
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
163.2k+ views
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.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Solutions Class 12 Notes: CBSE Chemistry Chapter 1

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

Electrochemistry Class 12 Notes: CBSE Chemistry Chapter 2
