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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
Most stable carbanion among the following isA. B. C. D. |
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Answer» Correct Answer - D Due to presence of electron withdrawing `NO_(2)` group, it is most stable. |
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| 152. |
Assertion (A): All the carbon atoms in `H_(2)C = C = CH_(2)` are `sp^(2)` hybridised. Reason (R) : In this molecule all the carbon atoms are attached to each other by double bonds.A. Both A and R are correct and R is the correct explanation of A.B. Both A and R are correct but R is not the correct explanation of A.C. Both A and R are not correct.D. A is not correct but R is correct. |
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Answer» Correct Answer - D Correct (A). In `CH_(2) = C = CH`, the central carbon is sp-hybridized while the terminal carbon atoms are `sp^(2)`-hybridized. |
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| 153. |
Assertion (A) : All the carbon atoms in `H_(2)C=C=CH_(2)` are `sp^(2)` hybridised Reason (R) : In this molecule, all the carbon atoms are attached to each other by double bonds.A. Both A and R are correct and R is the correct explanation of A.B. Both A and R are correct but R is not the correct explanation of A.C. Both A and R are not correctD. A is not correct but R is correct |
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Answer» Correct Answer - D Correct assertion. The hybridisation state of carbon atoms is : `overset((sp^(2)))(CH_(2))=overset((sp))(C)=overset((sp^(2)))(CH_(2))` |
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| 154. |
Hyperconjugation is most useful for stabilizing which of the following carbocations?A. neo-PnetylB. tert-BuylC. iso-PropylD. Ethyl |
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Answer» Correct Answer - B The stability of various alkyl carbocations on the basis of hyperconjugation is `3^(@)gt2^(@)gt1^(@)gt` methyl. In the tert-butyl cation, the cation bearing the positive charge is attached to three. methyl groups, hence it will give nine. hyperconjugative structures, thus imparting maximum stability. |
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| 155. |
Hyperconjugation is most useful for stabilizing which of the following carbocations ?A. neopentylB. tert-ButylC. isopropylD. ethyl |
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Answer» Correct Answer - B `(CH_(3))_(3)C^(+)` has the maximum number of 9 `alpha`-hydrogens and hence most useful in explaining its stability. |
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| 156. |
Match the carbocation in Column-I with the effect which is major. Stabilizing factor for it an column II |
| Answer» Correct Answer - A::B::C::D | |
| 157. |
The most unstable carbocation is:A. `CH_(3)overset(o+)(C)H_(2)`B. `Cl-CH_(2)-overset(o+)(C)H_(2)`C. `overset(o+)(C)H_(2)-CHO`D. `overset(o+)(C)H_(2)-O-CH_(3)` |
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Answer» Correct Answer - C `overset(o+)(C)H_(2)-underset(O)underset(||)C-H` most unstable due to strong `-I` effect of `-CHO` group. |
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| 158. |
Identify the most stable species in the following set of ions giving reasons : (i) `overset(+)(C)H_(3), overset(+)(C)H_(2)Br, overset(+)(C)HBr_(2), overset(+)(C)Br_(2)` (ii) `overset(Θ)(C)H_(3), overset(Θ)(C)H_(2)Cl, overset(Θ)(C)HCl_(2),overset(Θ)(C) Cl_(3)` |
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Answer» (i) `overset(C)H_(3)` is the most stable species because the -I-effect of Br intensifies the +ve charge and hence destability the species. Further, more the number of Br atoms, less stable is the species. Thus, the stability of the species decreases in the order : `overset(+)(C)H_(3) gt overset(+)(C)H_(2)Br gt overset(+)(C)HBr_(2) gt overset(+)(C)Br_(3)` (ii) -I effect of the Cl atom disperses the -ve charge and thus stabilizes the species. Further, more the number of Cl atoms, more is the dispersal of the -ve charge and hence more stable is the species. Thus, `.^(-)C Cl_(3)` is the most stable species. The stability of other species decreases in the order `.^(-)CHCl_(2) gt .^(-)CH_(2)Cl gt .^(-)CH_(3)`. |
