(a) Draw the structure of the given compound :
4-Bromo-3-methylpent-2-ene
(b) What happens when chloroethane is treated with aqueous potassium hydroxide ?
(a) Draw the structure of the given compound :
4-Bromo-3-methylpent-2-ene
(b) What happens when chloroethane is treated with aqueous potassium hydroxide ?
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Solution and Explanation
The parent chain is pent-2-ene, so it has 5 carbons with a double bond between C2 and C3.
A methyl group sits on C3.
A bromine atom sits on C4.
Step 2: Build the structure.
Number the chain C1 to C5 and attach the groups at their positions.
C1 is a CH3 group. C2 is a CH= group. C3 is a =C(CH3)- group. C4 is a CHBr group. C5 is a CH3 group.
The condensed formula is:
\[ \mathrm{CH_3-CH=C(CH_3)-CHBr-CH_3} \]
The molecular formula is \( \mathrm{C_6H_{11}Br} \).
Step 3: Answer part (b).
Chloroethane is a primary alkyl halide. With aqueous KOH, the hydroxide ion replaces the chlorine atom by nucleophilic substitution.
This is hydrolysis and the product is ethanol. Potassium chloride is formed as the by-product.
\[ \mathrm{C_2H_5Cl + KOH(aq) \rightarrow C_2H_5OH + KCl} \]
Alcoholic KOH would give elimination (ethene) instead, but aqueous KOH favours substitution.
Final Answer:
(a) The structure is \( \mathrm{CH_3-CH=C(CH_3)-CHBr-CH_3} \).
(b) Chloroethane undergoes hydrolysis to give ethanol and KCl.
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