This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the solutions to the questions from the image: Question 1: A carbocation is a species that contains a carbon atom bearing a positive charge. Carbocation Question 2: The product of a Friedel-Crafts acylation reaction is an acylbenzene (or aryl ketone). Acylbenzene Question 3: The Lewis structure for the tert-butyl radical (CH₃)₃C• is: [scale=0.8] (C) at (0,0) C; (CH3_top) at (0,1) CH_3; (CH3_left) at (-0.8,-0.5) CH_3; (CH3_right) at (0.8,-0.5) CH_3; (radical) at (0.3,0.3) ·; (C) -- (CH3_top); (C) -- (CH3_left); (C) -- (CH3_right); See diagram above Question 4: • Heterolysis (or heterolytic cleavage) is the unsymmetrical breaking of a covalent bond where one atom retains both electrons from the bond, forming a cation and an anion. • Homolysis (or homolytic cleavage) is the symmetrical breaking of a covalent bond where each atom retains one electron from the bond, forming two radicals. Question 5: A trivalent carbon atom that bears a positive charge is a carbocation. (Assuming the question refers to a carbocation based on common organic chemistry contexts). Positive charge Question 6: Markovnikov's Rule states that in the electrophilic addition of a protic acid (HX) to an alkene, the hydrogen atom adds to the carbon atom of the double bond that already has more hydrogen atoms, while the halogen (X) adds to the carbon atom with fewer hydrogen atoms. Question 7: a) CH₃CH₂CH(CH₃)₂ (2-methylbutane) [scale=0.8] (C1) at (0,0) CH_3; (C2) at (1,0) CH_2; (C3) at (2,0) CH; (C4) at (3,0) CH_3; (C5) at (2,-1) CH_3; (C1) -- (C2) -- (C3) -- (C4); (C3) -- (C5); at (0.5,0.3) 1^; at (1.5,0.3) 2^; at (2.5,0.3) 3^; at (2.5,-0.7) 1^; • CH₃ (C1) hydrogens: Primary (1°) • CH₂ (C2) hydrogens: Secondary (2°) • CH (C3) hydrogen: Tertiary (3°) • CH₃ (C4) hydrogens: Primary (1°) • CH₃ (C5) hydrogens: Primary (1°) b) (CH₃)₃CCH₂CH₃ (2,2-dimethylbutane) [scale=0.8] (C1) at (0,0) CH_3; (C2) at (1,0) C; (C3) at (2,0) CH_2; (C4) at (3,0) CH_3; (C5) at (1,1) CH_3; (C6) at (1,-1) CH_3; (C1) -- (C2) -- (C3) -- (C4); (C2) -- (C5); (C2) -- (C6); at (0.5,0.3) 1^; at (1.5,0.3) 4^; % Quaternary carbon, no hydrogens at (2.5,0.3) 2^; at (3.5,0.3) 1^; at (1.5,1.3) 1^; at (1.5,-1.3) 1^; • CH₃ (C1) hydrogens: Primary (1°) • C (C2) has no hydrogens (quaternary carbon) • CH₂ (C3) hydrogens: Secondary (2°) • CH₃ (C4) hydrogens: Primary (1°) • CH₃ (C5) hydrogens: Primary (1°) • CH₃ (C6) hydrogens: Primary (1°) c) (CH₃)₂CHCH(CH₃)CH₂CH₃ (2,3,4-trimethylpentane) [scale=0.8] (C1) at (0,0) CH_3; (C2) at (1,0) CH; (C3) at (2,0) CH; (C4) at (3,0) CH_2; (C5) at (4,0) CH_3; (C6) at (1,-1) CH_3; (C7) at (2,1) CH_3; (C1) -- (C2) -- (C3) -- (C4) -- (C5); (C2) -- (C6); (C3) -- (C7); at (0.5,0.3) 1^; at (1.5,0.3) 2^; % This is a secondary carbon, but