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| 159. |
Identify the most stable species in the following set of ions giving reasons (a) `overset(+)(C)H_(3), overset(+)(C)H_(2) Br, overset(+)(C)HBr_(2), overset(+)(C)Br_(3)` (b) `overset(Θ)(C)H_(3), overset(Θ)(C)H_(2)Cl, overset(Θ)(C)HCl_(2), overset(Θ)(C)Cl_(3)` |
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Answer» (i) `overset(o+)(C)H_(3)` ion the most stable : Replacement of H atoms in the carbocation with `Br` atoms discreases the magnitude of positive charge on the carbocation due to its - I effect and thus, destabilises the carbocation. (ii) `overset(Θ)(C)Cl_(2)` is the most stable because the `Cl` atom has a strong - I effect. by replacing H atoms in the carbanion with `Cl` atoms the magnitude of negative charge on the carbon atom is reduced and the anion is stabilised. |
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| 160. |
Assertion (A) : Energy of resonance hybrid is equal to the average of energies of all canonical forms. Reason (R) : Resonance hybrid cannot be represented by a single structure.A. Both A and R are correct and R is the correct explanation of A.B. Both A and R are correct but R is not the correct explanation of A.C. Both A and R are not correct.D. A is not correct but R is correct. |
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Answer» Correct Answer - D Correct A. The energy of the resonance hybrid is equal to the sum of the energies of the various canonical structures in proportion of their contribution towards the resonance hybrid. |
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| 161. |
Assertion (A) : Simple distillation can help in separating a mixture of propan-1-ol (boiling point `97^(@)C`) and propanone (boiling point `56^(@)C`). Reason (R) : Liquids with a difference of more than `20^(@)C` in their boiling points can be separated by simple distillation.A. Both A and R are correct and R is the correct explanation of A.B. Both A and R are correct but R is not the correct explanation of A.C. Both A and R are not correct.D. A is not correct but R is correct. |
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Answer» Correct Answer - A R is the correct explanation of A. |
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| 162. |
On complete combustion, 0.246 g of an organic compound gave 0.198g of carbon dioxide and 0.1014g of water. Determine the percentage composition of carbon and hydrogen in the compound |
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Answer» Percentage of carbon `=(12xx0.198xx100)/(44xx0.246)` `=21.95%` Percentage of hydrogen`=(2xx0.1014xx100)/(18xx0.246)` `=4.58%` |
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| 163. |
An organic compound gave the following results on analysis, 0.2496 g of the compound gave 0.3168 g of `CO_(2)` and 0.0864 g of `H_(2)O`. Calculate the percentage of carbo and hydrogen in the compound. |
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Answer» Correct Answer - `C = 34.61 %, H= 3.84 %` `%` of `C=12/44xx("Mass of "CO_(2))/("Mass of compound")xx100=12/44xx0.3168/0.2496xx100= 34.61%` `%` of `H=2/18xx("Mass of "H_(2)O)/("Mass of compound")xx100=2/18xx0.0864/0.2496xx100= 3.84 %` |
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| 164. |
0.2475g of an organic compound gave on combustion 0.4950 of `CO_(2)` and 0.2025g of `H_(2)O` calculate the % of C `&` H in it. |
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Answer» Wt of organic compound (w)=0.2475g Wt of `CO_(2)(b)=0.4950g` Wt. of `H_(2)O(a)=0.2025g` % of `C=((12)/(44)xxb)/(w)xx100` `=((12)/(44)xx0.4950)/(0.2475)xx100=54.54` % of `H=((2)/(18)xxa)/(w)xx100` `=((2)/(18)xx0.2025)/(0.2475)xx100=9.09` |
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| 165. |
0.465 g of an organic substance gave on combustion 1.32 g of `CO_(2)` and 0.315g of `H_(2)O`. Calculate the percentage of carbon and hydrogen in the compound. |
| Answer» Correct Answer - C = 77.42%, H = 7.53 % | |
| 166. |
An organic compound on analysis gave the following data: (i). 0.25 gm of the compound on complete combustion gave 0.37 gm of `CO_2` and 0.2 gm of water. (ii). 0.25 gm of the compound on analysis by Dumas method gave 32 ml of nitrogen gas at STP. Calculate the percentages of `C,H,N` and `O `in the organic compounds. |
