
Examine the structural formulas given below and identify number of compounds which are reduced by $$ $NaB{{H}_{4}}LiAl{{H}_{4}}-CHO-CO-O-N{{H}_{2}}-NH-$ $NaB{{H}_{4}}$.
Answer
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Hint: $NaB{{H}_{4}}$is a reducing agent which is used for the reduction of aldehydes, ketones and acid chlorides to alcohols. Although it is a weak reducing agent and less reactive than the reducing agent like Lithium Aluminium Hydride $LiAl{{H}_{4}}$. And that’s why other compounds like esters , amides , acids and nitriles are not reduced by this reagent.
Complete Step by Step Answer:
Sodium borohydride ($NaB{{H}_{4}}$) is an inorganic compound is a reducing agent that finds application in chemistry both in the lab and in the industrial scale. But it is a weak reducing agent in comparison to the strong reducing agents like $LiAl{{H}_{4}}$ and so it is only strong enough to reduce aldehydes, ketones and acid chlorides to alcohols but it does not affect other strong bonded compounds like esters, amides, acids and nitriles.
The compounds given here are respectively of the classes:
- Aldehyde (presence of $-CHO$ group)
- Amide(presence of $-CO-N{{H}_{2}}$ group)
- Ester(presence of $-COO-$ group)
- Ketone(presence of $-CO-$ group)
- Ketone(presence of $-CO-$ group)
- Aldehyde(presence of $-CHO$ group)
- Ester(presence of $-COO-$ group)
- Aldehyde (presence of $-CHO$ group)
- 2°Amine(presence of $-NH-$ group)
Since , $NaB{{H}_{4}}$ acts as reducing agents only on aldehydes, ketones and acid chlorides. Therefore, among the above compounds only aldehydes, ketones and acid chlorides will be considered as the correct answer .
Thus, the correct answers will be A,D,E,F,H.
Note: $NaB{{H}_{4}}$ acts as a reducing agent in those compounds also in which two groups are present like if ketone and ester both are present in a compound then it will reduce only ketone part and leave the rest as it is there present in the compounds. But in case of strong reducing agents it breaks the compound at both places and reduces it in corresponding acids and alcohols.
Complete Step by Step Answer:
Sodium borohydride ($NaB{{H}_{4}}$) is an inorganic compound is a reducing agent that finds application in chemistry both in the lab and in the industrial scale. But it is a weak reducing agent in comparison to the strong reducing agents like $LiAl{{H}_{4}}$ and so it is only strong enough to reduce aldehydes, ketones and acid chlorides to alcohols but it does not affect other strong bonded compounds like esters, amides, acids and nitriles.
The compounds given here are respectively of the classes:
- Aldehyde (presence of $-CHO$ group)
- Amide(presence of $-CO-N{{H}_{2}}$ group)
- Ester(presence of $-COO-$ group)
- Ketone(presence of $-CO-$ group)
- Ketone(presence of $-CO-$ group)
- Aldehyde(presence of $-CHO$ group)
- Ester(presence of $-COO-$ group)
- Aldehyde (presence of $-CHO$ group)
- 2°Amine(presence of $-NH-$ group)
Since , $NaB{{H}_{4}}$ acts as reducing agents only on aldehydes, ketones and acid chlorides. Therefore, among the above compounds only aldehydes, ketones and acid chlorides will be considered as the correct answer .
Thus, the correct answers will be A,D,E,F,H.
Note: $NaB{{H}_{4}}$ acts as a reducing agent in those compounds also in which two groups are present like if ketone and ester both are present in a compound then it will reduce only ketone part and leave the rest as it is there present in the compounds. But in case of strong reducing agents it breaks the compound at both places and reduces it in corresponding acids and alcohols.
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