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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.
| 301. |
In hexa-1, 3-dien-5yne, the number of C-C, `sigma`, C-C `pi` and C-H sigma` bonds respectively areA. 5, 4 and 6B. 6, 3 and 5C. 5, 3 and 6D. 6, 4 and 5 |
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Answer» Correct Answer - A `{:(H-overset(1)(C)=overset(2)(C)-overset(3)(C)=overset(4)(C)-overset(5)(C)-=overset(6)(C)-H),(" |"" |"" |"" |"),(" "H" "H" "H" "H),("Hexa-1,3-diene-5-yne"):}` C-C, `sigma`-bonds=5 C=C, `pi`-bonds = 4 C-H, `sigma`-bonds = 6 |
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| 302. |
The state of hybridization of `C_(2), C_(3), C_(5)` and `C_(6)` of the hydroccarbon, `{:(" "CH_(3)" "CH_(3)),(" |"" |"),(underset(7)(C)H_(3)-.^(6)C-underset(5)(C)H=underset(4)(C)H-underset(3)(C)H - underset(2)(C)-=underset(1)(C)H),(" |"),(" "CH_(3)):}` is in the sequenceA. `sp^(3), sp^(2), sp^(2)` and spB. `sp, sp^(2), sp^(2)` and `sp^(3)`C. `sp, sp^(2), sp^(3)` and `sp^(2)`D. `sp, sp^(3), sp^(2)` and `sp^(3)` |
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Answer» Correct Answer - D `{:(" "CH_(3)" "CH_(3)),(" |"" |"),(CH_(3)-overset(6)(C)-underset(sp^(2))overset(5)(C)H_(2)=overset(4)(C)H-underset(sp^(3))overset(3)(C)H-underset(sp)overset(2)(C)-=overset(1)(C)H),(" |"),(" "CH_(3)):}` Thus, option (d) is correct. |
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| 303. |
In allene `(C_(3)H_(4))`, the type(s) of hybridization of the carbon atoms is (are)A. sp and `sp^(3)`B. sp and `sp^(2)`C. only `sp^(2)`D. `sp^(2)` and `sp^(3)` |
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Answer» Correct Answer - B `overset(sp^(2))(C)H_(2)0 = overset(sp)(CH)=overset(cp^(2))(CH_(2))`. |
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| 304. |
The IUPAC name for the compound A. Ehtyl acrylateB. Ethyl methylbutenoateC. Ethyl acetoethenoateD. Ethyl 3-methylbut-3-enoate. |
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Answer» Correct Answer - D `underset("Ethyl 3-methylbut-3-enanoate")underset()(H_(2)overset(4)C=overset(CH_(3))overset(3|)C-overset(2)CH_(2)-overset(1)COOC_(2)H_(5))` |
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| 305. |
The inductive effectA. decreases with increase of distanceB. its extent increases with increase of distanceC. inidicates the transfer of `pi` pair of electrons from less electronegative atom to more electronegative atom in a moleculeD. shows the transfer of lone pair of electrons. |
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Answer» Correct Answer - A See electron displacement in covalent bond. |
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| 306. |
Which of the following compounds exhibits stereoisomerism?A. `2`-methyl butene-`1`B. `3`-methylbutyne`-1`C. `3`-methylbutanoic acidD. `2`-methylbutanoic acid |
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Answer» Correct Answer - D `2`-Methylbutanoic acid contains one asymmetric centre `CH_(3)CH_(2)-overset(H)overset(|)underset(CH_(3))underset(|)(C)-CO_(2)H` |
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| 307. |
Which of the following does not show electrical conductance?A. PotassiumB. GraphiteC. DiamondD. Sodium |
| Answer» Correct Answer - C | |
| 308. |
1.2 g of organic compound on Kjeldahilization liberates ammonia which consumes `30 cm^(3)` of 1 N HCl. The percentange of nitrogen in the organic compound isA. 30B. 35C. `46.67`D. `20.8` |
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Answer» Correct Answer - B `% N = (1.4 xx 30 xx 1 xx 1)/(1.2) = 35` |
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| 309. |
Give the IUPAC name for the amine. `{:(" "CH_(3)),(" |"),(CH_(3)-N-C-CH_(2)CH_(3)),(" |"" |"),(" "CH_(3)C_(2)H_(5)):}` |
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Answer» N, N, 3-Trimethylpentan-3-amine (Alphabets are given perference over numericals). (Please remember that capitals (i.e., N, S, O, etc) and lower case letters (i.e. n, o, m, p, etc.) have higher priority of citation than Greek letters, i.e., `alpha, beta, gamma`,.....etc.) which, in turn, have higher priority of citation than arabic numerals (i.e., 1, 2, 3, 4,........etc.). For example, N, `alpha`, 1, 2 have priority of citation than 1, 2, 3, 4,.......... . |
