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The reduction of 1,3-pentadience with sodium in liquid ammonia in the persence of an alcohol givesA. pentaneB. 2-penteneC. 1-pentaneD. propene+ethene |
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Answer» Correct Answer - B `Na//liq NH_(3)` is a reducing agent `CH_(2)=CH-CH=CH-CH_(3) underset("alcohol")overset(Na//liq.NH_(3))to underset("Pent-2-ene")(CH_(3)-CH_(2)-CH-CH_(3))` This is a brich reduction to reduce terminal double bond only. |
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