Question

Product (A) of above reaction is:
$ {{H}_{2}}C=CH-C(=O)-OC{{H}_{3}}+C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH\xrightarrow[TsOH]{\Delta }(A)+C{{H}_{3}}OH $
$ methylacrylate\left[ bpt\left( 81{}^\circ C \right) \right] $ $ n-butylalcohol\left[ bpt\left( 117{}^\circ C \right) \right] $ $ \left[ bpt\left( 145{}^\circ C \right) \right] $ $ \left[ bpt\left( 65{}^\circ C \right) \right] $
(A) $ {{H}_{2}}C=CH-C(=O)-OC{{H}_{2}}C{{H}_{2}}C{{H}_{3}} $
(B) $ {{H}_{2}}C=CH-C(=O)-O-CH{{\left( Me \right)}_{2}} $
(C) $ {{H}_{2}}C(OC{{H}_{3}})-C{{H}_{2}}-C(=O)-OC{{H}_{3}} $
(D) $ {{H}_{3}}C-{{\left( C{{H}_{2}} \right)}_{4}}-C(=O)-O-C{{H}_{3}} $

Answer 1

Hint: We know that normally the reaction would have been an oxidation reaction. But things change when a tertiary alcohol is present. Formation of double-bond will take place. The formation of the double bond takes place in such a way as to ensure maximum number of hyper conjugative structures.

Complete step by step solution:
Let us first know about the reaction that is mentioned in the question. Generally, when primary or secondary alcohol vapors are passed over heated copper at the dehydrogenation or in other words oxidation takes place. It uses methacrylate and n-butyl alcohol and the secondary one into compound A.
This is not the case with tertiary alcohols as they are difficult to oxidize. In their case, dehydration takes place rather than dehydrogenation. The dehydration takes a hydroxyl and hydrogen atom from adjacent carbons, which results in the formation of a double bond. Remember here $ TsOH $ is solvent so the corresponding compound with by product $ C{{H}_{3}}OH $
The major product will only be “A” because the number of hyper conjugative structures is while that in by product is $ C{{H}_{3}}OH $ . Number of hyper conjugative structures is equal to the number of hydrogen atoms that are attached to the carbon atoms adjacent to the double bond. Thus, here A is $ {{H}_{2}}C=CH-C(=O)-OC{{H}_{2}}C{{H}_{2}}C{{H}_{3}} $
Therefore, correct answer is option A i.e. $ {{H}_{2}}C=CH-C(=O)-OC{{H}_{2}}C{{H}_{2}}C{{H}_{3}} $

Note:
Note that the tertiary alcohols are not oxidized into carbonyl compounds or acids by mild oxidising agents such as chromic oxide. When subjected to strong oxidizing agents such as potassium permanganate, they convert into acids with lower numbers of carbon atoms. The heated copper process which is described above is an industrial one, and is seldom used in laboratories.