Preparation for Your ACS Examination in Organic Chemistry Study Guide Solutions

Preparation for Your ACS Examination in Organic Chemistry Answer Solutions Chapter 1 | Nomenclature - 1. The structure of the compound contains six carbons in the longest continuous chain and functional group is alcohol. So, the parent chain is named hexanol. Hence, choices A and D are eliminated. Choice B and C, both of which have parent chain name is hexanol. Option C can be rejected because functional group alcohol should get the lowest priority. The structure of the compound contains 6-membered carbon chain and functional groups is alcohol at C3. So, the parent chain is named as 3-hexanol. Two methyl groups attached on C5. Hence, the name of the compound is 5,5-dimethyl-3-hexanol. Therefore, the correct choice is B. - 2. Hydroxyl group is not included in the choice (A). Hence, choice (A) is eliminated. The terms ortho, meta, and para are used for benzene derivatives possess for two substituents. The provided compound has three substituents on benzene ring. Hence, choice (B) is eliminated. Choice (D) can be rejected because functional group phenol should get the lowest priority. Draw the 4-bromo-3-nitrophenol, add nitro group on C3 and bromine on C4 to the phenol. Therefore, the correct choice is (C). - 3. Choices (C) and (D) can be rejected because those are 1,2-dichloro-3-pentene. Choice (A) and (B) both of which have name is 1,2-dichloro-2-penetene except geometry of an alkene. If the two high priority groups are on the same side of the double bond, then it is known as Z isomer. If the two high priority groups are on the opposite side of the double bond then it is known as E isomer. In the choice (B) two higher-ranking alkyl groups are on same side of the double bond. Hence, it is Z-isomer. In the choice (A) two higher-ranking alkyl groups are on opposite side of the double bond. Hence, it is E-isomer. Therefore, the correct choice is (A). - 4. The structure of the compound contains seven carbons in the longest continuous chain and functional group is ketone. So, the parent chain is named as heptanone. Choices (A) and (B) are eliminated because the position of keto group is not mentioned. Choice (C) and (D), both of which have parent chain name is heptanone. Choice (D) can be rejected because functional group ketone should get the lowest priority. Hence, the correct name is 5bromo-4-methyl-3-heptanone. Therefore, the correct choice is C. - 5. To draw the structure from the name of the compound follows the following rules. First find the parent name and draw that number of carbon atoms then choose the suffix to identify the functional group. Arbitrarily number the carbon atoms in the chain. Add the substituents to the appropriate carbon atoms. Finally re-draw with hydrogen atoms to make carbon atom have four bonds. To draw the structure 3,5-dimethyl-3-nonene, draw a nine carbon chain and add methyl groups on C3 and C5 and double bond between C3 and C4. Hence, choice (A) can be rejected. To draw the structure 3,5-dimethyl-3-nonene, draw a nine carbon chain and add methyl groups on C3 and C5 and double bond between C3 and C4. In Z isomer two groups are same side. Therefore, the structure of (Z)-3,5-dimethyl-3-nonene is as follows: Therefore, the correct choice is (C). - 6. Alkyl groups attached to the nitrogen atom of an amide or amine are treated with as substituents. To avoid confusion with alkyl groups attached to the main chain of an amide or amine, N must precede the name of each alkyl group attached to nitrogen. Hence, choices (B) and (C) are eliminated. Choice (D) also eliminated because benzyl group contains CH2 group on benzene ring. Thus, the name of the provided structure is N-(4-bromophenyl) ethanamide. Therefore, the correct choice is (A). - 7. The structure of the compound contains seven carbons in the longest continuous chain, double bond and functional group is alcohol. So, the parent chain is named as heptenol. Hence, choice (A) can be rejected. Choice (B), (C), and (D) all of them have parent chain name is heptenol. Option (B) can be rejected because functional group alcohol should get the lowest priority. Choice (C) and (D) both of which have name is 5-methyl-4-heptene-2ol except geometry of an alkene. If the two high priority groups are on the same side of the double bond, then it is known as Z