Question

Ketones undergo a reduction when treated with sodium borohydride, NaBH4. The product of the above reaction has the following spectroscopic properties; propose a structure. MS: M+ = 86 IR: 3400 cm-1 1H NMR: 1.56 δ (4H, triplet); 1.78 δ (4H, multiplet); 3.24 δ (1H, quintet); 3.58 δ (1H, singlet) 13C NMR: 24.2, 35.5, 73.3 δ

Answer 1

Answer:

The product is cyclohexanol

Explanation:

Firstly,

A ketone undergo a borohydride reduction reaction to form an alcohol as below,

R-CO-R'  ⇒ R-CO(OH)-R'

1. IR Spectrum confirms that alcohol group is existed with the peak at 3400 cm⁻¹
2. From 1H-NMR, the product has 10 hydrogen atoms, the MS suggest that the formula is C₅H₁₀O (M = 86). With this formula, the alcohol is monosaturated. Since, the substance already underwent reduction reaction, the only way to suggest a monosaturated compound is a cyclic alcohol. So the compound is cyclopentanol.
3. Check with other spectroscopic properties,

• 3 signals of 13C NMR confirms the structure is symmetrical, δ 24.2, (-CH₂-CH₂-CH(CH₂-)-OH), δ 35.5 (-CH₂-CH₂-CH(CH₂-)-OH), δ 73.3 (-CH₂-CH₂-CH(CH₂-)-OH).

• 1H NMR confirms,

        1.56 δ (4H, triplet) - (-CH₂-CH₂-CH-OH) ; triplet as coupling with 2 H,

        1.78 δ (4H, multiplet)  - (-CH₂-CH₂-CH-OH); multiplet as coupling with 2H of CH₂, 1 H of CH

         3.24 δ (1H, quintet); - (-CH₂-CH₂-CH(CH₂-)-OH), coupling with4 H of 2 group of CH₂

         3.58 δ (1H, singlet); - (-CH₂-CH₂-CH(CH₂-)-OH), hydrogen of alcohol group, not tend to coupling with other hydrogen