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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.
| 1. |
Stereoisomers may be of different types. Stereosiomers that are radialy interconvertible by rotation around a `sigma-`bond are known as_______isomersA. geometricalB. opticalC. cis-transD. conformational |
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Answer» Correct Answer - D Conformational isomers differ by rotation about a `C-C` bond. |
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| 2. |
Which of the following is true for the given reaction? `overset( :overset(..)O:^-)overset(|)((S)-C_2H_5CHCH_3+CH_3I)rarroverset(OCH_3)overset(|)C_2H_5CHCH_3+I^(-)`A. The configuration is unchanged and the product is `S`B. The configuration is chnaged and the product is `R`C. The configuration is chnaged and product is `S`D. The configuration is unchanged and the product is `R` |
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Answer» Correct Answer - A Because the bonds to the chiral carbon atom are not distrubed, the configuration is unchanged. Since the priority order is unchanged, the product is `(S)`. Note that product (like reactant) is optically active. The products and reactants will have different specific rotations and their signs may differ. specific rotations of reactants and products are independent of each other. |
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| 3. |
Which of the folloiwng is true for the given reaction? `(S)-C_(3)H_(7)CH(OH)CH=CH_(2)+H_(2)overset(Pt)rarrC_(3)H_(7)CH(OH)CH_(2)CH_(3)`A. The configuration is unchanged and the product is `R`B. The configuration is chnaged and the product is `S`C. The configuration is chnaged and product is `R`D. The configuration is unchanged and the product is `S` |
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Answer» Correct Answer - A The configuration is unchanged because the bonds to the chiral carbon atom are undistured. The product is `R` since the priority sequncy changes. Like the reactant, the product is optically active. Note that the products and reactants will have different sepcific rotations and their signs may differ. specific rotations of reactants and products are independent of each other. |
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| 4. |
Consider the following organic compound `overset(1)CH_(3)overset(2)CH_(2)overset(3)CH_(2)overset(4)CH_(2)overset(5)CH_(2)overset(6)CH_(2)overset(7)CH_(3)` To make it a chiral compound, the attack should be on carbonA. `7`B. `3`C. `4`D. `1` |
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Answer» Correct Answer - B Compound becomes chiral if its molecule carries just one asymmetric carbon atom. Therefore, an attack on `C-3` will make it chiral: `CH_(3)CH_(2)overset(**)underset(X)underset(|)CHCH_(2)CH_(2)CH_(2)CH_(3)` |
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| 5. |
Isomers are different compounds that have the sameA. empirical formulaB. molecular formulaC. precentage compositionD. all of these |
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Answer» Correct Answer - B Isomers contain the same numbers of the same kinds of atoms and hence have tha same molecular formula. Ethyne `(C_(2)H_(2))`and benzene `(C_(6)H_(6))` have the same percentage composition and same emirical formula but they are not isomers as they differe in molecular formula. |
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| 6. |
Which of the following compounds possesses a chiral centre?A. B. C. D. |
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Answer» Correct Answer - C A chiral center is an atom (usually a `C` atom but may be `N,S,P,Si`, etc) bonded tetrahedrally to four dofferent atopms or groups of atoms, called ligands. In compound `(3)`, the carbon atom carrying the `-OH` group is a chiral center as the double bond in the ring is placed unsymmetrically. |
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| 7. |
If there is no rotation of plane polarized light by a compound in a specific solvent, through to be chiral, it may mean that:A. there is no compound in the solventB. the compound may be a racemic mixtureC. the compound is certainly a chiralD. the copound is certainly meso |
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Answer» Correct Answer - D A meso compound consists of molecule containing chiral carbon atoms but the compound is optically inactive as the molecule is superposable on its mirror image. |
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| 8. |
The device that is used for measuring the effect of plane polarized light on optically active compounds is known asA. polariscopeB. polarimeterC. both`(1)`and`(2)`D. periscope |
