Two liquids, A and B, were found in a box labeled only \"isomeric alkyl halides

Question

Two liquids, A and B, were found in a box labeled only "isomeric alkyl halides C5H11Br". You have been employed to deduce the structures of these compounds from the following data: Reaction of each compound with Mg in ether, followed by water, gives the same hydrocarbon. Compound A, when dissolved in warm ethanol, reacts to give an ethyl ether C and an acidic solution in a few minutes. Compound B reacts more slowly but eventually gives the SAME ether C and acidic solution. Both acidic solutions, when tested with AgNO3 solution, give a light yellow precipitate of AgBr. Reaction of compound B with sodium ethoxide in ethanol gives two alkenes, one of with reacts with O3, then aqueous H2O2, to give acetone as one product. Give the structures of compounds A, B, and C and explain reasoning.

Explanation / Answer

If we see the formula, C5H11Br , it is clear that there are only single bonds or is the form of alkane formula (CnH2n+2) first thing. Then after, goes on reactions we can deduce the structures of compounds clearly. Compound A will be 2-bromo-2-methylbutane, and compound B will be 3-bromo-2-methylbutane .The structure of C will be C5H11OC2H5 , As said compound A undergoing E1 elimination at the fastest rate than B so A would be a tertiary alkyl bromide ( 2-bromo-2-methylbuatane) . Reaction of Compound B as giving the mixture of two alkenes (2-methyl-1-butene and 2-methyl-2-butene(major)) is an E2 elimination reaction hence, it's structure is 2-bromo-3-methylbutane . 2-butene after on ozonolysis gives acetone as one product.

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