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2023 PEET Organic Chemistry

Recommended Post : 【Organic Chemistry】 PEET Organic Chemistry Solutions

1. Which pair of a compound’s structure and its IUPAC name is incorrectly matched?

⑴ Problem

⑵ Solution : ④

① Naming of Chain Alkenes and Alkanes, Naming of Alcohols : Since alcohols have the highest priority, they are named so that the alcohol gets the lower number.

② Naming of Cyclic Alkanes : When there is a complex substituent, the alphabetical order of the substituent names determines their priority over the functional group.

③ Naming of Bicycloalkanes : Number the bridgehead carbons as 1 and assign numbers following the longest path.

④ Naming of Alcohols : The main chain should have the most carbons, so it should be -oct- not -hept-.

⑤ Naming of Cyclic Alkenes : The two carbons forming the alkene should be numbers 1 and 2, and the direction determined so that substituent numbers are as low as possible.

2. Which comparison of the physicochemical properties of compounds is correct?

⑴ Problem

⑵ Solution : ①

① The angle between the C-N bonds in cyclic amines is smaller than in linear amines, increasing the vector sum of dipole moments

② The atomic radius of S in the 3rd period is larger than that of carbon in the 2nd period, so the ring strain in the right structure is smaller

③ The right structure has weaker bonds due to steric hindrance, leading to lower C-C bond dissociation energy compared to the left

④ Boiling points are influenced by molecular attractions, while melting points are more affected by lattice energy (≒ intermolecular overlap). Thus, cyclic hydrocarbons with alkenes have lower melting points due to decreased lattice energy, as seen with n-hexane and (E) hex-2-ene (ref)

⑤ The left structure has less steric hindrance, facilitating molecular interactions and leading to a higher boiling point

3. Which comparison of the reaction rate or selectivity of compounds in each reaction is incorrect?

⑴ Problem

⑵ Solution : ③

① Comparison of Leaving Group Reaction Rates : TfO- > TsO- > MsO- > I- > Br- > Cl- > H2O ≫ F- > OH- > OR- > NH2-. The right structure is similar to MsOMe, showing that TfO-, TsO-, MsO- are known as good leaving groups

② Solvent Conditions : H2O stabilizes the carbocation better due to more hydrogen bonds

③ [Friedel-Crafts Alkylation Reaction](https://jb243.github.io/pages/1370#:5.-,,-(FriedelCraftsalkylation) : Consider Cl-C-Ph-R → +C-Ph-R, then think of it as undergoing electrophilic addition to the aromatic ring. The position of the -R group does not host the carbocation, so -Cl only induces an electron-withdrawing effect without resonance stabilization due to electronegativity. Thus, the reaction rate decreases compared to when -H is present.

④ Halogen Addition Reaction involves the formation of a halonium ion intermediate, where Br- adds in a SN1-like manner to form dibromo alkane. This addition step is the rate-determining step (rds). The semi-carbocation is secondary on the left and tertiary on the right, making it more stable and faster to react.

⑤ With tert -BuOK, steric hindrance increases stereoselectivity, leading to nucleophilic attack on the less hindered side

4. Which pairing of structures for main products A and B is correct? (Note: Main products are obtained through appropriate separation and purification processes.)

⑴ Problem

⑵ Solution : ①

○ A. Ozonolysis. Splits the double bond and connects each end with a carbonyl group (=O). Considering the stereoarrangement, corresponds to structure A in ①, ②

○ B. Reaction Overview

○ The orientation of the hexagonal ring is determined by the alkenyl group

○ [Hydroboration-Oxidation](https://jb243.github.io/pages/1363#:23.-,,-(hydroborationoxidation) : anti -Markovnikov (i.e., addition of hydrogen to the less substituted side of the alkene), syn -addition

○ Reactions proceed faster with less substituted alkenes due to lower steric hindrance, starting with secondary alkenes

○ There’s little difference in steric hindrance depending on the direction of the reaction, but stereo-selectivity arises from the energy level differences of the products

5. When intending to synthesize the final main product B from starting material A through multi-step reactions, which reagents are most appropriate for steps ㈎ to ㈒? (Note: Main products are obtained through appropriate separation and purification processes.)

⑴ Problem

⑵ Solution : ③

○ ㈎ NBS, H2O results in [halohydrin](https://jb243.github.io/pages/1363#:13.-,% B0,-() formation

○ ㈏ PPh3 is used for ylide reagent preparation in Wittig reaction

○ ㈐ Alkane Method

○ ㈑ Alkane Acid-Base Reaction + S_N Reaction on Ketone Carbon

○ ㈒ Alkane_anti_-Addition Reaction

6. Which pairing of structures for main products A and B is correct? (Note: Main products are obtained through appropriate separation and purification processes.)

