Practice free Chemistry NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
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.
Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula $\text{C}_4\text{H}_8\text{O}$ is:
THF (1), 2-methyloxetane (2 enantiomers), 3-methyloxetane (1), 2-ethyloxirane (2), 2,2-dimethyloxirane (1), 2,3-dimethyloxirane (cis-meso 1 + trans pair 2 = 3): total $= 10$.
Identify the correct orders against the property mentioned: A. $\text{H}_2\text{O} > \text{NH}_3 > \text{CHCl}_3$ – dipole moment. B. $\text{XeF}_4 > \text{XeO}_3 > \text{XeF}_2$ – number of lone pairs on central atom. C. O–H > C–H > N–O – bond length. D. $\text{N}_2 > \text{O}_2 > \text{H}_2$ – bond enthalpy. Choose the correct answer:
A is correct (1.85 > 1.47 > 1.04 D). D is correct (N₂ 945 > O₂ 498 > H₂ 436 kJ/mol). B is wrong (lone pairs: XeF₂ 3 > XeF₄ 2 > XeO₃ 1). C is wrong (bond length N–O > C–H > O–H).
Higher yield of NO in $\text{N}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{NO(g)}$ can be obtained at [$\Delta H$ of the reaction $= +180.7\ \text{kJ mol}^{-1}$]: A. higher temperature, B. lower temperature, C. higher concentration of N₂, D. higher concentration of O₂. Choose the correct answer:
The reaction is endothermic, so higher temperature (A) favours the forward reaction. Increasing reactant concentrations N₂ (C) and O₂ (D) also shifts equilibrium toward NO. Lower temperature (B) does not.
If the rate constant of a reaction is $0.03\ \text{s}^{-1}$, how much time does it take for $7.2\ \text{mol L}^{-1}$ concentration of the reactant to get reduced to $0.9\ \text{mol L}^{-1}$? (Given: $\log 2 = 0.301$)
First order ($s^{-1}$): $t = \dfrac{1}{k}\ln\dfrac{C_0}{C} = \dfrac{1}{0.03}\ln 8 = \dfrac{2.303\times3\times0.301}{0.03} \approx 69.3$ s.
Which one of the following reactions does NOT belong to "Lassaigne's test"?
Lassaigne's test detects N, S and halogens by fusion with sodium. The reduction of CuO by carbon is a combustion-based carbon detection, not part of Lassaigne's test.
Ready to ace NEET?
Free access · No credit card required
Yes. You can attempt every Chemistry question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.
No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.
The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.