Alcohols Phenols And Ethers Question 92
Question: The product of the following reaction,
Options:
A) $ CH_3-\underset{CH_3}{\overset{CH_3}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}}}}-\underset{OH}{\mathop{\underset{|}{\mathop{C}}}}H-CH_3 $
B) $ CH_3-\underset{CH_3}{\overset{CH_3}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}}}}-\underset{{}}{\mathop{\underset{{}}{\mathop{C}}}}H_2-CH_2OH $
C) $ CH_3-\underset{CH_3}{\overset{OH}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}}}}-\underset{CH_3}{\mathop{\underset{|}{\mathop{C}}}}H-CH_3 $
D) $ HOCH_2-\underset{CH_3}{\overset{CH_3}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}}}}-\underset{{}}{\mathop{\underset{{}}{\mathop{C}}}}H_2-CH_3 $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Recall that oxymercuratoin-reduction of alkenes lead to hydration in Markovnikov’s way. Thus $ {{(CH_3)}_3}CCH=CH_2\xrightarrow[(ii)NaBH_4]{(i)Hg{{(OAc)}_2}.H_2O}{{(CH_3)}_3}C\overset{OH}{\mathop{\overset{|}{\mathop{C}}}}HCH_3 $
BETA