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Answer» (i) Calculation of percentage of carbon. Mass of the compound =0.25 g Mass of carbon dioxide formed =0.37 g Percentage of carbon `=12/44xx("Mass of "CO_(2))/("Mass of compound")xx100=12/44xx0.37/0.25xx100=40.36` (ii) Calculation of percentage of hydrogen Mass of the compound =0.25 g Mass of water formed =0.20 g Percentage of hydrogen `=2/18xx("Mass of "H_(2)O)/("Mass of compound")xx100=2/18xx0.20/0.25xx100=8.89` (iii) Calculation of persentage of nitrogen. Mass of the compound = 0.25 g Volume of `N_(2)` evolved at N.T.P. = 32 mL Percentage of nitrogen `=28/22400xx("Volume of "N_(2)" at N.T.P.")/("Mass of compound")xx100=28/22400xx32.0/0.25xx100=16.0` |
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| 167. |
0.2475 g of an organic substance gave on combustion 0.495 g of `CO_(2)` and 0.2025g of `H_(2)O`. Calculate the percentage of carbon and hydrogen in it. |
| Answer» Correct Answer - C = 54.54%, H = 9.09% | |
| 168. |
Name a suitable technique of the components from a mixture of calcium sulphate and comphor. |
| Answer» The process of sublimation is used to separate a mixture of camphor and calcium sulphate. In this process, the sublimable compound changes from solid to vapour state without passing through the liquid state. Camphor is a sublimable compound and calcium sulphate is a non-sublimable solid. Hence, on heating, camphor will sublime while calcium sulphate will be left behind. | |
| 169. |
Which is not true regarding conformers of ethane?A. Theoretically infinite conformations existB. By prices experimental setup, staggered conformer can be separeted out of systemC. Increasing temperature increase the percentage of eclipsed conformerD. Staggered conformer has lower torsional strain than eclipsed one |
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Answer» Correct Answer - B Although conformers differ in potential energy and stability, the difference is so small that it does not allow their practical separation. |
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| 170. |
The correct energy graduation of different conformers isA. Staggered `gt` Skew `gt` eclipsedB. Skew `gt` Staggered `gt` eclipsedC. Eclipsed `gt` skew `gt` staggeredD. Skew `gt` staggered `gt`eclipsed |
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Answer» Correct Answer - C Stability `prop1//`reactivity [Staggered `gt`skew`gt`eclipsed] stability |
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| 171. |
Which statement is correct about anti-conformation of `1`-Chloropropane?A. It is the most polar formB. It has maximum torsional strainC. It has minimum steric strainD. (a) and (b) both |
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Answer» Correct Answer - D Anti-conformation is stable and polarity is present in one directions. |
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| 172. |
The isomer of diethyl ether isA. `(CH_(3))_(2)CHOH`B. `(CH_(3))_(3)COH`C. `C_(3)H_(7)OH`D. `(C_(2)H_(5))_(2)CHOH` |
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Answer» Correct Answer - B Only this `(CH_(3))_(3)COH` compound has same molecular formula with diethylether. |
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| 173. |
Nucleophilicity order is correctly represented byA. `CH_(3)^(-)ltNH_(2)^(-)ltHO^(-)ltF^(-)`B. `CH_(3)^(-)=NH_(2)^(-)gtHO^(-)=F^(-)`C. `CH_(3)^(-)gtNH_(2)^(-)gtHO^(-)gtF^(-)`D. `NH_(2)^(-)gtF^(-)gtHO^(-)gtCH_(3)^(-)` |
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Answer» Correct Answer - C The correct order of nucleophilicity is `CH_(3)^(-)gtNH_(2)^(-)gtHO^(-)F^(-)` (Refer A-level information). |
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| 174. |
Electrophile in the case of chlorination of benzene in presence of `FeCl_(3)` isA. `Cl^(+)`B. `Cl^(-)`C. `Cl`D. `FeCl_(3)` |
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Answer» Correct Answer - A In the chlorination of benzene, the electrophile is chlorinium ion `(Cl^(+))`. |
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| 175. |