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Answer» Correct Answer - C Plane-polarized light (used in the analysis of enantiomers) is produced in an instrument called a polarimeter (polariscope) and passes thorugh a polarimeter tube holding either the liquid enantiomer or its solution. The angle through which the polarized light has been rotated is measured in an analyzer. The observed rotation `(alpha_(obs))` is expressed in degrees. If the polarized light plane is rotated to the right (clockwise), the anatiomer is said to be dextrorotatory. A rotation to the (counterclockwise) is called levorotatory. The symbols `(+)` and `(-)` desginate rotation to the right and left, respectively. Note that polarimeter consist of a light source, two lenses (polarizer and analyzer) and between the lenses a tube to hold the substance that is being examined for optical activity. A periscope is device for viewiing objets that are above the eye-level of the observer or are placed so that direct vision is obstructed. |
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| 9. |
The number of chiral centres present in `3,4`-dichloropentan`-2-ol` isA. `3`B. `4`C. `1`D. `2` |
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Answer» Correct Answer - A Draw the structure and identify the chiral carbon atoms: `CH_(3)-overset(CI)overset(|**)underset(H)underset(|)(C)-overset(CI)overset(|**)underset(H)underset(|)(C)-overset(CI)overset(|**)underset(H)underset(|)(C) -CH_(3)` |
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| 10. |
Which of the following generla rules for switching ligands or rotaing Fischer structures is incorrect?A. An even number of switches results in no change in the configurationB. An odd number of switches changes the configuration to that of the enantiomerC. The structure cannot be rotated out of the plane of the paperD. The structure can be rotated in the plane through `90^(@)` |
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Answer» Correct Answer - D Any rotation of Fisher projections that interconverts horizontal and vertical ligands is not allowed. Hence, the structure can be rotated in the plane through `180^(@)` or its multiples but not through `90^(@)` |
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| 11. |
The specific rotation of a pure enantiomer is `+16^(@)` its observed rotation if it is isolated from a reaction with 25% racemisation and 75% retention is,A. `-12^(@)`B. `+12^(@)`C. `+16^(@)`D. `-16^(@)` |
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Answer» Correct Answer - B % enantiomeric excess `=("observed specific rotation")/("Specific ration of pure enantiomer")xx100` observed specific rotation`=(3//4)/(100)xx(+16^(@))xx100=+12^(@)` |
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| 12. |
Dextrororatory bytan`-2-ol` has a specific rotation `[alpha]_(D)^(25) =+13.52^(@)`. A sample of butan`-2-ol` shows a specific rotation `[alpha]_(D)^(25)=+6.72^(@)`. Calculate the enantiomers excess (or optical purity) of the sample?A. `60%`B. `50%`C. `70%`D. `80%` |
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Answer» Correct Answer - B `%` Enantiomeric excess (optical purity) `=("observed specific rotaiton")/("specific rotation of the pure enantiomer") xx 100%` `= (+6.76^(@))/(+13.52^(@)) xx 100%` `= 50%` |
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| 13. |
The process of separating a racemic mixture into optically pure `(+)` and `(-)` enantiomers is known asA. asymmetric synthesisB. mutarotationC. epimerizationD. resolution |
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Answer» Correct Answer - D Resolution is the separation of two enantiomers from each other. A mixture of enantiomers is said to be resolvable. |
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| 14. |
The specific rotation of a pure enantiomers is `+12^(@)`. What will be its observed rotation if it is isolated form a reaction with `20%` racemization and `80%` retention.A. `+9.6^(@)`B. `-9.6^(@)`C. `+2.4^(@)`D. `-2.4^(@)` |
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Answer» Correct Answer - A `%`optical purity `= (alpha_(obs))/([alpha]_(D)) xx 100%` Thus, `alpha_(obs) = (80)/(100) (+12^(@))` `= +9.6^(@)` |
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| 15. |
`(+)-`Butan`-2-ol` has `[alpha]_(D)^(20) =+13.9^(@)`.a sample of butane`2-ol` contianing both the enantiomers was found to have a specific rotation value of `-3.5^(@)` under similar conditions. The percentages of the `(+)` and `(-)` enantiomers present in the mixture are, respectively,A. `37.4%` and `62.6%`B. `35.5%` and `64.5%`C. `42.2%` and `57.8%`D. `62.6%` and `37.4%` |