isomer. If the two high priority groups are on the opposite side of the double bond then it is known as E isomer. In the provided compound, two higher-ranking alkyl groups are on same side of the double bond. Hence, it is Z-isomer. Therefore, the correct choice is (D). - 8. Alkyl groups attached to the nitrogen atom of an amide or amine are treated with as substituents. To avoid confusion with alkyl groups attached to the main chain of an amide or amine, N must precede the name of each alkyl group attached to nitrogen. To draw the N-phenyl, add phenyl group to N atom. Hence, the choices (A), (B), and (C) are rejected. The structure of N-phenyl-4-nitrobenzamide is as follows: Therefore, the correct choice is (D). - 9. Terms ortho, meta, and para are used for benzene derivatives possess for two substituents. The provided compound has three substituents on benzene ring. Hence, choice (A) can be rejected. The IUPAC name for the hydroxyl benzene is phenol. Hence, the choices (C) and (D) can be rejected because functional group is phenol. The provided compound is a hydroxyl benzene derivative and two methyl groups on C3 and C5. Hence, the IUPAC name of the compound is 3,5-dimethylphenol. Therefore, the correct choice is (B). - 10. The structure of the compound contains six carbons in the longest continuous chain, double bond and functional group is ketone. So, the parent chain is named as cyclohexenone. Hence, choice (B) can be rejected. Choices (A), (C) and (D), all of which have parent chain name are cyclohexenone. Choices (A) and (C) can be rejected because functional group ketone should get the lowest priority. The correct name of the provided compound is 3-metyhyl-6-methylethyl-2 cyclohexenone. Therefore, the correct choice is (D). - 11. A nitrogen atom bonded to the adjacent alkyl carbon is known as amines. An oxygen atom bonded to the adjacent carbonyl group is known as esters. A nitrogen atom bonded to adjacent carbonyl carbon is known as amides. Hence, choices (A) and (D) can be rejected. An oxygen atom attached with two alkyl groups is known as ethers. Hence, choice (C) can be rejected. The functional group inside the triangle is amide; rectangle is ketone, pentagonal is ester, and circle is ether. Therefore, the correct choice is (B). Chapter 2 | Structure, Hybridization, Resonance, Aromaticity - 1. The Lewis structure of A of the oxygen atom has only six electrons around it and oxygen atom not satisfies the octet rule. Hence, choice (A) can be rejected. The formal charge of an atom in a molecule or ion is defined as the charge it would have if the bonded electrons are equally shared by the bonded atoms. The formal charge of an atom either in a molecule or in an ion will be smaller. The charge separation between the atoms should be less. The total formal charge of a molecule will be the sum of the individual formal charges of all the atoms present and will be always zero. Similarly the total formal charge of an ion will be the sum of the individual formal charges of all the atoms present and will always be equal to the charge of the ion. The formal charge of each of the atoms in choice (B) is calculated in the table. The lowest energy Lewis structure is one that has negative charge on more electronegative atom. The formal charge of each of the atoms in choice (D) is calculated in the table. Oxygen is more electronegative atom so it cannot grab the positive charge. The Lewis structure (D) has more energy. Hence, choice (D) can be rejected. The formal charge of each of the atoms in choice (C) is calculated in the table. The lowest energy Lewis structure is one that has negative charge on more electronegative atom like fluorine and nitrogen. Therefore, the correct choice is (C). - 2. The carbon atoms of alkanes are sp3 hybridized. The normal tetrahedral bond angle of an sp3-hybridized atom is 109.5. Hence, choices (A), (B), and (D) can be rejected. The internal C-C-C bond angle in the cyclohexane is sp3 hybrid orbitals, which have tetrahedral symmetry. Therefore, the angles between bonds of a tetravalent carbon atom have a preferred value 109.5 which is very close of choice (C). Therefore, the correct choice is (C). - 3. The molecular formula of diazomethane is CH2N2. To form a bond required two electrons. Calculate the total number of valency electrons in diazomethane as follows: Lewis structure