A) Na-Ethanol
B) \[NaB{{H}_{4}}\]
C) Catalytic hydrogenation
D) \[LiAl{{H}_{4}}\]
Correct Answer: D
Solution :
: \[C{{H}_{3}}-{{(C{{H}_{2}})}_{14}}-COOH\xrightarrow[{{H}_{3}}{{O}^{+}}]{LiAl{{H}_{4}},ether}\] \[C{{H}_{3}}{{(C{{H}_{2}})}_{14}}C{{H}_{2}}OH\]You need to login to perform this action.
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