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Answer» Correct Answer - A The specific rotation of pure `(-)-`butane`-2-`ol is `-13.9^(@)`. Therefore, the optical purity `(OP)` of the sample si `%` optical purity `= (-3.5^(@))/(-13.9^(@)) xx100% = 25%` This implies that `25%` of the sample consists of `(-)-`enantiomers in excess and `75%` is the racemate. the total `%` of `(-)` is thus `25% +1//2 (75%) = 62.5%`, and the remaining `37.5%` is `(+)`. |
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| 16. |
The process of separating of a racemic mixture into `d` and `l` enantiomers is calledA. dehydrogenationB. revolutionC. dehydrationD. resolution |
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Answer» Correct Answer - D Resolution refers to method of separating a racemic mixture into its enantiomeric constituents. One of the most common methods is to allow a racemic mixture of react with an enantiomer of some other compounds. This is changed a racemic form into a mixture of diastereomers which have dofferent melting points, boiling points and solubilities and hence can be separated by conventional methods. |
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| 17. |
`R` and `S` paris of enantiomers differ form one another inA. optical rotation of polarized lightB. solubility in racemic mixtureC. reaction will racemic mixtureD. none of these |
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Answer» Correct Answer - A Enantiomers have identical physical properties apart form their differing influence on polarized light which they rotate in oppoiste deirection but to an equal extent. |
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| 18. |
Which of the following pairs of compounds are enantiomers?A. B. C. D. |
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Answer» Correct Answer - A If we rotate the second compound out of the plane of paper by `180^(@)` (clockwise or counterclockwise), we notice that the two are nonsuperposable mirror images. Compounds in `(2)` and `(3)` are not enantomers as their molecules posses a plane of symmetry (i.e. of plane perpendicular to the paper and containing both `CH_(3)` and `OH`, the plane theat passers between the two Br atoms). In `(4)`, the second compound becomes identical to the first one if we rotate it (out of the plane paper) by `180^(@)` (clockwise or counterclockwise). |
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| 19. |
Which one of the following pairs represnets steroeoisomeris?A. Chain isomerism and rotational isomerismB. structual isomerism and geomtrical isomerismC. Kinkage isomerism and geomertical isomerismD. optical isomerism and geometrical isomericm |
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Answer» Correct Answer - D Optical isomerism corresponds to enantiomerism while geometrical isomerism corresponds to diastereomerism. |
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| 20. |
`Cis` and trans isomers generallyA. contain a triple bondB. contain double bonded carbon atomsC. are enantiomersD. rotate the plane of plane-polarized light |
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Answer» Correct Answer - B Cis-trans isomerism occurs in any molecule if four approprite ligands are held in one plane by some element of rigidity with in the molcule. The element of rigidity may be a double bond (such as `C =C,C =N`, or `N = N`) or a ring |
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| 21. |
The compound `1,3`-ibromo`-2-`methy`1` butane, `CH_(3)CHBrCH(CH_(3))CH_(2)Br, `has two dissimilar chiral carbons `(C-2` and `C-3)`. Its four stereoisomers are shown below I Which of the following has `(2R, 3R)` designation?A. `I`B. `II`C. `III`D. `IV` |
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Answer» Correct Answer - D In order of priority, the four ligands bonded to `C-2` are `CH_(3)CHBr gtCH_(2)Br gtCH_(3) gtH`. On `C-3` they are `Br gt CH(CH_(3))CH_(2)Br gtCH_(3) gtH`. |
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| 22. |
Racemic modification can be resolved byA. the use of enzymesB. fractional crystailisationC. fractional distillationD. none of the above |
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Answer» Correct Answer - A The separation of racemic mixture back into d and l isomers is known as resolution. It can be done by (I). Mechanical method (II). Biochemical method using enzymes (III). Chemical method (salt formation). |
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| 23. |