of (C) contains six bonds and one lone pair. Hence, it has total 14 electrons. Therefore, choice (C) can be rejected. Lewis structure of (D) contains six bonds and three lone pair. Hence, it has total 18 electrons. Therefore, choice (D) can be rejected. Lewis structure of (A) and (B) contains 16 valency electrons. To decide the acceptable Lewis structure, check the formal charges on each atom.The formal charge of an atom is calculated by using the following formula: To calculate the formal charge, count the number of valence electrons, lone pair electrons and shared bonding electrons. The formal charge of each of the atoms in Lewis structure of A is calculated in the table. Therefore, choice (A) can be rejected. To calculate the formal charge, count the number of valence electrons, lone pair electrons and shared bonding electrons. The formal charge of each of the atoms in Lewis structure of B is calculated in the table. Therefore, the correct choice is (B). - 4. The formal charge of an atom is calculated by using the following formula: To calculate the formal charge, count the number of valence electrons, lone pair electrons and shared bonding electrons. The formal charge of sulfur atom in Lewis structure of A is calculated in the table. The formal charge of sulfur atom in Lewis structure of C is calculated in the table. Hence choice (C) can be rejected. The formal charge of sulfur atom in Lewis structure of D is calculated in the table. Hence choice (D) can be rejected. The formal charge of sulfur atom in Lewis structure of B is calculated in the table. The formal charge on sulfur atom is +1 in the Lewis structure (B). Therefore, the correct choice is (B). - 5. Carbon atoms form four bonds, oxygen atom forms two bonds, nitrogen forms three bonds and hydrogen and halogens atoms each form single bonds. Choices (B) and (C) can be rejected because carbon does not form five bonds. The provided condensed formula has parentheses around the CH3 and CH2. The subscript are 3 and 2 . This means that carbon atom contains three CH3 groups and adjacent two CH2 groups. Hence, the choice (A) can be rejected. The structure for the condensed formula is as follows: (CH3)3C(CH2)2CH3. Therefore, the correct choice is (D) - 6. Hybridization: The process of mixing the atomic orbitals in an atom (usually the central atom) to generate a set of new atomic orbital is called hybridization. To determine the orbitals used in bonding, count the number of groups (atoms + lone pairs): The hybridization of the orbital containing the unshared pair of electron on nitrogen atom is sp2 because it has two bond pairs and one lone pair. Therefore, the correct choice is (C). - 7. Hybridization: The process of mixing the atomic orbitals in an atom (usually the central atom) to generate a set of new atomic orbital is called hybridization.To determine the orbitals used in bonding, count the number of groups (atoms + lone pairs): The hybridization of atom marked I is sp2 because it has two bond pairs. Hence, choices (C) and (D) can be rejected. The hybridization of atom marked II is sp3 because it has four bond pairs. Hence, choice (A) can be rejected. The hybridizations of the atoms marked I, II, III and IV in the molecule are shown the below diagram. Therefore, the correct choice is (B). - 8. The heat of combustion of cycloalkanes is tabulated as follows: The most stable compound produces less heat of combustion. From the heat of combustion table, the least heat of combustion produced by cyclopentadecane. Therefore, the correct choice is (D). - 9. Resonance structure represents the electron distribution in a molecule. Choices (A) and (B) can be rejected because electron distribution does not take place in those molecules. Resonance structure represents the electron distribution in a molecule. The following are the possible resonance structures for the compound C. Most stable compounds form more resonance structures. The following are the possible resonance structures for the compound D. Therefore, the correct choice is (D). - 10. Resonance structure represents the electron distribution in a molecule. The following are the possible resonance structures for the provided compound. From the resonance structures, the negative charge is located at 2, 4, 6, and 8. Therefore, the correct choice is (C). - 11. Resonance structures differ only in the