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| 310. |
In the estimation of sulphur by Carius method, 0.480 f og an organic comppund gives 0.699 g of barium sulphate. The percentage of sulphur in this compound is (atomic masses : Ba = 137, S = 32 and O = 16)A. 0.2B. 0.15C. 0.35D. 0.3 |
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Answer» Correct Answer - A `%S = (32)/(233) xx (0.699)/(0.480) xx 100 = 20` |
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| 311. |
For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M/10 sulphuric acid. The unreacted acid required 20 Ml of M/10 sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound isA. 0.05B. 0.06C. 0.1D. 0.03 |
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Answer» Correct Answer - C Volume of the acid taken = 60 mL of M/10 `H_(2)SO_(4)` Let the acid left unused = v mL of M/10 `H_(2)SO_(4)` Volume of M/10 NaOH used to neutralize the unreacted acid = 20mL Applying molarity equation, `underset("(Acid)")(n_(a)M_(a)V_(a)) = underset("(Base)")(n_(b)M_(b)V_(b))` where `n_(a), M_(a)` and `V_(a)` are the basicity, molarity and volume of the acid and `n_(b), M_(b)` and `V_(b)` are the acidity, molarity and volume of the base or `2 xx (1)/(10) xx v = 1 xx (1)/(10) xx 20` `:. v = 10 mL` `:.` Volume of the acid used = 60 - 10 = 50 mL of M/10 `H_(2)SO_(4)` Now % of N `= (1.4 xx "Molarity" xx "Volume" xx "Basicity of the acid used")/("Weight of the substance taken")` `= (1.4 xx 1 xx 50 xx 2)/(10 xx 1.4) = 10.0` |
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| 312. |
`(pm)`-2 Butanol is optically inactive. Give reasons. |
| Answer» `(pm)`-2 Butanol is a racemic mixture and hence contains equal amounts of the two enantiomers, i.e, (+)-2- butanol and (-)-2- butanol. Whereas (+)-2-butanol rotates the plane of polarized light toward right, (-)-2-butanol rotates the plane of polarized light towards left but to the same extent. The overall rotation is zero and hence `(pm)`-2-butanol is optically inactive. | |
| 313. |
The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound isA. acetamideB. benzamideC. ureaD. thiourea |
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Answer» Correct Answer - C Le the vol. of acid left unused = v mL of 0.1 `M H_(2)SO_(4)` Applying molarity equation `n_(a)M_(a)V_(a) = n_(b)M_(b)V_(b)`, we have, `2 xx 0.1 xx v = 1 xx 0.5 xx 20` or v = 50 mL `:.` Vol. of acid used = 100 - 50 = 50 mL of `0.1 M H_(2)SO_(4)` `%N = (1.4 xx n_(1)M_(a)V_(a))/("wt. of substance taken") = (1.4 xx 2 xx 0.1 xx 50)/(0.3)` `= 46.6` % N in urea `(NH_(2)CONH_(2)) = (28//60) xx 100 = 46.6%`, in acetamide `(CH_(3)CONH_(2)) = (14//59) xx 100 = 23.72%`, in benzamide `(C_(6)H_(5)CONH_(2)) = (14//21) xx 100 = 11.57%` and in thiourea `(NH_(2)CSNH_(2)) = (28//76) xx 100 = 36.84%` Thus, option (c) is correct. |
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| 314. |
0.35 g of an organic susbtance was Kjeldahilsed and the ammonia obtained was passed into 100ml of M/`10H_(2)SO_(4)` The excess acid required 154 ml of `M//10 NaOH` for neurtralisation, calculate the % of nitrogen in the compound. |
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Answer» Volume of `(M)/(10)H_(2)SO_(4)` taken=100ml Excess volume of `(M)/(10)H_(2)SO_(4)` is 154 ml of `(M)/(10)NaOH=(154)/(2)ml` of `(M)/(10)H_(2)SO_(4)` `therefore` Volume of `(M)/(10)H_(2)SO_(4)` left unused=77ml Volume of `(M)/(10)H_(2)SO_(4)` used for neutralisation of `NH_(3)=100-77=23ml`. |
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| 315. |
0.185 g of an organic substance when treated with conc `HNO_(3)` gave 0.32 g of silver bromide. Calculate the % of bromine in the compound. |
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Answer» W=0.185g wt. of AgBr=0.320g % of `Br=(80xx0.320xx100)/(188xx0.185)=73.60` |
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| 316. |
Which of the following compounds will give negative Lassaigne is test for Nitrogen?A. B. C. D. |
| Answer» Correct Answer - B | |
| 317. |
In Lassaigne test for nitrogen, conc `HNO_(3)` is used to destroy..................and ................. . |
| Answer» Correct Answer - NaCN, `Na_(2)S` | |