Incorrect statement isA. ethane can have an infinite number of conformationsB. cyclopropane molecule has considerable angle strainC. eclipsed form of ethane is less stable than staggered conformationD. staggered conformation possesses maximum energy. |
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Answer» Correct Answer - D Staggered conformation is most stable due to its minimum energy. |
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| 24. |
At room temperature, the eclipsed and the staggered forms of ethane cannot be isolated becauseA. both the conformers are equally stableB. they interconvert rapidlyC. There is a large energy barrier of rotation about the `sigma`-bondD. the energy difference between the conformers is large. |
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Answer» Correct Answer - B Staggered and eclipsed conformers cannot be physically separated because energy difference between them is so small that they most readily interconvert at room temperature. |
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| 25. |
According to `CIP` sequence rule, the correct arrangement in order of decreasing priority isA. `OH gt COOH gt CH (OH)CH_(3) gtCH_(2)OH`B. `COOH gtOH gtCH_(2)OH gt CH(OH)CH_(3)`C. `OH gt CH_(2)OH gt COOH gtCH(OH)CH_(3)`D. `CH(OH)CH_(3) gtCOOH gtOH gtCH_(2)OH` |
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Answer» Correct Answer - A `OH` has the highest priority because the directly attacjed oxygen atom has the highest atomic number. `-overset(O)overset(||)C-OH`equals`-overset(O)overset(|)underset((O))underset(|)C-OH`,i.e.`C(COO)`and takes priority over other ligands while `-overset(H)overset(|)underset(OH)underset(|)C-CH_(3)`,i.e.`C(OCH)` takes priority over `-overset(H)overset(|)underset(H)underset(|)C-OH`,i.e.`C(OHH)` |
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| 26. |
Which of the following are diastereomers ? A. I and IIIB. II and IVC. I and IID. None of these |
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Answer» Correct Answer - C Superimposable compounds which are also not the mirror images of each other, are called diastereomers. Thus, I and II are diastereomers. |
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| 27. |
Which statement is true?A. A compound with R configuration is the (+) enantiomerB. If configuration changes from + to -, that essentially means inversion of configuration takes placeC. An achiral molecule reacts always with racemic forms, to give a chiral molecule.D. By breaking two bonds on the chiral centre, configuration changes. |
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Answer» Correct Answer - D We know that by breaking two bonds on the chiral centre, cionfiguration changes. |
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| 28. |
A molecule is said to chiral if itA. contains no chiral carbonsB. contains chiral carbonsC. cannot be superposed on its mirrot imageD. contains a plane of symmetry |
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Answer» Correct Answer - C Chiral molecule is nonsuperosable on its mirror image. |
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| 29. |
Which of the following is the chiral molecule?A. `CH_(3)CI`B. `CH_(2)CI_(2)`C. `CHBr_(3)`D. `CHCIBrI` |
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Answer» Correct Answer - D It contains just one chiral carbon atom `CI-overset(H)overset(**|)underset(I)underset(|)C-Br` |
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| 30. |
How many carbon atoms in the molecule `HOOC-(CHOH)_(2)-COOH` are asymmetric?A. 1B. 2C. 3D. none of these |
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Answer» Correct Answer - B A carbon atom which is attached to four diferent groups is called as asymmetric carbon atom or chiral centre. HOOC`(CHOH)_(2)COOH` has two asymmetric carbon `(overset(**)(C))` atoms. `HOOC-underset(OH)underset(|)overset(**)(C)H-underset(OH)underset(|)overset(**)(C)H-COOH` |
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| 31. |
Which of the following is a chiral compound?A. `2,3,4-`trimethylhexaneB. `n-`hexaneC. methaneD. `n-`butane |
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Answer» Correct Answer - A `underset(2,3,4-"trimethylhexane")(H_(3)overset(1)C-overset(CH_(3))overset(2|)underset(H)underset(|)C-overset(CH_(3))overset(3|**)underset(H)underset(|)C-overset(CH_(3))overset(4|)underset(H)underset(|)C-overset(5)CH_(2)overset(6)CH_(3))` This compound is chiral beacuse its moelcule has two chiral carbon atoms `(C-3` and `C-4)`. Pther molecules are chiral due to the presence of a plane of symmetry. |
