Aldehydes Ketones Question 49
Question: A substance $ C_4H _{10}O $ yields on oxidation a compound, $ C_4H_8O $ which gives an oxime and a positive iodoform test. The original substance on treatment with cone. $ H_2SO_4 $ gives $ C_2H_8 $ The structure of the compound is
Options:
A) $ CH_3CH_2CH_2CH_2OH $
B) $ CH_3CHOHCH_2CH_3 $
C) $ {{( CH_3 )}_3}COH $
D) $ CH_3CH_2-O-CH_2CH_3 $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ C_4H_8\xrightarrow[(-H_2O)]{conc.H_2SO_4}C_4H _{10}O\xrightarrow{oxidation} $
$ C_4H_8O( R-COCH_3 ) $ Thus $ C_4H_8O $ should be $ CH_3CH_2COCH_3 $ hence $ C_4H _{10}O $ should be $ CH_3CH_2CHOHCH_3 $
BETA
