(a)
Interpretation:
Draw the structural formula the compound formed by the reaction of given compound with sodium borohydride.
Concept Introduction:
Answer to Problem 57P
Explanation of Solution
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Therefore, the product formed will be as follows:
(b)
Interpretation:
Draw the structural formula the compound formed by the reaction of given compound with sodium borohydride.
Concept Introduction:
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Answer to Problem 57P
Explanation of Solution
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Therefore, the product formed will be as follows:
(c)
Interpretation:
Draw the structural formula the compound formed by the reaction of given compound with sodium borohydride.
Concept Introduction:
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Answer to Problem 57P
Explanation of Solution
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Therefore, the product formed will be as follows:
(d)
Interpretation:
Draw the structural formula the compound formed by the reaction of given compound with sodium borohydride.
Concept Introduction:
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Answer to Problem 57P
Explanation of Solution
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Therefore, the product formed will be as follows:
(e)
Interpretation:
Draw the structural formula the compound formed by the reaction of given compound with sodium borohydride.
Concept Introduction:
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Answer to Problem 57P
Explanation of Solution
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Therefore, the product formed will be as follows:
(f)
Interpretation:
Draw the structural formula the compound formed by the reaction of given compound with sodium borohydride.
Concept Introduction:
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Answer to Problem 57P
Explanation of Solution
Aldehydes are reduced to primary alcohols and ketones are reduced in secondary alcohols. The most commonly used reagent for reduction of aldehydes and ketone is sodium borohydride
Therefore, the product formed will be as follows:
Want to see more full solutions like this?
Chapter 16 Solutions
Introduction to General, Organic and Biochemistry
- Nonearrow_forwardJON Determine the bund energy for UCI (in kJ/mol Hcl) using me balanced chemical equation and bund energies listed? का (My (9) +36/2(g)-(((3(g) + 3(g) A Hryn = -330. KJ bond energy и-н 432 bond bond C-1413 C=C 839 N-H 391 C=O 1010 S-H 363 б-н 467 02 498 N-N 160 N=N 243 418 C-C 341 C-0 358 C=C C-C 339 N-Br 243 Br-Br C-Br 274 193 614 (-1 214||(=olin (02) 799 C=N 615 AALarrow_forwardDetermine the bond energy for HCI ( in kJ/mol HCI) using he balanced cremiculequecticnand bund energles listed? also c double bond to N is 615, read numbets carefully please!!!! Determine the bund energy for UCI (in kJ/mol cl) using me balanced chemical equation and bund energies listed? 51 (My (9) +312(g)-73(g) + 3(g) =-330. KJ спод bond energy Hryn H-H bond band 432 C-1 413 C=C 839 NH 391 C=O 1010 S-1 343 6-H 02 498 N-N 160 467 N=N C-C 341 CL- 243 418 339 N-Br 243 C-O 358 Br-Br C=C C-Br 274 193 614 (-1 216 (=olin (02) 799 C=N 618arrow_forward
- Differentiate between single links and multicenter links.arrow_forwardI need help on my practice final, if you could explain how to solve this that would be extremely helpful for my final thursday. Please dumb it down chemistry is not my strong suit. If you could offer strategies as well to make my life easier that would be beneficialarrow_forwardNonearrow_forward
- Introduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage LearningOrganic And Biological ChemistryChemistryISBN:9781305081079Author:STOKER, H. Stephen (howard Stephen)Publisher:Cengage Learning,General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage Learning