When neopentyl alcohol is treated with strong acid, such as concentrated sulfuric acid (H2SO4), it undergoes an elimination reaction to form a mixture of two alkenes. The elimination reaction is known as the E1 mechanism, and in the case of neopentyl alcohol, it results in the formation of a mixture of two isomeric alkenes: 2-methyl-2-butene and 2-methyl-1-butene.
Now let's evaluate the given statements:
A. Both gave the same major products on treatment with HBr.
This statement is not necessarily true. The alkenes formed from neopentyl alcohol may give different products upon treatment with HBr.
B. Both give different major products on treatment with HBr in the presence of peroxide.
This statement is likely to be true. The presence of peroxide (such as in a radical reaction) typically leads to anti-Markovnikov addition, resulting in different major products compared to the reaction without peroxide.
C. The alkene which is formed in 85% concentration has a higher heat of hydrogenation than the other alkene.
This statement is not necessarily true. The concentration of the alkene does not provide information about its heat of hydrogenation. The heat of hydrogenation depends on the stability of the alkene, and without specific data, it's not possible to determine which alkene has a higher heat of hydrogenation.
D. Both alkenes on ozonolysis give the same products.
This statement is likely to be true. Ozonolysis typically results in the cleavage of carbon-carbon double bonds, forming carbonyl compounds. Since both alkenes are isomers of each other, they would likely yield the same products upon ozonolysis.
In summary, statements B and D are likely to be correct based on typical reactions, while statements A and C cannot be determined without additional information.
