Write equations for the reactions that occur. Put conditions and catalysts over the arrow.
1.CH3COOH + CH3(CH2)7OH
2.C6H14 + KMNO4 + H
3. CH3H12 + KMNO4 + H
4. CH3CHOHCH3 + KMNO4 + H
5. (CH3)3COH + KMNO4 + H
6. (CH3)2C=O + KMNO4 + H
7. CH3COOH + KMNO4 + H
8. How could you distinguish CH3(CH2)8CH=CH(CH2)4COOH from CH3(CH2)4COOH ??? is it bc the first molecule has a double bond perhaps ?
Also, for the above equations I am unsure how to do the ones with potassium permanganate ?? PLEASE HELP and Thank you !!
1. CH3COOH + CH3(CH2)7OH
Equation: CH3COOH + CH3(CH2)7OH → CH3COO(CH2)7CH3 + H2O
Conditions/Catalysts: Acidic conditions (e.g., sulfuric acid) are often used as a catalyst.
2. C6H14 + KMNO4 + H
Equation: C6H14 + KMNO4 + H2O → C6H13CHO + KOH + MnO2
Conditions/Catalysts: Acidic conditions are required for this reaction, such as dilute sulfuric acid (H2SO4).
3. CH3H12 + KMNO4 + H
Equation: CH3H12 + KMNO4 + H2O → CH3CHO + KOH + MnO2
Conditions/Catalysts: Acidic conditions are required (e.g., dilute sulfuric acid) for this reaction.
4. CH3CHOHCH3 + KMNO4 + H
Equation: CH3CHOHCH3 + KMNO4 + H2O → CH3COCH3 + KOH + MnO2
Conditions/Catalysts: Acidic conditions, with a catalyst such as dilute sulfuric acid, are necessary.
5. (CH3)3COH + KMNO4 + H
Equation: (CH3)3COH + KMNO4 + H2O → (CH3)2CO + H2CO3 + KOH + MnO2
Conditions/Catalysts: Acidic conditions are required (e.g., dilute sulfuric acid) for this reaction.
6. (CH3)2C=O + KMNO4 + H
Equation: (CH3)2C=O + KMNO4 + H2O → (CH3)2CO + KOH + MnO2
Conditions/Catalysts: Acidic conditions are necessary, with a catalyst such as dilute sulfuric acid.
7. CH3COOH + KMNO4 + H
Equation: CH3COOH + KMNO4 + H2O → CH3COOH + KOH + MnO2
Conditions/Catalysts: Acidic conditions are required (e.g., dilute sulfuric acid) for this reaction.
8. To distinguish CH3(CH2)8CH=CH(CH2)4COOH from CH3(CH2)4COOH, you can use a simple chemical test. Bromine water test can be used to differentiate between saturated and unsaturated hydrocarbons. CH3(CH2)8CH=CH(CH2)4COOH being unsaturated, the carbon-carbon double bond reacts with bromine water, whereas CH3(CH2)4COOH being saturated, does not react. If CH3(CH2)8CH=CH(CH2)4COOH is present, a decolorization of bromine water will occur as the double bond is added to bromine. On the other hand, if CH3(CH2)4COOH is present, the bromine water remains unchanged.