position of non-bonded electrons and pi bonds whereas isomers differ in the arrangement of both the atoms and electrons. Choices (B), (C), and (D) can be rejected because in those pair of molecules are differ in the arrangement of atoms. Resonance structures differ only in the position of non-bonded electrons and pi bonds. To draw another resonance structure move pi bond electrons as follows: Therefore, the correct choice is (A) - 12. Lone pair on nitrogen atom in compound II combined with four pi electrons of the carbon-carbon - double bonds gives six pi electrons. Hence, choices (B) and (D) can be rejected. Lone pair on sulfur atom in compound III combined with four pi electrons of the carbon-carbon double bonds gives six pi electrons. Hence, choice (C) can be rejected. Lone pair on nitrogen atoms in compound I combined with four pi electrons of the carbon-carbon double bonds gives eight pi electrons. Therefore, the correct choice is (A). - 13. Structures with a maximum of octets are most important. Hence, choices (A) and (B) can be rejected because in structure (A) and (B) carbon atom attached with nitrogen has only six electrons. Charges should be preferentially located on atoms with compatible electronegativity. Oxygen is more electronegative than nitrogen atom. So, the most contribute structure must have a negative charge on oxygen atom. Hence, choice (D) can be rejected. The most contribute structure is as follows: Therefore, the correct choice is (C). - 14. Aromatic compound contains (4n+2)pi electrons. Compound I contains four pi electrons so it is anti-aromatic. Hence, choices (A) and (C) can be rejected. Compound II contains ten pi electrons (electrons) so it is an aromatic. Both choices (B) and (D) saying that compound II is an aromatic compound. To decide the correct choice, find the nature of the compound III. Compound III contains ten electrons ((4n+2)pi electrons) so it is an aromatic. Therefore, the correct choice is (B). - 15. Lone pair on nitrogen atom in compound B combined with four pi electrons of the carbon-carbon double bonds gives six pi electrons. Aromatic compound contains (4n+2)pi electrons. Hence, choice (B) is an aromatic compound. Aromatic compound contains (4n+2)pi electrons. Compound C contains six pi electrons so it is an aromatic compound. Hence, choice (C) can be rejected. Aromatic compound contains (4n+2)pi electrons. Compound D contains two pi electrons so it is an aromatic compound. Hence, choice (D) can be rejected. Compound A contains four pi electrons so it is not an aromatic compound. Therefore, the correct choice is (A). - 16. The order of stability of carbocation is as follows: Benzylic carbocation > allylic carbocation > tertiary carbocation > secondary carbocation > primary carbocation > methyl carbocation Choice (B) is primary carbocation. Choice (D) is secondary carbocation. Choices (A) and (C) are benzylic carbocations. Hence, choices (B) and (D) can be rejected because benzylic carbocation is more stable than primary and secondary carbocations. Presence of alkyl groups increases the stability of carbocation. Hence, choice (C) can be rejected. The most stable carbocation is choice (A). Therefore, the correct choice is (A). Chapter 3 | Acids and Bases - 1. Solutions are identified as acidic, neutral or basic according to the pH values: Amino acids contain both an acidic group (—COOH) and a basic group (—NH2). These two groups can undergo an Intramolecular acid-base reaction, a reaction within the amino acid itself. The result is a loss of the hydrogen ion from the —COOH group and a gain of a hydrogen ion to the —NH2 group. This H+ transfer within the amino acid forms a dipolar ion, an ion that has one positive charge and one negative charge and is thus electrically neutral because the sum of the charges on the molecule is zero. Thus, the structure of provided compound at pH is as follows: Therefore, the correct choice is (C). - 2. Aromatic amines are much weaker bases than alkyl amine. We can account for this effect, on the basis of resonance contributions to the overall hybrid of an aryl amine. Choices (B) , ( C) , and (D) can be rejected because those are aryl amines. Choice (A) is a alkyl amine. Therefore, the correct choice is (A). - 3. Electron withdrawing groups decreases the basic nature of negative species. Due to presence of electron withdrawing