For the reaction of phenol with `CHCl_(3)` in presence of KOH, the electrophile isA. `overset(+)CHCl_(2)`B. `:C"C"l_(2)`C. `overset(cdot)CHCl_(2)`D. `"C"Cl_(4)` |
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Answer» Correct Answer - B In the reaction of phenol with `CHCI_(3)` in the presence of KOH (Reimer-Tiemann reaction) the electrophile is a carbene `(:CCl_(2))` |
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| 176. |
The reaction: `CH_(3)CH_(2)I + KOH(aq)toCH_(3)CH_(2)OH+KI` is classified as : (a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition |
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Answer» `CH_(3)CH_(2)I + KOH(aq)toCH_(3)CH_(2)OH+KI` It is an example of nucleophilic substitution reaction. The hydroxyl group of KOH (`OH^(–)`) with a lone pair of itself acts as a nucleophile and substitutes iodide ion in `CH_(3)CH_(2)I` to form ethanol. |
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| 177. |
`KOH` can be used as drying agent forA. carboxylic acidsB. phenolsC. aminesD. esters |
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Answer» Correct Answer - C Amines are basic, do not react with base `KOH`, Phenols, acid forms salts with `KOH` while esters undergo hydrolysis in presence of `KOH`. |
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| 178. |
Which of the following behaves both as a nucleophile and as an electrophile ?A. `H_(3)C-CequivN`B. `H_(3)COH`C. `H_(2)C=CH-CH_(3)`D. `H_(3)C-NH_(2)` |
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Answer» Correct Answer - A `H_(3)C-Cequivoverset(cdotcdot)N`can act both as an electrophile and a nucleophile compounds with a multiple bond between carbon and a more electronegative atom can act both as an electrophile and nucleophile. |
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| 179. |
Carbanion is iso-structural withA. Free radicalB. Carbonium ionC. AmmoniaD. Carbene |
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Answer» Correct Answer - C Carbanion is isostructural with ammonia. Both are pyramidal. |
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| 180. |
The compounds `CH_(3)NH_(2)" and "CH_(3)CH_(2)NH_(2)` are :A. isomersB. isobarsC. homologuesD. allotropes |
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Answer» Correct Answer - C `CH_(3)CH_(2)NH_(2)" is the next to "CH_(3)NH_(2)` |
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| 181. |
The kind of delocalisation involving sigma bond orbitals is called………..A. Mesomeric effectB. Tautomeric effectC. Electromeric effectD. Hyperconjugative effect. |
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Answer» Correct Answer - D Delocalisation of electrons involving `alpha`-bond is hyperconjugative effect. |
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| 182. |
The kind of delocalization invloving `sigma`-bond orbitals is called............... . |
| Answer» Correct Answer - hyperconjugation | |
| 183. |
Statement-1. Nitrenes cannot be isolated. Statement-2. Nitrenes are the nitrogen analogous of carbene.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is False |
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Answer» Correct Answer - B Correct explanation. Nitrenes are too reactive to be isolated. |
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| 184. |
Write the structures of the following : (i) Acetone (ii) Ethylene glycol (iii) Sec-butyl alcohol. |
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Answer» (i) `CH_(3)-overset(O)overset(||)(C)-CH_(3)` (ii) `HO-CH_(2)-CH_(2)-OH` (iii) `CH_(3)-underset(OH)underset(|)(CH)-C_(2)H_(5)` |
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| 185. |
Assertion (A) . Pent-1-ene and pent-2-ene are position isomers. Reason (R). Position isomers differ in the position of functional group or a substituent.A. Both A and R are correct and R is the correct explanation of AB. Both A and R are correct but R is not the correct explanation of AsC. both A and R are not correctD. A is not correct but R is correct |
