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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.
| 451. |
How many optically active stereomers are possible for butane-2,3-diolA. 3B. 4C. 1D. 2 |
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Answer» Correct Answer - A `{:(" "CH_(3)),(" |"),(H-C_(**)-OH),(" |"),(H-C^(**)-OH),(" |"),(" "CH_(3)),("butane-2,3-diol"):}` No. of asymmetric of chiral carbon =2 , Therefore no. of optical isomers `= 2^(n) = 2^(2) = 4`, No. of optically active stereomers = 4-1=3, because meso compound is optically inactive due to internal compensation. |
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| 452. |
IUPAC name of `CH_(2)OH-CH_(2)OH` isA. 1, 2 - dihydroxy ethaneB. Ethylene glycolC. Ethane - 1, 2- diolD. Ethane - 1, 2 - dial |
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Answer» Correct Answer - C |
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| 453. |
The IUPAC name of `CH_(3) -CH(OH) - CH_(2) - C(OH) ( CH_(3))_(2)` is `:`A. 1,1-dimethyl - 1, 3-butanediolB. 2-methyl - 2, 4-pentanediolC. 4-methyl-2, 4-pentanediolD. 1, 3, 3-trimethyl-1, 3-propanediol |
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Answer» Correct Answer - B `{:(" "OH" "CH_(3)),(" |"" |"),(underset(5)(C)H_(3)-underset(4)(C)H-underset(3)(C)H_(2)-underset(2|)(C)-OH),(" "1CH_(3)),(" 2-methyl-2,4-pentane-diol"):}` |
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| 454. |
What is the IUPAC name of the following A. 3 - chloro cyclo hexa - 1, 5 - dieneB. 5 - chloro cyclo hexa - 1, 3 - dieneC. 1 - chloro cyclo hexa - 2 , 5 - dieneD. 2 - chloro cyclo hexa - 1, 4 - diene |
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Answer» Correct Answer - B |
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| 455. |
Empirical formula of compound is `CH_(2)O`.If its molecular weight is 180 then the molecular formula of the compound isA. `C_(3)H_(6)O_(3)`B. `C_(4)H_(8)O_(4)`C. `C_(6)H_(12)O_(6)`D. `C_(5)H_(10)O_(5)` |
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Answer» Correct Answer - C Molecular formula `= ("Empirical formula")_(n)` `n = ("Molecular weight")/("Empirical formula wt.") = (180)/(30) = 6` `= (CH_(2)O)_(6) = C_(6)H_(12)O_(6)`. |
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| 456. |
The compound having only primary hydrogen atoms isA. IsobuteneB. 2,3-dimethylbuteneC. CyclohexaneD. Propyne |
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Answer» Correct Answer - A::D Propyne have the structure `CH_(3)-C-=CH`. It consist 2 primary carbon (a carbon to which single carbon is bonded) and one secondary carbon. Its structure show that it contain only primary hydrogen. Isobutene have the structure `{:(CH_(3)-C=CH_(2)),(" |"),(" "CH_(3)):}` which also contain only primary Hydrogen atom. |
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| 457. |
The IUPAC name of `{:(" "CH_(3)O),(" |" " ||"),(CH_(3)-C-C-CH_(2)-CH_(2)OH):}` isA. 1 - hydroxy - 4 - methyl pentan - 3 - oneB. 2-methyl-5-hydroxy pentane-3-oneC. 4-methyl-3-oxopentan-1-olD. Hexan-1-ol-3-one |
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Answer» Correct Answer - A `{:(" "CH_(3)" "O),(" |"" ||"),(underset("1hydroxy-4-methyl pentan-3-one")(CH_(3)-CH-C-CH_(2)-CH_(2)OH)):}` |
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| 458. |
What is the IUPAC name of the following A. Ethenyl cyclo pentaneB. cyclopentyl etheneC. cyclopentyl ethyleneD. vinyl cyclopentane |
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Answer» Correct Answer - C |
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| 459. |
The correct structure of 4 - bromo - 3 - methyl but - 1 - ene isA. `Br-CH = C(CH_(3))_(2)`B. `CH_(2)=CH-CH(CH_(3))-CH_(2)Br`C. `CH_(2)=C(CH_(3))CH_(2)CH_(2)Br`D. `CH_(3)=C(CH_(3))=CHCH_(2)-Br` |
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Answer» Correct Answer - B `{:(overset(1)(C)H_(2)=overset(2)(C)H-overset(3)(C)H-overset(4)(C)H_(2)Br),(" |"),(" "underset("4-bromo-3-methyl but-1-ene")(CH_(3))):}` |
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| 460. |
