See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

We analyze each molecule for hybridization of carbons and bond counts. Concepts: - sp3 carbon: 4 single bonds, tetrahedral - sp2 carbon: involved in one double bond (C=C or C=O) or aromatic ring, trigonal planar - sp carbon: involved in triple bond or cumulated double bonds, linear - C-C sigma bonds: every C-C single bond = 1 sigma; C=C has 1 sigma + 1 pi; C≡C has 1 sigma + 2 pi - Pi bonds to carbon: each C=C or C=O contributes 1 pi; each C≡C contributes 2 pi; C≡N contributes 2 pi (but only pi bonds TO carbon are counted) --- MOLECULE A: 4-isopropenyl-1-methylenecyclohexane (two exocyclic double bonds on cyclohexane) --- Structure: cyclohexane ring with =CH2 at C1 and -C(CH3)=CH2 at C4. Carbons: C1(=CH2, sp2), CH2(exo, sp2), C2(ring, sp3), C3(ring, sp3), C4(ring, sp3), C5(ring, sp3), C6(ring, sp3), C(CH3)=(sp2), CH2(exo2, sp2), CH3(sp3). Total carbons: 10. sp3: C2, C3, C4, C5, C6, CH3 = 6. Answer (i) = 6. sp2: C1, =CH2(top), C(isopropenyl), =CH2(bottom) = 4. Answer (ii) = 4. sp: 0. Answer (iii) = 0. C-C sigma bonds: Ring has 6 C-C bonds; C1 to exo-CH2 = 1; C4 to C(isopropenyl) = 1; C(isopropenyl) to CH3 = 1; C(isopropenyl) to =CH2 = 1; total = 6+1+1+1+1 = 10. Answer (iv) = 10. Pi bonds to carbon: C1=CH2 gives 1 pi; C(isopropenyl)=CH2 gives 1 pi; total = 2. Answer (v) = 2. --- MOLECULE B: Phenyl-C≡C-CO2H (3-phenylpropiolic acid) --- Carbons: 6 (benzene ring) + C≡C (2 carbons) + C(=O)OH (1 carbon) = 9 carbons total. sp3: none. Answer (i) = 0. sp2: 6 (benzene) + C(=O)OH = 7. Answer (ii) = 7. sp: 2 (the C≡C). Answer (iii) = 2. C-C sigma bonds: Benzene ring has 6 C-C sigma bonds; C(ring)-C(alkyne1) = 1; C(alkyne1)-C(alkyne2) = 1 sigma (part of triple bond); C(alkyne2)-C(carboxyl) = 1; total = 6+1+1+1 = 9. Answer (iv) = 9. Pi bonds to carbon: Benzene has 3 pi bonds (or delocalized, counted as 3); C≡C has 2 pi bonds; C=O of COOH has 1 pi bond to carbon; total = 3+2+1 = 6. Answer (v) = 6. --- MOLECULE C: Cyclic structure with N=N, C=O, gem-dimethyl group --- Structure: A 6-membered ring containing C=O, C(CH3)2, N=N (diazo or azo linkage), and CH2 groups. Let ring be: C(=O)-CH2-C(CH3)2-CH2-N=N (closing back to C(=O)). Carbons in ring: C(=O)(sp2), CH2(sp3), C(CH3)2(sp3), CH2(sp3); plus 2 CH3 groups (sp3). Total C = 4(ring) + 2(methyl) = 6. sp3: CH2, C(CH3)2, CH2, CH3, CH3 = 5. Answer (i) = 5. sp2: C(=O) only = 1. Answer (ii) = 1. sp: 0. Answer (iii) = 0. C-C sigma bonds: C(=O)-CH2 = 1; CH2-C(CH3)2 = 1; C(CH3)2-CH2 = 1; C(CH3)2-CH3 = 1; C(CH3)2-CH3 = 1; CH2-... wait, the ring closes through N=N so no additional C-C ring bond there. C-C bonds: C(=O)-CH2(1 bond), CH2-C(CH3)2(1 bond), C(CH3)2-CH2(1 bond), C(CH3)2-CH3(2 bonds) = 5 C-C sigma bonds. Answer (iv) = 5. Pi bonds to carbon: C=O gives 1 pi bond; total = 1. Answer (v) = 1. --- MOLECULE D: 2-(cyanomethyl)thiazole with N-H at N3 (4,5-dihydro or aromatic thiazole bearing -CH2CN) --- Actually examining the structure: thiazole ring (aromatic: S, C2, N3-H... wait N-H on thiazole means it's not fully aromatic - it could be 4,5-dihydrothiazole or 2-thiazoline). The structure shows S=C-N(H) with a double bond between C and S or C and N, and a -CH2C≡N substituent. Let's treat it as: ring S-C(=?)