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| 32. |
How many carbon atoms in the molecule `HOOC-(CHOH)_(2)-COOH` are asymmetric?A. `1`B. `2`C. `3`D. None of these |
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Answer» Correct Answer - B Asymmetric carbon atom is always bonded to four different ligands. Therefore, there are two asymmetric acrbon atoms in the given molecule: `HOOC-overset(H)overset(|**)underset(OH)underset(|)C-overset(H)overset(|**)underset(OH)underset(|)C-COOH` |
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| 33. |
Which of the following statements is not correct?A. A meso compound has chiral centres but exhibits no optical activityB. A meso compound has no chiral centre. Thus it is optically inactive.C. A meso compound has molecule in which one-half of molecule is superimposable on the other even through chiral centre is present in themD. A meso compound is optically inactive because the rotation caused by one-half of molecule is cancelled by the rotation produced by another half |
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Answer» Correct Answer - D Compounds which do not show optical activity inspite of the presence of chiral carbon atoms are called meso-compounds. |
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| 34. |
Which of the following objects is chiral?A. ForkB. SpoonC. HammerD. Ear |
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Answer» Correct Answer - D A chiral object is one whose mirror image is not superposable as it it lacks a plane or centre of symmetry. On the other hand, an chiral object has a superposable mirror image. Ear is chiral as it does not possess a plane of symmetry or a cnetre of symmetry. |
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| 35. |
Which of the following atoms can be a chiral center in a given molecular species? (i) Carbon (ii) Nitrogen (iii) Phosphorus (iv) SiliconA. (i),(ii),(iii),(iv)B. (ii),(iii),(iv)C. (iii),(iv)D. Only (i) |
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Answer» Correct Answer - A A chiral center is an atom bonded tetrahedrally to four different atoms or groups of atoms (ligands).A chiral center is usually a `C` atom but may be `N,P,Si,` etc. Example of `C,Si` and `P` serving as chiral centers (stereocenters) are: `H_(2)C=CH-underset(Br)underset(|)overset(H)overset(|)(C^(**))-CH_(2)CH_(3)` `CH_(3)CH_(2)-.^**underset(CI)underset(|)overset(CH_(3))overset(|)(Si)-CH_(2)CH_(2)CH_(3)` `CH_(3)CH_(2)-overset(CH_(3))underset(underset(..)(overset(-)overset(.).Ooverset(.).))underset(|)overset(+|**)(N)-CH_(2)CH_(2)CH_(3)` `CH_(3)CH_(2)-overset(CH_(3))overset(+|**)underset(H)underset(|)(P)-CH_(2)CH_(2)CH_(3)` |
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| 36. |
A carbon atom is chiral whenA. it has tetrahdral orientationB. it is `sp^(3)` hybridizedC. it has four different ligandsD. it forms multiple bonds |
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Answer» Correct Answer - C A carbon atom to which four different atoms (or groups) are bonded is a chiral centre. |
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| 37. |
The necessary and sufficient condition for a compound to be optically active is thatA. it must contain asymmetric carbon atomsB. its molecule must be nonidentical with its mirroe imageC. its must be symmetricD. its molecule must be identical with its mirror image |
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Answer» Correct Answer - B Optical activity is not the property of a single molecule but rather of a collection of moolecules in which one serves as the nonsuperposable mirror image of the other. |
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| 38. |
The optical activity is the characterstic ofA. gaseous stateB. liquid stateC. crystalline stateD. molecules |
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Answer» Correct Answer - D Optical activity is associated with the three dimensional arrangement of atoms (or groups of atoms) in molecules |
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| 39. |
Calculate the specific rotation of coniine (the toxic component of poison hemlock), if a solution containing `0.75g `in `10 mL` is placed in a `1-dm` polarimeter tube and its observed rotation at `25^(@)C(D` line) is `+1.2^(@)`.A. `+1.6^(@)`B. `+16^(@)`C. `+3.2^(@)`D. `+32^(@)` |
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Answer» Correct Answer - B Specific rotation `= ("observed rotation"(degress))/("length"(dm)xx "concetration or density" (g//mL^(-1)))` `[alpha]_(D)^(25^(@)) = (+1.2^(@))/((1)(0.75)/(10))=+16^(@)` Note that the specific rotation of the enantiomer of coniine is `-16^(@)` Also note that the specific rotation is a constant and is independent of concetration and path length. |