| 318. |
In which of the following reactions equilibrium will shift towards right?A. B. C. D. |
| Answer» Correct Answer - D | |
| 319. |
Compound having molecular formula `C_(5)H_(12)O` cannot showA. TautomerismB. Position isomerismC. MetamerismD. Functional isomerism |
| Answer» Correct Answer - A | |
| 320. |
Keto-enol Tautomerism is observed inA. `C_(6)H_(5)-CHO`B. `C_(6)H_(5)-CO-CH_(3)`C. D. `C_(6)H_(5)-CO-CH_(2)-CO-CH_(3)` |
| Answer» Correct Answer - B::C::D | |
| 321. |
0.46g of an organic compound was analysed. The increase in mass of `CaCl_(2)` U-tube was 0.54g and potash bulb was 0.88g. The percentage composition of the compound isA. `C=52.17%,H=13.04%,O=34.79%`B. `C=50%,H=50%`C. `C=32.19%,H=18.01%,O=49.8%`D. `C=72%,H=28%` |
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Answer» Correct Answer - A Mass of `H_(2)O=0.54g` Mass of `CO_(2)=0.88g` % of `C=(12)/(44)xx(0.88)/(0.46)xx100=52.17%` % of `H=(2)/(18)xx(0.54)/(0.46)xx100=13.04%` % of `O=100-(52.17+13.04)=34.79`% |
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| 322. |
The correct order priority for the `-CONH_(2)CN" and "-COOR` isA. `-CONH_(2),-COOR,-CN`B. `-COOR,-CONH_(2),-CN`C. `-CN,-COOR,-CONH_(2)`D. `-CN,-CONH_(2),-COOR` |
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Answer» Correct Answer - B See the nomenclature. |
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| 323. |
No bond resonance explains the stability of the following :A. benzyneB. carbanionsC. free radicalsD. carbenes. |
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Answer» Correct Answer - C Hyperconjugation explains the stability of free radicals. |
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| 324. |
In Hoffinann bromamide reaction, the reactive intermediate involved is aA. CarbocationB. CarbanionC. CarbeneD. Nitrine. |
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Answer» Correct Answer - D See A-Level Information. |
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| 325. |
Total number of alkyl groups having the molecular formula `C_(4)H_(9)` is |
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Answer» Correct Answer - 4 Four, i.e., n-butyl,isobutyl, sec-butyl and tert-butyl. |
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| 326. |
Consider thiol anion `(RS^(Θ))` and alkoxy anion `(RO^(Θ))`. Which of the following statement is correct ?A. `RS^(Θ)` is less basic and less nucleophilic than `RO^(Θ)`B. `RS^(Θ)` is less basic but more nucleophilic than `RO^(Θ)`C. `RS^(Θ)` is more basic and more nucleophilic than `RO^(Θ)`D. `RS^(Θ)` is more basic but less nucleophilic than `RO^(Θ)` |
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Answer» Correct Answer - B Since O-H bond is stronger than S-H bond, therefore, `RO^(-)` has a greater tendency to accept a proton than `RS^(-)` and hence `RO^(-)` is a stronger base than `RS^(-)`. Conversely, since S is less electro-negative than O, therefore, `RS^(-)` is more willing to donate a pair of electrons than `RO^(-)` and hence `RS^(-)` is more nucleophilic than `RO^(-)`. Thus, option (b) is correct. |
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| 327. |
As per IUPAC nomenclature, the name of the complex `[Co(H_(2)O)_(4) (NH_(3))_(2)] Cl_(3)` isA. tetraaquadiaminecobalt(III) chlorideB. tetraaquadiamminecobalt(III) chlorideC. diamineteraaquacobalt(III) chlorideD. diamminetetraaquacobalt(III) chloride |
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Answer» Correct Answer - D The name of the given complex is : diamminetetraaquacobalt(III) chloride. Special care should be taken in spelling the name of `NH_(3)` ligand as it is ammine and not amine. |
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| 328. |
The `IUPAC` name of the following compound `Cl_(3)C-CH_(2)CHO` isA. 3,3,3-TrichloropropanalB. 1,1,1-TrichloropropanalC. 2,2,2-TrichloropropanalD. Chloral. |
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Answer» Correct Answer - A `underset("3,3,3-Trichloropropanal")underset()(Cl_(3)overset(3)Coverset(2)CH_(2)overset(1)CHO)` |
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| 329. |
Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound ? |
