$Write the chemical equations for Williamson synthesis of 2 - Ethoxy - 2 - methyl - pentane starting from$ $ethanol and 2 - Methylpentan - 2 - ol .$

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Solution

In williamson synthesis, alkyl halide must always be less hindered because the reaction takes place by SN2 mechanism. Hence, base remove proton from alcohol to form alkoxide which reacts with alkyl halide to form ether.

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Similar questions

Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.

Write structures of the compounds whose IUPAC names are as follows:

(i) 2-Methylbutan-2-ol

(ii) 1-Phenylpropan-2-ol

(iii) 3,5-Dimethylhexane −1, 3, 5-triol

(iv) 2,3 − Diethylphenol

(v) 1 − Ethoxypropane

(vi) 2-Ethoxy-3-methylpentane

(vii) Cyclohexylmethanol

(viii) 3-Cyclohexylpentan-3-ol

(ix) Cyclopent-3-en-1-ol

(x) 3-Chloromethylpentan-1-ol.

Write structures of the compounds whose IUPAC names are as follows.
(a). 2-methylbutan-2-ol

(b). 1-phenylpropan-2-ol

(c). 3,5-dimethylhexane-1,3,5-triol

(d). 2,3-diethylphenol

(e). 1-ethoxypropane

(f). 2-ethoxy-3-methylpentane

(g). Cyclohexylmethanol

(h). 3-cyclohexylpentan-3-ol

(i). Cyclopent-3-en-1-ol

(j). 3-chloromethylpentan-1-ol

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