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| 40. |
Consider the reaction `{:((R)-CH_(3)CHICH_(2)CH_(3)+^(137)T^(-),rarrCH_(3)CHICH_(2)CH_(3),,,),(alpha_(obs)=-15.90^(@),"containing" 2%137T^(-),,,),(,alpha_(obs)=-15.26^(@),,,):}` The percentage of racemic from in the product isA. `10%`B. `8%`C. `4%`D. `12%` |
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Answer» Correct Answer - C Note that replacement of `I` with `.^(137)I` does not change he sign or value of the rotation. The optical purity of the recovered iodide is `(15.26//15.90)(100%) = 96%`. Hence `4%` of the iodide is racemic. This means thet `2%`, (the same percentage that contains isostopic `I)`. has been converted to the `(S)` configuraiton. It is concluded that every reaction with `.^(137)T^(-)` resulted in an inversion at chiral carbon. |
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| 41. |
Which of the following reactions occurs wiyh reacemization?A. `CH_(3)CH_(2)CHH(OH)CH_(2)Br+OH^(-)rarrCH_(3)CH_(2)CH(OH)CH_(2)OH+Br^(-)`B. `CH_(3)CH_(2)CH=CH =CH_(2)+D_(2)overset(catalyst)rarrCH_(3)CH_(2)CH(D)CH_(2)D`C. `CH_(3)CH_(2)CH(OH)CH(CH_(3))_(2)+CH_(3)COCIrarrCH_(3)CH_(2)CH(OCOCH_(3))CH(CH_(3))_(2)+HCI`D. `CH_(3)CH(NH_(2))COOH+NaOHrarrCH_(3)CH_(2)(NH_(2))COO^(-)Na^(+)+H_(2)O` |
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Answer» Correct Answer - B The product has a chiral center `(D` being sufficiently different from `H)`, both enantiomers are formed in equal amounts because `D_(2)` adds to the enantiotopic faces of the alkene equally. Reactions `(1),(3)` and `(4)` occur with retention because no bond-breacking occurs at the chiral carbon. To answers this, we need to draw out the structurnes of the reactants and the products and identify and chiral centers. |
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| 42. |
The prefixes syn and anti are used to denoteA. structural isomersB. conformationalC. geonmertical isomersD. optical isomers |
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Answer» Correct Answer - C The prefixes syn and anti are used to denote geometrical isomers of oximers, azo compounds, imines and hydrozones. |
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| 43. |
An enantiomerically pure acid is treated with racemic mixture of an alcohol having one chiral carbon. The ester formed will be :A. optically active mixtureB. pure enantiomerC. meso compoundD. racemic mixture. |
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Answer» Correct Answer - A If an enantiomerically pure acid is treated with racemic mixture of an alcohol having chiral carbon, the product formed will be optically active mixture. |
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| 44. |
The alkene that axhibits geometrical isomerism isA. but`-2-`eneB. propeneC. `2-`methy`1`but`-2`-eneD. `2-`methy`1`propene |
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Answer» Correct Answer - A `{:(underset(But-2-en e)(CH_(3)CH=CHCH_(3)),,underset(Prpen e)(CH_(3)CH=CH_(2)),,),(underset(2-Methy1but-2-en e)((CH_(3))_(2)C=CHCH_(3)),,underset(2-Methy1pr open e)((CH_(3))_(2)C=CH_(2)),,):}` Alkenes with two identical ligands on the same double-bonded `C` have no geometric isomers. only in but`-2-`ene each doubly-bonded carbon has two different ligands |
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| 45. |
What type of isomerism is possible for `1-`chloro`-2-`nitroethene?A. Optical isomerismB. `E//Z` isomerismC. Position isomerismD. Functional group isomerism |
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Answer» Correct Answer - B `underset(1-"Chloro-2-nitroethen")(CICH=CHNO_(2))` It exhibits `E//Z` (or geometrical) isomerism because each of the doubly bonded carbon atom is linked to different ligands. |
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| 46. |
Which fo the following stereosimers represent meso compounds?A. B. C. D. All of these |
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Answer» Correct Answer - D In `(1)` and `(2)`, a plane of symmetry cuts through the central carbon and the two horizontal ligands revealing that the top half of the molecule is the mirror image of the bottom half. It does not matter which ligands ate bonded to this carbon because they are within the symmetry plane. This is also true in `(3)` if `n =1` or an odd number. However, if `n` is an even number, the symmetry plane cuts through the central `C-C` bond. |
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