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This deck includes 5 flashcards for Chemistry. Use it to review key material, focus on weak cards, and prepare for your next exam with StudyLess.

1. 01

   What is the type of reaction when 2-bromo-2-phenylbutane is treated with NaOH at high temperature, and why?

   Elimination reaction (E2 or E1) because the strong base (NaOH) and high temperature favor elimination over substitution.

2. 02

   What are the major and minor products when 2-bromo-2-phenylbutane is treated with NaOH at high temperature?

   Major product: Mixture of B and a minor product C. (The question implies B is a major product, and is a minor product).

3. 03

   Detail the mechanism of formation of product B from 2-bromo-2-phenylbutane with NaOH.

   This requires drawing out the E2 or E1 mechanism, showing the abstraction of a beta-hydrogen by the base and the departure of the bromide leaving group, leading to alkene formation. (Specifics depend on whether E1 or E2 is favored and regioselectivity).

4. 04

   How can compound A (2-bromo-2-phenylbutane) be prepared from compound D under UV light?

   This suggests a radical bromination reaction, likely involving an alkane or alkene precursor (D) reacting with bromine under UV light to form A.

5. 05

   Deduce the structure of compound D, given that it can be used to prepare 2-bromo-2-phenylbutane under UV light.

   Compound D is likely 2-phenylbutane, which would undergo radical bromination at the tertiary benzylic position under UV light to yield 2-bromo-2-phenylbutane (A).