Alcohols Phenols And Ethers Question 108
Question: $ H_2C=CH-CH_2-CH_2-\underset{OH}{\mathop{\underset{|}{\mathop{C}}}}H-CH_3 $ $ \xrightarrow[Pyridine]{SOCl_2}(A)\xrightarrow[(H_2O)]{O_3/Zn}\underset{C_5H_9ClO}{\mathop{(B)}}\xrightarrow{NaBH_4}(C) $ Compound (C) is
Options:
A) $ CH_3-\overset{OH}{\mathop{\overset{|}{\mathop{C}}}}H-CH_2-\underset{Cl}{\mathop{\underset{|}{\mathop{C}}}}H-CH_3 $
B) $ HOCH_2-CH_2-CH_2-CH_2-CH_2-CH_2-Cl $
C) $ HO-CH_2-CH_2-CH_2-\overset{Cl}{\mathop{\overset{|}{\mathop{C}}}}H-CH_3 $
D) $ HO-CH_2-CH_2-\underset{Cl}{\mathop{\underset{|}{\mathop{C}}}}H-CH_2-CH_3 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ H_2C=CH-CH_2-CH_2-\underset{OH}{\mathop{\underset{|}{\mathop{C}}}}H-CH_3\xrightarrow[Pyridine]{SOCl_2} $
BETA