group Cl, choices (B), (C), and (D) can rejected. Due to lack of electron withdrawing group, compound A is strongest base among the provided species. Therefore, the correct choice is (A). - 4. Protonated alcohols easily donate its proton. Hence, protonated alcohols are more acidic than protonated amines. Thus, choices (B) and (C) can be rejected. The acidity of H–Y increases across a row and down a column of the periodic table. Oxygen is farther to the right in the periodic table than nitrogen. Hence, methanol is stronger acid than methyl amine. Hence, the order from the strongest acid to the weakest acid is as follows: Therefore, the correct choice is (A). - 5. The electron-withdrawing power of the two carbonyl groups increases the acidity of the active methylene group in compound II. So, the most acidic compound is II. Hence, choices (A) and (D) can be rejected. The electronwithdrawing power of the carbonyl group of ester decreases by the adjacent mesomeric effect of oxygen lone pair of electrons. Thus, acidity of the ketones is more than esters. Hence, the order from the most acidic to the least acidic is as follows: Therefore, the correct choice is (C). - 6. The electron-withdrawing power of the nitro-group increases the acidity of p-nitro-phenol than phenol. The best resonance form after lose of proton from p-nitro-phenol is as follows: Hence, choices (A) ( B) and (D) can be rejected. Due to formation of resonance form after lose of proton from p-nitro-phenol; it is more acidic than phenol. Therefore, the correct choice is (C). - 7. Proton A attached with oxygen whereas proton B, C and D attached with C. The acidity of H–Y increases across a row of the periodic table. Therefore, the most acidic proton is A. Therefore, the correct choice is (A). - 8. The sp3 hybridized carbon atom of hydrogen has lower percent s-character than sp2 hybridized carbon atom of hydrogen. Hence, the least acidic hydrogen is labeled (B). Strong acid has lower pKa value. The sp hybridized carbon atom of hydrogen has higher percent s-character hence it is most acidic. Therefore, the correct choice is (C). - 9. The nitrogen labeled (2) is least basic nature because delocalized electron pair on nitrogen for aromatic nature. Nitrogen labeled (1) is more basic due to the formation of negative charge on nitrogen. Therefore, the correct choice is (B). - 10. The electron-withdrawing power of the oxygen atom increases the acidity of compound. Thus, increases number of oxygen atoms increases its acidic nature of compound. So, the most acidic compound is II. Hence, choices (A), (C) and (D) can be rejected. Compound II is strongest acid because it contains more oxygen atoms than other. Hence, the order of acidic from strongest to weakest is as follows: Therefore, the correct choice is (B). Chapter 4 | Stereoisomerism - 1. To assign R or S to the molecule, first rank the groups. The lowest priority group must be present at below the plane. If drawing a circle from (1) (2) (3) proceeds in the clockwise direction, the stereogenic center is labeled R; if the circle is counterclockwise, it is labeled S. (A)First rank the groups using Cahn-ingold-Prelog system as follows: Trace a circle (1) (2) (3) proceeds in the counterclockwise. Therefore, the configuration at stereogenic center is S. (B) First rank the groups using Cahn-ingold-Prelog system as follows: Trace a circle (1) (2) (3) proceeds in the counterclockwise. Therefore, the configuration at stereogenic center is S. (C) First rank the groups using Cahningold-Prelog system as follows: Trace a circle (1) (2) (3) proceeds in the clockwise. Therefore, the configuration at stereogenic center is R. (D) First rank the groups using Cahn-ingold-Prelog system as follows: Trace a circle (1) (2) (3) proceeds in the counterclockwise. Therefore, the configuration at stereogenic center is S. Molecule (C) has the R configuration. Therefore, the correct choice is (C). - 2. To assign R or S to the molecule, first rank the groups. The lowest priority group must be present at below the plane. If drawing a circle from (1) (2) (3) proceeds in the clockwise direction, the stereogenic center is labeled R; if the circle is counterclockwise, it is labeled S. First rank the groups using Cahn-ingold-Prelog system as follows: Trace a circle (1) (2) (3) proceeds in the clockwise in the right carbon. Hence, its