the hydrogen is tertiary. Let's clarify. at (2.5,0.3) 3^; at (3.5,0.3) 2^; at (4.5,0.3) 1^; at (1.5,-0.7) 1^; at (2.5,1.3) 1^; Let's re-evaluate the structure for (c) (CH₃)₂CHCH(CH₃)CH₂CH₃. This is 2,3-dimethylpentane. [scale=0.8] (C1) at (0,0) CH_3; (C2) at (1,0) CH; (C3) at (2,0) CH; (C4) at (3,0) CH_2; (C5) at (4,0) CH_3; (C6) at (1,1) CH_3; % Methyl on C2 (C7) at (2,-1) CH_3; % Methyl on C3 (C1) -- (C2) -- (C3) -- (C4) -- (C5); (C2) -- (C6); (C3) -- (C7); at (0.5,0.3) 1^; at (1.5,0.3) 3^; % H on C2 is tertiary at (2.5,0.3) 3^; % H on C3 is tertiary at (3.5,0.3) 2^; at (4.5,0.3) 1^; at (1.5,1.3) 1^; at (2.5,-0.7) 1^; • CH₃ (C1) hydrogens: Primary (1°) • CH (C2) hydrogen: Tertiary (3°) • CH₃ (C6) hydrogens: Primary (1°) • CH (C3) hydrogen: Tertiary (3°) • CH₃ (C7) hydrogens: Primary (1°) • CH₂ (C4) hydrogens: Secondary (2°) • CH₃ (C5) hydrogens: Primary (1°) Question 8: Reaction of methylcyclopentane with bromine under irradiation (h) is a free radical halogenation. Bromine is selective for the most substituted carbon. The methyl group is on a tertiary carbon in methylcyclopentane. The product is 1-bromo-1-methylcyclopentane. [scale=0.8] % Methylcyclopentane (0,0) -- (1,0) -- (1.5,0.866) -- (0.5,1.732) -- (-0.5,0.866) -- (0,0); (0.5,1.732) -- (0.5,2.232) node[above] CH_3; at (2.5,1) + Br_2 h; % 1-bromo-1-methylcyclopentane [xshift=5cm] (0,0) -- (1,0) -- (1.5,0.866) -- (0.5,1.732) -- (-0.5,0.866) -- (0,0); (0.5,1.732) -- (0.5,2.232) node[above] CH_3; (0.5,1.732) -- (1,1.732) node[right] Br; 1-bromo-1-methylcyclopentane Question 9: To prepare cyclohexene from cyclohexane: Step 1: Free radical halogenation of cyclohexane with bromine and light (h) to form bromocyclohexane. [scale=0.8] % Cyclohexane (0,0) -- (1,0) -- (1.5,0.866) -- (1,1.732) -- (0,1.732) -- (-0.5,0.866) -- (0,0); at (2,0.866) + Br_2 h; % Bromocyclohexane [xshift=4cm] (0,0) -- (1,0) -- (1.5,0.866) -- (1,1.732) -- (0,1.732) -- (-0.5,0.866) -- (0,0); (0.5,1.732) -- (0.5,2.232) node[above] Br; Step 2: Dehydrohalogenation (E2 elimination) of bromocyclohexane using a strong base (e.g., potassium hydroxide in ethanol) to form cyclohexene. [scale=0.8] % Bromocyclohexane (0,0) -- (1,0) -- (1.5,0.866) -- (1,1.732) -- (0,1.732) -- (-0.5,0.866) -- (0,0); (0.5,1.732) -- (0.5,2.232) node[above] Br; at (2.5,0.866) KOH, EtOH, heat; % Cyclohexene [xshift=5cm] (0,0) -- (1,0); [double] (1,0) -- (1.5,0.866); (1.5,0.866) -- (1,1.732) -- (0,1.732) -- (-0.5,0.866) -- (0,0); Question 10: A nucleophile is a chemical species that donates an electron pair to an electrophile to form a chemical bond. Nucleophiles are typically electron-rich and can be neutral or negatively charged. Question 11: The relative stabilities of alkyl radicals from least stable to most stable are: Methyl radical < Primary radical < Secondary radical < Tertiary radical Methyl radical < Primary radical < Secondary radical < Tertiary radical Question 12: Assuming the question refers to (2S,3S)-2,3-dibromobutane, as "1,2-dibromobutane" with an (S) configuration is ambiguous for a compound with two chiral centers. The Fischer projection for (2S,3S)-2,3-dibromobutane is: [scale=0.8] at (0,1.5) CH_3; at (0,0.5) C; at (0,-0.5) C; at (0,-1.5) CH_3; (0,0.5) -- (0,-0.5); % Vertical bond (-0.5,0.5) -- (0.5,0.5); % Horizontal bond for C2 (-0.5,-0.5) -- (0.5,-0.5); % Horizontal bond for C3 at (-0.5,0.5) H; at (0.5,0.5) Br; % For (2S) at (-0.5,-0.5) H; at (0.5,-0.5) Br; % For (3S) See diagram above for (2S,3S)-2,3-dibromobutane Question 13: An asymmetric carbon (or chiral carbon) is a carbon atom that is bonded to four different groups. Question 14: The compound is CH₃-CH(Cl)-CH₂-CH₃, which is 2-chlorobutane. Step 1: Identify asymmetric centers. The carbon at position 2 (CH(Cl)) is bonded to four different groups: -CH₃, -H, -Cl, and -CH₂CH₃. Therefore, it is an asymmetric (chiral) center. The other carbons are not asymmetric. Number of asymmetric centers = 1. Step 2: Calculate the number of stereoisomers. For a molecule with n chiral centers and no meso compounds, the number of stereoisomers is 2^n. Number of stereoisomers = 2^1 = 2. 1 asymmetric center, 2 stereoisomers Question 15: The compound is 1,3-dimethylcyclohexane. 1,3-dimethylcyclohexane Section B: Answer All Questions Question 16 (a): Functional groups are specific groups of atoms within molecules that are responsible for the characteristic chemical reactions of those molecules. They determine the chemical properties and reactivity of organic compounds. Question 16 (b): Grignard reagents (RMgX) react with carbonyl compounds (aldehydes and ketones) to form alcohols. The type of alcohol (primary, secondary, or tertiary) depends on the starting carbonyl compound: • To synthesize a primary alcohol: React a Grignard reagent with formaldehyde (methanal, HCHO). [scale=0.8] (RMgX) at (0,0) RMgX; (HCHO) at (2,0) HCHO; at (1,0) +; at (3.5,0) 1. Ether, 2. H_3O^+; (RCH2OH) at (6,0) RCH_2OH; Example: CH₃MgBr + HCHO → CH₃CH₂OH (ethanol) • To synthesize a secondary alcohol: React a Grignard reagent with any aldehyde other than formaldehyde (R'CHO). [scale=0.8] (RMgX) at (0,0) RMgX; (R'CHO) at (2,0) R'CHO; at (1,0) +; at (3.5,0) 1. Ether, 2. H_3O^+; (RCHOHR') at (6,0) RCH(OH)R'; Example: CH₃MgBr + CH₃CHO → CH₃CH(OH)CH₃ (2-propanol) • To synthesize a tertiary alcohol: React a Grignard reagent with a ketone (R'COR''). [scale=0.8] (RMgX) at (0,0) RMgX; (R'COR'') at (2,0) R'COR''; at (1,0) +; at (3.5,0) 1. Ether, 2. H_3O^+; (RC(OH)R'R'') at (6,0) RC(OH)R'R''; Example: CH₃MgBr + (CH₃)₂CO → (CH₃)₃COH (2-methyl-2-propanol) Got more? Send 'em 📸