The structure of 3-bromoprop-1 -ene isA. `CH_(3)-overset(Br)overset(|)(C)=CH_(2)`B. `CH_(3)-CH=CH-Br`C. `CH_(3)-underset(Br)underset(|)(C)=CH-Br`D. `Br-CH_(2)-CH=CH_(2)` |
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Answer» Correct Answer - D |
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| 461. |
Which of the following pentoses will be optically activeA. AllB. II and IIIC. ID. II |
| Answer» Correct Answer - A | |
| 462. |
Hydrolysis of d-sucrose results in change of optical rotation becauseA. number of moles of d-Glucose are less than number of moles of l-Fructose produced during hydrolysisB. Angle of rotation of l-Fructose is greater than AOR of d-GlucoseC. Both glucose and fructose and levorotatory.D. Glucose is l and fructose is d and number of moles of glucose formed during reaction are greater than number of moles of fructose. |
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Answer» Correct Answer - B Hydrolysis of d-sucrose reducts in change of optical rotation because angle of rotation of l-fructose is greater than angle of rotation of d-glucose. |
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| 463. |
Hydrolysis of sucrose is called:A. Thermal decompositionB. SaponificationC. InversionD. Hydration |
| Answer» Correct Answer - C | |
| 464. |
The number of racemic mixture obtained by optical isomers of 2, 3-dihydroxy butanal is/areA. ThreeB. TwoC. OneD. Zero |
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Answer» Correct Answer - B `{:(CH_(3)-overset(**)(C)H-overset(**)(C)H-CHO),(" |"" |"),(" "OH" "OH):}` Number of stereoisomers `= 2^(n) = 4`. Here 2 racemic mixtures are obtained. |
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| 465. |
An organic compound (A) `(C_(10)H_(20))` on reductive ozonolysis gives `2-` methyl butanal. Based on this information, answer the following question. If the stereochemistry about the double bond in (A) is trans, the total number of enantiomers for (A) is :A. NoneB. 2C. 4D. 6 |
| Answer» Correct Answer - B | |
| 466. |
In which of the following lone-pair indicated is not involved in resonance:A. `CH_(2)=CH-CH=underset(..)overset(..)(O)`B. `CH_(2)=CH-underset(..)overset(..)(O)-CH=CH_(2)`C. `CH_(2)=CH-C-=N:`D. |
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Answer» Correct Answer - A::C::D (B). A double bonded oxygen will not releases electron density. So its I.p. will not involve in resonance. (D). `CH_(2)=CH-C-=overset(..)(N)` Carbon having only four valencies (E). Not a conjugates system. |
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| 467. |
Arrange following increasing order of stability A. 1gt4gt3gt2B. 4gt1gt3gt2C. 4gt1gt2gt3D. 2gt1gt4gt3 |
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Answer» Correct Answer - B 4(extended conjugation) 1(Back-bonding) 3(Low-I effect of oxygen) 2(High-I effect of oxygen) |
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| 468. |
Which of the following resonating structures of 1-methoxy-1,3-butadiene is least stable ?A. `overset (Ө) C H_2 - CH = CH - CH = overset (oplus) O - CH_3`B. `CH_2 = CH- overset (Ө) CH - CH = overset (oplus) O - CH_3`C. `overset (Ө) CH_2 - overset (oplus) CG - CH = CH - O - CH_3`D. `CH_2 = CH - overset (Ө) C H - overset (oplus) C H - O - CH_3` |
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Answer» Correct Answer - C ( c) The octete of all atoms are complete in structures `a` and `b` The molecule in which all the atoms have complete octet is more stable than the atom having incomplete octet. Larger the number of resonating structures, larger will be the stability. Thus, structures `a` and `b` are stable. Instructure `(d)`, the electron deficiency of positive charged carbon is duly compensated by lone pair electrons of adjacent oxygen atoms, while such neihbour group support is not available in structure `(c)`. Hence, structure `(c)` is less stable in comparion to structure `(d)`. |
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| 469. |
An electron releasing group will not stabilize which of the following groups ?A. CarbocationB. CarbanionC. Free radicalD. None of these |
