In the following reaction: H_3C-C^{|↖{ \CH_3}}_{|↙{ \CH_3}}-CH=CH_2\stackrel{\H_2\O{/}{──────▶}\H^{+}}\A_{\text{Major Product}}+\B_{\text{Minor Product}} The major product is :

Correct Answer is : H3C−CH3|C|OH−C|CH3H=CH2

H_3C-C^{|↖{ \CH_3}}_{|↙{ \OH}}-C_{|↙{ \CH_3}}H=CH_2

C_{|↙{ \OH}}H_2-C^{|↖{ \CH_3}}_{|↙{ \CH_3}}-CH_2-CH_3

H_3C-C^{|↖{ \CH_3}}_{|↙{ \CH_3}}-C_{|↙{ \OH}}H-CH_3

H_3C-C^{|↖{ \CH_3}}_{|↙{ \CH_3}}-CH_2-C_{|↙{ \OH}}H_2

AIPMT 2012 Physics and Chemistry Paper

Question : 100 of 100

Marks: +1, -0

In the following reaction:

H3C−

‌‌‌CH3

−CH=CH2

H2O/H+

──────▶

Major Product

Minor Product

The major product is :

Go to Question:

Prev Question