20. A+Barrow HCequiv CHxrightarrow [-Z](NaH)arrow A(CH_(3)I)/(-272)arrow B(NaNH_(3))/(-272)arrow Cxrightarrow [-2](CH_(2)I)arrow (H_(2)O)/(H_(2)SO_(4))E

To determine the correct answer, let's analyze the given chemical reactions step by step:<br /><br />1. $HC\equiv CH \xrightarrow{NaH} A$<br /> - This reaction involves the addition of sodium hydride (NaH) to acetylene (HC≡CH), resulting in the formation of acetylide ion (C2H−).<br /><br />2. $A \xrightarrow{CH_{3}I} B$<br /> - In this reaction, the acetylide ion (A) reacts with methyl iodide (CH3I) to form propyne (CH3C≡CH).<br /><br />3. $B \xrightarrow{NaNH_{3}} C$<br /> - The propyne (B) reacts with sodium amide (NaNH3) to form propyne sodium salt (C).<br /><br />4. $C \xrightarrow{CH_{2}I} D$<br /> - The propyne sodium salt (C) reacts with methyl iodide (CH2I) to form 1-methylpropyne (D).<br /><br />5. $D \xrightarrow{H_{2}O/H_{2}SO_{4}} E$<br /> - The 1-methylpropyne (D) undergoes hydrolysis in the presence of water (H2O) and sulfuric acid (H2SO4) to form 2-methylpropene (E).<br /><br />Therefore, the correct answer is E.