| Answer» Correct Answer - B | |
| 470. |
Number of isomers of `C_(4)H_(10)` isA. 2B. 3C. 4D. Isomerism not exist |
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Answer» Correct Answer - A `{:(underset("n-Butane")(CH_(3)-CH_(2)-CH_(2)-CH_(3))",",,CH_(3)-CH-CH_(3)),(,," |"),(,," "CH_(3)),(,," Isobutane"):}` |
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| 471. |
The number of optical isomers for the compound, `CH_(3)-CH(Br)-CH(Br)C_(2)H_(5)` isA. 1B. 2C. 4D. 3 |
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Answer» Correct Answer - C The compound `CH_(3)-CH(Br)CH(Br)C_(2)H_(5)` cannot be divided into two identical halves. It has 2 chiral centres. The formula for number of optically active form is `2^(n)`. Where n is the number of chiral centres. Hence `2^(2) = 4` Hence the number of isomers are 4 so choice (a), (b) and (d) are incorrect while (c) is correct. |
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| 472. |
The number of constitutional isomers of the formula `C_(5)H_(11)` Br isA. 4B. 8C. 6D. 10 |
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Answer» Correct Answer - B The compound `C_(5)H_(11)` Br possessess 8 constitutional isomers |
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| 473. |
3-Phenyl propenoic acid is the IUPAC name of which of the following compoundA. Cinnamic acidB. Mendaleic acidC. Picric acidD. Citric acid |
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Answer» Correct Answer - A |
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| 474. |
The correct IUPAC name of A. 1-Chloro-2-butylcyclopropaneB. 1-Chloro-2-cyclopropybutaneC. 1-Chloro-methyl-1-ethyl cyclo propaneD. 3-Chloro metyl-1,2-methylenepentane |
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Answer» Correct Answer - B |
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| 475. |
IUPAC name of the compound A. N-ethyl-N-methyl isopropaneB. N-ethyl-N-methyl amino propaneC. N-ethyl-N-methyl-1-amino propaneD. N-ethyl-N-methyl-2-propanamine |
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Answer» Correct Answer - D |
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| 476. |
Which of the following compounds is strogest base?A. B. C. D. |
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Answer» Correct Answer - B Alipatic amines are more basic than aromatic amines. |
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| 477. |
Which of the following reactions does not proceed in forward direction?A. `CH_(3)CH_(2)SH+NaOHrarrCH_(3)CH_(2)SNa+H_(2)O`B. `CH_(3)CH_(2)OH+CH_(3)MgBrrarrCH_(3)CH_(2)OMgBr+CH_(4)`C. `CH_(3)CH_(2)OH+NaOHrarrCH_(3)CH_(2)ONa+H_(2)O`D. `CH_(3)CH_(2)OH+NaHrarrCH_(3)CH_(2)ONa+H_(2)` |
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Answer» Correct Answer - C Alipatic amines are more basic than armatic amines. |
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| 478. |
Reason for geometrical isomerism shown by 2- butene isA. Chiral carbonB. Free rotation about single bondC. Free rotation about double bondD. Restricted rotation about double bond |
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Answer» Correct Answer - D Restricted rotation is essential condition for geometrical isomerism. |
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| 479. |
Which one of the following pairs represents stereoisomerismA. Chain isomerism and rotational isomerismB. Structural isomerism and geometric isomerismC. Linkage isomerism and geometric isomerismD. Optical isomerism and geometric isomerism |
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Answer» Correct Answer - D When isomers have the structural formula but differ in the relative arrangement of atoms of groups in space with in the molecule, these are known as stereoisomers and the phenomenon as stereisomerism. Stereoisomerism is of three types (i) Geometrical isomerism (ii) Optical isomerism (iii) Conformational isomerism. |
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| 480. |
