Explain Williamson ether synthesis mechanism. Also, predict the major product formed and write the complete reaction when 2-chloro-2-methylpropane reacts with sodium methoxide.

Q: Explain Williamson ether synthesis mechanism. Also, predict the major product formed and write the complete reaction when 2-chloro-2-methylpropane reacts with sodium methoxide.

Williamson Ether Synthesis Mechanism\nWilliamson ether synthesis is an important laboratory method for the preparation of symmetrical and unsymmetrical ethers. It involves the reaction of an alkyl halide with a sodium alkoxide. General Reaction: $R-\text{X} + R'-\text{ONa} \rightarrow R-\text{O}-R' + \text{NaX}$ Mechanism: The reaction generally follows an $SN2$ mechanism for primary alkyl halides. The alkoxide ion acts as a strong nucleophile and attacks the carbon atom bonded to the halogen atom in the alkyl halide, displacing the halide ion ($X^-$) in a concerted single-step process with inversion of configuration. Reaction of 2-chloro-2-methylpropane with Sodium Methoxide\nWhen tertiary alkyl halides like 2-chloro-2-methylpropane react with a strong nucleophile/base like sodium methoxide, elimination competes strongly over substitution due to steric hindrance. Reaction Equation: $$\text{(CH}3)3\text{C}-\text{Cl} + \text{CH}3\text{ONa} \rightarrow \text{(CH}3)2\text{C}=\text{CH}2 + \text{CH}3\text{OH} + \text{NaCl}$$ Explanation of Product: Instead of forming an ether via $SN2$ substitution (which is hindered by the three bulky methyl groups on the tertiary carbon), sodium methoxide acts as a strong base and abstracts a proton from one of the methyl groups. This triggers an $E2$ elimination reaction, leading to the formation of 2-methylpropene (an alkene) as the major product, along with methanol and sodium chloride.

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Williamson Ether Synthesis Mechanism
Williamson ether synthesis is an important laboratory method for the preparation of symmetrical and unsymmetrical ethers. It involves the reaction of an alkyl halide with a sodium alkoxide. General Reaction: $R-	ext{X} + R'-	ext{ONa} ightarrow R-	ext{O}-R' + 	ext{NaX}$ Mechanism: The reaction generally follows an $SN2$ mechanism for primary alkyl halides. The alkoxide ion acts as a strong nucleophile and attacks the carbon atom bonded to the halogen atom in the alkyl halide, displacing the halide ion ($X^-$) in a concerted single-step process with inversion of configuration. Reaction of 2-chloro-2-methylpropane with Sodium Methoxide
When tertiary alkyl halides like 2-chloro-2-methylpropane react with a strong nucleophile/base like sodium methoxide, elimination competes strongly over substitution due to steric hindrance. Reaction Equation: $$	ext{(CH}3)3	ext{C}-	ext{Cl} + 	ext{CH}3	ext{ONa} ightarrow 	ext{(CH}3)2	ext{C}=	ext{CH}2 + 	ext{CH}3	ext{OH} + 	ext{NaCl}$$ Explanation of Product: Instead of forming an ether via $SN2$ substitution (which is hindered by the three bulky methyl groups on the tertiary carbon), sodium methoxide acts as a strong base and abstracts a proton from one of the methyl groups. This triggers an $E2$ elimination reaction, leading to the formation of 2-methylpropene (an alkene) as the major product, along with methanol and sodium chloride.

CBSE · Class 12 · Chemistry · Alcohols, Phenols and EthersExplain Williamson ether synthesis mechanism. Also, predict the major product formed and write the complete reaction when 2-chloro-2-methylpropane reacts with sodium methoxide.

Williamson Ether Synthesis Mechanism\nWilliamson ether synthesis is an important laboratory method for the preparation of symmetrical and unsymmetrical ethers. It involves the reaction of an alkyl halide with a sodium alkoxide.

Reaction of 2-chloro-2-methylpropane with Sodium Methoxide\nWhen tertiary alkyl halides like 2-chloro-2-methylpropane react with a strong nucleophile/base like sodium methoxide, elimination competes strongly over substitution due to steric hindrance.