See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material. The starting material is 1,2-dimethylcyclohexane-1,2-diol. From the wedge-dash notation, both OH groups are on wedge bonds and both methyl groups are on dash bonds at C1 and C2, making this the trans-1,2-diol (the two OH groups are on the same face, and the two methyl groups are on the same face — actually this is the cis-diol with respect to OH groups both being wedge). Looking carefully: both OH on wedge (same side), both CH3 on dash (same side) — this is actually a meso or racemic trans diol. Given the symmetrical substitution (C1 and C2 both bear OH and CH3), with both OH wedge and both CH3 dash, this is the compound where the two OH groups are cis to each other. Step 2: Reaction with HIO4 (periodic acid cleavage). HIO4 cleaves vicinal diols. The cyclohexane-1,2-diol is cleaved to give a dialdehyde (ring-opened). Since C1 and C2 each bear one OH and one CH3 (plus being part of the ring), cleavage of the C1-C2 bond gives a linear dialdehyde: OHC-C(CH3)(CH2CH2CH2CH2)-CHO, i.e., 2-methylheptanedial — wait, let me reconsider the structure. The ring has 6 carbons; C1 and C2 each have OH and CH3. Cleavage gives: CHO-C(CH3)-(CH2)4-C(CH3)-CHO, which is 2,7-dimethyloctanedial... No: the ring opening of cyclohexane-1,2-diol gives OHC-(CH2)4-CHO (hexanedial) for unsubstituted, but with methyl groups at C1 and C2: product A = 2-oxopentanedial... Actually: each carbon bearing OH becomes an aldehyde (since each C has one OH and one H... but here each C has OH, CH3, and two ring carbons). So each carbon with OH and no H becomes a ketone? No — C1 has: OH, CH3, and two ring C-C bonds. Upon cleavage, C1 becomes CH3-CO- (ketone end). So product A is: CH3CO-(CH2)4-COCH3, i.e., octane-2,7-dione. Step 3: Product A is octane-2,7-dione (CH3-CO-(CH2)4-CO-CH3). This is a symmetric diketone. Step 4: Reaction with excess LiAlH4 then H+ reduces both ketone groups to secondary alcohols: CH3-CH(OH)-(CH2)4-CH(OH)-CH3, which is octane-2,7-diol. Step 5: Count stereoisomers of octane-2,7-diol. The molecule has two stereocenters at C2 and C7, each bearing OH, CH3, H, and a chain. The molecule is symmetric (C2 and C7 are equivalent). The possible stereoisomers are: (2R,7R), (2S,7S), and (2R,7S)=(2S,7R) meso. So there are 3 stereoisomers: one pair of enantiomers (RR and SS) and one meso compound. Step 6: Why other options fail: (a) 2 ignores the meso form; (c) 4 would apply to a molecule with two non-equivalent stereocenters (2^2=4); (d) 5 is too many. The answer is 3 because of the internal symmetry creating a meso compound, reducing the total from 4 to 3. Therefore, the correct answer is B.