CUET Preparation Today
Read the passage carefully and answer the Questions. Ethers with the general formula $C_nH_{2n+2}O$ and functional group C-O-C may be divided as symmetrical and unsymmetrical. Lower symmetrical ether may be obtained by the dehydration of alcohols at low temeratures in the presence of protic acids, such as $H_2SO_4$ and $H_3PO_4$ wheras at high temperature intramolecular dehydration takes place. Formation of ether from primary alcohols is also accompanied by the formation of small amount of alkenes. Due to the stearic hinderance, secondary and tertiary alcohols undergo elimination reaction rather than substitution reaction yielding alkene as major product. Ethers can also be obtained by heating alkyl halides with sodium or potassium alkoxides. Substituted Ethers (secondary & tertiary) can also be prepared using this method. If primary halides are replaced with secondary and tertiary halides, alkenes are the major products. |
Major product in the following reaction is
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The correct answer is Option (1) →
The reactants are tert-butyl bromide $(CH_3)_3C-Br$ and sodium methoxide $(CH_3ONa)$. Key concept from the passage If secondary or tertiary alkyl halides react with alkoxides, elimination reaction occurs instead of substitution because of steric hindrance. Therefore alkenes become the major products. Reasoning
Reaction $$(CH_3)_3CBr + CH_3ONa \rightarrow CH_2 = C(CH_3)_2 + NaBr + CH_3OH$$ The product formed is 2-methylpropene (isobutene). Final answer (as per question format) The correct answer is Option (1) $\rightarrow$ $CH_2 = C(CH_3)_2$ (2-methylpropene). Summary of Options Option 1 (Correct): 2-methylpropene (Alkene formed via E2 elimination). Option 2: Tert-butyl alcohol (Usually formed in aqueous conditions, not here). Option 3: Tert-butyl methyl ether (Would be the product if $S_N2$ could occur). Option 4: 2,2-dimethylpropane (Neopentane, not possible in this mechanism).
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