Which of the following compounds may not exist as enantiomersA. `CH_(3)CH(OH)CO_(2)H`B. `CH_(3)CH_(2)CH(CH_(3))CH_(2)OH`C. `C_(6)H_(5)CH_(2)CH_(3)`D. `C_(6)H_(5)CHClCH_(3)` |
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Answer» Correct Answer - C In `C_(6)H_(5)CH_(2)CH_(3)` chiral centre is absent. |
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| 481. |
Which of the following pairs of compounds are enantiomersA. B. C. D. |
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Answer» Correct Answer - B Structures are not mirror of each other which are non superimposable so they are enantiomers. |
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| 482. |
exhibit which isomerismA. Position isomerismB. Geometrical isomerismC. Optical isomerismD. Functional isomerism |
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Answer» Correct Answer - B Geometrical isomerism. |
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| 483. |
Which of the following pairs of compounds are enantiomers?A. B. C. D. |
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Answer» Correct Answer - 1 Given compound in I-option are non superimposable mirror images. |
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| 484. |
Among the following compounds which exhibits optical isomerismA. PropanolB. 2-PropanolC. 1-ButanolD. 2-Butanol |
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Answer» Correct Answer - 4 2-butanol has chiral carbon. |
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| 485. |
An alkane forms isomers if the number of least carbon atom isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - D `underset("n-butane")(CH_(3)-CH_(2)-CH_(2)-CH_(3))` and `{:(CH_(3)-CH-CH_(3)),(" |"),(" "underset("Iso-butan")(CH_(3))):}` |
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| 486. |
Which of the following compounds exhibits optical isomerismA. `CH_(3)CH_(2)COOH`B. `CH_(3)CHOHCOOH`C. `CH_(3)CH_(2)CH_(2)OH`D. `CH_(3)CHOHCH_(3)` |
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Answer» Correct Answer - B `{:(" "H),(" |"),(CH_(3)-C^(**)-COOH^(**)),(" |"),(" "OH):}` chiral centre is present. |
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| 487. |
The conversion of a keto form into enol is enolisation. It depends on structural factor, temperature and nature of solvent. Resonance and hydrogen bonding increases enol content. Enolic form of phenol is more stable than keto form by-13 kcal/mol of energy hence phenol exist exclusively as an enol. Enolic tautomer is less polar due to intramolecular hydrogen bonds than the corresponding keto form. Any polar solvent would decrease the enolisation and fovour keto form intramolecular hydrogen bonding stabilized enol form by 7kcal/mol and resonance stabilizes enol form by 15 kcal/mole Which one of the following compounds has maximum enol content?A. `CH_(3)COCH_(2)COOCH_(3)`B. `CH_(3)COCH_(2)COOC_(2)H_(5)`C. `CH_(3)COCH_(2)COCH_(3)`D. `C_(6)H_(5)COCH_(2)COC_(6)H_(5)` |
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Answer» Correct Answer - D More acidic active metylene group is found in compound D. |
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| 488. |
The conversion of a keto form into enol is enolisation. It depends on structural factor, temperature and nature of solvent. Resonance and hydrogen bonding increases enol content. Enolic form of phenol is more stable than keto form by-13 kcal/mol of energy hence phenol exist exclusively as an enol. Enolic tautomer is less polar due to intramolecular hydrogen bonds than the corresponding keto form. Any polar solvent would decrease the enolisation and fovour keto form intramolecular hydrogen bonding stabilized enol form by 7kcal/mol and resonance stabilizes enol form by 15 kcal/mole The enol content of ethyl acetoacetate is highest inA. WaterB. acetic acidC. ethanolD. hexane |
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Answer» Correct Answer - A Water facilitate Hydrogen bonding. |
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| 489. |
