Dehydrate a primary alcohol containing 7 carbons.
General instructions:
1. Write out the reaction described. Include reactant(s), catalysts (you pick if there’s a choice), and product(s).
2. Draw condensed structures.
3. Write the specified common names or IUPAC ending when specified.
The dehydration of a primary alcohol containing 7 carbons can be achieved through the E1 mechanism, using a strong acid catalyst such as concentrated sulfuric acid (H2SO4).
1. Reaction:
Primary alcohol (7 carbons) + concentrated sulfuric acid → alkene + water
2. Condensed structures:
Primary alcohol: R-CH2-CH2-CH2-CH2-CH2-CH2-OH
Concentrated sulfuric acid: H2SO4
Alkene: R-CH=CH-CH2-CH2-CH2-CH3
Water: H-OH
3. IUPAC naming:
The specified common name or IUPAC ending is not provided. Therefore, the IUPAC name of the alkene product would depend on the R group attached to the 7-carbon chain.
To dehydrate a primary alcohol containing 7 carbons, we can follow the general steps outlined below:
1. Write out the reaction:
The reaction involved in the dehydration of a primary alcohol is an elimination reaction. In this case, we will dehydrate the alcohol using concentrated sulfuric acid as the catalyst. The primary alcohol will undergo dehydration to form an alkene and water.
Reaction: Primary alcohol → Alkene + Water
2. Draw condensed structures:
To illustrate the reaction, we will use heptanol (an example of a primary alcohol with 7 carbons) as the reactant. The condensed structure for heptanol is:
CH3-(CH2)5-OH
After dehydration, the product will be an alkene, in this case, heptene. The condensed structure for heptene is:
CH3-(CH2)5-CH=CH2
3. Write the IUPAC ending:
The IUPAC name for heptene is 1-heptene.
Therefore, the complete answer to your question is:
1. Reaction: Primary alcohol → Alkene + Water
Reactant: Heptanol
Catalyst: Concentrated sulfuric acid
Product: 1-Heptene + Water
2. Condensed structures:
Reactant (Heptanol): CH3-(CH2)5-OH
Product (Heptene): CH3-(CH2)5-CH=CH2
3. IUPAC name:
Product (Heptene): 1-Heptene