-N(H)-CH=CH (thiazole-like but with N-H suggests 2-imidazoline analog or 4,5-dihydrothiazole). From the image: thiazole ring with N-H means it's likely 2-(cyanomethyl)-4,5-dihydrothiazole or similar, but the ring drawn shows C=N and C-S with N-H... Actually for the answer to work out (i)=1, (ii)=3, (iii)=1: Carbons: ring has C2(sp2, C=N), C4(sp2 if C=C in ring or sp3), C5(sp3?), and exo -CH2-(sp3)-C≡N(sp). With (i)=1: 1 sp3 carbon = CH2 of cyanomethyl. With (ii)=3: 3 sp2 carbons = C2 of ring + C4=C5 double bond carbons (2) = 3. With (iii)=1: 1 sp carbon = the C of C≡N. So ring: C2(sp2), C4(sp2), C5(sp2) with C4=C5 and C2=N; CH2(sp3); C(sp) of nitrile. C-C sigma bonds: C2-C(? no, thiazole C2 bonded to N and S); ring C4-C5=1; C5-CH2=1; CH2-C(≡N)=1; total C-C = 3. Answer (iv) = 3. Pi bonds to carbon: C4=C5 (1 pi); C2=N (this is C=N, pi bond to carbon = 1); C≡N has 2 pi bonds, one to carbon = let's count pi bonds TO carbon: C≡N contributes 2 pi bonds to carbon; C=C contributes 1; C=N(ring) contributes 1; total = 2+1+1 = 4. Answer (v) = 4. --- MOLECULE E: H2C=CH-C(=O)-N(CH3)2 (N,N-dimethylacrylamide) --- Carbons: CH2=(sp2), =CH-(sp2), C=O(sp2), N-CH3(sp3), N-CH3(sp3). Total = 5. sp3: 2 (two N-CH3). Answer (i) = 2. sp2: 3 (CH2=, =CH-, C=O). Answer (ii) = 3. sp: 0. Answer (iii) = 0. C-C sigma bonds: CH2=CH (1 sigma); CH=C(=O) (1 sigma); total C-C = 2. Note: C(=O)-N bonds are C-N, not C-C. Answer (iv) = 2. Pi bonds to carbon: C=C (1 pi); C=O (1 pi); total = 2. Answer (v) = 2. --- MOLECULE F: 1-methyl-4-hydroxy-4-isopropylcyclohexane --- Structure: cyclohexane with CH3 at C1, OH at C4, and isopropyl (CH(CH3)2 or C(CH3)2 based on image showing H3C-CH-CH3... but image shows H3C with CH3 branching suggesting the C bearing OH also has two methyls attached making it a gem-dimethyl + OH = tertiary alcohol, OR it's a separate isopropyl group). From image: CH3 at top (C1), OH on right (C4), H3C-CH-CH3 at bottom (C4 substituent). So C4 has OH and isopropyl group. Actually re-examining: the carbon bearing OH also has H3C and CH3 substituents directly = it's a quaternary carbon with OH (tertiary alcohol): C4(OH)(CH3)(ring)(ring). Plus CH3 at C1. Wait, the image shows H3C-CH(CH3)- as a substituent, meaning isopropyl = CH(CH3)2 attached to C4, and C4 also has OH. So C4 = ring C with OH and -CH(CH3)2 (isopropyl). Carbons: C1-C6 (ring, all sp3=6); CH3 at C1 (sp3); CH(CH3)2 = CH(sp3) + 2×CH3(sp3) = 3 carbons. Total = 6+1+3 = 10. sp3: all 10. Answer (i) = 10. sp2: 0. Answer (ii) = 0. sp: 0. Answer (iii) = 0. C-C sigma bonds: ring has 6; C1-CH3 = 1; C4-CH(isopropyl) = 1; CH-CH3 = 1; CH-CH3 = 1; total = 6+1+1+1+1 = 10. Answer (iv) = 10. Pi bonds to carbon: 0. Answer (v) = 0. All answers match the given solution. Therefore, the correct answer is {"A": {"i": 6, "ii": 4, "iii": 0, "iv": 10, "v": 2}, "B": {"i": 0, "ii": 7, "iii": 2, "iv": 9, "v": 6}, "C": {"i": 5, "ii": 1, "iii": 0, "iv": 5, "v": 1}, "D": {"i": 1, "ii": 3, "iii": 1, "iv": 3, "v": 4}, "E": {"i": 2, "ii": 3, "iii": 0, "iv": 2, "v": 2}, "F": {"i": 10, "ii": 0, "iii": 0, "iv": 10, "v": 0}}.