In an E2 elimination using a bulky base, which product is the Hofmann product?

In an E2 elimination using a bulky base, which product is the Hofmann product?

• A. The less substituted alkene.
• B. The more substituted alkene.
• C. A mixture with equal amounts.
• D. A cyclic alkene.

Answer: (A)

The main idea here is how base size affects E2 regioselectivity. In an E2, the base removes a beta-hydrogen anti to the leaving group in a concerted step. A bulky base is sterically hindered, so it prefers the less hindered, less substituted beta-hydrogen. Abstracting that hydrogen leads to the less substituted double bond—the Hofmann product. Smaller bases, by contrast, can access the more hindered, more substituted beta-hydrogens more easily, giving the Zaitsev (more substituted) alkene. So, using a bulky base in E2 favors the less substituted alkene, which is why the Hofmann product is formed.

The main idea here is how base size affects E2 regioselectivity. In an E2, the base removes a beta-hydrogen anti to the leaving group in a concerted step. A bulky base is sterically hindered, so it prefers the less hindered, less substituted beta-hydrogen. Abstracting that hydrogen leads to the less substituted double bond—the Hofmann product. Smaller bases, by contrast, can access the more hindered, more substituted beta-hydrogens more easily, giving the Zaitsev (more substituted) alkene. So, using a bulky base in E2 favors the less substituted alkene, which is why the Hofmann product is formed.