⑴ Problem

⑵ Solution : ②

○ A. Acid-Catalyzed Alkene Hydration follows the Markovnikov rule. H+ adds to allow carbocation formation at the most stabilized position by the alkene. Then tautomerization occurs. Thus, a structure with a ketone and alkene attached is required (②, ③, ④)

○ B. In competitions between alkene and alkane, the alkene reacts first (ref) ( ∵ the halonium ion intermediate formed by addition to the alkane is too unstable to proceed well)

7. Which compounds have the carbon marked with an asterisk (*) as a chiral center?

⑴ Problem

⑵ Solution : ⑤

① Since the carbon is on a plane of symmetry within the molecule, it is not a chiral center

② The carbon is a pivot that allows the epoxide to rotate like a top, hence it is not a chiral center

③ Since the carbon is on a plane of symmetry within the molecule (i.e., the plane containing two carbons attached to an alcohol), it is not a chiral center

④ The two carbons attached to an alcohol are symmetrical to each other, hence not chiral centers

8. Which of the following is the most appropriate structure for the final major product of each reaction? (Note, the major product is obtained through appropriate separation and purification processes.)

⑴ Problem

⑵ Solution : ③

① Addition of halogen to alkene + S_N reaction (cf. Et3N prefers S_N nucleophile)

② Hydrogen and i-Pr determine the stereo-orientation of the compound, with the orientation being less hindered the further away from i-Pr and above rather than below

③ The acid-base reaction occurs most rapidly, hence leading to an intramolecular S_N reaction quickly

④ After a carbocation forms on the side that can resonate with the ketone, Br- approaches considering steric hindrance

⑤ Two main products are possible, but the direction towards forming a conjugated alkene is more stable and thus the main product

9. Which of the following is not the most appropriate structure for the main product of each reaction? (Note, the main product is obtained through appropriate separation and purification processes.)

⑴ Problem

⑵ Solution : ②

① Since -NO₂ is the strongest EWG in the molecule, a reaction that generates a negative charge on the carbon substituted by -NO₂ is preferred. Thus, F- can decide on which of the two Cl positions to substitute

② While -NO₂ being the strongest EWG means a reaction generating a negative charge on the substituted carbon by -NO₂ is preferred, the substitution by F is correct, but an amine group is a stronger electron donor than an alkoxy group, hence the -NH-CH2-CH2-OH group moves as the substituent

③ Considering steric hindrance, the middle -Br or -F group will not be substituted

⑤ Two main considerations:

○ Multiple substitution effect: Given two benzene structures, with -NH2 (strong activating substituent), -Me (weak activating substituent), -Cl (weak deactivating substituent), -NO₂ (strong deactivating substituent) on one benzene and -OH (strong activating substituent), -NH2 (strong activating substituent), -NO₂ (strong deactivating substituent) on the other, the latter is more reactive, thus acting as the electron donor for the reaction

○ In the latter benzene ring, although -NH2 is a better electron donor than -OH, the strong EWG -NO₂ leads to carbocation formation near -NH2, hence -OH acts as the electron donor for the reaction

10. Which of the following correctly pairs the structure of the main products A and B? (Note, the main product is obtained through appropriate separation and purification processes.)

⑴ Problem

⑵ Solution : ④

○ A : TsOH acts as an acid. The acid-catalyzed epoxide opening reaction occurs in the least sterically hindered orientation

○ B : Stereoarray inversion + determination of stereo-orientation

11. The following describes the synthesis of main product B from reactant A, including the 〈reaction formula〉 and 〈experimental process〉. The most inappropriate description is:

⑴ Problem

⑵ Solution : ②

① Friedel-Craft acylation

② AlCl₃ acts as a catalyst. Heating oxolane-2,5-dione ( A ; 100 g/ml) and 4-oxo-4-phenylbutanoic acid ( B ; 178 g/ml) induces decarboxylation from esters or carboxylic acids

③ Since the boiling point of benzene is 80.1 ℃, benzene can be distilled from a mixture with water through fractional distillation

④ Adding a base, the carboxylic acid dissolves as carboxylate (i.e., R-CO2Na)

⑤ Yield = 0.28 mol / (35 g × 1 mol / 100 g) × 100 = 80 (%)

12. The following describes the halogenation reaction between reactants A and B. The correct explanations as per the options are:

⑴ Problem

⑵ Solution : ⑦

○ ㄱ. The reactivity ratio is the ratio of amount produced to the number of reaction possibilities. There are 6 primary hydrogens and 4 secondary hydrogens, hence the reactivity ratio is 70 / 4 : 30 / 6 = 17.5 : 5 = 3.5 : 1.

○ ㄴ. 26 / 10 = 2.6

○ ㄷ. As the actual production ratio shows, the radical intermediate becomes relatively more stable the further it forms from the EWG -Cl group, hence increasing the speed of formation

13. The following describes the synthesis process from starting material A to main products B ~ D. (Note, the main product is obtained through appropriate separation and purification processes.) The correct explanations as per the options are:

⑴ Problem

⑵ Solution : ①

○ ㄱ. Unsaturation refers to how many rings or double bonds need to be broken to convert a molecule into a linear structure

○ ㄴ. Hydroboration-oxidation considers steric hindrance to add -H and -OH groups in a syn-addition (here, 9-BBN approaches in a less hindered 5-membered ring orientation, determining the orientation of the -OH group). Followed by Pd/C hydrogenation reaction. The number of chiral centers (*) in B is 4

○ ㄷ. D is produced from the hydrogenation of an alkene, with the alkene being on the t-Bu side ( ∵ the hydrogenation reaction occurs on the less hindered side). C is produced from an E2 elimination reaction, with the alkene being on the opposite side to t-Bu

14. Which of the following options correctly identifies the structure of the final main product? (Note, the main product is obtained through appropriate separation and purification processes.)

