Alcohols Phenols And Ethers Question 15
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 conc. $ H_2SO_4 $ gives $ C_4H_8 $ . The structure of the compound is
[SCRA 2000]
Options:
A) $ CH_3CH_2CH_2CH_2OH $
B) $ CH_3CH(OH)CH_2CH_3 $
C) $ {{(CH_3)}_3}COH $
D) $ CH_3CH_2-O-CH_2CH_3 $
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Answer:
Correct Answer: B
Solution:
$ CH_3CH(OH)CH_2CH_3\xrightarrow{Conc\text{.}H_2SO_4} $
$ \underset{(C_4H_8)}{\mathop{CH_3CH=CHCH_3}} $
$ \underset{2-Butanol}{\mathop{CH_3CHOHCH_2CH_3}}\xrightarrow{[O]}\underset{Butanone}{\mathop{CH_3COCH_2CH_3}} $ Butanone gives both an oxime and positive iodoform test, therefore, the original compound is 2-butanol.
BETA