configuration of stereogenic center is R. Trace a circle (1) (2) (3) proceeds in the counterclockwise in the left carbon. The least group at above the plane present is in the left carbon. So the configuration at left stereogenic center is R. Hence, left carbon and right carbons has R configuration. Therefore, the correct choice is (A). - 3. The wedge-dash form the provided Newmann projection is as follows: To assign R or S to the molecule, first rank the groups. The lowest priority group must be present at below the plane. If drawing a circle from (1) (2) (3) proceeds in the clockwise direction, the stereogenic center is labeled R; if the circle is counterclockwise, it is labeled S. First rank the groups using Cahn-ingold-Prelog system as follows: Trace a circle (1) (2) (3) proceeds in the counterclockwise in the right carbon. The least group is not present at below the plane. So the configuration at stereogenic center is R. The configuration at chiral carbon of compounds (A), (B), and (C) are R-configuration. Stereoisomers of a chiral substance that have a mirror image relationship. Enantiomers have opposite configurations at all chirality centers. Provided Newmann projection and compound D are enantiomers. Therefore, the correct choice is (D). - 4. The potential energy vs angle of rotation for one complete rotation of compounds A, C, and D shows different peaks of unequal heights and depths due to different gauche and eclipsed interactions of different atoms. Hence, choices (A) , (C) and (D) can be rejected. The potential energy vs angle of rotation for one complete rotation of compound B shows different peaks of equal heights and depths due to different gauche and eclipsed interactions of different atoms. Therefore, the correct choice is (B). - 5. To exhibit meso, compound must have at least two chiral carbon atoms. Due to lack of chiral carbon atoms in compound I, choices (A) and (D) can be rejected. Meso compound cuts into two equal parts. Due to lack of plane of symmetry in the compound III, it does not exist as meso compound. Meso compound is achiral due to exhibit the plane of symmetry. The plane of symmetry for the compound II is as follows: Therefore, the correct choice is (B). - 6. The most stable conformations have two methyl groups anti to each other. Choices (A), (B) , and (D) has only one anti conformation between two methyl groups. Choice (C) has two anti conformations. Therefore, the correct choice is (C). - 7. To assign R or S to the molecule, first rank the groups. The lowest priority group must be present at vertical position. If drawing a circle from (1) (2) (3) proceeds in the clockwise direction, the stereogenic center is labeled R; if the circle is counterclockwise, it is labeled S. (A) First rank the groups using Cahn-ingold-Prelog system as follows: Trace a circle (1) (2) (3) proceeds in the counterclockwise. The least group is present at horizontal position. Therefore, the configuration at each stereogenic center is R. The configuration of the two stereo centers is R. Therefore, the correct choice is (A). - 8. Bothe the compounds have same molecular formula and relationship between groups / atoms are different. Hence, pair of compounds is not identical. Hence, choice (A) can be rejected. Both compounds have same molecular formula C4H8O but not different connectivity of substituents. Hence, the relationship between A and B is not constitutional isomers. Hence, choice (B) can be rejected. Non-mirror image stereoisomers are diastereomers. They have the same configuration at one or more chirality centers but differ at other chirality centers. Hence, choice ( D) can be rejected. Stereoisomers of a chiral substance that have a mirror image relationship. Enantiomers have opposite configurations at all chirality centers. Compound A and B are enantiomers. Therefore, the correct choice is (C) - 9. Treatment of 2-butene with HBr gives (R)-2-bromobutane and (S)-2-bromobutane. Thus, the reaction is as follows: The mechanism of the provided reaction is as follows: Addition of HBr to the double bond forms a secondary carbocation. Now, the bromide ion (nucleophile) attacks the either side of carbocation to form the recemate. Therefore, the correct choice is (D) - 10. The most stable chair conformation has less 1,3-diaxial interactions. 