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When `CH_(3)CH_(2)CHCl_(2)` is treated with `NaNH_(2)` the product formed isA. `CH_(3)-CH=CH_(2)`B. `CH_(3)-C-=CH`C. `CH_(3)CH_(2)CH(NH_(2))(Cl)`D. `CH_(3)CH_(2)C(NH_(2))_(2)` |
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Answer» Correct Answer - B `CH_(3)-CH_(2)-CHCl_(2)+2NaNH_(2)underset(196K)overset("liq. "NH_(3))toCH_(3)C-=CH+2NaCl+2NH_(3)` |
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