kcal/mol, standard heat of formation of $\text{BaSO}_4(s) = -349$ kcal/mol]
$\Delta H_f(\text{BaSO}_4) = \Delta H_f(\text{Ba}^{2+}) + \Delta H_f(\text{SO}_4^{2-}) + \Delta H_{cryst}$. $-349 = x + (-216) + (-4.5)\Rightarrow x = -128.5$ kcal/mol.
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