Conversion of `CH_(4)` to `CH_(3)Cl` is an example of which of the following reactionA. Electrophilic substitutionB. Free radical additionC. Nucleophilic substitutionD. Free radical substitution |
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Answer» Correct Answer - D When methane gas is treated with chlorine in the presence by sunlight, one hydrogen of methane replaced by the chlorine atom and forms methyl chloride. The mechanism involved in this reaction is free radical mechanism. So it is an example of free radical substitution reaction. |
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| 490. |
The order of decreasing basicity in the four halide ions is :A. `I^(-) gt Br^(-) gt Cl^(-) gt F^(-)`B. `Cl^(-) gt Br^(-) gt I^(-) gt F^(-)`C. `F^(-) gt Cl^(-) gt Br^(-) gt I^(-)`D. `Cl^(-) gt F^(-) gt Br^(-) gt I^(-)` |
| Answer» Correct Answer - C | |
| 491. |
Carbocation areA. Arrhenius acidsB. Bronsted acidsC. Lowry acidsD. Lewis acids |
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Answer» Correct Answer - D Due to the tendency to complete the octet of carbon, the carbocation is an exceedingly reactive particle, Since it takes a pair pf electrons to complete the ocet, the carbocation is a Lewis acid, and an extremely powerful one. |
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| 492. |
ElectrophilesA. positive reagentsB. netural reagentsC. Lewis acidsD. all of these |
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Answer» Correct Answer - D Electrophiles are electron seeking reagents. They include positive reagents such as protons, neutral reagents such as bromine (because it can be polarized so that one end is positive), and the Lewis acids `BF_(3) and AlCl_(3)`. |
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| 493. |
Which of the following metal ions cannot act as an electrophile?A. `Ag^(+)`B. `Hg^(2+)`C. `Pt^(2+)`D. `Na^(+)` |
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Answer» Correct Answer - D Metal ions that contain vacant orbitals of lower energy-the silver ion `(Ag^(+))`, the mercuric ion `(Hg^(2+))`, and the platinum ion `(Pt^(2+))`-act as electrophiles. Sodium ion ia not electrophilic as it lacks an empty orbital of lower enrgy. |
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| 494. |
Glucose and fructose formA. Same osazoneB. Same acid on oxidationC. Same alcohol when reducedD. Different osazone |
| Answer» Correct Answer - A | |
| 495. |
The change of optical rotation of glucose solution with time is refered to as:A. MutarotationB. InversionC. Specific rotationD. Autorotation |
| Answer» Correct Answer - A | |
| 496. |
Give the correct order of initials T or F for following statements Use T if statements is true and F if it is false. Statement-I: p-chloro phenol is more acidic than p-flouro phenol. Statement-II: `T_(1//2)` time period of `(C_(6)H_(5))_(3)overset(o+)(C)` is less than `T_(1//2)` of `C_(6)H_(5)overset(o+)(C)H_(2)` Statement-III: Solubility of `Br^(-)` salt of alkali metals in water increases down the groupA. TFFB. FFFC. FTTD. TFT |
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Answer» Correct Answer - A Statement-I: p-chlorophenol is more acidic then p-flouro phenol |
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| 497. |
The number of sigma `(sigma)` and pi `(pi)` bonds present in acetylene are _____ respectively.A. 1,2B. 2,1C. 5,2D. 3,2 |
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Answer» Correct Answer - D |
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| 498. |
C-H and 7 C-C sigma bonds are inA. cyclohexaneB. 3,3-dimethylphentaneC. 2,2,3-trimethylpentaneD. n-heptane |
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Answer» Correct Answer - C |
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| 499. |
The `IUPAC` name of viny1 acetylene is:A. Pent - 1 - ene-4-yneB. Pent-4-yne-1-eneC. But-1-en-3-yneD. But-1-yn-3-ene |
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Answer» Correct Answer - C |
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| 500. |
Explain why A has lower point than B ? |
| Answer» Correct Answer - A::B::C::D | |