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Answer» `CO_(2)` is acidic in nature, therefore, it reacts with the strong base KOH to form `K_(2)CO_(3)`. `2KOH + CO_(2) rarr K_(2)CO_(3) + H_(2)O` The increase in the mass of U-tube containing KOH then gives the mass of `CO_(2)` produced and from its mass, the percentage of carbon in the organic compound can be estimated by using the equation, `% C = (12)/(44) xx ("Mass of " CO_(2) " formed")/("Mass of substance taken") xx 100` |
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| 330. |
Total number of stereoisomers of compound is: `CH_(3)-underset(OH)underset(|)CH-underset(Br)underset(|)CH-CH_(3)`A. `2`B. `4`C. `6`D. `8` |
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Answer» Correct Answer - B Unsymmetrical compound with `2` centres has `2^(2)=4` stereoisomers. |
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| 331. |
How many cyclic and acyclic isomers (including tautomers) can be made by the formula `C_(3)H_(6)O` ?A. 4B. 5C. 9D. 10 |
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Answer» Correct Answer - C 9 isomers are possible. |
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| 332. |
The total number of acyclic structural isomers possible for compound with molecular formula `C_(4)H_(10)O` isA. 9B. 7C. 5D. 6 |
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Answer» Correct Answer - B Seven, out of which four are alcohols : (i) `CH_(3)CH_(2)CH_(2)CH_(2)OH`, (ii) `(CH_(3))_(2)CHCH_(2)OH` (iii) `CH_(3)CH_(2)CHOHCH_(3)`, (iv) `(CH_(3))_(3)COH` and three are ethers : (v) `CH_(3)CH_(2)OCH_(2)CH_(3)`, (vi) `CH_(3)OCH_(2)CH_(2)CH_(3)` and (vii) `CH_(3)OCH(CH_(3))_(2)` |
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| 333. |
Select the correct statements/s:A. Eclipsed and staggered ethanes give different products on relation with chlorine in presence of light.B. The conformational isomers can be isolated at room temperature.C. Torsional strain in ethane is minimum at dihedrated angles `60^(@),180^(@)` and `300^(@)`D. Steric strain is minimum is staggered gauche form of `n`-butane. |
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Answer» Correct Answer - C Torisional strain is minimum in staggered form. |
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| 334. |
Which among the following is the most stable carbocation ?A. `CH_(3)Coverset(+)H_(2)`B. `overset(+)CH_(3)`C. `(CH_(3))_(3)C^(+)`D. `(CH_(3))_(2)CH^(+)` |
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Answer» Correct Answer - C The names of alkyl substituents should be in alphabetical order. |
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| 335. |
Why is it necessary to use acetic acid and not suplhuric acid for the acidification of sodium extract for testing suplhur by lead acetate test? |
| Answer» Although the addition of sulphuric acid will precipitate lead sulphate, the addition of acetic acid will ensure a complete precipitation of sulphur in the form of lead sulphate due to common ion effect. Hence, it is necessary to use acetic acid for acidification of sodium extract for testing sulphur by lead acetate test. | |
| 336. |
The most satisfactory method to separate mixture of sugars is :A. Fractional crystallisationB. SublimationC. ChromatographyD. None of these |
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Answer» Correct Answer - C Chromatography is the best method to deparate the sugars. |
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| 337. |
A miscible mixture of benzene and chloroform can be separated by :A. sublimationB. distillationC. filtrationD. Crystallisation |
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Answer» Correct Answer - B Benzene boils at `81^(@)C` while chloroform at about `61^(@)C`. The separation can be done by distillation. |
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| 338. |
Chloroform and benzene form a pair of miscible liquids. These can be separated by:A. SublimationB. FiltrationC. Separating funnelD. Fractional distillation |
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Answer» Correct Answer - D Fractional distillation can be used because they differ in their boiling point. |
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| 339. |