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Answer» Correct Answer - A Both A and R are correct and R is the correct explanation of A. `underset("Pent-1-ene")(overset(5)(C)H_(3) - overset(4)(C)H_(2) - overset(3)(C)H_(2) - overset(2)(C)H = overset(1)(C)H_(2))` `underset("Pent-2-ene")(overset(5)(C)H_(3) - overset(4)(C)H_(2) - H overset(3)(C) = overset(2)(C)H - overset(1)(C)H_(3))` When two or more compounds differ in the position of substituent atom or functional group on the carbon skeleton then it is position isomerism. Double bond is a functional group whose position varies. |
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| 186. |
Assertion : But-1-ene amd 2-methylprop-1-ene are position isomers. Reason : Position isomers have same molecular formula but different arrangement of carbon atoms.A. If both assertion and Reason are true, and reason is the true explanation of the assertion.B. If both assertion and reason are true, but reason is not the true explanation of the assertion.C. If assertion is true, but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - D Correct assertion. But-1-ene and 2-methylprop-1-ene are chain isomers. Correct reason. Chain isomers have same molecular formula but different arrangement of carbon atoms. |
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| 187. |
`H_(3)O^(+)` or `RN_(4)^(+)` neither acts as an electrophile nor as a nucleophile. Explain why ? |
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Answer» `H_(3)overset(..)(O^(+))` has a lone pair of electrons bu due to the presence of +ve charge, it cannot donate its electron pair and hence it does act as a nucleophile. `R_(4)N^(+)`, however, does not have a lone pair of electrons, therefore, it does not act as a nucleophile. `H_(3)overset(..)(O^(+))` has 8 electrons in the valence shell. It cannot expand its valence shell beyond 8 due to the absence of d-orbitals. Therefore, it does not act as an electrophile. Similarly, `R_(4)overset(+)(N)` also has 8 electrons in the valence shell. Like O, N also cannot expand its valence shell beyond 8 and hence it also does not act as an electrophile. Thus, `H_(3)O^(+)` or `RN_(4)^(+)` neither acts as a nucleophile nor as an electrophile. |
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| 188. |
The IUPAC name for the formula `CH_(3) - overset(CH_(3))overset(|)(C)= CH - COOH` isA. 2-Methylbut-2-enoic acidB. 3-Methylbut-3-enoic acidC. 3-Methylbut-2-enoic acidD. 2-Methylbut-3-enoic acid |
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Answer» Correct Answer - C `{:(" "CH_(3)),(" |"),(overset(4)(C)H_(3)overset(3)(-C)=overset(2)(C)H-overset(1)(C)OOH):}` 3-Methylbut-2-enoic acid |
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| 189. |
The IUPAC name for the formula `CH_(3) - overset(CH_(3))overset(|)(C)= CH - COOH` isA. 2-Methylbut-2-enoic acidB. 3-Methylbut-3-ionic-acidC. 3-Methylbut-2-enoic acidD. 2-Methylbut-3-enoic acid |
| Answer» Correct Answer - C | |
| 190. |
Which of the following has neither secondary nor tertiary hydrogen?A. IsobutaneB. IsopentaneC. PentaneD. Neopentane |
| Answer» Correct Answer - D | |
| 191. |
The IUPAC name of the compound is `CH_(3)-underset(CH_(3))underset(|)(CH)-CH_(2)-CH(OH)-CH_(2)`A. 4-Methylhexan-3-olB. Heptan-2-olC. 4-Methylhexan-2-olD. None of these |
| Answer» Correct Answer - C | |
| 192. |
0.3080 gm of and organic chloro compound gave 0.5740 gm of siver chloride in carius estimation. Calculate the percentage of chloride presents in the compound |
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Answer» Given that, Mass of organic compound is 0.3780 g. Mass of AgCl formed = 0.5740 g 1 mol of AgCl contains 1 mol of Cl. Thus, mass of chlorine in 0.5740 g of AgCl `(35.5xx05740)/143.32` =0.1421g `therefore` Percentage of chlorine=`0.1421/0.3780xx100=37.59%` Hence, the percentage of chlorine present in the given organic chloro compound is 37.59% |
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| 193. |
A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. `H_(2)SO_(4)`. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be............. . |
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Answer» Correct Answer - 2.8 `underset(underset(46 g)("Formic acid"))(HCOOH) overset(conc.H_(2)SO_(4))rarrunderset(28g)(CO)+ H_(2)O` Now 46 g of COOH evolve CO = 28 g `:. 2.3 g` of HCOOH will evolve `CO = (28)/(46) xx 2.3` = 1.4 g `{:(""COOH),("| " overset(conc.H_(2)SO_(4))rarr CO+CO_(2)+H_(2)O),(COOH" "28 g),("Oxalic acid"),(" "90g):}` When the gaseous mixture of `(CO+CO_(2)` is passed through KOH pellets, `CO_(2)` is absorbed while CO phase out `2KOH + CO_(2) rarr K_(2)CO_(3) + H_(2)O` Now, 90g of oxalic acid evolve `CO = 28 g` `:. 4.5 g` of oxalic acid will evolve CO `= (28)/(90) xx 4.5 = 1.4 g` Total amount of CO evolved = 1.4 + 1.4 = 2.8 g |
