Question

When neopentyl alcohol is treated with , a mixture of two alkenes is formed. Which statements are correct about these alkenes?
A. Both gave the same major products on treatment with HBr.
B. Both give different major products on treatment with HBr in presence of peroxide.
C. The alkene which is formed in 85% concentration has higher heat of hydrogenation than other alkene.
D. Both alkenes on ozonolysis give the same products.

Answer 1

Hint: To answer this question, you must recall the reactions of alkenes. On treatment with sulphuric acid, alcohols undergo dehydration and an alkene is formed after the elimination of a water molecule.

Complete step by step solution:
On treatment with HBr, both alkenes undergo additional reactions. The product formed will be in accordance to the Markovnikov’s rule which suggests that during an addition of a protic acid to an alkene, the hydrogen atom of the acid adds on to that carbon atom of the alkene which has the greater number of hydrogen atoms. In both the alkenes the tertiary carbon atom has the least number of hydrogen atoms and thus will be the additional site for the bromine ion. Thus, the same product is obtained.

- On treatment with HBr in presence of peroxide, the reaction proceeds through the peroxide effect or the Anti- Markovnikov’s rule. The bromine atom bonds to that carbon atom which has the greater number of hydrogen atoms than the starting alkene. Hence, different products are obtained.
- So after rearrangement, the two products obtained are 2-methyl but-1-ene and 2- methyl but-2-ene
- The di substituted alkene (2-methyl but-1-ene) is less substituted and thus minor product whereas the tri substituted alkene (2- methyl but-2-ene) is the major product.
- Since the major product (2- methyl but-2-ene) is tri substituted, it is more stable than the minor product (2-methyl but-1-ene). Thus, heat of hydrogenation is more for the minor (15%) product.
- In ozonolysis, the double bond is cleaved and an oxygen atom bonds to each carbon atom forming a carbonyl compound. Different products are formed in both cases.

The correct answers are A and B.

Note: When alkenes are formed after an elimination reaction the major product is formed on the basis of Saytzeff’s rule. According to the Saytzeff rule, more substituted alkene is preferably formed out of two possible alkene products. In other words, the hydrogen atom is released from the carbon atom attached to less number of hydrogen atoms.