H H | | 1) CH-CH + KOH = | | Br Br 2) Br H | | CH3- C-CH + KOH = | | Br H

To solve the reactions provided, we need to analyze how KOH (potassium hydroxide) will react with the organic compounds presented. 1. **CH3-CH(Br)-Br + KOH**: This is an example of a bromoalkane (specifically, a dibromopropane derivative). KOH can promote an elimination reaction (usually an E2 mechanism), resulting in the formation of an alkene. The reaction is as follows: - The base (KOH) will abstract a proton (H) from the β-carbon that is adjacent to the carbon bearing the bromine leaving group (Br). - This leads to the formation of a double bond between the α-carbon and β-carbon, with the bromine atoms leaving as bromide ions (Br-). The overall reaction: **CH3-CH(Br)-Br + KOH → CH3-CH=CH2 + KBr + H2O** The product is propene (CH2=CH-CH3). 2. **CH3-C(Br)-H + KOH**: Here we have a bromoalkane with a bromine substituent on a secondary carbon atom. KOH can again promote elimination (E2) to produce an alkene. In this case, the strongest potential for a double bond formation is between the carbons adjacent to the carbon with the bromo group. The leaving group (Br) leaves and forms a double bond in the process. The overall reaction: **CH3-C(Br)-H + KOH → CH3-CH=CH2 + KBr + H2O** In this case, the product is again propene (CH2=CH-CH3), assuming that KOH abstracts the hydrogen from the β-carbon adjacent to the bromo-group. ### Summary of Products: 1. For **CH$3$-C(Br)-C(Br)** + KOH, the product is alkene: **CH2=CH-CH3 (Propene)**. 2. For **CH$3$-C(Br)-H** + KOH, the product is also: **CH2=CH-CH3 (Propene)**. This means in both equations, under the elimination condition with KOH, the reaction leads to the formation of propene.