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| 194. |
What is the IUPAC name of the compound `underset(SH)underset(|)CH_(2)-underset(SH)underset(|)CH-underset(OH)underset(|)CH_(2)`A. 1-Hydroxyethane-2, 3-dithiolB. 3-Hydroxyethane-2, 3-dithiolC. 2, 3-Disulphanyl propan-1-o1D. 2, 3-hydrosulphidopropan-1-o1. |
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Answer» Correct Answer - C Factual questions. |
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| 195. |
Classify the following reactions in one of the reaction type studied in this unit (i) `CH_(3)CH_(2)Br+SH^(-) rarr CH_(3)CH_(2)SH+Br^(-)` (ii) `(CH_(3))_(2)C=CH_(2)+HCl rarr (CH_(3))_(2)C(Cl)CH_(3)` (iii) `(CH_(3))_(3)C CH_(2)OH+HBr rarr (CH_(3))_(2)CBrCH_(2)CH_(3)` (iv) `CH_(3)CH_(2)Br+HO^(-) rarr CH_(2)=CH_(2)+H_(2)O+Br^(-)` |
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Answer» (i) Nucleophilic substitution (ii) Electrophilic addition (iii) Rearrangement of carbocation intermediate formed followed by nuclephilic substitution. (iv) Elimination. |
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| 196. |
In the estimation of sulphur by carius method, 0.468 gm of an organic sulphur compound afforded 0.668 gm of barium sulphate. Find out the percentage of sulphur in the given compound. |
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Answer» Total mass of organic compound = 0.468 g [Given] Mass of barium sulphate formed = 0.668 g [Given] 1 mol of `BaSO_(4)` = 233 g of `BaSO_(4)` = 32 g of sulphur Thus, 0.668 g of `BaSO_(4)` contains `(32xx0.668)/233g` of sulphur = 0.0917 g of sulphur Therefore, percentage of sulphur = `0.0197/0.468xx100=19.59%` Hence, the percentage of sulphur in the given compound is 19.59 %. |
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| 197. |
Classify the following reactions in one of the reaction type studied in this unit. (a) `CH_(3)CH_(2)Br +HS^(-) rarr CH_(3)CH_(2)SH + Br^(-)` (b) `(CH_(3))_(2)C = CH_(2) + HCl rarr (CH_(3))_(2)C Cl - CH_(3)` (c) `CH_(3)CH_(2)Br + HO^(-) rarr CH_(2) + H_(2)O + Br^(-)` (d) `(CH_(3))_(3)C-CH_(2)OH + HBr rarr (CH_(3))_(2)CBrCH_(2)CH_(3) + H_(2)O` |
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Answer» (a) Nucleophilic substitution (b) Electrophilic addition (c) Bimolecular elimination (d) Nucleophilic subtitution with rearrangement. |
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| 198. |
Assertion : Alkyl carbanions like ammonia have pyramidal shape. Reason : The carbon atom carrying negative charge has an octet of electrons.A. If both assertion and reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect. |
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Answer» Correct Answer - B Corrrect explanation : Alkyl carbonions assume pyramidal structure due to presence of lone pair responsible for negative charge on the carbon atom. |
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| 199. |
Statement-1. Simple carbanions are usually pyramidal but allyl carbanion is a planar species. Statement-2. All the carbon atoms in allyl carbanion are `sp^(2)`-hybridized.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is False |
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Answer» Correct Answer - B Correct explanation. Allyl carbanion is stabilized by resonance. |
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| 200. |
Assertion : Alkyl carbanion like ammonia have pyramidal shape. Reason : The carbon atom carrying negative charge has an octet of electrons.A. If both assertion and Reason are true, and reason is the true explanation of the assertion.B. If both assertion and reason are true, but reason is not the true explanation of the assertion.C. If assertion is true, but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - B Correct explanation. Due to greater lp-bp over bp-bp repulsions, carbanions assume pyramidal shape. |
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