1,3-diaxial interactions are shown in the compounds (A), (C) and (D) as follows: Due to lack of 1,3-diaxial interactions in compound (B) it is most stable. Therefore, the correct choice is (B). - 11. 1-methylcyclohexene reacts with HCl to give 1-chloro-1-methylcyclohexane. To exhibit meso or diastereomers, compound must have at least two chiral carbon atoms. Due to lack of chiral carbon atoms in1-chloro-1methylcyclohexane, choices (B) and (C) can be rejected. To form a recemate, product must have one chiral carbon but 1-chloro-1-methylcyclohexane does not contain any chiral carbon atom. Hence, choice (D) can be rejected. 1methylcyclohexene reacts with HCl to give 1-chloro-1-methylcyclohexane. Due to lack of chiral carbon in the product is a achiral but not meso. Therefore, the correct choice is (A). - 12. Bothe the compounds have same molecular formula and relationship between groups / atoms are different. Hence, pair of compounds is not identical. To exhibit the meso compound it should process the plane of symmetry. Due to lack of plane of symmetry in the provided compounds rejected the choice (B). The c Non-mirror image stereoisomers are diastereomers. They have the same configuration at one or more chirality centers but differ at other chirality centers. Hence, choice (C) can be rejected. Stereoisomers of a chiral substance that have a mirror image relationship. Enantiomers have opposite configurations at all chirality centers. Thus, the provided pair is enantiomers. Therefore, the correct choice is (D). - 13. The most stable conformations have two groups are anti to each other. Conformer II is most stable due to Intramolecular hydrogen bonding between gauche interactions of hydroxyl groups and methyl and hydroxyl groups are anti to each other. Hence, choices (A), (B) , and (C) can be rejected. Conformer II is most stable due to Intramolecular hydrogen bonding between as shown in below. - 14. The maximum number of stereoisomers is 2” where n is the number of stereogenic centers. Compound 1,2dimethylcyclopropane has two stereogenic centers. Hence, the maximum number of stereoisomers is four. Hence, choices (C) and (D) can be rejected. The possible stereo isomers for 1,2-dimethylcyclopropane are as follows: Compound 3 and 4 are identical. Hence, The maximum number of stereoisomers for the 1,2- dimethylcyclopropane is 3. Therefore, the correct option is (A) - 15. Incase identical both compounds have same configuration at stereogenic centre. Hence, choice (A) can be rejected. Meso compounds exhibits plane of symmetry. (1S, 2S)-1, 2-cyclohexanediol does not contains plane of symmetry. Hence, choice (C) can be rejected. Racemates contains equal mixture of enantiomers. Hence, choice (D) can be rejected. Compounds (1S, 2S)-1, 2-cyclohexanediol and (1R, 2S)-1, 2-cyclohexanediol are diastereomers because configuration at one stereogenic centre is same while it is different at other stereogenic centre. Therefore, the correct choice is (B). - 16. To assign R or S to the molecule, first rank the groups. The lowest priority group must be present at below the plane. If drawing a circle from (1) (2) (3) proceeds in the clockwise direction, the stereogenic center is labeled R; if the circle is counterclockwise, it is labeled S. There is no relationship between plane polarizing light and configuration of chiral carbon. Hence, choice (A) can be rejected. One: one ratio mixture of (+) and (-) enantiomers shows no rotation and is therefore optically inactive. Such a mixture is called a racemic mixture. Hence, choice (C) can be rejected. Meso compound has two chiral carbons. Optical rotation of meso compounds is zero due to cancel of each rotation of chiral carbon. Hence, choice (D) can be rejected. The statement (B) is correct. Therefore, the correct choice is (B). - 17. A tetrahedral stereogenic center has a carbon bonded to four different groups. Provided compound has one stereogenic center, stereogenic center is as shown below: The provided compound reacts with phenyllithium to give (1R,3S)-3-methyl-1-phenylpentan-1-ol, which has two stereogenic centers. Therefore, the correct choice is (C). - 18. Two stereoisomers that are related to each other as image-nonsuperimposable mirror images are called enantiomers. A compound containing of stereocenters is chiral and exists as a pair of enantiomers. A 1:1 mixture of enantiomers is racemate. Enantiomers and racemate