Outline all steps in the conversion of: (a) ethylbenzene into phenylacetylene; (b) trans-1-phenylpropene into cis-1-phenylpropene.

First, we need to transform ethylbenzene into styrene by removing two hydrogen atoms from the CH3 side chain. This process is known as dehydrogenation. The reaction requires a transition metal catalyst such as Iron(III) oxide and Potassium oxide mixture at a temperature of 500-600 °C. The balanced chemical reaction can be written as follows: \[ C_6H_5CH_2CH_3 \xrightarrow[500-600\,\text{°C, KO}]{Fe_2O_3} C_6H_5CH=CH_2 + H_2 \]

Now, we need to transform the styrene into phenylacetylene. This can be done by hydrogenating the double bond in the presence of a poisoned catalyst (Lindlar's catalyst). Lindlar's catalyst consists of Palladium on Calcium carbonate with lead acetate and quinoline as a poison to avoid full hydrogenation to ethylbenzene. This catalyst is very selective and only reduces alkynes into cis-alkenes without further reduction to alkanes. The balanced chemical reaction can be written as follows: \[ C_6H_5CH=CH_2 + H_2 \xrightarrow[\text{Pd/CaCO}_3]{\text{Lindlar's catalyst}} C_6H_5C\equiv CH \] So, our final product is phenylacetylene, as desired. \( \) (b) Conversion of trans-1-phenylpropene to cis-1-phenylpropene

First, we need to brominate the trans-1-phenylpropene to form a vicinal dibromide. This can be done using Br2 in the presence of an inert solvent like dichloromethane (CH2Cl2). The balanced chemical reaction can be written as follows: \[ C_6H_5CH=CHCH_3 \xrightarrow[\text{CH}_2\text{Cl}_2]{Br_2} C_6H_5CHBrCHBrCH_3 \] Next, we need to eliminate the vicinal dihalide formed by using a strong base such as NaNH2 in anhydrous ammonia (NH3) solvent. The base will remove two hydrogens in a two-step elimination process, converting into an alkyne, which is 1-phenylpropyne. The balanced chemical reaction can be written as follows: \[ C_6H_5CHBrCHBrCH_3 + 2\, NaNH_2 \xrightarrow[NH_3]{\Delta} C_6H_5C\equiv CCH_3 + 2\, NaBr + 2\, NH_3 \]

Now, we will convert the 1-phenylpropyne into cis-1-phenylpropene using Lindlar’s catalyst. The reaction is similar to the one in first part of the exercise. The balanced chemical reaction can be written as follows: \[ C_6H_5C\equiv CCH_3 + H_2 \xrightarrow[\text{Pd/CaCO}_3]{\text{Lindlar's catalyst}} C_6H_5CH=CHCH_3 \] So, our final product is cis-1-phenylpropene, as desired.