Which is not used to purify organic solids ?A. DistillationB. SublimationC. CrystallisationD. None of these |
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Answer» Correct Answer - A Distillation is used to purify organic liquids only. |
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| 340. |
Hybridization of nitrogen atom in pyridine is A. `sp^(3)`B. `sp^(2)`C. `sp`D. `sp^(3)d` |
| Answer» Correct Answer - B | |
| 341. |
Which one of the following is not correct in respect to hybridization of orbitals ?A. The orbitals present in the valence shell only are hybridized.B. The orbitals undergoing hybridization have almost equal energyC. Promotion of electron is not essential condition for hybrizationD. It is not always that only partially filled orbitals participate in hybridization, in some cases even filled orbitals in valence shell take part |
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Answer» Correct Answer - D (E) Pure atomic orbitals are less effective in forming stable bonds than hybrid orbitals. |
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| 342. |
For which of the following compound tautomerization reaction is very slow?A. B. C. D. |
| Answer» Correct Answer - A | |
| 343. |
Which of the following statements is correct ?A. Desmotropism is another name for tautomerismB. Allyl carbocation is more stable than isopropyl carbocationC. `+I` effect is exhibited by `-overset(+)NH_(3)`D. The formula `CH_(2)Cl_(2)` is nonpolar |
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Answer» Correct Answer - A Desmotropism is another name for tautomerism. |
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| 344. |
Which of the following can act as nucleophilcA. Diethyl etherB. Anilinium ionC. Acylium ionD. Dichloromethylene carbene. |
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Answer» Correct Answer - A `C_(2)H_(5)-overset(cdotcdot)underset(cdotcdot)O-C_(2)H_(5)` is a nucleophile due to the presence of lone pairs of electrons on oxygen. |
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| 345. |
Out of the following, the one containing only nucleophiles isA. `AlCl_(3),BF_(3),NH_(3)`B. `NH_(3),CN^(-),CH_(3)OH`C. `AlCl_(3),NH_(2)^(-),H_(2)O`D. `RNH_(2), :CX_(2),H^(-)` |
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Answer» Correct Answer - B `NH_(3),CN^(-),CH_(3)OH` are nucleophiles. |
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| 346. |
Which of the following species is not electrophile in nature ?A. `Cl^(+)`B. `BH_(3)`C. `H_(3)O^(+)`D. `overset(+)(N)O_(2)` |
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Answer» Correct Answer - C `H_(3)O^(+)` cannot act as an electrophile since its octet is already complete and hence has no room to accept an additional pair of electrons. |
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| 347. |
Nucleophiles areA. Nucleus loving speciesB. Electron loving speciesC. Nucleus hating speciesD. Electron hating species. |
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Answer» Correct Answer - A Nucleophiles are nucleus loving species. |
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| 348. |
A nucleophile must necessarily haveA. an unpaired electronB. two lone pairs of electronsC. an overall positive chargeD. tendency to donate electron pair. |
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Answer» Correct Answer - D A nucleophile must necessarily have a tendency to donate electron pair. |
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| 349. |
Which of the following species is an electrophile ?A. `H_(2)O`B. `NH_(3)`C. `C_(2)H_(5)OH`D. `SO_(3)` |
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Answer» Correct Answer - D `SO_(3)` is an electrophile . |
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| 350. |
A nucleophilic reagent must necessarily haveA. An overall positive chargeB. An overvall negative chargeC. An unpaired electronD. A species with complete octet and lone pair of electrons. |
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Answer» Correct Answer - D A micleophile must necessarily have a lone pair of electrons. |
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