Identify the products of the reaction between 2KOH and H₂SO4- OH.
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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'
- IR Spectrum confirms that alcohol group is existed with the peak at 3400 cm⁻¹
- 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.
- Check with other spectroscopic properties,
- 3 signals of 13C NMR confirms the structure is symmetrical, δ 24.2, (-<u>C</u>H₂-CH₂-CH(CH₂-)-OH), δ 35.5 (-CH₂-<u>C</u>H₂-CH(CH₂-)-OH), δ 73.3 (-CH₂-CH₂-<u>C</u>H(CH₂-)-OH).
- 1H NMR confirms,
1.56 δ (4H, triplet) - (-C<u>H</u>₂-CH₂-CH-OH) ; triplet as coupling with 2 H,
1.78 δ (4H, multiplet) - (-CH₂-C<u>H</u>₂-CH-OH); multiplet as coupling with 2H of CH₂, 1 H of CH
3.24 δ (1H, quintet); - (-CH₂-CH₂-C<u>H</u>(CH₂-)-OH), coupling with4 H of 2 group of CH₂
3.58 δ (1H, singlet); - (-CH₂-CH₂-CH(CH₂-)-O<u>H</u>), hydrogen of alcohol group, not tend to coupling with other hydrogen