have different physical properties like melting points. Hence, choice (B) can be rejected. A compound containing of stereocenters is chiral and exists as a pair of enantiomers. A 1:1 mixture of enantiomers is racemate. A molecule containing stereocenters and a mirror plane is identical with its mirror image is called meso compound. Enantiomers and meso compounds have different physical properties like melting points. Hence, choices (C) and (D) can be rejected. Two stereoisomers that are related to each other as image-nonsuperimposable mirror images are called enantiomers. Most of the physical properties of enantiomers are the same. A major exception is their interaction with plane-polarized light. Therefore, the correct choice is (A). - 19. Two stereoisomers that are related to each other as image-nonsuperimposable mirror images are called enantiomers. A compound containing of stereocenters is chiral and exists as a pair of enantiomers. A 1:1 mixture of enantiomers is racemate. Enanotimer and racemate are diastereomers because configuration at one stereogenic centre is same while it is different at other stereogenic centre. Therefore, the correct choice is (B). - 20. Cyclohexene reacts with bromine to give racemate of 1,2-dibromocyclohexane through the formation of bromonium ion intermediate. The most stable 1,2-dibromocyclohexane having two bromine atoms at equatorial because lack of 1,3-diaxial H and Br interactions. Therefore, the correct choice is (A). - 21. Identical molecules are having the same connectivity of atoms. Hence, choice (A) can be rejected. Stereoisomers are different in the arrangement of atoms. Hence, choice (D) can be rejected. Conformational isomers arise by the rotation of carbon – carbon bond. Hence, choice (B) can be rejected. Both compounds have same molecular formula C6H12O2 but different connectivity of substituents. Hence, the relationship between the pair is constitutional isomers. Therefore, the correct choice is (C). - 22. If the two groups are on the same side or either above or below the plane, then it is known as cis isomer. If the two groups are on the opposite side or one is above the plane other one is below the plane it is known as trans isomer. Cycloalkanes and alkenes exhibit cis-trans isomerism. Alkynes and allenes do not show the cis- trans isomerism. Therefore, the correct choice is (B). - 23. To exhibits optical active (chiral) compound must have chiral carbon. Chiral carbon contains four different atoms or groups. Due to lack of chiral carbon atoms in compound I and II, those are not optical active compounds. Hence, choices (A), (B) and (D) can be rejected. A molecule to be chiral it must not be super imposable on its mirror image. The mirror image of allene does not super imposable. Hence, allene is a chiral molecule. Therefore, the correct choice is (C). - 24. Hydrogenation of the double bond with hydrogen in the presence of Pt is syn addition. In syn addition both the hydrogen atoms attaches same side of the double bond. Hydrogenation of compound B, C and D does not produce any chiral carbon atoms. Hence, choices (B), (C) and (D) can be rejected. A compound containing of stereocenters is chiral and exists as a pair of enantiomers. A 1:1 mixture of enantiomers is racemate. Hydrogenation of compound A produces a racemate. The chemical reaction is as follows: Therefore, the correct choice is (A). - 25. To exhibits the meso compound, it must have at least two chiral carbon atoms. Reactions of A and C produces one chiral carbon containing products. Hence, choices (A) and (C) can be rejected. Alkenes react with bromine to give trans-1,2-dibromoalkane. Thus, cyclobutene reacts with bromide in the presence of carbon tetrachloride to give trans-1,2-dibromocyclobutane. Meso compound bisects the molecule in to two halves. Hence, choice (B) can be rejected. Alkenes react with D2 to give cis-1,2-dideutueroalkane. Thus, cyclobutene reacts with D2in the presence of Pt to give cis-1,2-dideutueroalkane, which bisects the molecule in to two halves. Therefore, the correct choice is (D). - 26. To assign R or S to the molecule, first rank the groups. The lowest priority group must be present at below the plane. If drawing a circle from (1) (2) (3) proceeds in the clockwise direction, the stereogenic center is labeled R; if the circle is counterclockwise, it is labeled S.