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.