⑴ Problem

⑵ Solution : ⑥

○ ㄱ. Initially, a halonium ion intermediate is formed, avoiding steric hindrance and causing stereoselectivity

○ ㄴ. Epoxidation (considering steric hindrance) → two S_N reactions

○ ㄷ. [Alkene halohydrin formation reaction](https://jb243.github.io/pages/1363#:13.-,%B0,-() (-CO2Me group forms a plane with the six-membered ring, thus reducing steric hindrance) → conducts an intramolecular S_N reaction

15. The compounds that are diastereomers of each other are:

⑴ Problem

⑵ Solution : ②

○ RS nomenclature and vector cross product

○ ㄱ. Enantiomers

○ ㄴ. Diastereomers

○ ㄷ. Nitrogen inversion does not consider the stereo arrangement of amines. The stereo arrangements of the two compounds are completely identical

16. The following are the structures of heterocyclic compounds A to C. Which of the following statements are correct?

⑴ Problem

⑵ Solution : ④

○ ㄱ. Aromaticity

○ ㄴ. The lone pair of electrons on the nitrogen at position 1 does not participate in resonance and thus does not act as a base. Conversely, the lone pair of electrons on the nitrogen at position 3 can act as a base, donating electrons to a proton.

○ ㄷ. The basicity order of 1H-imidazole (A), 1,3-oxazole (B), and 1,3-thiazole (C) is as follows. In the case of 1H-imidazole, the nitrogen at position 1 acts as a very good EDG, moving electrons to the nitrogen at position 3 through electron resonance effect, increasing its basicity. On the other hand, when comparing 1,3-oxazole and 1,3-thiazole, sulfur (S), having a larger atomic radius, contributes more to electron resonance effect than oxygen (O) (∵ lowering the positive charge density on the carbocation, making it less unstable), making 1,3-thiazole more basic than 1,3-oxazole.

17. The following are the structures of natural products A and B. Which of the following statements are correct?

⑴ Problem

⑵ Solution : ⑦

○ ㄱ. The C-C bonds connecting the segolene compounds in A can rotate freely, so the plane bisecting these bonds is a plane of symmetry.

○ ㄴ. A is not optically active, while B is optically active. Thus, they are diastereomers.

○ ㄷ. Correct explanation (However, the next number follows the CIP rule

18. Only select the structures of the main products that are correct from the options. (Note: The main products are obtained through appropriate separation and purification processes.)

⑴ Problem

⑵ Solution : ⑤

○ ㄱ. The addition of NBS, NIS can be considered similar to bromination in EAS. It’s a question of which position on the alkene the electrophile substitutes, as in the problem above, so that the carbocation intermediate can be stabilized by the electron-donating resonance effect of the amine group.

○ ㄴ. The -Br group must substitute where the carbocation intermediate is stabilized by the strong EDG, -OMe group.

○ ㄷ. The issue is which aromatic carbon in the pyrrole ring to substitute with an -I group. As in the problem above, it must proceed so that the carbocation intermediate can be stabilized by the electron-donating resonance effect of the amine group. If the -I group is substituted at a different position (as shown below), it cannot be assumed that the pyridine ring would break just to provide resonance stabilization.

19. Only select the structures of the main products that are correct from the options. (Note: The main products are obtained through appropriate separation and purification processes.)

⑴ Problem

⑵ Solution : ⑥

○ ㄱ. Alkene epoxidation

○ ㄴ. It is presumed that HClO4 attaches to the oxygen of the C=O bond, and two -OH groups attack the ketone carbon, removing one oxygen followed by a concerted reaction caused by heat.

○ ㄷ. Reverse reaction of tautomerization. A 1,2-methyl shift also occurs.

20. Only select the structures of the final main products that are correct from the options. (Note: The main products are obtained through appropriate separation and purification processes.)

⑴ Problem

⑵ Solution : ①

○ ㄱ. Carbon-carbon coupling reaction by Grignard reagent

○ ㄴ. Hydroboration-oxidation of alkenes. The -OH group substitutes on the side with less steric hindrance. If there were no substituents on the

hetero ring, the -OH group would be correct on the oxygen side, but since there is a very large substituent, the -OH group must substitute as far away from that substituent as possible.

○ㄷ. Alcohol protection reaction preferentially occurs with primary alcohols that have less steric hindrance.

Input: 2023.12.07 20:57

Revised: 2023.12.24 23:17