Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
Reduction of acetone in the presence of sodium borohydride gives
Q. Reduction of acetone in the presence of sodium borohydride gives
2816
223
AMU
AMU 2012
Aldehydes Ketones and Carboxylic Acids
Report Error
A
1-propanol
0%
B
2-propanol
0%
C
propene
100%
D
n-propane
0%
Solution:
Correct answer is (b) 2-propanol