The major product of the following reaction:
NaBH4 reduces both carbonyl group and imine.
This reaction involves the addition of HBr to an alkene. The key concept here is Markovnikov's Rule, which states that when a protic acid (like HBr) adds to an unsymmetrical alkene, the hydrogen atom (H⁺) attaches to the carbon with the greater number of hydrogen atoms, and the halide (Br⁻) attaches to the carbon with the fewer hydrogen atoms (the more substituted carbon).
Let's analyze the given alkene: . This is 2-butene, an unsymmetrical alkene.
Step 1: Identify the more and less substituted carbons of the double bond.
The double bond is between carbon 2 and carbon 3.
Step 2: Apply Markovnikov's Rule (which gives the same result from either direction for a symmetrical alkene).
The proton (H⁺) will add to one carbon, and the bromide ion (Br⁻) will add to the other. The product will be 2-bromobutane.
The structure of 2-bromobutane is: or .
Step 3: Match the product to the options.
Looking at the provided images, the correct structure is the one that shows a carbon chain with a Br atom on a secondary carbon, which is 2-bromobutane.
Final Answer: The major product is 2-bromobutane.
Markovnikov's Rule: A rule used to predict the regiochemistry of electrophilic addition reactions of alkenes and alkynes. It states that the electrophile adds to the sp² carbon that is bonded to the most hydrogens.
General Reaction: is incorrect. The correct product is (for an alkene of the type R-CH=CH₂).
Anti-Markovnikov Addition: In the presence of peroxides (ROOR), HBr adds to alkenes in an anti-Markovnikov fashion. This is known as the peroxide effect or Kharasch effect. This effect is specific to HBr; HCl and HI do not show this behavior.
In the given transformation, which of the following is the most appropriate reagent ?
